Class 12 Commerce Maths - Determinants

If A is $2\times 2$ matrix, $det A=4$ then find the product of $det(3A)$ and $det(A^{-1})$

Given

$A$ is

$2 \times 2$ matrix,

$\left| A \right|$

i) det(3A)

using properties of determinat

$\left| KA \right| ={ K }^{ n }\left| A \right|, n=$ order of determinant

$det(3A)={ 3 }^{ 2 }\left| A \right| \\ det(3A)=9\times 4=36\\ ii)det({ A }^{ -1 })$

using properties of determinant

$det({ A }^{ -1 })=\dfrac { 1 }{ det(A) } =\dfrac { 1 }{ 4 } .$

Write the value of the determinant $\begin{bmatrix} p & p+1 \\ p-1 & p\end{bmatrix}$
when $p=1342$

We have,

$\begin{bmatrix} p & p+1 \\ p-1 & p\end{bmatrix}$

$=p^2-(p+1)(p-1)$

$=p^2-(p^2-1)$

$=1$

Using the properties of determinants, find the value of

$\begin{vmatrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{vmatrix}$

Let

$\Delta = \begin{vmatrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{vmatrix}$
Expanding along first column,

$\Delta=a(0+bc)+b(-ac-0)=abc-abc=0$

Evaluate the determinants
$\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$

The value of

$\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$

$=2(-1)-4(-5)$

$=20-2$

$=18$

If $A=\begin{bmatrix} 1 & 0 & 0 \\ 2 & 3 & 4 \\ 5 & -6 & x \end{bmatrix}$ and $\det A = 45$ ; then find $x$ .

$\text{det(A)} =\begin{vmatrix} 1&0&0\\2&3&4\\5&-6&x\end{vmatrix}=1(3x+24)$ , expand along first row

$\Rightarrow 45=3x+24$

$\Rightarrow 3x=45-24=21$

$\therefore x=7$

Find adjoint of matrix $\ \ \ A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

Let

$A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

We know that

$adj A= C^{T}$

So, let's find cofactors of each of the elements of A.

$C_{11}=(-1)^{1+1} 4 =4$

$C_{12}=(-1)^{1+2} 3= -3$

$C_{21}=(-1)^{2+1} 2 =-2$

$C_{22}=(-1)^{2+2} 1 =1$

So, cofactor matrix is

$C=\begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix}$

$\Rightarrow C^{T}=\begin{bmatrix} 4 &-2 \\ -3 & 1 \end{bmatrix}$

Hence,

$adjA=\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$

Show that the points $A(2,3),B(4,0)$ and $C(6,-3)$ are collinear.

Area of the

$\triangle {ABC}$ is

$=\cfrac{1}{2}[(0-12+18)-(12+0-6)]$

$=\cfrac{1}{2}(6-6)=0$

$\therefore$ The given points are collinear.

Write the value of $\triangle = \begin{vmatrix}x+y&y+z&z+x\\z&x&y\\-3&-3&-3\end{vmatrix}$ .

On expansion, we get,

$\begin{vmatrix} x+y & y+z & z+x \\ z & x & y \\ -3 & -3 & -3 \end{vmatrix}=(x+y)(-3x+3y)-(y+z)(-3z+3y)+(z+x)(-3z+3x)\\=-3({ x }^{ 2 }-{ y }^{ 2 }+{ y }^{ 2 }-{ z }^{ 2 }+z^2-{ x }^{ 2 })\\ =0$

The area of a triangle is zero, then the three points are said to be ______ points.

Three or more points are said to be collinear if they lie on a single straight line.

Therefore, area formed by them will be zero.

Find the value of the determinant:
$\begin{vmatrix}5 & -2\\ -3 & 1\end{vmatrix}$

$\begin{vmatrix}5 & -2\\ -3 & 1\end{vmatrix} = (5\times 1) - [(-2) \times (-3)] = 5 - 6 = -1$

$\therefore$ Value of the determinant is

$-1$ .

If for any $2 \times 2$ square matrix A, $A (adj A) = \begin{bmatrix} 8& 0\\ 0 & 8\end{bmatrix}$ , them write the value of det[A].

let A =

$\\ \begin{bmatrix} a & b \\ c & d \end{bmatrix}\\$

Adj(A) =

$\\ \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\\$

Now

$A(AdjA)= \begin{bmatrix} a & b \\ c & d \end{bmatrix} * \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\\$

$\Rightarrow \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} * \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\\$

$\Rightarrow ad-bc=8,-ab+ab=0,dc-dc=0,-bc+ad=8$

$det(A)=\begin{vmatrix} a & b \\ c & d \end{vmatrix}=ad-bc=8$

Find a value of $x$ if $\begin{vmatrix}x& 2 \\18&x\end{vmatrix} = \begin{vmatrix}6& 2 \\18&6\end{vmatrix}$

$\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$

${ x }^{ 2 }-18\times 2=6\times 6-2\times 18$

${ x }^{ 2 }-36=36-36$

$x=\pm 6$

$\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ { a }^{ 2 } & { b }^{ 3 } & { c }^{ 3 } \end{vmatrix}$
Prove that : = $(a - b) (b - c) (c - a) (a + b + c).$

Now,

$\begin{vmatrix}1&a&a^3\\1&b&b^3\\1&c&c^3\end{vmatrix}$

[

$R_2'=R_2-R_1$ and

$R_3'=R_3-R_1$ ]

$=\begin{vmatrix}1&a&a^3\\0&b-a&b^3-a^3\\0&c-a&c^3-a^3\end{vmatrix}$

$=(b-a)(c-a)\begin{vmatrix}1&a&a^3\\0&1&b^2+ab+a^2\\0&1&c^2+ac+a^2\end{vmatrix}$

$=-(b-a)(c-a)(b^2-c^2+a(b-c))$

$=-(b-a)(c-a)(b-c)(a+b+c)$

$=(a-b)(b-c)(c-a)(a+b+c)$ .

Find the values of $x$ for which $\begin{vmatrix}3&x\\x&1\end{vmatrix}=\begin{vmatrix}3&2\\4&1\end{vmatrix}$ .

Given

$\begin{vmatrix} 3 & x \\ x & 1 \end{vmatrix}=\begin{vmatrix} 3&2 \\ 4&1 \end{vmatrix}$

Solving the determinants, we get:

$3-{ x }^{ 2 }=3-8$

${ x }^{ 2 }=8$

$x=\pm 2\sqrt { 2 }$

Evaluate the following deteminants :
i) $\begin{vmatrix} x& -7\\ x &5x+1 \end{vmatrix}$

ii) $\begin{vmatrix} \cos 15^{\circ} & \sin 15^{\circ}\\ \sin 75^{\circ} & \cos 75^{\circ} \end{vmatrix}$

$\begin{vmatrix}x&-7\\x&5x+1\end{vmatrix}=5x^2+x+7x=5x^2+8x=x(5x+8)$

ii)

$\begin{vmatrix}\cos 15^o&\sin 15^o\\\sin 75^o&\cos 75^o\end{vmatrix}$

$=\begin{vmatrix}\cos 15^o&\sin 15^o\\\cos 15^o&\sin 15^o\end{vmatrix}=0$ [Since

$\sin 15^o=\cos 75^o$ and

$\cos 15^o=\sin 75^o]$

If $3^{n}$ is a factor of the determinant
$\left| \begin{matrix} 1 & 1 & 1 \\ _{ }^{ n }{ { C }_{ 1 } } & _{ }^{ n+3 }{ { C }_{ 1 } } & ^{ n+6 }{ { C }_{ 2 } } \\ ^{ n }{ { C }_{ 2 } } & ^{ n+3 }{ { C }_{ 2 } } & ^{ n+6 }{ { C }_{ 2 } } \end{matrix} \right| ,$
then the maximum value of $n$ is $____$ .

$^{ n }C_{ 1 }=n$ ,

$^{ n }C_{ 2 }=\dfrac { n(n-1) }{ 2 }$
Making two zeros and expanding

$\Delta=27=3^{3}$
If

$3^{n}$ is to be factor of

$\Delta$ , then value of

$n$ is

$3$ .

$^{ n }C_{ 1 }=n$ ,

$^{ n }C_{ 2 }=\dfrac { n(n-1) }{ 2 }$
Making two zeros and expanding

$\Delta=27=3^{3}$
If

$3^{n}$ is to be factor of

$\Delta$ , then value of

$n$ is

$3$ .

Find the determinant in following case:
$A=\left [ \begin{array}{c}5\hspace{0.5cm}{20}\\0\hspace{0.5cm}{2}\end{array} \right ]$

Given

$A=\begin{bmatrix}5&20\\0&2\end{bmatrix}$ .

So,

$|A|=\begin{vmatrix}5&20\\0&2\end{vmatrix}=10.$

Find if the points (0, 8/3), (1/3) , and (82, 30) are collinear.

Collinear. If points are

$P \, (0, 8/3), \, Q \, (1,3), \, R \, (82,30)$ , then,

$m_1$ = slopes of PQ =

$\dfrac{3 \, - \, (8/3)}{1 \, - \, 0} \, = \, \dfrac{1}{3}$

$m_2$ = slopes of QR =

$\dfrac{30 \, - \, 3}{82 \, - \, 1} \, = \, \dfrac{1}{3}$
Since slope of PQ is the same as the slope of QR, the three points P, Q , R are collinear.

Prove the following :
$\left| \begin{matrix} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{matrix} \right| =4xyz$

Apply

${ C }_{ 1 }-\left( { C }_{ 2 }+{ C }_{ 3 } \right)$ to make one zero we get

$\triangle =\begin{vmatrix} 0 & z & y \\ -2x & z+x & x \\ -2x & x & x+y \end{vmatrix}$
Take out

$-2x$ from

${C}_{1}$ and then apply

${R}_{2}-{R}_{3}$

$\Delta =-2x\left| \begin{matrix} 0 & z & y \\ 0 & z & -y \\ 1 & x & x+y \end{matrix} \right|$

$=-2x\left( -zy-zy \right) =4xyz$ .

Prove that the points (a , b), (c , d) and (a - c, b - d) are collinear if ad = bc. Also show that the straight line passing through these points passes through origin.

If the given points be A,B,C then as they are collinear therefore slopes of AC and BC are same.

$\therefore \, \, \, \dfrac{d}{c} \, = \, \dfrac{b}{a}$ or

$ad = bc$ .....(1)
Since AC is

$y - b = \dfrac{d}{c} (x - a)$
It will pass through (0, 0) if

$- b = \dfrac{d}{c} (- a)$
or

$bc = ad$ which is true by (1).

Show that
$\begin{vmatrix} \sin 10^{\circ}& -\cos 10^{\circ}\\ \sin 80^{\circ}&\cos 80^{\circ} \end{vmatrix}=1$

$\begin{vmatrix}\sin 10^o&-\cos 10^o\\ \sin 80^o& \cos 80^o\end{vmatrix}$

$=\begin{vmatrix}\sin 10^o&-\cos 10^o\\ \cos 10^o& \sin 10^o\end{vmatrix}$

$=\sin^2 10^o+\cos^2 10^o=1$ .

Find the value of $K$ , for which the given points are collinear:
$\begin{matrix} A \\ \left( 7,-2 \right) \end{matrix}\begin{matrix} B \\ \left( 5,1 \right) \end{matrix}\begin{matrix} C \\ \left( 3,K \right) \end{matrix}$

As the three points are collinear:

$AB+BC=CA$ and

$AB$ and

$AC$ will have the same slope.

Slope of

$AB=\dfrac{-2-1}{7-5}$ (1)

Slope of

$AB=\dfrac{-2-K}{7-3}$ (2)

Equating (1) and (2) gives

$\dfrac{-3}{2}=\dfrac{-2-K}{4}$

$3=\dfrac{2+K}{2}$

$6=2+K$

Hence

$K=4$

In each of the following find the value of $k$ , for which the points are collinear..
(ii) $(8, 1), (k, -4), (2, -5)$

Let the points are

$A\left( {8,1} \right)\,\,B\left( {K, - 4} \right)\,\,C\left( {2, - 5} \right)$

For collinearity

$Area\,of\,\Delta ABC$

$\frac{1}{2}\left| {8\left( { - 4 + 5} \right) + K\left( { - 5 - 1} \right) + 2\left( {1 + 4} \right)} \right| = 0$

$\left| {8 - 6k + 10} \right| = 0$

$18 - 6k = 0$

$6k = 18$

$k = 3$

Let $A=\begin{bmatrix}5&8\\8&13\end{bmatrix}$ then show that $A$ satisfies the equation $x^2-18x+1=0$ .

Given,

$A=\begin{bmatrix}5&8\\8&13\end{bmatrix}$ .

Now the characteristic equation corresponding to the matrix

$A$ is

$|A-aI|=0$ [Where

$a$ is a scalar and

$I$ is the identity matrix of order

$2$ ]

$\begin{vmatrix}5-a&8\\8&13-a\end{vmatrix}=0$

or,

$a^2-18a+1=0$ .

According to the Caley-Hamilton's theorem we have " Every square matrix satisfies its own characteristic equation.".

Then

$A$ will satisfy the equation

$x^2-18x+1=0$ .

In each of the following find the value of $k$ , for which the points are collinear.(i) $(7, -2), (5, 1), (3, k)$

Let the points are

$A\left( {7, - 2} \right)\,B\left( {5,1} \right)\,C\left( {3,K} \right)$

For collinearity

$Area\,of\,\Delta ABC$

$\Rightarrow \dfrac{1}{2}\left| {7\left( {1 - K} \right) + 5\left( {K + 2} \right) + 3\left( { - 2 - 1} \right)} \right| = 0$

$\Rightarrow \left| { - 7 - 7K + 5K + 10 - 9} \right| = 0$

$\Rightarrow \left| { - 2K + 8} \right| = 0$

$\Rightarrow - 2K + 8 = 0$

$\Rightarrow 2K = 8$

$K = 4$

Find the value of $x$ if the points $(x,5),(2,3)$ and $(-2,-11)$ are collinear.

Points are collinear , so slope is equal

$\displaystyle \left [\frac{y_2 - y_1}{x_2 - x_1}\right] = slope.$

$\displaystyle \frac{3 -5}{2-x} = \frac{-11-3}{-2-2}$

$\displaystyle \frac{-2}{2-x} = \frac{14}{4} \Rightarrow -8 = 28 - 14x$

$\displaystyle 14x = 36 \Rightarrow x = \frac{18}{7}$

1042410_1038032_ans_b2bb9ccf5edc49baa3abdf15cabec8d5.jpg

If the minor of three-one element (i.e $M_{31}$ ) in the determinant $\left| \begin{matrix} 0 & 1 & \sec { \alpha } \\ \tan { \alpha } & -\sec { \alpha } & \tan { \alpha } \\ 1 & 0 & 1 \end{matrix} \right|$ is $1$ then find the value of $\alpha$ . $(0 \le \alpha \le \pi)$

$\begin{vmatrix} 0 & 1 & \sec\alpha \\ \tan\alpha & -\sec\alpha & \tan\alpha \\ 1 & 0 & 1 \end{vmatrix}$

${M_{31}} = \tan \alpha + {\sec ^2}\alpha$

$\Rightarrow \tan \alpha + {\sec ^2}\alpha$

$\Rightarrow \tan \alpha + 1 + {\tan ^2}\alpha = 1$

$\Rightarrow \tan \alpha + {\tan ^2}\alpha = 0$

$\Rightarrow \tan \alpha \left( {1 + \tan \alpha } \right) = 0$

$\Rightarrow \tan \alpha = 0\,\,\,\,or\,\,\,\,1 + \tan \alpha = 0$

$\Rightarrow \alpha = 0\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\tan \alpha = - 1$

$\Rightarrow \alpha = 0\,\,\,\,or\,\,\alpha = \dfrac{{3\pi }}{4}$

Show that $\triangle$ ABC is an isosceles triangle, if the determinant
$\begin{vmatrix} 1 & 1 & 1 \\ 1+\cos A & 1+\cos B & 1+\cos C \\ \cos^2A +\cos A & \cos^2B+\cos B & \cos^2C+\cos C\end{vmatrix}=0$

$\begin{vmatrix} 1 & 1 & 1 \\ 1+\cos A & 1+\cos B & 1+\cos C \\ \cos^2A+\cos A & \cos^2B+\cos B & \cos^2C +\cos C \end{vmatrix}=0$

Applying

$C_{1}\rightarrow C_{1}-C_{3}, C_{2}\rightarrow C_{2}-C_{3}$ , we get,

$\begin{vmatrix} 0 & 0 & 1 \\ \cos A -\cos C & \cos B -\cos C & 1+\cos C \\ \cos^2A +\cos A -\cos^2C -\cos C & \cos^2B+\cos B -\cos^2C -\cos C & \cos^2C+\cos C\end{vmatrix}=0$

$\begin{vmatrix} 0 & 0 & 1 \\ \cos A -\cos C & \cos B -\cos C & 1 +\cos C \\ (\cos A+\cos C +1)(\cos A -\cos C) & (\cos B+\cos C +1)(\cos B-\cos C) & \cos^2C+\cos C \end{vmatrix}=0$

Taking

$(\cos A -\cos C)$ common from

$C_{1}$ and

$(\cos B-\cos C)$ common from

$C_{2}$ , we get

$(\cos A-\cos C)(\cos B-\cos C) \begin{vmatrix} 0 & 0 & 1 \\ 1 & 1 & 1+\cos C \\ \cos A+\cos C+1 & \cos B +\cos C +1 & \cos^2C+\cos C \end{vmatrix}=0$

Expanding along

$R_{1}$ ,

$(\cos A -\cos C)(\cos B -\cos C)[(\cos B +\cos C+1)-(\cos A+\cos C+1)]=0$

$(\cos A -\cos C)(\cos B-\cos C)(\cos B-\cos A)=0$

$\cos A =\cos C \ or \ \cos B=\cos C \ or \ \cos B=\cos A$

$A=C \ or \ B=C \ or \ B=A$

Hence, ABC is an isosceles triangle.

Using properties of determinants, prove the following
$\begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{vmatrix}=(1+a^2+b^2)^3$ .

LHS :

$\begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{vmatrix}$

Applying

$C_{1}\rightarrow C_{1}-bC_{3},C_{2}\rightarrow C_{2}+aC_{3}$ , we get,

$=\begin{vmatrix} 1+a^2+b^2 & 0 & -2b \\ 0 & 1+a^2+b^2 & 2a \\ b(1+a^2+b^2) & -a(1+a^2+b^2) & 1-a^2-b^2 \end{vmatrix}$

Taking out

$(1+a^2+b^2)$ common from

$C_{1}$ and

$C_{2}$ , we get,

$=(1+a^2+b^2)^2\begin{vmatrix} 1 & 0 & -2b \\ 0 & 1 & 2a \\ b & -a & 1-a^2-b^2\end{vmatrix}$

Applying

$R_{3}\rightarrow R_{3}-bR_{1}+aR_{2}$ , we get,

$=(1+a^2+b^2)^2\begin{vmatrix} 1 & 0 & -2b \\ 0 & 1 & 2a \\ 0 & 0 & 1+a^2+b^2 \end{vmatrix}$

Expanding along

$C_{1}$ , we get,

$=(1+a^2+b^2)^2(1+a^2+b^2)=(1+a^2+b^2)^3$ = RHS

Solve:
$\begin{vmatrix} 0& -3 &x \\ x+1& 3 &1 \\ 4& 1 & 5\end{vmatrix}$ = $0$

$\begin{vmatrix} 0& -3 &x \\ x+1& 3 &1 \\ 4& 1 & 5\end{vmatrix}$ =

$0$

$\begin{array}{l}0(3 \times 5 - 1) + 3(5 \times (x + 1) - 4) + x(x + 1 - 12) = 0\\0 + 3\left[ {5x + 5 - 4} \right] + x\left[ {x - 11} \right] = 0\\15x + 3 + {x^2} - 11x = 0\\{x^2} + 4x + 3 = 0\\{x^2} + 3x + x + 3 = 0\\x(x + 3) + 1(x + 3) = 0\\(x + 1)(x + 3) = 0\\\therefore \,x = - 1, - 3\end{array}$

If $A = \left| {\begin{array}{*{20}{c}} 1&{ - 1} \\ { - 1}&1 \end{array}} \right|$ , then show that ${A^2} = 2A\,\,\& \,\,{A^3} = 4A$

Given that:

$A=\begin{vmatrix}1 &-1 \\ -1 &1 \end{vmatrix}$

Now,

$A^2=A\times A$

$A\times A=\begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix}\times \begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix}$

$A\times A=\begin{vmatrix} 1\times 1+(-1)\times (-1) & -1 \times 1+1\times (-1)\\ -1\times 1+1\times (-1) & 1\times 1+(-1)\times (-1) \end{vmatrix}$

$A^2=\begin{vmatrix} 2 & -2 \\ -2 & 2 \end{vmatrix}$

$A^2=2\begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix}$

$A^2=2A$

Hence proved.

Now,

$A^3=A^2\times A$

$=2A\times A$ (Since we proved

$A^2=2A$ )

$=2A^2$

$=2\times 2A$

$A^3=4A$

Hence proved.

Find $\begin{vmatrix}4&6&2\\3&0&9\\7&1&5\end{vmatrix}$

$\begin{vmatrix}4&6&2\\3&0&9\\7&1&5\end{vmatrix}=4(0-9)-6(15-63)+2(3-0)\\-36+288+6=258$

Show that the point $P (a,b+c)$ , $Q(b,c+a)$ and $R(c,a+b)$ are collinear.

$\begin{array}{l} P(a,b+c)=({ x_{ 1 } },{ y_{ 1 } }) \\ Q=(b,c+a)=({ x_{ 1 } },{ y_{ 2 } }) \\ R=(c,a+b)=({ x_{ 3 } },{ y_{ 3 } }) \\ { m_{ pq } }=slopeofPQ=\cfrac { { { y_{ 2 } }-{ y_{ 1 } } } }{ { { x_{ 2 } }-{ x_{ 1 } } } } \\ =\cfrac { { c+a-b-c } }{ { b-a } } \\ =\cfrac { { a-b } }{ { -(a-b) } } \\ =-1 \\ \\ { m_{ QR=\, } }slope\, of\, QR=\, \cfrac { { { y_{ 3 } }-{ y_{ 2 } } } }{ { { x_{ 3 } }-{ x_{ 2 } } } } \\ =\cfrac { { a+b-c-a } }{ { c-b } } \\ =\cfrac { { b-c } }{ { -(b-c) } } \\ =-1 \\ SlopeofPQ=SlopeofQRso,pointsA,B,Carecollinear. \end{array}$

Find the adjoint of $\begin{vmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{vmatrix}$

Let

$A=\left[ \begin{matrix} 1 & -3 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right] \\ adj\quad A=\left[ \begin{matrix} 0 & -11 & 0 \\ 9 & 1 & -3 \\ 6 & 8 & 9 \end{matrix} \right] \\ =\left[ \begin{matrix} 0 & 9 & 6 \\ -11 & 1 & 8 \\ 0 & -3 & 9 \end{matrix} \right]$

Find the value of following determinant.
$\begin{vmatrix} \dfrac{7}{3} & \dfrac{5}{3}\\ \dfrac{3}{2} & \dfrac{1}{2}\end{vmatrix}$ .

We know that

$\begin{vmatrix} a&b\\c&d \end{vmatrix}=[ad-bc]$

$\begin{vmatrix} \dfrac{7}{3} & \dfrac{5}{3}\\ \dfrac{3}{2} & \dfrac{1}{2}\end{vmatrix}=\dfrac73\times\dfrac12-\dfrac53\times\dfrac32=\dfrac76-\dfrac52=\dfrac{7-15}{6}=\dfrac{-8}{6}=-\dfrac43$

If $A=\left[ \begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix} \right],$ find $|A|$

Given,

$A=\left[ \begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix} \right],$

$|A|=1(1)-2(1)=-1$

If A is a square matrix of order $3$ such $|A|=5$ , then find the value of $|adjA|$ ?

Property:

$|\text{adj }A| = |A|^{n-1}$ where

$n$ is the order of the square matrix

$A$ .

Here,

$n=3$ and

$|A| = 5$
Therefore,

$|\text{adj }A| = 5^2 = \boxed{25}$

If three points $(x_1 , y_1) , (x_2 , y_2 )$ and $(x_3 , y_3 )$ lie on the same line, prove that
$\dfrac {y_2 - y_3}{x_2 x_3} + \dfrac {y_3 - y_1}{x_3 x_1}+ \dfrac {y_1 - y_2}{x_1 x_2} = 0$

$(x_{1},y_{1}),(x_{2},y_{2}),(x_{3},y_{3})$ lie on same line

To prove

$\displaystyle \frac{_{2}-y_{3}}{x_{2}x_{3}}+\frac{y_{3}-y_{1}}{x_{3}x_{1}}+\frac{y_{1}-y_{2}}{x_{1}x_{2}}=0$

Let the line be

$y=mx+c$

$(x_{1},y_{1})$ is on the line

$\Rightarrow y_{1}=mx_{1}+c$

$(x_{2},y_{2})$ is on the line

$\Rightarrow y_{2}=mx_{2}+c$

$(x_{3},y_{3})$ is on the line

$\Rightarrow y_{3}=mx_{3}+c$

LHS

$\displaystyle \Rightarrow \frac{y_{2}-y_{3}}{x_{2}x_{3}}+\frac{y_{3}-y_{1}}{x_{3}x_{1}}+\frac{y_{1}-y_{2}}{x_{1}x_{2}}$

$\displaystyle \Rightarrow \frac{(mx_{2}+c-mx_{3}-c)}{x_{2}x_{3}}+\frac{(mx_{3}+c-mx_{1}-c)}{x_{1}x_{3}}+\frac{(mx_{1}+c-mx_{2}-c)}{(x_{1}x_{2})}$

$\Rightarrow \displaystyle m\left[\frac{(x_{2}-x_{3})}{x_{2}x_{3}}+\frac{(x_{3}-x_{1})}{x_{3}x_{2}}+\frac{(x_1-x_2)}{x_1x_2}\right]$

$\Rightarrow \displaystyle m\left[\frac{(x_{1}x_{2}-x_{1}x_{3})}{x_{1}x_{2}x_{3}}+\frac{(x_{2}x_{3}-x_{2}x_{1})}{x_{1}x_{2}x_{3}} + \dfrac{(x_1x_3-x_2x_3)}{x_1x_2x_3}\right]$

$\Rightarrow m\left(\dfrac{0}{x_{1}x_{2}x_{3}}\right)\Rightarrow 0\Rightarrow RHS$

$\Rightarrow LHS=RHS$

Evaluate $\begin{vmatrix} a & b & c \\ -1 & 1 & -1 \\ 1 & -1 & 1 \end{vmatrix}$ .

$A=\begin{vmatrix} a & b & c \\ -1 & 1 & -1 \\ 1 & -1 & 1 \end{vmatrix}$

$A=a(1-1)-b(-1+1)+c(1-1)$

$A=a\times 0-b\times 0+c \times 0=0$

Find the values of the following determinants
$\begin{vmatrix} 2i & -3i \\ { i }^{ 3 } & -2{ i }^{ 5 } \end{vmatrix}$ where $i=\sqrt{-1}$ .

$\begin{vmatrix} 2i & -3i\\ i^3 & -2i^5\end{vmatrix}$

$=(2i\times 2i^5)-(i^3\times -3i)$

$=-4i^6-(3i^2)$

$=4+3$

$=7$ .

1072120_1160080_ans_42f0620479cb4af9b60031601ffa9b31.png

$A = \begin{bmatrix} 2&4 \\ 6&8 \end{bmatrix}$ find det(A)

The given matrix can be solved as

$A=2\times 8-4\times 6\Rightarrow A=16-24\Rightarrow A=-8.$

Compute the following determinant :

$\begin{vmatrix} { a }^{ 2 } & ab \\ ab & { b }^{ 2 } \end{vmatrix}$

Let A =

$\begin{vmatrix} { a }^{ 2 } & ab \\ ab & { b }^{ 2 } \end{vmatrix}$

$\because$ determinant of a matrix

$\begin{vmatrix} { a }_{11} & a_{12} \\ a_{21} & { a_{22} } \end{vmatrix}$ is

given by

$a_{n} a_{22}-a_{21}a_{12}$

$\Rightarrow |A| = a^{2}b^{2}-(ab)(ab)$

$= a^{2}b^{2}-a^{2}b^{2} = 0$

$\therefore |A| = 0$

1120301_1140153_ans_4098d85f26f142e286933a110486870a.jpg

Show that $\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ { a }^{ 2 } & b^{ 2 } & c^{ 2 } \end{matrix} \right| =(a-b)(b-c)(c-a)$ .

$\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ {a}^{2} & {b}^{2} & {c}^{2} \end{vmatrix} = \left( a - b \right) \left( b - c \right) \left( c - a \right)$

Solving determinant, we have

$\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ {a}^{2} & {b}^{2} & {c}^{2} \end{vmatrix}$

Applying

${C}_{2} \rightarrow {C}_{2} - {C}_{1}$ and

${C}_{3} \rightarrow {C}_{3} - {C}_{1}$

$= \begin{vmatrix} 1 & 0 & 0 \\ a & b - a & c - a \\ {a}^{2} & {b}^{2} - {a}^{2} & {c}^{2} - {a}^{2} \end{vmatrix}$

Taking

$\left( c - a \right)$ and

$\left( b - a \right)$ as common from

${C}_{3}$ and

${C}_{2}$ respectively.

$= \left( b - a \right) \left( c - a \right) \begin{vmatrix} 1 & 0 & 0 \\ a & 1 & 1 \\ {a}^{2} & b + a & c + a \end{vmatrix}$

Expanding the determinant along

${R}_{1}$ , we have

$= \left( b - a \right) \left( c - a \right) \left( 1 \left( c + a - b - a \right) - 0 + 0 \right)$

$= \left( b - a \right) \left( c - a \right) \left( c - b \right)$

$= \left( a - b \right) \left( b - c \right) \left( c - a \right)$

$=$ R.H.S.

Hence proved.A

Compute the following determinant :

$\begin{vmatrix} 3 & 5 \\ -2 & 6 \end{vmatrix}$

Let A =

$\begin{vmatrix} 3 & 5 \\ -2 & 6 \end{vmatrix}$

$[\because \,determinate = a_{4}a_{22}-a_{12}a_{21}]$

$|A|$ = determinant

$= 3\times 6-5\times (-2)$

$= 18+10 = 28$

1120285_1140151_ans_a632e4da51b946398cd488c4cc1e3fb4.jpg

Find value of the determinant
$A = \begin{vmatrix}3 & 2\\ 5 & 7\end{vmatrix}$

$A=\begin{vmatrix} 3 & 2\\ 5 & 7\end{vmatrix} =3(7)-2(5)=21-10=11$ .

Prove that $\left| \begin{matrix} 1 & \omega & { \omega }^{ 2 } \\ \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & { \omega } \end{matrix} \right| =0$

Solving given determinant, we get

$[1(\omega^3-1)-\omega(\omega^2-\omega^2)+\omega^2(\omega-\omega^4)]$ ....... as

$(\omega^3=1)$

$\implies [(1-1)-\omega(0)+\omega^2(\omega-\omega)]$

$\implies 0+0+0$

$\implies 0$

Hence proved

For what value of $p$ are the points $(2,1),(p,-1)$ and $(-1,3)$ collinear.

$^m{A_C}{ = ^m}{A_B}$

$\Rightarrow \frac{{3 - 1}}{{ - 1 - 2}} = \frac{{ - 1 - 1}}{{P - 2}}$

$\Rightarrow \frac{{ - 2}}{3} = \frac{{ - 2}}{{P - 2}}$

$\Rightarrow 3 = P - 2$

$\therefore P = 5$

Hence the value of

$P=5$ .

1199859_1300618_ans_932b914c4a264861b1f3c0840a93f499.JPG

Using determinants show that points $A(a,b+c),B(b,c+a)$ and $C(c,a+b)$ are col-linear.

For the points to be collinear the area of the triangle formed by the points must be zero.

Therefore,

$\begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{vmatrix}$ must be equal to

$0$ .

L.H.S. =

$(a+b+c)\begin{vmatrix} 1 & b+c & 1 \\ 1 & c+a & 1 \\ 1 & a+b & 1 \end{vmatrix}$ =

$0$ . (Since, elements of two columns are equal.)

Hence, the given points are collinear.

Prove that the adjoint of a symmetric matrix is also a symmetric matrix.

1339981_1281791_ans_eaffe2101f484815b86f61504acd2692.jpg

Find deteminant $\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$

Let the matrix be

$\Delta$

$\begin{array}{l} \Delta =\left[ { \begin{array} { *{ 20 }{ c } }1 & 0 \\ 0 & 2 \end{array} } \right] \\ Now, \\ \Delta =\left| { \begin{array} { *{ 20 }{ c } }1 & 0 \\ 0 & 2 \end{array} } \right| \\ \Delta =\left( { 2-0 } \right) \\ Hence, \\ \Delta =2 \end{array}$

Prove that: $2\left| \begin{matrix} 8 & -5 \\ -2 & 6 \end{matrix} \right|=\left| \begin{matrix} 14 & -2 \\ -4 & 6 \end{matrix} \right|$

$2\begin{vmatrix} 8 & -5\\ -2 & 6\end{vmatrix}$

$=2[8\times 6-(-5\times -2)]$

$=2[48-(10)]$

$=2[48-10]$

$=2\times 38$

$=76$

$\begin{vmatrix} 14 & -2\\ -4 & 6\end{vmatrix}$

$=14\times 6-(-2\times -4)$

$=84-(8)$

$=76$

Hence, proved.

1209666_1301476_ans_a0fd931c94e34659b894a6c6fbadc3b9.jpg

In the diagram on a lunar eilpse, if the positions od sun,Earth and moon are shown by $(-4,6), (k,-2)$ and $(5,-6)$ respectively, then find the value of k.

As the sun, earth and moon lie on the same axis. Thus the points

$\left( -4, 6 \right), \left( k, -2 \right)$ and

$\left( 5, -6 \right)$ are collinear.

Therefore,

$\begin{vmatrix} {x}_{1} & {y}_{1} & 1 \\ {x}_{2} & {y}_{2} & 1 \\ {x}_{3} & {y}_{3} & 1 \end{vmatrix} = 0$

$\begin{vmatrix} -4 & 6 & 1 \\ k & -2 & 1 \\ 5 & -6 & 1 \end{vmatrix} = 0$

$-4 \left( -2 - \left( -6 \right) \right) - 6 \left( k - 5 \right) + 1 \left( -6k - \left( -10 \right) \right) = 0$

$-16 - 6k + 30 - 6k + 10 = 0$

$24 - 12k = 0$

$\Rightarrow k = \cfrac{24}{12} = 2$

Hence the value of

$k$ is

$2$ .

If the points $(3,-2), (x,2)$ and $(8,8)$ are collinear find $10x$ using determinant.

$(3, -2), (x, 2), (8, 8)$ are collinear

$\therefore$ Area of triangle formed by point

$= 0$

$\dfrac{1}{2}\begin{vmatrix} 3 & -2 & 1 \\ x & 2 & 1 \\ 8 & 8 & 1 \end{vmatrix}=0$

$3(2-8)+2(x-8)+1(8x-16)=0$

$(3)(-4)+2x-16+8x-16=0$

$-12-16-16+10x=0$

$10x=44$

$x=4.4$

If the value of determinant $\left| \begin{matrix} m & 6 \\ -5 & 4 \end{matrix} \right|$ is $58$ , then find the value of $m$ .

A] Given

$\left| \begin{matrix} m & 6 \\ -5 & 4 \end{matrix} \right|$ is

$58$ ,

$=4m-(-5)\times 6 = 58$

$=4m+30 = 58$

$=4m = 28$

$\therefore m = 7$

1177900_1292289_ans_ad39ab0353b8408487c67d51c6fbe18e.jpg

If $A(x_{1},y_{1}),B(x_{2},y_{2}),C(x_{3},y_{3})$ are vertices of an equilateral triangle whose each side is equal to a, then prove that $\begin{vmatrix} x_{1} & y_{1} & 2 \\ x_{2} & y_{2} & 2 \\ x_{3} & y_{3} & 2 \end{vmatrix}=3a^4$ .

1347066_1279104_ans_61277206d32a405bb18930d392f3ae2c.jpg

Find the value of $k$ if det $\begin{vmatrix} K+3 & 1 & 2 \\ 3 & -2 & 1 \\ -4 & -3 & 3 \end{vmatrix}=0$

$\begin{vmatrix} k+3 & 1 & 2 \\ 3 & -2 & 1 \\ -4 & -3 & 3 \end{vmatrix}=0$

$C_1\rightarrow C_{1}-3C_{3}$

$C_2\rightarrow C_{2}+2C_{3}$

$=\begin{vmatrix} k-3 & 5 & 2 \\ 0 & 0 & 1 \\ -13 & 3 & 3 \end{vmatrix}=0$

Expanding along row

$2$ :

$=(-1)\left[(k-3)(3)-5(-13)\right]=0$

$=(-1)\left[3k-9+65\right]=0$

$=3k+56=0$

$= k=-\dfrac{56}{3}$

Using the method of slope, show that the following points are collinear :
$( i ) A \left( 4,8\right ) , B \left( 5,12 \right) , C \left( 9,28 \right)$
$(ii) A \left( 16 , - 18 \right) , B \left( 3 , - 6 \right) , C \left( - 10,6\right )$

(i) Slope of AB

$=\dfrac{12-8}{5-4}=\dfrac{4}{1}=4$

Slope of BC

$=\dfrac{28-12}{9-5}=\dfrac{16}{4}=4$

Hence AB

$||$ BC, but B is a point of intersect.

Hence A, B, C are collinear.

(ii) Slope of AB

$=\dfrac{-6+18}{3-16}=\dfrac{12}{-13}=-\dfrac{12}{13}$

Slope of BC

$=\dfrac{6+6}{-10-3}=\dfrac{12}{-13}=-\dfrac{12}{13}$

Slope of AB

$=$ slope of BC

Hence A, B, C are collinear.

1216017_1314072_ans_f232307919684892a4f624e35d7f193b.jpg

Show that the points $P(-2,3,5), Q(1,2,3)$ and $R(7,0,-1)$ are collinear.

$\textbf{Step 1: Find the distances between points.}$

$\text{Let point }(-2,3,5),(1,2,3) \text{ and } (7,0,-1)$

$\text{be denoted by P , Q , and R respectively.}$

$\text{Point }P, Q$

$\text{ and }R$

$\text{are collinear if they lie on a line.}$

$PQ = \sqrt {{{\left( {1 + 2} \right)}^2} + {{\left( {2 - 3} \right)}^2} + {{\left( {3 - 5} \right)}^2}}$

$\left[\because\boldsymbol{d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_1} - {z_1}} \right)}^2}} }\right]$

$= \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 2} \right)}^2}}$

$= \sqrt {9 + 1 + 4}$

$= \sqrt {14}$

$QR = \sqrt {{{\left( {7 - 1} \right)}^2} + {{\left( {0 - 2} \right)}^2} + {{\left( { - 1 - 3} \right)}^2}}$

$\left[\because\boldsymbol{d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_1} - {z_1}} \right)}^2}} }\right]$

$= \sqrt {{{\left( 6 \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( { - 4} \right)}^2}}$

$= \sqrt {36 + 4 + 16}$

$= 2\sqrt {14}$

$PR = \sqrt {{{\left( {7 + 2} \right)}^2} + {{\left( {0 - 3} \right)}^2} + {{\left( { - 1 - 5} \right)}^2}}$

$\left[\because\boldsymbol{d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_1} - {z_1}} \right)}^2}} }\right]$

$= \sqrt {{{\left( 9 \right)}^2} + {{\left( { - 3} \right)}^2} + {{\left( { - 6} \right)}^2}}$

$= \sqrt {81 + 9 + 36}$

$= 3\sqrt {14}$

$\textbf{Step 2: Show that points lies on a straight line.}$

$\text{Clearly,}$

$PQ + QR = RP$

$=\sqrt {14} + 2\sqrt {14}$

$= 3\sqrt {14}$

$\textbf{Hence, the point are collinear.}$

Show that the points $P(-2, 3, 5), Q(1, 2, 3)$ and $R(7, 0, -1)$ are collinear.

Let point

$(-2,3,5),(1,2,3) and (7,0,-1)$ be denoted by P , Q , and R respectively.

by using distance formula

$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_1} - {z_1}} \right)}^2}}$

Point

$P, Q$ and

$R$ are collinear if they lie on a line.

$PQ = \sqrt {{{\left( {1 + 2} \right)}^2} + {{\left( {2 - 3} \right)}^2} + {{\left( {3 - 5} \right)}^2}}$

$= \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 2} \right)}^2}}$

$= \sqrt {9 + 1 + 4}$

$= \sqrt {14}$

$QR = \sqrt {{{\left( {7 - 1} \right)}^2} + {{\left( {0 - 2} \right)}^2} + {{\left( { - 1 - 3} \right)}^2}}$

$= \sqrt {{{\left( 6 \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( { - 4} \right)}^2}}$

$= \sqrt {36 + 4 + 16}$

$= 2\sqrt {14}$

$PR = \sqrt {{{\left( {7 + 2} \right)}^2} + {{\left( {0 - 3} \right)}^2} + {{\left( { - 1 - 5} \right)}^2}}$

$= \sqrt {{{\left( 9 \right)}^2} + {{\left( { - 3} \right)}^2} + {{\left( { - 6} \right)}^2}}$

$= \sqrt {81 + 9 + 36}$

$= 3\sqrt {14}$

Hence,

$PQ + QR = RP$

$=\sqrt {14} + 2\sqrt {14} = 3\sqrt {14}$

the point are collinear

Solve:
$\left[ \begin{matrix} 1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & 6 \end{matrix} \right]$

we have

$\left[ { \begin{array} { *{ 20 }{ c } }1 & 4 & 2 \\ 2 & { -1 } & 4 \\ { -3 } & 7 & 6 \end{array} } \right]$

now

$\Delta =\left| { \begin{array} { *{ 20 }{ c } }1 & 4 & 2 \\ 2 & { -1 } & 4 \\ { -3 } & 7 & 6 \end{array} } \right|$

Expanding the given determinant by

$1st$ row, we get

$\begin{array}{l} \Delta =1\left| { \begin{array} { *{ 20 }{ c } }{ -1 } & 4 \\ 7 & 6 \end{array} } \right| -4\left| { \begin{array} { *{ 20 }{ c } }2 & 4 \\ { -3 } & 6 \end{array} } \right| +\left| { \begin{array} { *{ 20 }{ c } }2 & { -1 } \\ { -3 } & 7 \end{array} } \right| \\ =1(-6-28)-4(12+12)+2(14-3) \\ =-34-96+22 \\ =-130+22 \\ =-108 \end{array}$

Hence,

$\Delta = - 108$

If $A= \left[ {\matrix{10 & 0 \cr 4 & 10\cr } } \right]$ , then find $\left| A \right|$ .

Consider the given matrix.

$A= \left[ {\matrix{10 & 0 \cr 4 & 10\cr } } \right]$

$|A|=(10\times 10)-(0\times 4)$

$|A|=100-0$

$|A|=100$

Hence, this is the answer.

Construct a $3\times 2$ matrix whose elements are given by $a_{1}=\dfrac{1}{2}|i-3j|$ .

$a_{11}=\dfrac{1}{2}|-2|=1$ ;

$a_{12}=\dfrac{1}{2}|1-6|=\dfrac{5}{2}$

$a_{21}=\dfrac{1}{2}|2-3|=\dfrac{1}{2}$ ;

$a_{22}=\dfrac{1}{2}|2-6|=2$

$a_{31}=\dfrac{1}{2}|3-3|=0$ ;

$a_{32}=\dfrac{1}{2}|3-6|=\dfrac{3}{2}$

$\Rightarrow M=\begin{bmatrix} 1 & 5/2\\ 1/2 & 2\\ 0 & 3/2\end{bmatrix}$ .

1214807_1396411_ans_90bb807742974308ad95bb9158ff49a3.jpg

Write the value of the determinant $\left| \begin{array}{cc}{3} & {-1} \\ {2} & {1}\end{array}\right|$

Let

$A=\begin{vmatrix}3&-1\\2&1\end{vmatrix}$

$=3(1)-(2)(-1)=3+2=5$

$\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { b }^{ 2 } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| =(a-b)(b-c)(c-a)$

$R_1\rightarrow R_1 - R_3$

$R_2\rightarrow R_2 - R_3$

$\begin{vmatrix} 0 & a-c & a^2-c^2\\ 0 & b-c & b^2-c^2\\ 1 & c & c^2\end{vmatrix} =(a-b)(b-c)$

$\begin{vmatrix} 0 & 1 & a+c\\ 0 & 1 & b+c\\ 1 & c & 0^2\end{vmatrix}$

Expanding along

$C_1$ ,

$(a-c)(b-c)(b+c-a-c)$

$=(a-c)(b-c)(b-a)=(a-b)(b-c)(c-a)$ .

1208003_1392621_ans_c8d76f2c119d47be997c494399dcf0c7.jpg

Solve:
$\left| \begin{matrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{matrix} \right| =0$

$\left| \begin{matrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{matrix} \right| =0$

using matrix propertis

$\Rightarrow o(o+c^{2})-a(-ao+bc)-b[-ac-o\times b]$

$o=-abc+abc+o$

1212311_1363227_ans_09c57ca716f646cd887a1964fb354d54.JPG

Find the value of x for which $|_{ x }^{ 3 }\quad _{ 1 }^{ x }|=|_{ 8 }^{ 3 }\quad _{ 1 }^{ 2 }|$ .

$\left|\begin {matrix} 3&2\\8&1\end{matrix}\right |=3-16=-13$

$\left|\begin {matrix} 3&x\\x&1\end{matrix}\right |=3-x^2$

From Question

$\implies -13=3-x^2\\x^2=13+3\\x^2=16\\x=\pm4$

If the points (-3,6), (-9,a) and (0,15) are collinear, then find a.

Let the given points be A(-3,6), B(-9,a) and C(0,15).

Slope of AB =

$\dfrac{a-6}{-9+3}=\dfrac{6-a}{6}$

Slope of BC =

$\dfrac{a-15}{-9-0}=\dfrac{15-a}{9}$

Since the points A, B and C are collinear.

Slope of AB = Slope of BC

$\dfrac{6-a}{6}=\dfrac{15-a}{9}$

$18-3a=30-2a$

$a=-12$

Convert $\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$ into an identify matrix by suitable row transformations.

$R_2\rightarrow R_2-2R_1$

$\begin{bmatrix} 1 & -1\\ 0 & 5\end{bmatrix}$

$R_2\rightarrow \dfrac{R_2}{5}$

$\begin{bmatrix} 1 & -1\\ 0 & 1\end{bmatrix}$

$R_1\rightarrow R_1+R_2$

$\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix} \rightarrow I$ .

1214504_1396330_ans_ad3c8d927a6f4cfeac42458899349f69.jpg

If $A = \left| \begin{array} { c c } { 2 } & { 3 } \\ { - 1 } & { 2 } \end{array} \right|$ Find $A$ .

$A = \left| \begin{array} { c c } { 2 } & { 3 } \\ { - 1 } & { 2 } \end{array} \right|$

$\implies A=2\times 2-3\times (-1)$

$=4+3=7$

Evaluate $\Delta = \begin{vmatrix} \cos 2\alpha & \sin 2\alpha \\-\sin 3\alpha & \cos 3\alpha \end{vmatrix}$

$\Delta = \begin{vmatrix} \cos 2\alpha & \sin 2\alpha \\-\sin 3\alpha & \cos 3\alpha \end{vmatrix}$

$=\cos{2\alpha}\cos{3\alpha}+\sin{3\alpha}\sin{2\alpha}$

$=\cos{\left(2\alpha-3\alpha\right)}$

$=\cos{\left(-\alpha\right)}$

$=\cos{\alpha}$

The value of $\begin{vmatrix} \cos15^{\circ} & \sin15^{\circ} \\ \sin75^{\circ} & \cos75^{\circ} \end{vmatrix}$ is

$\begin{vmatrix} \cos15^{\circ} & \sin15^{\circ} \\ \sin75^{\circ} & \cos75^{\circ} \end{vmatrix}$

$=\cos 15^{\circ}\cos 75^{\circ}-\sin {75}^{\circ}\sin 15^{\circ}$

$=\cos(75+15)$

$=\cos 90=0$

Find the value of x, if
$\begin{vmatrix} x+2 & x \\ x-4 & x+3 \end{vmatrix}=\begin{vmatrix} 4 & 2 \\ -2 & 5 \end{vmatrix}$

$\begin{vmatrix} x+2 & x \\ x-4 & x+3 \end{vmatrix}=\begin{vmatrix} 4 & 2 \\ -2 & 5 \end{vmatrix}$

Since both have the same order, we can eqaute the corresponding rows and columns respectively.

$x+2=4\Rightarrow\,x=4-2=2$

$x=2$

$\Rightarrow\,x-4=-2\Rightarrow\,x=-2+4=2$

$\Rightarrow x+3=5\Rightarrow\,x=5-3=2$

$\therefore\,x=2$

Evaluate: $\begin{vmatrix} \cos30^{\circ} & \sin30^{\circ} \\ \sin105^{\circ} & \cos105^{\circ} \end{vmatrix}$

$\begin{vmatrix} \cos30^{\circ} & \sin30^{\circ} \\ \sin105^{\circ} & \cos105^{\circ} \end{vmatrix}$

$=\cos{{30}^{\circ}}\cos{{105}^{\circ}}-\sin{{30}^{\circ}}\sin{{105}^{\circ}}$

$=\cos{\left({30}^{\circ}+{105}^{\circ}\right)}$ using

$\cos{A}\cos{B}-\sin{A}\sin{B}=\cos{\left(A+B\right)}$

$=\cos{\left({135}^{\circ}\right)}$

$=\cos{\left({90}^{\circ}+{45}^{\circ}\right)}$

$=-\sin{{45}^{\circ}}=-\dfrac{1}{\sqrt{2}}$

Evaluate the determinant :
$\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}$

$\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}$

Expanding the determinant, we get

$\cos^2\theta-(-\sin ^2\theta)$

$\cos^2\theta+\sin ^2\theta$

$=1$ ..............

$\because \cos^2\theta+\sin ^2\theta=1$

Find the determinant of the matrix $A$
$A=\begin{bmatrix} -1 & 4 \\ 2 & 3 \end{bmatrix}$

$A=\begin{bmatrix} -1 & 4 \\ 2 & 3 \end{bmatrix}$

$\left|A\right|=\begin{vmatrix} -1 & 4 \\ 2 & 3 \end{vmatrix}$

$=-1\times 3-2\times 4$

$=-3-8=-11$

Show that the points $A(-7, 4, -2), B(-2, 1, 0)$ and $C(3, -2, 2)$ are collinear.

Given,

$A(-7, 4, -8), B(-2, 1, 0), C(-3, -2, 2)$

Since,

$\vec {AB}=\vec {OB}-\vec {OA}$

$\vec {AB}=(-2i+j+0k)-(-7i+4j-2k)$

$\vec {AB}=5i-3j+2k$

Since,

$\vec {AC}=\vec {OC}-\vec {OA}$

$\vec {AC}=(3i-2j+2k)-(-7i+4j-2k)$

$\vec {AC}=10i-6j+4k$

$\vec {AC}=2(5i-3j+2k)=2\vec {AB}$

Therefore,

$A, B, C$ are collinear.

Find the coordinates of a point $A$ , where $AB$ is diameter of a circle whose centre is $(2, -3)$ and $B$ is the point $(1, 4)$ .

Center of circle is mid point of Diameter

Let

$A$ be

$(x,y)$

$\implies 2=\dfrac{x+1}{2},-3=\dfrac{y+4}{2}$

$\implies x=3,y=-10$

So, Coordinates of

$A$ are

$(3,-10)$

Evaluate the following determinants :
$\begin{vmatrix} \cos15^{\circ} & \sin15^{\circ} \\ \sin75^{\circ} & \cos75^{\circ} \end{vmatrix}$

$\begin{vmatrix} cos15^{\circ} & sin15^{\circ} \\ sin75^{\circ} & cos75^{\circ} \end{vmatrix}$

$=\begin{vmatrix} cos15^{\circ} & sin15^{\circ} \\ sin\left({90}^{\circ}-{15}^{\circ}\right) & cos\left({90}^{\circ}-{15}^{\circ}\right) \end{vmatrix}$

$=\begin{vmatrix} cos15^{\circ} & sin15^{\circ} \\ cos15^{\circ} & sin15^{\circ} \end{vmatrix}$

$=\cos{{15}^{\circ}}\sin{{15}^{\circ}}-\cos{{15}^{\circ}}\sin{{15}^{\circ}}$

$=0$

Find the values of $x$ , if
$\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}$ = $\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$

$\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}$ =

$\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$

Expanding the determinant, we get

$10-12=5x-6x$

$\Rightarrow x=2$

Are the following pairs of sets equal? Give reasons.
$A=\{x : x$ is a letter of the word "WOLF" $\}$ ,
$B=\{x : x$ is a letter of the word "FOLLOW" $\}$ .

Yes.

We have,

$A=\{x : x$ is a letter of the word "WOLF"

$\}=\{$ W, O, L, F

$\}$

$B=\{x : x$ is a letter of the word "FOLLOW"

$\}=\{$ W, O, L, F

$\}$

Clearly,

$A=B$ .

The value of the determinant $\begin{vmatrix} \cos30^{\circ} & \sin60^{\circ} \\ \sin30^{\circ} & \cos60^{\circ} \end{vmatrix}$ is

Given,

$\begin{vmatrix} \cos30^{\circ} & \sin60^{\circ} \\ \sin30^{\circ} & \cos60^{\circ} \end{vmatrix}$

Expanding the determinant, we get

$\Rightarrow \cos 30^{\circ}\cos 60^{\circ}-\sin {30}^{\circ}\sin 60^{\circ}$

$\Rightarrow \cos(30+60)$

$\Rightarrow \cos 90=0$

Evaluate $\begin{bmatrix} 14 & 9 \\ -8 & -7 \end{bmatrix}$ .

Let

$A= \begin{bmatrix} 14&9\\-8&-7 \end{bmatrix}$

Determinant of A,

$|A|=\begin{vmatrix}14&9\\-8&-7\end{vmatrix}$

$|A|=[(14\times(-7))-((-8)\times9)]$

$|A| =-98+72 =-26$

Find the adjoint of the following matrice:
$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$

Given the matrix is

$\begin{bmatrix} a & b \\ c & d \end{bmatrix} =A$ (Let).

Now in

$A$ co-factors of the elements

$a,b,c,d$ are

$d,-c,-b,a$ respectively.

Then adjoint of

$A$ will be

$\begin{bmatrix} d & -c \\ -b & a \end{bmatrix} ^t$

$= \begin{bmatrix} d & -b \\ -c & a\end{bmatrix}$ .

The value of the determinant $\begin{vmatrix} \cos15^{\circ} & \sin15^{\circ} \\ \sin75^{\circ} & \cos75^{\circ} \end{vmatrix}$ is______

Given,

$\begin{vmatrix} \cos15^{\circ} & \sin15^{\circ} \\ \sin75^{\circ} & \cos75^{\circ} \end{vmatrix}$

Expanding the determinant, we get

$\Rightarrow \cos 15^{\circ}\cos 75^{\circ}-\sin {75}^{\circ}\sin 15^{\circ}$

$\Rightarrow \cos(75+15)$

$\Rightarrow \cos 90=0$

If ${ e }^{ i\theta }=\cos { \theta } +i\sin { \theta }$ , find the value of $-\begin{vmatrix} 1 & { e }^{ i\pi /3 } & { e }^{ i\pi /4 } \\ { e }^{ -i\pi /3 } & 1 & { e }^{ i2\pi /3 } \\ { e }^{ -i\pi /4 } & { e }^{ -i2\pi /3 } & 1 \end{vmatrix}\quad-{2}^{\frac{1}{2}}$

Let

$\Delta = \begin{vmatrix} 1 & { e }^{ i\pi /3 } & { e }^{ i\pi /4 } \\ { e }^{ -i\pi /3 } & 1 & { e }^{ i2\pi /3 } \\ { e }^{ -i\pi /4 } & { e }^{ -i2\pi /3 } & 1 \end{vmatrix}$
Expanding along first column we get,

$\displaystyle \Delta = (1)\begin{vmatrix} 1 & { e }^{ i2\pi /3 } \\ { e }^{ -i2\pi /3 } & 1 \end{vmatrix}-{ e }^{ -i\pi /3 }\begin{vmatrix} { e }^{ i\pi /3 } & { e }^{ i\pi /4 } \\ { e }^{ -i2\pi /3 } & 1 \end{vmatrix}+{ e }^{ -i\pi /4 } \begin{vmatrix} { e }^{ i\pi /3 } & { e }^{ i\pi /4 } \\ 1 & { e }^{ i2\pi /3 } \end{vmatrix}$

$\Rightarrow \Delta = 1(1-1)-{ e }^{ -i\pi /3 }({ e }^{ i\pi /3 }-{ e }^{ -5i\pi /12 })+{ e }^{ -i\pi /4 } ({ e }^{ \pi } -{ e }^{ i\pi /4 } )$

$\Rightarrow \Delta =-2+{ e }^{ 3i\pi /4 }+{ e }^{ -3i\pi /4 }=-2+2\cos { (3\pi /4) }=-2-\sqrt{2}$
Hence required value is 2.

Evaluate $\begin{vmatrix} \cos { \alpha } \cos { \beta } & \cos { \alpha } \sin { \beta } & -\sin { \alpha } \\ -\sin { \beta } & \cos { \beta } & 0 \\ \sin { \alpha } \cos { \beta } & \sin { \alpha } \sin { \beta } & \cos { \alpha } \end{vmatrix}$

$\begin{vmatrix} \cos { \alpha } \cos { \beta } & \cos { \alpha } \sin { \beta } & -\sin { \alpha } \\ -\sin { \beta } & \cos { \beta } & 0 \\ \sin { \alpha } \cos { \beta } & \sin { \alpha } \sin { \beta } & \cos { \alpha } \end{vmatrix}$

$=\cos { \alpha } \cos { \beta } \left( \cos { \alpha } \cos { \beta } \right) +\cos { \alpha } \sin { \beta } \left( \cos { \alpha } \sin { \beta } \right) +\sin { \alpha } \left( \sin { \alpha } \sin ^{ 2 }{ \beta } +\sin { \alpha } \cos ^{ 2 }{ \beta } \right)$

$=\cos ^{ 2 }{ \alpha } \cos ^{ 2 }{ \beta } +\cos ^{ 2 }{ \alpha } \sin ^{ 2 }{ \beta } +\sin ^{ 2 }{ \alpha } \left( 1 \right)$

$=\cos ^{ 2 }{ \left( \alpha \right) } \left[ \cos ^{ 2 }{ \beta } +\sin ^{ 2 }{ \beta } \right] +\sin ^{ 2 }{ \alpha }$

$=\cos ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \alpha }$

$=1$

If $A,B$ and $C$ are the angles of non-right angles triangle $ABC$ , then find the value of $\begin{vmatrix} \tan { A } & 1 & 1 \\ 1 & \tan { B } & 1 \\ 1 & 1 & \tan { C } \end{vmatrix}$

$A+B=\pi -C$

$\tan { (A+B) } =-\tan { C }$

$\displaystyle \frac { \tan { A } +\tan { B } }{ 1-\tan { A } \tan { B } } =-\tan { C }$

$\tan { A } +\tan { B } +\tan { C } =\tan { A } \tan { B } \tan { C }$ ....(1)

Consider,

$\begin{vmatrix} \tan { A } & 1 & 1 \\ 1 & \tan { B } & 1 \\ 1 & 1 & \tan { C } \end{vmatrix}$

$\Delta =\tan { A } \tan { B } \tan { C } -\tan { A } -\tan { C } +1+1-\tan { B }$

$\Delta=2$ (by (1))

If the value of $\begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{vmatrix}$ is $k$ , then find $\dfrac{-k}{13}$ .

$k=\begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{vmatrix}$

$k=1(0)-2(0)+4(-1-12)$

$k=-4(13)$

Now,

$\cfrac { -k }{ 13 } =\cfrac { -(-4)(13) }{ 13 }$

$=4$

$\therefore$ There the correct answer is

$4$

The absolute value of the determinant $\begin{vmatrix} -1 & 2 & 1 \\ 3+2\sqrt { 2 } & 2+2\sqrt { 2 } & 1 \\ 3-2\sqrt { 2 } & 2-2\sqrt { 2 } & 1 \end{vmatrix}$ is $\displaystyle\frac { k }{ \sqrt { 2 } }$ then $k=$

Let

$\displaystyle\Delta =\begin{vmatrix} -1 & 2 & 1 \\ 3+2\sqrt { 2 } & 2+2\sqrt { 2 } & 1 \\ 3-2\sqrt { 2 } & 2-2\sqrt { 2 } & 1 \end{vmatrix}$

Applying

${ C }_{ 1 }\rightarrow { C }_{ 1 }-\left( { C }_{ 2 }+{ C }_{ 3 } \right) ,$ then

$\Delta =\begin{vmatrix} -4 & 2 & 1 \\ 0 & 2+\sqrt { 2 } & 1 \\ 0 & 2-\sqrt { 2 } & 1 \end{vmatrix}=\left( -4 \right) \left( 4\sqrt { 2 } \right) \\ =2\left[ -4\left( 2+2\sqrt { 2 } -2 \right) \right] =-16\sqrt { 2 }$

Hence absolute value

$=16\sqrt{2}$

Therefore

$k=16$

If $\begin{vmatrix} x & x+y & x+y+z \\ 2x & 3x+2y & 4x+3y+2z \\ 3x & 6x=3y & 10x+6y+3z \end{vmatrix}=64$ , then the real value of $x$ is .........

$\triangle =x\begin{vmatrix} 1\quad & x+y\quad & x+y+z \\ 2 & 3x+2y\quad & 4x+3y+2z \\ 3 & 6x+3y & 10x+6y+3z \end{vmatrix}$
Using

$\begin{vmatrix} { C }_{ 3 }\rightarrow & { C }_{ 3 }\quad & -z{ C }_{ 1 } \\ { C }_{ 2 }\rightarrow & { C }_{ 2 } & -y{ C }_{ 1 } \end{vmatrix}$

$=x^{ 2 }\begin{vmatrix} 1\quad & 1 & x+y \\ 2\quad & 3\quad & 4x+3y \\ 3\quad & 2\quad & 10x+6y \end{vmatrix}$

$=x^{ 3 }\begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 3 & 6 & 10 \end{vmatrix}\left[ { C }_{ 3 }{ C }_{ 3 }-y{ C }_{ 2 } \right] \\ ={ x }^{ 3 }\left( 6-8+3 \right) ={ x }^{ 3 }$
Given

$x^{ 3 }=64\Rightarrow x=4,4\omega ,4{ \omega }^{ 2 }$

The value of $\begin{vmatrix} 2{ x }_{ 1 }{ y }_{ 1 } & { x }_{ 1 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 1 } & { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } \\ { x }_{ 1 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 1 } & 2{ x }_{ 2 }{ y }_{ 2 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } \\ { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } & 2{ x }_{ 3 }{ y }_{ 3 } \end{vmatrix}$ is ........

2020779_131233_ans_57753a7fdfc54897851cd33fd50a8e2a.jpg

The constant $k$ is such that the following system of equations posses a non-trivial(i.e., not all zero) solution over the set of rationals $Q$
$x+ky+3z=0,\quad 3x+ky-2z=0\quad 2x+3y-4z=0$
Then $\displaystyle \frac{2k}{11}$ is equal to

For non-trivial solution

$\Delta = \begin{vmatrix} 1&k&3 \\ 3 & k & -2 \\2&3&-4 \end{vmatrix}=0$
Expanding along first column,

$1(-4k+6)-3(-4k-9)+2(-2k-3k)=0\Rightarrow -2k+33=0 \Rightarrow k = 33/2$

If $A + B + C = \pi$ , then $\begin{vmatrix} sin (A + B + C)& sin B & cos C\\ - sin B & 0 & tan A\\ cos (A + B) &- tan A &0 \end{vmatrix} =$ .............

$A+B+C=\pi$

$\Rightarrow \sin { 180 } =0\quad \cos { \left( A+B \right) } =\cos { \left( 180-C \right) } =-\cos { C }$

$\Rightarrow \begin{vmatrix} 0 & \sin { B } & \cos { C } \\ \sin { B } & 0 & \tan { A } \\ -\cos { C } & -\tan { A } & 0 \end{vmatrix}$

The obtained matrix is a skew symmetric

$det=0$

If $(1,2)$ , $\displaystyle \left(\frac{1}{2} , 3 \right)$ and $(0, k)$ are collinear points, find the value of $k$ .

When three points A, B and C are collinear,

slope of line joining any two points (say AB) $=$ slope of line joining any other two

points (say BC).

Slope of the line passing through points $\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and $\left( { x }_{ 2 },{ y }_{ 2 } \right)$ $=$ $\cfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-x_{ 1 } }$

Slope of the line passing through $(1, 2)$ and $( \cfrac {1}{2}, 3) =$ Slope of line passing through $(\cfrac {1}{2}, 3)$ and $(0,k)$

$\cfrac { 3-2 }{ \frac {1}{2} -1 } = \cfrac { k - 3 }{ 0 -\frac {1}{2} }$

$\cfrac { 1}{ - \frac {1}{2} } =\cfrac { k-3 }{ -\frac {1}{2} }$

On solving, $k = 4$

State whether True or False
(-1, 8), (9, -2), (3,4) are collinear points.

When three points

$A, B$ and

$C$ are collinear, slope of line joining any two points (say

$AB$ )

$=$ slope of line joining any other two points (say

$BC$ ).
Slope of the line passing through points

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and

$\left( { x }_{ 2 },{ y }_{ 2 } \right)$

$=$

$\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-x_{ 1 } }$
Slope of the line passing through

$(-1, 8)$ and

$(9, -2) = \dfrac { -2-8 }{ 9-(-1) } =\dfrac { -10 }{ 10 }= -1$ .
Slope of the line passing through

$(9, -2)$ and

$(3,4) = \dfrac { 4-(-2)}{ 3-9 } =\dfrac { 6 }{ -6 }= -1$
Slope of both lines are same. Hence the given points are collinear.

Let
$f\left ( x \right )=\begin{vmatrix} 2\cot x & -1 & 0\\ 1 & \cot x & -1\\ 0 & 1 & 2\cot x \end{vmatrix}$
then

$f\left( x \right) =\begin{vmatrix} 2\cot x & -1 & 0 \\ 1 & \cot x & -1 \\ 0 & 1 & 2\cot x \end{vmatrix}\\ { R }_{ 1 }\rightarrow { R }_{ 1 }+{ R }_{ 3 }\\ f\left( x \right) =\begin{vmatrix} 2\cot x & 0 & 2\cot x \\ 1 & \cot x & -1 \\ 0 & 1 & 2\cot x \end{vmatrix}\\ C_{ 1 }\rightarrow { C }_{ 1 }-{ C }_{ 3 }\\ f\left( x \right) =\begin{vmatrix} 0 & 0 & 2\cot x \\ 2 & \cot x & -1 \\ 2\cot x & 1 & 2\cot x \end{vmatrix}=2\cot x\left( 2-2\cot ^{ 2 }{ x } \right) =4\cot { x } \csc ^{ 2 }{ x }$
A)

$\int _{ \pi /6 }^{ \pi /3 }{ f\left( x \right) dx } =\int _{ \pi /6 }^{ \pi /3 }{ 4\cot { x } \csc ^{ 2 }{ x } dx } =-2{ \left[ \cot ^{ 2 }{ x } \right] }_{ \pi /6 }^{ \pi /3 }=-\cfrac { 16 }{ 3 }$
B)

$f\left( \cfrac { \pi }{ 4 } \right) =4\cot { \cfrac { \pi }{ 4 } } \csc ^{ 2 }{ \cfrac { \pi }{ 4 } } =4.1.2=8$
C)

$f'\left( x \right) =-4\csc ^{ 4 }{ x } -8\cot ^{ 2 }{ x } \csc ^{ 2 }{ x }$

$f'\left( \cfrac { \pi }{ 2 } \right) =-4\csc ^{ 4 }{ \cfrac { \pi }{ 2 } } -8\cot ^{ 2 }{ \cfrac { \pi }{ 2 } } \csc ^{ 2 }{ \cfrac { \pi }{ 2 } } =-4$
D)

$f\left( \cfrac { \pi }{ 2 } \right) =4\cot { \cfrac { \pi }{ 2 } } \csc ^{ 2 }{ \cfrac { \pi }{ 2 } } =0$

Let $f\left ( \theta \right )=\begin{vmatrix} \cos ^{2}\theta & \cos \theta \sin \theta & -\sin \theta \\ \cos \theta \sin \theta & \sin ^{2}\theta & \cos \theta \\ \sin \theta & -\cos \theta & 0 \end{vmatrix}$
find $f\left ( \pi /3 \right )$ .

Given

$f\left( \theta \right) =\begin{vmatrix} \cos ^{ 2 } \theta & \cos \theta \sin \theta & -\sin \theta \\ \cos \theta \sin \theta & \sin ^{ 2 } \theta & \cos \theta \\ \sin \theta & -\cos \theta & 0 \end{vmatrix}$

$\Rightarrow f(\theta)=\cos^{4}\theta+2\sin^2\theta\cos^{2}\theta+\sin^4\theta$

$\Rightarrow f(\theta)=(\cos^2\theta+\sin^2\theta)^2 =1$

$\therefore f(\pi/3)=1$

$A_{3 \times 3}$ is a matrix such that $|A|=a, \:B = (adj \:A)$ such that $|B|= b$ . Find the value of $\dfrac{(ab^2 + a^2b + 1)S}{25}$ where $\displaystyle \frac{1}{2}S=\frac{a}{b}+\frac{a^2}{b^3}+\frac{a^3}{b^5} + \:......... \:up \:to \:\infty$ , and $a = 3$ .

$|B|=|adj \:A|= |A|^2 = 9$

$\displaystyle \frac{S}{2}=\frac{a/b}{1-a/b^2}=\frac{ab}{b ^2-a}=\frac{27}{81-3}=\frac{27}{78}=\frac{9}{26}$

$\therefore (ab^2 + a^2b + 1)S= 225$

Find the value of $x$ for which the points $(x, -1), (2,1)$ and $(4, 5)$ are collinear.

If point

$A(x,-1), B(2,1)$ , and

$C(4,5)$ are collinear, then slope of

$AB$

$\displaystyle =$ slope of

$BC$

$\displaystyle\Rightarrow \frac{1-(-1)}{2-x}= \frac{5-1}{4-2}$

$\displaystyle\Rightarrow \frac{1+1}{2-x}= \frac{4}{2}$

$\displaystyle\Rightarrow \frac{2}{2-x}= 2$

$\displaystyle\Rightarrow 2= 4-2x$

$\displaystyle\Rightarrow 2x= 2$

$\displaystyle \Rightarrow x= 1$

Find the values of $x$ , if
(i) $\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}$ (ii) $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}=\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$

(i)

$\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}$

$\Rightarrow 2-20 =2x^2-24$

$\Rightarrow 2x^2=6$

$\Rightarrow x^2=3$

$\Rightarrow x=\pm \sqrt{3}$

(ii)

$\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}=\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$

$\Rightarrow 10-12=5x-6x$

$\Rightarrow -2=-x$

$\Rightarrow x=2$

Using the properties of determinants, show that:

(i) $\begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}=(a+b+{ c) }^{ 2 }$

(ii) $\begin{vmatrix} x+y+2z & x & y \\ z & y+z+2x & y \\ z & x & z+x+2y \end{vmatrix}=2(x+y+{ z) }^{ 3 }$

(i)

$\begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$

$C_1\rightarrow C_1-C_2, C_2\rightarrow C_2-C_3$

$=\begin{vmatrix} -(a+b+c) & 0 & 2a \\ (a+b+c) & -(a+b+c) & 2b \\ 0 & (a+b+c) & c-a-b \end{vmatrix}$

Taking

$a+b+c$ common from

$C_1, C_2$

$={ (a+b+c) }^{ 2 }\begin{vmatrix} -1 & 0 & 2a \\ 1 & -1 & 2b \\ 0 & 1 & c-a-b \end{vmatrix}$

$R_2\rightarrow R_2+R_1$

$={ (a+b+c) }^{ 2 }\begin{vmatrix} -1 & 0 & 2a \\ 0 & -1 & 2a+2b \\ 0 & 1 & c-a-b \end{vmatrix}$

Expanding along first column,

$={ (a+b+c) }^{ 2 }[-1(-c+a+b-2a-2b)]$

$={ (a+b+c) }^{ 3 }$

(ii)

$\begin{vmatrix} x+y+2z & x & y \\ z & y+z+2x & y \\ z & x & z+x+2y \end{vmatrix}$

$R_1\rightarrow R_1-R_2, R_2\rightarrow R_2-R_3$

$=\begin{vmatrix} x+y+z & -(x+y+z) & 0 \\ 0 & x+y+z & -(x+y+z) \\ z & x & z+x+2y \end{vmatrix}$

Taking

$(x+y+z)$ common from

$R_1, R_2$

$={ (x+y+z) }^{ 2 }\begin{vmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ z & x & z+x+2y \end{vmatrix}$

$C_2 \rightarrow C_2+C_3$

$={ (x+y+z) }^{ 2 }\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ z & x+z & z+x+2y \end{vmatrix}$

Expanding along first row,

$={ (x+y+z) }^{ 2 }\begin{vmatrix} 1 & -1 \\ x+z & x+z+2y \end{vmatrix}$

$={ (x+y+z) }^{ 2 }(x+z+2y+x+z)$

$=2(x+y+{ z) }^{ 3 }$

Using the properties of determinants, show that:

(i) $\begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix}=(5x+4)(4-{ x) }^{ 2 }$

(ii) $\begin{vmatrix} y+k & y & y \\ y & y+k & y \\ y & y & y+k \end{vmatrix}={ k }^{ 2 }(3y+k)$

(i) Consider,

$LHS= \begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix}$

${ C }_{ 1 }\rightarrow { C }_{ 1 }+{ C }_{ 2 }+{ C }_{ 3 }$

$=\begin{vmatrix} 5x+4 & 2x & 2x \\ 5x+4 & x+4 & 2x \\ 5x+4 & 2x & x+4 \end{vmatrix}$

Taking

$(5x+4)$ as a common factor from

$C_1$

$=(5x+4)\begin{vmatrix} 1 & 2x & 2x \\ 1 & x+4 & 2x \\ 1 & 2x & x+4 \end{vmatrix}$

$R_2\rightarrow R_2-R_1, R_3\rightarrow R_3-R_1$

$=(5x+4)\begin{vmatrix} 1 & 2x & 2x \\ 0 & 4-x & 0 \\ 0 & 0 & 4-x \end{vmatrix}$

Taking

$(4-x)$ as common factor from

$R_2$ and

$R_3$

$=(5x+4){ (4-x) }^{ 2 }\begin{vmatrix} 1 & 2x & 2x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$

Expanding along first column, we get

$=(5x+4){ (4-x) }^{ 2 }\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}$

$=(5x+4){ (4-x) }^{ 2 }$

$=RHS$

(ii)

$LHS= \begin{vmatrix} y+k & y & y \\ y & y+k & y\\ y & y & y+k \end{vmatrix}$

${ C }_{ 1 }\rightarrow { C }_{ 1 }+{ C }_{ 2 }+{ C }_{ 3 }$

$LHS= \begin{vmatrix} 3y+k & y & y \\ 3y+k & y+k & y\\ 3y+k & y & y+k \end{vmatrix}$

Taking

$(3y+k)$ as a common factor from

$C_1$

$=(3y+k)\begin{vmatrix} 1 & y & y \\ 1 & y+k & y \\ 1 & y & y+k \end{vmatrix}$

$R_2\rightarrow R_2-R_1, R_3\rightarrow R_3-R_1$

$=(3y+k)\begin{vmatrix} 1 & y & y \\ 0 & k & 0 \\ 0 & 0 & k \end{vmatrix}$

Taking

$k$ as common factor from

$R_2$ and

$R_3$

$=(3y+k)k^2\begin{vmatrix} 1 & 2x & 2x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$

Expanding along first column, we get

$=(3y+k)k^{ 2 }\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}$

$=k^{2}(3y+k)$

$=RHS$

$\therefore \begin{vmatrix} y+k & y & y \\ y & y+k & y \\ y & y & y+k \end{vmatrix}={ k }^{ 2 }(3y+k)$

Find the value of determinant.
(i) $\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}$
(ii) $\begin{vmatrix} { x }^{ 2 }-x+1 & x-1 \\ x+1 & x+1 \end{vmatrix}$

(i)

$\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}$

$=(\cos\theta )(\cos\theta )-(-\sin\theta )(\sin\theta )={ \cos }^{ 2 }\theta +{ \sin }^{ 2 }\theta =1$
(ii)

$\begin{vmatrix} { x }^{ 2 }-x+1 & x-1 \\ x+1 & x+1 \end{vmatrix}$

$=({ x }^{ 2 }-x+1)(x+1)-(x-1)(x+1)$

$={ x }^{ 3 }-{ x }^{ 2 }+x+{ x }^{ 2 }-x+1-({ x }^{ 2 }-1)$

$={ x }^{ 3 }+1-{ x }^{ 2 }+1$

$={ x }^{ 3 }-{ x }^{ 2 }+2$

(i) Find equation of line joining $(1, 2)$ and $(3, 6)$ using determinants
(ii) Find equation of line joining $(3, 1)$ and $(9, 3)$ using determinants.

(i) Let

$P(x,y)$ be any point on the line joining the points

$(1,2)$ and

$(3,6)$

So, these points are collinear.

$\Rightarrow \begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix}=0$

$\Rightarrow x(2-6)-y(1-3)+1(6-6)=0$

$\Rightarrow -4x+2y=0$

So, the equation of line joining the points

$(1,2)$ and

$(3,6)$ is

$-4x+2y=0$

(ii) Let

$P(x,y)$ be any point on the line joining the points

$(3,1)$ and

$(9,3)$

So, these points are collinear.

$\Rightarrow \begin{vmatrix} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{vmatrix}=0$

$\Rightarrow x(1-3)-y(3-9)+1(9-9)=0$

$\Rightarrow -2x+6y=0$

So, the equation of line joining the points

$(3,1)$ and

$(9,3)$ is

$-2x+6y=0$

Find co-factors of the matrix,
$A=\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$

$\displaystyle A=\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$

matrix of minor of

$A$ is

$\displaystyle \begin{bmatrix} \left( 2 \right) \left( 4 \right) -\left( -3 \right) \left( -2 \right) & \left( 0 \right) \left( 4 \right) -\left( -3 \right) \left( 3 \right) & \left( 0 \right) \left( -2 \right) -\left( 2 \right) \left( 3 \right) \\ \left( -1 \right) \left( 4 \right) -\left( 2 \right) \left( -2 \right) & \left( 1 \right) \left( 4 \right) -\left( 2 \right) \left( 3 \right) & \left( 1 \right) \left( -2 \right) -\left( -1 \right) \left( 3 \right) \\ \left( -1 \right) \left( -3 \right) -\left( 2 \right) \left( 2 \right) & \left( 1 \right) \left( -3 \right) -\left( 0 \right) \left( 2 \right) & \left( 1 \right) \left( 2 \right) -\left( 0 \right) \left( -1 \right) \end{bmatrix}$

$\displaystyle =\begin{bmatrix} 2 & 9 & -6 \\ 0 & -2 & 2 \\ -1 & -3 & 2 \end{bmatrix}.$

$\therefore$ Matrix of co-foctar of

$A$ is

$\displaystyle \begin{bmatrix} 2 & -9 & -6 \\ 0 & -2 & -2 \\ -1 & +3 & 2 \end{bmatrix}$

Write minors and cofactors of the elements of following determinants
(i) $\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}$
(ii) $\begin{vmatrix} a & c \\ b & d \end{vmatrix}$

(i)

$\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}$

Minors:

$M_{11}=3$ ;

$M_{12}=0$

$M_{21}=-4$ ;

$M_{22}=2$

Cofactors:

$C_{11}=(-1)^{1+1}\,M_{11}$

$\Rightarrow C_{11}=3$

$C_{12}=(-1)^{1+2}\,M_{12}$

$\Rightarrow C_{12}= 0$

$C_{21}=(-1)^{2+1}\,M_{21}$

$\Rightarrow C_{21}=-(-4)=4$

$C_{22}=(-1)^{2+2}\, M_{22}$

$\Rightarrow C_{22}=2$

(ii)

$\begin{vmatrix} a & c \\ b & d \end{vmatrix}$

Minors:

$M_{11}=d$ ;

$M_{12}=b$

$M_{21}=c$ ;

$M_{22}=a$

Cofactors:

$C_{11}=(-1)^{1+1}\,M_{11}$

$\Rightarrow C_{11}=d$

$C_{12}=(-1)^{1+2}\,M_{12}$

$\Rightarrow C_{12}= -b$

$C_{21}=(-1)^{2+1}\,M_{21}$

$\Rightarrow C_{21}=-c$

$C_{22}=(-1)^{2+2}\, M_{22}$

$\Rightarrow C_{22}=a$

Find the Minors and Cofactors of the elements of the following determinants:
(i) $\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$
(ii) $\begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -2 \\ 0 & 1 & 2 \end{vmatrix}$

(i)

$A =\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$

The elements of the minor of the matrix

$A$ will be given by

$M_{1,1} = \begin{vmatrix} 1&0\\0&1\end{vmatrix} = 1j\\$

$M_{1,2} = \begin{vmatrix} 0&0\\0&1\end{vmatrix} = 0\\$

$M_{1,3} = \begin{vmatrix} 0&1\\0&0\end{vmatrix} = 0\\$

$M_{2,1} = \begin{vmatrix} 0&0\\0&1\end{vmatrix} = 0\\$

$M_{2,2} = \begin{vmatrix} 1&0\\0&1\end{vmatrix} = 1\\$

$M_{2,3} = \begin{vmatrix} 1&0\\0&0\end{vmatrix} = 0\\$

$M_{3,1} = \begin{vmatrix} 0&0\\1&0\end{vmatrix} = 0\\$

$M_{3,2} = \begin{vmatrix} 1&0\\0&0\end{vmatrix} = 0\\$

$M_{3,3} = \begin{vmatrix} 1&0\\0&1\end{vmatrix} = 1$

Hence,

$M = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$

The cofactor matrix will be given by

$C_{i,j} = (-1)^{i+j} M_{i,j}$

$\therefore C = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$

(ii)

$B = \begin{bmatrix}1&0&4\\3&5&-2\\0&1&2 \end{bmatrix}$

$M_{1,1} = \begin{vmatrix}5&-2\\1&2 \end{vmatrix} = 12\\$

$M_{1,2} = \begin{vmatrix}3&-2\\0&2 \end{vmatrix} = 6\\$

$M_{1,3} = \begin{vmatrix}3&5\\0&1 \end{vmatrix} = 3\\$

$M_{2,1} = \begin{vmatrix}0&4\\1&2 \end{vmatrix} = -4\\$

$M_{2,2} = \begin{vmatrix}1&4\\0&2 \end{vmatrix} = 2\\$

$M_{2,3} = \begin{vmatrix}1&0\\0&1 \end{vmatrix} = 1\\$

$M_{3,1} = \begin{vmatrix}0&4\\5&-2 \end{vmatrix} = -20\\$

$M_{3,2} = \begin{vmatrix}1&4\\3&-2 \end{vmatrix} = -14\\$

$M_{3,3} = \begin{vmatrix}1&0\\3&5 \end{vmatrix} = 5$

Hence,

$M = \begin{bmatrix}12&6&3\\-4&2&1\\-20&-14&5 \end{bmatrix}$

The cofactor matrix will be given by

$C_{i,j} = (-1)^{i+j} M_{i,j}$

$\therefore C = \begin{bmatrix} 12&-6&3\\4&2&-1\\-20&14&5\end{bmatrix}$

Find the adjoint of matrix $\ \ \ \ A= \begin{bmatrix} 1 & 1 & 2 \\ 2 & 3 & 5 \\ 2 & 0 & 1 \end{bmatrix}$

Let

$A= \begin{bmatrix} 1 & 1 & 2 \\ 2 & 3 & 5 \\ 2 & 0 & 1 \end{bmatrix}$

We know that

$adj A=C^{T}$

So, we will find out co-factors of each element of

$A$ .

$C_{ 11 }=(-1)^{ 1+1 }\begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix}$

$\Rightarrow C_{11}=3-0=3$

$C_{ 12 }=(-1)^{ 1+2 }\begin{vmatrix} 2 & 5 \\ 2 & 1 \end{vmatrix}$

$\Rightarrow C_{12}=-(2-10)=8$

$C_{ 13 }=(-1)^{ 1+3 }\begin{vmatrix} 2 & 3 \\ 2 & 0 \end{vmatrix}$

$\Rightarrow C_{13}=0-6=-6$

$C_{ 21 }=(-1)^{ 2+1 }\begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix}$

$\Rightarrow C_{21}=-(1-0)=-1$

$C_{ 22 }=(-1)^{ 2+2 }\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}$

$\Rightarrow C_{22}=1-4=-3$

$C_{ 23 }=(-1)^{ 2+3 }\begin{vmatrix} 1 & 1 \\ 2 & 0 \end{vmatrix}$

$\Rightarrow C_{23}=-(0-2)=2$

$C_{ 31 }=(-1)^{ 3+1 }\begin{vmatrix} 1 & 2 \\ 3 & 5 \end{vmatrix}$

$\Rightarrow C_{31}=5-6=-1$

$C_{ 32 }=(-1)^{ 3+2 }\begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix}$

$\Rightarrow C_{32}=-(5-4)=-1$

$C_{ 33 }=(-1)^{ 3+3 }\begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix}$

$\Rightarrow C_{33}=3-2=1$

So, the cofactor matrix

$C=\begin{bmatrix} 3 & 8 & -6 \\ -1 & -3 & 2 \\ -1 & -1 & 1 \end{bmatrix}$

$\Rightarrow { C }^{ T }=\begin{bmatrix} 3 & -1 & -1 \\ 8 & -3 & -1 \\ -6 & 2 & 1 \end{bmatrix}$

Hence,

$adj A=\begin{bmatrix} 3 & -1 & -1 \\ 8 & -3 & -1 \\ -6 & 2 & 1 \end{bmatrix}$

Find co-factors of the matrix,
$A= \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$

Let

$A= \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$

Here,

$|A|=-\cos^2\alpha -\sin^2 \alpha$

$\Rightarrow |A|=-1$ (

$\because \sin^2\alpha+\cos^2\alpha=1$ )

Since,

$|A|\ne 0$

Hence,

$A^{-1}$ exists.

We have

${ A }^{ -1 }=\cfrac { adjA }{ |A| }$

and

$adj A=C^{T}$

So, we will find the co-factors of each element of

$A$ .

$C_{11}=(-1)^{1+1} \begin{vmatrix} \cos \alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha \end{vmatrix}$

$\Rightarrow C_{11}=-(\sin^2\alpha+\cos^2\alpha) =-1$

$C_{12}=(-1)^{1+2} \begin{vmatrix} 0 & \sin\alpha \\ 0 & -\cos\alpha \end{vmatrix}$

$\Rightarrow C_{12}=-(0-0)=0$

$C_{13}=(-1)^{1+3} \begin{vmatrix} 0&\cos \alpha \\ 0 & \sin\alpha \end{vmatrix}$

$\Rightarrow C_{13}=0-0=0$

$C_{21}=(-1)^{2+1} \begin{vmatrix} 0 & 0 \\ \sin\alpha & -\cos\alpha \end{vmatrix}$

$\Rightarrow C_{21}=-(0-0) =0$

$C_{22}=(-1)^{2+2} \begin{vmatrix} 1 & 0 \\ 0 & -\cos\alpha \end{vmatrix}$

$\Rightarrow C_{22}=(-\cos\alpha-0)=-\cos\alpha$

$C_{23}=(-1)^{2+3} \begin{vmatrix} 1 & 0 \\ 0 & \sin\alpha \end{vmatrix}$

$\Rightarrow C_{23}=-(\sin\alpha-0) =-\sin\alpha$

$C_{31}=(-1)^{3+1} \begin{vmatrix} 0 & 0 \\ \cos\alpha & \sin\alpha \end{vmatrix}$

$\Rightarrow C_{31}=(0-0) =0$

$C_{32}=(-1)^{3+2} \begin{vmatrix} 1 & 0 \\ 0 & \sin\alpha \end{vmatrix}$

$\Rightarrow C_{32}=-(\sin\alpha-0) =-\sin\alpha$

$C_{33}=(-1)^{3+3} \begin{vmatrix} 1 & 0 \\ 0& \cos\alpha \end{vmatrix}$

$\Rightarrow C_{33}=(\cos\alpha-0) =\cos \alpha$

Hence, the co-factor matrix of A is

$C= \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix}$

$\Rightarrow adj A=C^T= \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix}$

Now,

${ A }^{ -1 }=\cfrac { adjA }{ |A| } =\dfrac { 1 }{ -1 } \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{bmatrix}$

$\displaystyle \Rightarrow { A }^{ -1 }=\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{bmatrix}$

Using properties of determinants, prove that:
$\begin{vmatrix} \alpha & { \alpha }^{ 2 } & \beta +\gamma \\ \beta & { \beta }^{ 2 } & \gamma +\alpha \\ \gamma & { \gamma }^{ 2 } & \alpha +\beta \end{vmatrix}$ $=(\alpha- \beta)(\beta- \gamma)(\gamma-\alpha)(\alpha +\beta +\gamma )$

Consider,

$\begin{vmatrix} \alpha & { \alpha }^{ 2 } & \beta +\gamma \\ \beta & { \beta }^{ 2 } & \gamma +\alpha \\ \gamma & { \gamma }^{ 2 } & \alpha +\beta \end{vmatrix}$

$C_3\rightarrow C_1+C_3$

$\begin{vmatrix} \alpha & { \alpha }^{ 2 } & \alpha +\beta +\gamma \\ \beta & { \beta }^{ 2 } & \alpha +\beta +\gamma \\ \gamma & { \gamma }^{ 2 } & \alpha +\beta +\gamma \end{vmatrix}$

Taking

$\alpha+\beta+\gamma$ common from

$C_3$

$=(\alpha +\beta +\gamma )\begin{vmatrix} \alpha & { \alpha }^{ 2 } & 1 \\ \beta & { \beta }^{ 2 } & 1 \\ \gamma & { \gamma }^{ 2 } & 1 \end{vmatrix}$

$R_1\rightarrow R_1-R_2 , R_2\rightarrow R_2-R_3$

$=(\alpha +\beta +\gamma )\begin{vmatrix} \alpha -\beta & { \alpha }^{ 2 }-{ \beta }^{ 2 } & 0 \\ \beta -\gamma & { \beta }^{ 2 }-{ \gamma }^{ 2 } & 0 \\ \gamma & { \gamma }^{ 2 } & 1 \end{vmatrix}$

Taking

$\alpha-\beta$ common from

$R_1$ and

$\beta-\gamma$ from

$R_2$

$=(\alpha +\beta +\gamma )(\alpha -\beta )(\beta -\gamma )\begin{vmatrix} 1 & \alpha +\beta & 0 \\ 1 & \beta +\gamma & 0 \\ \gamma & { \gamma }^{ 2 } & 1 \end{vmatrix}$

Expanding along the third column, we get

$=(\alpha +\beta +\gamma )(\alpha -\beta )(\beta -\gamma )[1(\beta +\gamma -\alpha -\beta )]$

$=(\alpha +\beta +\gamma )(\alpha -\beta )(\beta -\gamma )(\gamma -\alpha )$

Evaluate $\begin{vmatrix} \cos\alpha \cos\beta & \cos\alpha \sin\beta & -\sin\alpha \\ -\sin\beta & \cos\beta & 0 \\ \sin\alpha \cos\beta & \sin\alpha \sin\beta & \cos\alpha \end{vmatrix}$

$\begin{vmatrix} \cos\alpha \cos\beta & \cos\alpha \sin\beta & -\sin\alpha \\ -\sin\beta & \cos\beta & 0 \\ \sin\alpha \cos\beta & \sin\alpha \sin\beta & \cos\alpha \end{vmatrix}$

Expanding along the first row, we get

$=\cos\alpha\cos\beta (\cos\alpha\cos\beta-0)-\cos\alpha\sin\beta (-\sin\beta\cos\alpha-0)-\sin\alpha(\sin\alpha\sin^2\beta-\sin\alpha\cos^2\beta)$

$=\cos ^{ 2 }{ \alpha } \cos ^{ 2 }{ \beta } +\cos ^{ 2 }{ \alpha } \sin ^{ 2 }{ \beta } +\sin ^{ 2 }{ \alpha } (\sin ^{ 2 }{ \beta } +\cos ^{ 2 }{ \beta } )$

$=\cos ^{ 2 }{ \alpha } (\cos ^{ 2 }{ \beta } +\sin ^{ 2 }{ \beta } )+\sin ^{ 2 }{ \alpha }$ (Using

$\sin^2 \theta+\cos^2\theta=1$ )

$=\cos ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \alpha }$ (Using

$\sin^2 \theta+\cos^2\theta=1$ )

$=1$ (Using

$\sin^2 \theta+\cos^2\theta=1$ )

If true Enter $'1'$ else $'0'.$
$\begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix}=-2(x+y)(x^2+y^2-xy)$

Let

$\Delta=\begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix}$

Using

$C_1=C+C_2, C_2=C_2+C_3$

$\Delta =\begin{vmatrix} 2(x+y) & y & x+y \\ 2(x+y) & x+y & x \\ 2(x+y) & x & y \end{vmatrix}=2(x+y)\begin{vmatrix} 1 & y & x+y \\ 1 & x+y & x \\ 1 & x & y \end{vmatrix}$

Using

$R_1-R_2, R_2=R_2-R_3$

$\Delta =2(x+y)\begin{vmatrix} 0 & -x & y \\ 0 & y & x-y \\ 1 & x & y \end{vmatrix}$

Now expanding along first column we get,

$\Delta =2(x+y)[-x(x-y)-y^2]=-2(x+y)(x^2+y^2-xy)$

Evaluate $\begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix}$

Let

$\Delta=\begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix}$
Using

$R_1=R_1-R_2, R_2=R_2-R_3$

$\Delta =\begin{vmatrix} 0 & -y & 0 \\ 0 & y & -x \\ 1 & x & x+y \end{vmatrix}$
Now expanding along first column we get,

$\Delta =1[-x(-y)-0]=xy$

Find cofactors of $A=\begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix}$

$A$ is a square matrix, then the minor of the entry in the

$i^{th}$ row and

$j^{th}$ column (also called the

$(i,j)$ minor, or a first minor) is the determinant of the submatrix formed by deleting the

$i^{th}$ row and

$j^{th}$ column. This number is often denoted

$M_{i,j}$ .

Minor is

$\begin{vmatrix} -1&4 &1 \\ -5 & 23 & 11 \\ 3 &-12 &-6 \end{vmatrix}$
Cofactors is
A different type of cofactor, sometimes called a cofactor matrix, is a signed version of a minor

$M_{(ij)}$ defined by

$C_{ij}=(-)^{i+j} M_{ij}$

$\begin{vmatrix} -1&-4 &1 \\ 5 & 23 & -11 \\ 3 &12 &-6 \end{vmatrix}$

If $A_{ij}$ is the cofactor of the element $a_{ij}$ of the determinant $\begin{bmatrix}2&-3&5\\6&0&4\\1&5&-7 \end{bmatrix}$ then write the value of $a_{32}.A_{32}.$

Given a square matrix, the cofactor of $a_{ij}$ is denoted by $A_{ij}=(-1)^{i+j}M_{ij}$ , where $M_{ij}$ is the minor of the entry $a_{ij}$

$M_{ij}=\begin{vmatrix}2& 5\\6& 4\end{vmatrix}$

From the matrix, $a_{32}=5$

Determinant of $A_{32}=(-1)^{3+2}\begin{vmatrix}2& 5\\ 6& 4\end{vmatrix}=-(-8-30)=22$

Therefore, $a_{32}.A_{32}=5\times 22=110$ .

If $\triangle = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}$ , white the cofactor of the element $a_{32}$ .

$\triangle =\begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}$

${ a }_{ 32 }=2$

Minor

$=5(1)-2(8)$

$=-11$

Co factor

$={ \left( -1 \right) }^{ 3+2 }\left( -11 \right)$

$=-1\left( -11 \right) =11$

Answer

$=11$

If $\Delta = \begin{vmatrix} 5& 3 & 8\\ 2 & 0 & 1\\ 1 & 2 & 3\end{vmatrix}$ , write the cofactor of the element $a_{32}$

Cofactor of

$a_{32}=(-1)^{3+2}\begin{vmatrix} 5& 8\\2 &1 \end{vmatrix}=-(5-16)=11$

If $\Delta = \begin{vmatrix}1 & 2 & 3\\ 2 & 0 & 1\\ 5 & 3 & 8\end{vmatrix}$ , write the minor of the elements $a_{22}$

Minor of

$a_{22}=\begin{vmatrix}1&3 \\ 5& 8\end{vmatrix}=8-15=-7$

If $\Delta = \begin{vmatrix} 5 & 3 & 8\\ 2 & 0 &1 \\ 1 & 2 & 3\end{vmatrix}$ , write the minor of the element $a_{23}.$

In linear algebra, a minor of a matrix is the determinant of some smaller square matrix, cut down from the original matrix by removing one or more of its rows or columns.

Since we need the minor of element

$a_{23}$ , the row II and column III would get cut.

The determinant of the remaining square matrix, i.e.

$det\begin{bmatrix} 5 & 3 \\ 1 & 2 \end{bmatrix}$ is the required minor.

$\therefore$ minor of

$a_{23} = 5 \times 2 - 1 \times 3 = 10 - 3 = 7.$

Find the relation between $x$ and $y$ if the points $A(x, y), B(-5, 7)$ and $C(-4, 5)$ are collinear.

Area of triangle

$=\dfrac{1}{2}\left [ x_{1}(y_{2}-y_{1})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{3}) \right ]$

$=\dfrac{1}{2}\left [ x(7-5)+5(5-y)-4(4-7) \right ]$

Given

$A,B$ and

$C$ are collinear, then area of triangle must be zero.

$\therefore \dfrac{1}{2}\left [ x(7-5)+5(5-y)-4(4-7) \right ]=0$

$\Rightarrow \dfrac{1}{2}\left [ 2x-25+5y-4y+25 \right ]=0$

$\Rightarrow \dfrac{1}{2}(2x+y+3)=0$

Then, relation between

$x$ and

$y$ is

$2x+y+3=0$

For how many real values of $'m'$ the points $A (m + 1, 1), B (2m + 1, 3)$ and $C (2m + 2, 2m)$ are collinear.

The points will be collinear if

$\quad \\ \dfrac { 3-1 }{ 2m+1-m-1 } =\dfrac { 2m-3 }{ 2m+2-2m-1 }$

$\dfrac { 2 }{ m } =\dfrac { 2m-3 }{ 1 }$

$2=2{ m }^{ 2 }-3m$

$2{ m }^{ 2 }-3m-2=0$

$m\quad has\quad two\quad possitive\quad real\quad roots$

Show that the adjoint of $A = \begin{bmatrix}-4 & -3 & -3\\ 1 & 0 & 1\\ 4 & 4 & 3\end{bmatrix}$ is $A$ itself

Given

$A = \begin{bmatrix}-4 & -3 & -3\\ 1 & 0 & 1\\ 4 & 4 & 3\end{bmatrix}$
The co-factor matrix of

$A$ is

$[A_{ij}]$

$= \begin{bmatrix}\begin{vmatrix}0 & 1\\ 4 & 3\end{vmatrix} &- \begin{vmatrix} 1& 1\\ 4 & 3\end{vmatrix} &+ \begin{vmatrix}1 & 0\\ 4 & 4\end{vmatrix} \\ -\begin{vmatrix} -3& -3\\ 4 & 3\end{vmatrix} & + \begin{vmatrix} -4& -3\\ 4 & 3\end{vmatrix} & - \begin{vmatrix} -4& -3\\ 4 & 4\end{vmatrix}\\ \begin{vmatrix}-3 & -3\\ 0 & 1\end{vmatrix} & + \begin{vmatrix} -4& -3\\ 1 & 1\end{vmatrix} &- \begin{vmatrix} -4& -3\\ 4 & 4\end{vmatrix} \end{bmatrix}$

$= \begin{bmatrix}-4 & 1 & 4\\ -3 & 0 & 4\\ -3 & 1 & 3\end{bmatrix}$

$\therefore \text{adj}\ A = [A_{ij}]^{T} = \begin{bmatrix}-4 & -3 & -3\\ 1 & 0 & 1\\ 4 & 4 & 3\end{bmatrix} = A$

$\therefore \text{adj}\ A=A$ itself.

Find the value of $k$ if $A (4, 11), B (2, 5), C (6, k)$ are collinear points.

Given: Points

$A (4, 11), B (2, 5)$ and

$C (6, k)$ are collinear.
Value of

$k$ :

$A (4, 11) = (x_1, y_1), B(2, 5) = (x_2, y_2), C (6, k) = (x_3, y_3)$
Slope of line

$AB = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{5 - 11}{2 - (4)} = \dfrac{-6}{-2} = 3$

$\therefore$ Slope of line

$AB = 3$ ........ (1)
Slope of line

$BC = \dfrac{y_3 - y_2}{x_3 - x_2} = \dfrac{k - 5}{6 - 2} = \dfrac{k - 5}{4}$

$\therefore$ Slope of line

$BC = \dfrac{k - 5}{4}$ ......... (2)

$\therefore$ Slope of line

$AB =$ Slope of line

$BC$

$\therefore 3 = \dfrac{k - 5}{4}$ (From (1) and (2))

$\therefore \dfrac{k - 5}{4} = 3$

$\therefore k - 5 = 12$

$k = 12 + 5$

$= 17$

Find the value of $a$ for which the following points $A (a, 3), B(2, 1)$ and $C(5, a)$ are collinear. Hence find the equation of the line.

Given:

$A(a, 3), B(2, 1)$ and

$C(5, a)$ are collinear.

$\therefore$ Slope of

$AB =$ Slope of

$BC$

$\Rightarrow \cfrac {1 - 3}{2 - a} = \dfrac {a - 1}{5 - 2}$

$\Rightarrow \cfrac {-2}{2 - a} = \dfrac {a - 1}{3}$

$\Rightarrow -6 = (2 - a)(a - 1)$

$\Rightarrow -6 = 2a - 2 - a^{2} + a$

$\Rightarrow a^{2} - 3a - 4 = 0$

$\Rightarrow a^{2} - 4a + a - 4 = 0$

$\Rightarrow (a - 4)(a + 1) = 0$

$a = 4, -1$

For

$a = 4$
Slope of

$BC = \dfrac {a - 1}{5 - 2} = \dfrac {4 - 1}{3} = \dfrac {3}{3} = 1$
Equation of

$BC; (y - 1) = 1(x - 2)$

$\Rightarrow y - 1 = x - 2$

$\Rightarrow x - y = 1$

For

$a = -1$

Slope of

$BC = \dfrac{a-1}{5-2} = \dfrac{-1-1}3 = -\dfrac23$

Equation of

$BC: (y-1) = -\cfrac23 (x-2) \Rightarrow 3y -3 = 4-2x$

$\Rightarrow 2x+3y = 7$

A, B, and C are three collinear points. The coordinates of A and B are (3, 4) and (7, 7) respectively and AC =10 units. Find sum of co-ordinates of C.

A (3,4) and B(7,7) and C(x,y) or C'(x,y) are collinear points.

AB = √[(7-3)²+(7-4)²] = 5

AC = 10, given.

Slope of AC = slope of AB = (7-4)/(7-3) = 3/4

=> (y-4)/(x-3) = 3/4 --- (1)

=> 4 y - 3 x = 7 --- (2)

AC² = 10² = (y - 4)² + (x - 3)²

= [ 3/4 * (x - 3) ]² + (x-3)²

= (x-3)² * 25/16

=> x - 3 = + 8

=> x = +11 or -5

=> y = (7+3x)/4

= 10 or -2

C = (11, 10) and C' = (-5, -2)

Using determinants show that points $A (a, b + c), B(b, c + a)$ and $C (c, a + b)$ are collinear.

Given

$A\left( a,b+c \right) ,\quad B\left( b,c+a \right) ,\quad C\left( c,a+b \right)$

Three points

$\left( { x }_{ 1 },y_{ 1 } \right) ,\left( { x }_{ 2 },y_{ 2 } \right) ,\left( { x }_{ 3 },y_{ 3 } \right)$ are collinear

$\begin{vmatrix} 1 & { x }_{ 1 } & y_{ 1 } \\ 1 & { x }_{ 2 } & y_{ 2 } \\ 1 & { x }_{ 3 } & y_{ 3 } \end{vmatrix}=0$

$\Rightarrow \begin{vmatrix} 1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b \end{vmatrix}$

$\Rightarrow \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ b+c & c+a & a+b \end{vmatrix}$

${ R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 3 }$

$\Rightarrow \begin{vmatrix} 1 & 1 & 1 \\ a+b+c & a+b+c & a+b+c \\ b+c & c+a & a+b \end{vmatrix}$

$\Rightarrow (a+b+c)\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ b+c & c+a & a+b \end{vmatrix}$

${ R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 }$

$\Rightarrow (a+b+c)\begin{vmatrix} 0 & 0 & 0 \\ 1 & 1 & 1 \\ b+c & c+a & a+b \end{vmatrix}=0$

find the value of $k$ for which the points $(k,-1)(2,1)$ and $(4,5)$ are collinear.

Consider the given points.

$(k, -1), (2, 1)$ and

$(4, 5)$

Since, these points are collinear means that the area of triangle must me zero.

So,

$\dfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|=0$

where

$(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are the points

Therefore,

$k(1-5)+2(5+1)+4(-1-1)=0$

$k(-4)+2(6)+4(-2)=0$

$-4k+12-8=0$

$-4k+4=0$

$4k=4$

$k=1$

Hence, this is the answer.

Show that the points $A(-3,3)$ , $B(0,0)$ , $C(3,-3)$ are collinear.

Let the points

$(-3, 3), (0, 0)$ , and

$(3,-3)$ be representing the vertices A, B, and C of the given triangle respectively.
Let

$A = (-3, 3)$ ,

$B = (0, 0)$ and

$C = (3,-3)$
AB= =

$\sqrt { ({x}_{1}-{x}_{2} )^2+({y}_{1}-{y}_{2})^2 }$
=

$\sqrt { (-3-0 )^2+(3-0)^2 }$ =

$\sqrt {18}$
BC =

$\sqrt { (0-3 )^2+(0-(-3))^2 }$ =

$\sqrt {18}$
AC =

$\sqrt { (-3-3 )^2+(3-(-3))^2 }$ =

$\sqrt {36}$
AB + BC is equal to AC Therefore, the points

$(-3, 3), (0, 0)$ , and

$(3,-3)$ are collinear.

If $A=\begin{bmatrix}1&2\\3&4\end{bmatrix},$ find $|2A|$ .

Given that

$A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

We have to find the determinant of matrix

$2A$

We have

$2A=2 \times \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}=\begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix}$

Therefore the value of

$|2A|=\begin{vmatrix} 2 & 4 \\ 6 & 8 \end{vmatrix}=2 \times 8-6 \times 4=16-24=-8$

Hence,

$|2A|=-8$ .

Find the equation of line joining (1, 2) and (3, 6) using determinants.

We have to find the equation of line joining

$(1,2)$ and

$(3,6)$

Let the third point on the line be

$(x,y)$

The area of triangle with vertices

$(x,y),(1,2),(3,6)$ is

$\frac{1}{2}\begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix}$

Since the three points are collinear, the area formed will be zero

$\implies \begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix}=0$

$\implies x(2-6)-y(1-3)+1(6-6)=0$

$\implies -4x+2y=0$

$\implies 2x-y=0$ .

Hence, the equation of line joining

$(1,2)$ and

$(3,6)$ is

$2x-y=0$ .

Calculate the values of the determinants:
$\begin{vmatrix}a & h & g\\ h& b & f\\ g & f & c\end{vmatrix}$ .

$\Delta =\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \end{matrix} \right|$

Expanding along

${R}_{1}$

$\Rightarrow \Delta =a\{ bc-{ f }^{ 2 }\} -h\{ hc-gf\} +g\{ hf-bg\} \\ \Rightarrow \Delta =abc-a{ f }^{ 2 }-c{ h }^{ 2 }+fgh+fgh-b{ g }^{ 2 }\\ \Rightarrow \Delta =abc+2fgh-a{ f }^{ 2 }-b{ g }^{ 2 }-c{ h }^{ 2 }$

For what values of $m$ will the expression ${y}^{2}+2xy+2x+my-3$ be capable of resolution into two rational factors?

For a expression in the form

$ax^2+by^2+2hxy+2gx+2fy+c=0$ to have two rational factors, then

$\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}=0$

$\Rightarrow \begin{vmatrix} 0 & 1 & 1 \\ 1 & 1 & m \\ 1 & m & -3 \end{vmatrix}=0$

$\Rightarrow-1(-3-m)+1(m-1)=0$

$\Rightarrow m+3+m-1=0$

$\Rightarrow m = -1$

Calculate the values of the determinants:
$\begin{vmatrix}1 & 1 & 1\\ 1& 1 + x & 1\\ 1 & 1 & 1 + y\end{vmatrix}$ .

$\Delta =\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{matrix} \right| \\ { R }_{ 1 }={ R }_{ 1 }-{ R }_{ 2 }\\ \Rightarrow \Delta =\left| \begin{matrix} 0 & -x & 0 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{matrix} \right|$

Expanding along

${R}_{1}$

$\Rightarrow \Delta =0-(-x)\{ 1+y-1\} +0\\ \Rightarrow \Delta =xy$

Calculate the values of the determinants:
$\begin{vmatrix}1 & z & -y\\ -z& 1 & x\\ y & -x & 1\end{vmatrix}$ .

$\Delta =\left| \begin{matrix} 1 & z & -y \\ -z & 1 & x \\ y & -x & 1 \end{matrix} \right|$

Expanding along

${R}_{1}$

$\Rightarrow \Delta =1\{ 1+{ x }^{ 2 }\} -z\{ -z-xy\} +(-y)\{ xz-y\} \\ \Rightarrow \Delta =1+{ x }^{ 2 }+{ z }^{ 2 }+xyz-xyz+{ y }^{ 2 }\\ \Rightarrow \Delta =1+{ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }$

Solve for $\lambda$ if $\begin{vmatrix} a^2+\lambda & ab & ac\\ ab & b^2+\lambda & bc\\ ac & bc & c^2+\lambda\end{vmatrix}=0$ .

$\begin{vmatrix}a^2+\lambda&ab&ac\\ab&b^2+\lambda&bc\\ac&bc&c^2+\lambda\end{vmatrix}=0$

$\therefore (a^2+\lambda)((b^2+\lambda)(c^2+\lambda)-b^2c^2)-ab(ab(c^2+\lambda)-abc^2)+ac(ab^2c-ac(b^2+\lambda))=0$

$\therefore (a^2+\lambda)(b^2c^2+b^2\lambda+c^2\lambda+\lambda^2-b^2c^2)-ab(abc^2+ab\lambda-abc^2)+ac(ab^2c-ab^2c-ac\lambda))=0$

$\therefore (a^2+\lambda)(b^2\lambda+c^2\lambda+\lambda^2)-a^2b^2\lambda-a^2c^2\lambda=0$

$\therefore a^2(b^2\lambda+c^2\lambda+\lambda^2)+\lambda(b^2\lambda+c^2\lambda+\lambda^2)-a^2(b^2\lambda+c^2\lambda)=0$

$\therefore a^2(\lambda^2)+\lambda(b^2\lambda+c^2\lambda+\lambda^2)=0$

$\therefore \lambda(a^2\lambda+b^2\lambda+c^2\lambda+\lambda^2)=0$

$\therefore \lambda^2(a^2+b^2+c^2+\lambda)=0$

$\therefore \lambda^2=0 \text{ or }a^2+b^2+c^2+\lambda=0$

$\therefore \lambda=0 \text{ or }\lambda=-(a^2+b^2+c^2)$

These are the required values of

$\lambda$ .

Calculate the values of the determinants:
$\begin{vmatrix}b + c & a & a\\ b& c + a & b\\ c & c & a + b\end{vmatrix}$ .

$\Delta =\left| \begin{matrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{matrix} \right| \\ { R }_{ 1 }={ R }_{ 1 }-{ R }_{ 2 }-{ R }_{ 3 }\\ \Rightarrow \Delta =\left| \begin{matrix} 0 & -2c & -2b \\ b & c+a & b \\ c & c & a+b \end{matrix} \right|$

Expanding along

${R}_{1}$

$\Rightarrow \Delta =0-(-2c)\{ b(a+b)-bc)\} +(-2b)\{ bc-c(c+a)\} \\ \Rightarrow \Delta =2bc(a+b-c)-2bc(b-c-a)\\ \Rightarrow \Delta =2bc(a+b-c-b+c+a)\\ \Rightarrow \Delta =4abc$

Prove that $\begin{vmatrix} 1+a & 1 & 1\\ 1& 1+b & 1\\ 1& 1& 1+c\end{vmatrix} =abc\displaystyle (1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ .

$\begin{vmatrix} 1+a & 1 & 1\\ 1& 1+b & 1\\ 1& 1& 1+c\end{vmatrix}$

$=(1+a)[(1+b)(1+c)-1]-(1+c-1)+(-1-b+1)$

$=(1+a)[1+b+c+bc-1]-c-b$

$=(1+a)[b+c+bc]-c-b$

$=b+c+bc+ab+ac+abc-c-b$

$=bc+ab+ac+abc$

$=abc\left(1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)$

Hence, proved.

Find the area of $\triangle PQR$ whose vertices are $P(2,1),Q(3,4)$ and $R(5,2)$ .

Area of $\Delta =\displaystyle\frac{1}{2}\begin{vmatrix} 1 & 1 & 1\\ 2 & 3 & 5\\ 1 & 4 & 2\end{vmatrix}$

$=\displaystyle\frac{1}{2}\left[(6-10)+(5-4)+(8-3)\right]$

$=\displaystyle\frac{1}{2}\left[-4+1+5\right]=1$ sq.units.

Find the equation of the line joining $A(1,3)$ and $B(0,0)$ using determine and find $k$ of $D(k,0)$ is a point such that area of $\triangle$ $ABD$ is $3$ sq. units.

$A\left( 1,3 \right) B\left( 0,0 \right) D\left( k,0 \right)$

$\triangle =\begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & k \\ 3 & 0 & 0 \end{vmatrix}$

$\triangle =1\left( 0-0 \right) -1\left( 0-3k \right) +1\left( 0-0 \right)$

$\triangle =3k$

Area

$=\cfrac { 1 }{ 2 } \left| \triangle \right| =3$

$3k=\pm 6$

$\Rightarrow k=\pm 2$

If points $\left( a,0 \right) ,\left( 0,b \right)$ and $\left( x,y \right)$ are collinear, prove that $\cfrac { x }{ a } +\cfrac { y }{ b } =1$ .

$\textbf{Step 1: Apply condition of three points being collinear }$

$\text{ If three points }$

$(x_1,y_1),\, (x_2,y_2)$

$\text{ and }$

$(x_3,y_3)$

$\text{are collinear, then}$

$\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\ x_3&y_3 &1\end{vmatrix}=0$

$\text{Here, we have }$

$(x_1,y_1)=(x,y), (x_{2},y_{2})=(a,0)$

$\text{ and }$

$(x_{3},y_{3})=(0,b)$

$\therefore \begin{vmatrix}x&y&1\\a&0&1\\0&b&1\end{vmatrix}=0$

$\textbf{Step 2: Simplify and get the required result }$

$\Rightarrow x(0-b)-y(a-0)+1(ab-0)$

$\Rightarrow -bx-ay+ab=0$

$\Rightarrow bx+ay=ab$

$\Rightarrow \dfrac{x}{a}+\dfrac{y}{b}=1$

$\text{ }$

$\text{............[Divide by 'ab'= throughout]}$

$\textbf{Hence Proved }$

If the points $A(-2, 1), B(a, b)$ and $C(4, -1)$ are collinear and $a-b=1$ , find the values of $a$ and $b$ .

If three points

$A,B,C$ are collinear

Slope of AB

$=$ Slope of BC

Slope of AB

$=\dfrac{b-1}{a+2}$

Slope of AC

$=\dfrac{-1-1}{4+2}=-\dfrac{2}{3}$

Since

$A,B,C$ are collinear

$\dfrac{b-1}{a+2}=-\dfrac{2}{3}\Rightarrow 3b-3+2a-4=0$

$3b+2a=7$ .............

$(1)$

Given

$a-b=1\Rightarrow 3a-3b=3$ ...........

$(2)$

Adding both

$(1)$ and

$(2)$ , we get

$3b-3b+3a+2a=7+3$

$5a=10\Rightarrow a=2$

Substituting it in equation

$(2)$ , we get

$2-b=1\Rightarrow b=1$

Hence

$a=2,b=1$ .

Find the ratio in which the line segment joining the points $P(x, 2)$ divides the line segment joining the points $A (12, 5)$ and $B(4,-3)$ . Also find the value of $x$ .

Given that, points

$P,A,B$ are collinear

So slope of

$PA=$ Slope of

$AB$

$\Rightarrow$

$\dfrac{5-2}{12-x}=\dfrac{-3-5}{4-12}$

$\Rightarrow$

$\dfrac{3}{12-x}=\dfrac{-8}{-8}=1$

$\Rightarrow$

$3=12-x\Rightarrow x=9$

$\Rightarrow$ So point

$P(9,2)$

Ratio of

$\dfrac{AP}{PB}=\dfrac{\sqrt{(12-9)^2+(5-2)^2}}{\sqrt{(4-9)^2+(-3-2)^2}}=\dfrac{\sqrt{18}}{\sqrt{50}}=\dfrac{3\sqrt{2}}{5\sqrt{2}}=\dfrac{3}{5}$

The point

$P$ divides AB in the ration

$3:5$ internally

Evaluate :
$\begin{vmatrix} 2& 3 & -5\\ 7 &1 & -2\\ -3 & 4 &1 \end{vmatrix}$

$\begin{vmatrix}2&3&-5\\7&1&-2\\-3&4&1\end{vmatrix}=2(1+8)-3(7-6)-5(28+3)=18-3-155=-140$ .

If three points $\left( { x }_{ 1 },{ y }_{ 1 } \right) ,\left( { x }_{ 2 },{ y }_{ 2 } \right) ,\left( { x }_{ 3 },{ y }_{ 3 } \right)$ lie on the same line, prove that
$\cfrac { { y }_{ 2 }-{ y }_{ 3 } }{ { x }_{ 2 }{ x }_{ 3 } } +\cfrac { { y }_{ 3 }-{ y }_{ 1 } }{ { x }_{ 3 }{ x }_{ 1 } } +\cfrac { { y }_{ 1 }-{ y }_{ 2 } }{ { x }_{ 1 }{ x }_{ 2 } } =0$

Points

$\left( { x }_{ 1 },{ y }_{ 1 } \right) ,\left( { x }_{ 2 },{ y }_{ 2 } \right) ,\left( { x }_{ 3 },{ y }_{ 3 } \right)$ are collinear

$\left( { x }_{ 1 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } \right) -\left( { x }_{ 2 }{ y }_{ 1 }+{ x }_{ 3 }{ y }_{ 2 }+{ x }_{ 1 }{ y }_{ 1 } \right) =0$

$\Rightarrow \left( { x }_{ 1 }{ y }_{ 2 }-{ x }_{ 1 }{ y }_{ 3 } \right) +\left( { x }_{ 2 }{ y }_{ 3 }-{ x }_{ 2 }{ y }_{ 1 } \right) +\left( { x }_{ 3 }{ y }_{ 1 }-{ x }_{ 3 }{ y }_{ 2 } \right) =0$

$\Rightarrow { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) =0$

$\Rightarrow \cfrac { { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) }{ { x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 } } +\cfrac { { x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) }{ { x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 } } +\cfrac { { x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) }{ { x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 } } =0$

$\Rightarrow \cfrac { \left( { y }_{ 2 }-{ y }_{ 3 } \right) }{ { x }_{ 2 }{ x }_{ 3 } } +\cfrac { \left( { y }_{ 3 }-{ y }_{ 1 } \right) }{ { { x }_{ 1 }x }_{ 3 } } +\cfrac { \left( { y }_{ 1 }-{ y }_{ 2 } \right) }{ { x }_{ 1 }{ x }_{ 2 } } =0$

Show that the following set of points are collinear.
$(2,5), (4,6)$ and $(8,8)$

Formula:

Area of triangle

$=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

If the area of a triangle is zero, then the points are collinear.

Given,

$A(2,5),B(4,6),C(8,8)$

Area of

$\triangle ABC=\dfrac{1}{2}[2(6-8)+4(8-5)+8(5-6)]$

$=\dfrac{1}{2}[-4+12-8]$

$=0$ sq.units

Hence the given points are collinear

If $a\ne b\ne c$ , prove that the points $\left( a,{ a }^{ 2 } \right) ,\left( b,{ b }^{ 2 } \right) ,\left( c,{ c }^{ 2 } \right)$ can never be collinear.

We know that the area of a triangle 0 when

$a=b=c,$

Let Δ be the area of the triangle formed by the points

$\Delta = \dfrac{1}{2}\mid y_1(x_2-x_3)+y_2(x_3-x_1)+y_3(x_1-x_2)\mid$

Here, $x_1=a,y_1=a^2, \ x_2=b,y_2=b^2, \ x_3=c, y_3=c^2$

$\therefore \Delta = \cfrac { 1 }{ 2 } \left| (ab^{ 2 }+bc^{ 2 }+ca^{ 2 })−(a^{ 2 }b+b^{ 2 }c+c^{ 2 }a) \right| \\ \Delta =\cfrac { 1 }{ 2 } ∣(a^{ 2 }c−a^{ 2 }b)+(ab^{ 2 }−ac^{ 2 })+(bc^{ 2 }−b^{ 2 }c)∣\\ \Delta =\cfrac { 1 }{ 2 } ∣−a^{ 2 }(b−c)+a(b^{ 2 }−c^{ 2 })−bc(b−c)∣\\ \Delta =\cfrac { 1 }{ 2 } \left| \left( b-c \right) \left\{ -a^{ 2 }+a(b+c)+bc \right\} \right| \\ \Delta =\cfrac { 1 }{ 2 } \left| \left( b-c \right) \left\{ -a^{ 2 }+ab+ac+bc \right\} \right|$

Hence, it is given that the area of triangle 0 when

$a=b=c,$ but

$a\neq b\neq c\quad$ so the area of triangle can't be zero. That's why all the given three points can never be collinear.

Find the value of $k$ if points $(k,3),(6,-2)$ and $(-3,4)$ are collinear.

Given points are

$A(k,3)\quad B(6,-2)\quad C(-3,4)$
Three points are said to be collinear if their slopes are equal

Collinearity:
Slope of

$AB=$ Slope of

$BC.$

$\cfrac { -2-3 }{ 6-k } =\cfrac { 4-(-2) }{ -3-6 } \\ \cfrac { -5 }{ 6-k } =\cfrac { 6 }{ -9 } \\ 36-6k=45\\ k=-\cfrac { 9 }{ 6 } \\ k=-\cfrac { 3 }{ 2 }$

For what value of $a$ the point $(a,1),(1,-1)$ and $(11,4)$ are collinear?

For what the value of a

$(a,1)\quad (1,-1)\quad and\quad (11,4)$ are collinear.

Slope of

$AB=$ Slop of

$BC$

$\cfrac { -1-1 }{ 1-a } =\cfrac { 4+1 }{ 11-1 } \\ \cfrac { -2 }{ 1-a } =\cfrac { 5 }{ 10 } \\ 1-a=-4\\ a=5\quad$

For this value, the points will be collinear.

Evaluate :
$\Delta=\begin{vmatrix} \cos \alpha \cos \beta& \cos \alpha \sin \beta & -\sin \alpha\\ -\sin \beta &\cos \beta & 0\\ \sin \alpha \cos \beta & \sin \alpha \sin \beta &\cos \alpha \end{vmatrix}$

$\begin{vmatrix}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0\\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{vmatrix}$

[

$R_1'=R_1-(\cot \alpha) R_3]$

$=\begin{vmatrix}0 & 0& -\mbox{cosec} \alpha \\ -\sin \beta & \cos \beta & 0\\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{vmatrix}$

$=(-\mbox{cosec}\alpha).(-\sin \alpha). (\sin^2\beta+\cos^2\beta)$

[Expanding with respect to first row]

$=1$

Prove that the points $\left( ab, \right) ,\left( { a }_{ 1 },{ b }_{ 1 } \right)$ and $\left( a-{ a }_{ 1 },b-{ b }_{ 2 } \right)$ are collinear if $a{ b }_{ 1 }={ a }_{ 1 }b$

Consider the following points

$A(a,b), B(a1,b1), C(a−a1,b−b1)$

Since the given points are collinear, we have

$area(\triangle ABC)=0$

First find the area of

$area(\triangle ABC)$ as follows:

$area(\triangle ABC)=\dfrac{1}{2} |x_1(y_1−y_3)+x1(y_3−y_1)+x_3(y_1−y_2)|$

$=|a(b_1−(b−b_1))+a_1((b−b_1)−b)+(a−a_1)(b−b_1)|$

$= |a(b_1−b+b_1)+a_1(b−b_1−b)+a(b−b_1)−a1(b−b_1)|$

$=|−ab−a_1b_1+ab−ab_1+a_1b+a_1b_1|$

$=|−(ab_1−a_1b)|$

$= (ab_1−a_1b)$

This gives,

$ab_1−a_1b=0$

Prove that the points $(a,0),(0,b)$ and $(1,1)$ are collinear if $\left ( \cfrac { 1 }{ a } +\cfrac { 1 }{ b } =1 \right )$

Formula:

Area of triangle

$=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

If the area of a triangle is zero, then the points are collinear.

Given,

$A(a,0),B(0,b),C(1,1)$

Area of

$\triangle ABC=\dfrac{1}{2}[a(b-1)+0(1-0)+1(0-b)]$

$=\dfrac{1}{2}[ab-a+0-b]$

from given condition, we have,

$\dfrac{1}{a}+\dfrac{1}{b}=1$

$\Rightarrow ab=a+b$

$=(a+b)-a-b=0$ sq.units

Hence the given points are collinear

Prove that the points $(2,3), (-4,-6)$ and $1,3/2)$ do not form a triangle.

Let the point be

$A=(-4,6),\ B=(1, 3/2),\ C=(2,3)$

if we can show that

$AB+BC=AC$

Then it's proved that these

$3$ points can't from a triangle

$AB=\sqrt {(-4 -1)^2 + (-6 -3/2)^2}=\sqrt {25+\dfrac {225}{4}}$

$=\sqrt {\\dfrac {325}{4}}$

$BC=\sqrt {(1-2)^2 +(3/2 -3)^2}= \sqrt {1+9/4}=\sqrt {13/4}$

$AC=\sqrt {(-4 -2)^2 + (-6 -3)^2}=\sqrt {36+81}=\sqrt {117}$

$AB+BC=\sqrt {\dfrac {325}{4}}+\sqrt {13/4}$

$= \left [\left (\sqrt {\dfrac {325}{4}}+\sqrt {13/4}\right)^2\right]^{1/2}$

$=\left (\dfrac {325}{4}+\dfrac {13}{4}+2 \sqrt {\dfrac {325}{4}}.\sqrt {13/4} \right)^{1/2}$

$=\left (\dfrac {338}{4} +2\sqrt {\dfrac {13^2 . 25}{4^2}} \right)^{1/2}$

$=\left (\dfrac {169}{2} + \dfrac {2.13.5}{4} \right)^{1/2}$

$=\left (\dfrac {234}{2}\right)^{1/2}= (117)^{1/2}$

$=\sqrt {117}$

$=AC$ [proved]

Find the value of $\lambda$ if the following equations are consistent
$x + y - 3 = 0$
$(1 + \lambda) x + (2 + \lambda) y - 8 = 0$
$x - (1 + \lambda) y + (2 + \lambda) = 0$

For the equations to be consistent,

$D = 0.$

$\therefore D = \begin{vmatrix} 1 & 1 & -3 \\ 1 + \lambda & 2 + \lambda & -8 \\ 1 & -1 - \lambda & 2 + \lambda \end{vmatrix} = 0$
Here the equations are in two variables x and y. If they are consistent then the values of x and y obtained from first two should satisfy the third and hence D = 0
Apply

$C_2 - C_1, C_3 + 3C_1$

$D = \begin{vmatrix} 1 & 0 & 0 \\ 1 + \lambda & 1 & -5 + 3\lambda \\ 1 & -2 - \lambda & 5 + \lambda \end{vmatrix} = 0$
or

$(5 + \lambda) + (2 + \lambda) (-5 + 3 \lambda) = 0$
or

$5 + \lambda + (-10 + \lambda + 3 \lambda^2) = 0$

$3\lambda^2 + 2 \lambda - 5 = 0$ or

$(\lambda - 1) (3 \lambda + 5) = 0$

$\therefore \lambda = 1, -5/3.$

Find cofactors of the elements of the matrix $A = \begin{bmatrix} -1 & 2 \\ -3 & 4 \end{bmatrix}$

Given

$A=\begin{bmatrix} -1 & 2 \\ -3 & 4 \end{bmatrix}$

Cofactor of

$(-1)$ is

$4$

Cofactor of

$(2)$ is

$3$

Cofactor of

$(-3)$ is

$-2$

Cofactor of

$(4)$ is

$-1$

$\therefore A=\begin{bmatrix} 4 & 3 \\ -2 & -1 \end{bmatrix}$

Find the value of $k$ , if the points $A(8,1),B(3,-4)$ and $C(2,k)$ are collinear.

$A(8,1),\quad B(3,-4)\quad and\quad C(2,k)$

Two points are said to be collinear if their slopes are equal

$\Rightarrow$ Slope of

$AB$

$=$ Slope of

$BC$

We have,

Slope between two points

$=\left (\dfrac{y_2-y_1}{x_2-x_1} \right)$

$\Rightarrow$ Slope of AB

$=$ Slop of BC

$\cfrac { -4-1 }{ 3-8 } =\cfrac { k+4 }{ -1 }$

$\cfrac { -5 }{ -5 } =\cfrac { k+4 }{ -1 }$

$k+4=-1$

$k=-5$

If the points $A(-1,-4),B(b,c)$ and $C(5,-1)$ are collinear and $2b+c=4$ , find the values of $b$ and $c$ .

Given,

$A(-1,-4),\quad B(b,c)\quad and\quad C(5,-1)$

Two points are said to be collinear if their slopes are equal

$\Rightarrow$ Slope of $AB$ $=$ Slope of $BC$

We have,

Slope between two points $=\left (\dfrac{y_2-y_1}{x_2-x_1} \right)$

$\Rightarrow$

$\cfrac { c+4 }{ b+1 } =\cfrac { -1-c }{ 5-b }$

$\Rightarrow$

$(c+4)(5-b)=(-1-c)(b+1)$

As we know,

$2b+c=4$

$\Rightarrow$

$c=4-2b$

Substituting this value, we get

$\Rightarrow$

$(4-2b+4)(5-b)=(-1-4+2b)(b+1)$

$\Rightarrow$

$(8-2b)(5-b)=(2b-5)(b+1)$

$\Rightarrow$

$40-8b-10b+2{ b }^{ 2 }=2{ b }^{ 2 }+2b-5b-5$

$\Rightarrow$

$10-18b=-3b-5$

$\Rightarrow$

$-18b+3b=-5-40$

$\Rightarrow$

$-15b=-45$

$\Rightarrow$

$b=3$ and

$c=-2$

Find the value(s) of $k$ for which the points $(3k-1,k-2),(k,k-7)$ and $(k-1,-k-2)$ are collinear.

Given,

$A(3k-1,k-2),\quad B(k,k-7)\quad and\quad C(k-1,-k-2)$

Two points are said to be collinear if their slopes are equal

$\Rightarrow$ Slope of

$AB$

$=$ Slope of

$BC$

We have,

$\Rightarrow$ Slope between two points $=\left (\dfrac{y_2-y_1}{x_2-x_1} \right)$

$\Rightarrow$

$\cfrac { k-7-k+2 }{ k-3k+1 } =\cfrac { -k-2-k+7 }{ k-1-k }$

$\Rightarrow$

$\cfrac { -5 }{ -2k+1 } =\cfrac { -2k+5 }{ -1 }$

$\Rightarrow$

$(5-2k)(1-2k)=5$

$\Rightarrow$

$5-10k-2k+4{ k }^{ 2 }=5$

$\Rightarrow$

$4{ k }^{ 2 }-12k=0$

$\Rightarrow$

$4k(k-3)=0$

$\Rightarrow$

$k=0,3$

Find the value of $k$ , if the points $A(7,-2),B(5,1)$ and $C(3,2k)$ are collinear.

$A(7,−2),\quad B(5,1)\quad and\quad C(3,2k)$

Two points are said to be collinear if their slopes are equal

$\Rightarrow$ Slope of

$AB$

$=$ Slope of

$BC$

We have,

Slope between two points

$=\left (\dfrac{y_2-y_1}{x_2-x_1} \right)$

$\Rightarrow$

$\cfrac { 1+2 }{ 5-7 } =\cfrac { 2k-1 }{ 3-5 }$

$\Rightarrow$

$\cfrac { 3 }{ -2 } =\cfrac { 2k-1 }{ -2 }$

$\Rightarrow$

$-2(2k-1)=-6$

$\Rightarrow$

$2k-1=3$

$\Rightarrow$

$k=\cfrac { 4 }{ 2 }$

$\Rightarrow$

$k=2$

Let $a > 0,\ d > 0.$ Find the value of the determinant
$\left| \begin{matrix} \dfrac { 1 }{ a } & \dfrac { 1 }{ a\left( a+d \right) } & \dfrac { 1 }{ \left( a+d \right) \left( a+2d \right) } \\ \dfrac { 1 }{ \left( a+d \right) } & \dfrac { 1 }{ \left( a+d \right) \left( a+2d \right) } & \dfrac { 1 }{ \left( a+2d \right) \left( a+3d \right) } \\ \dfrac { 1 }{ \left( a+2d \right) } & \dfrac { 1 }{ \left( a+2d \right) \left( a+3d \right) } & \dfrac { 1 }{ \left( a+3d \right) \left( a+4d \right) } \end{matrix} \right|$

Multiplying

${R}_{1}$ by

$a(a+d)(a+2d)$ to remove fraction and hence divide

$\Delta$ by the same.
Similarly clean fractions in

${R}_{2}$ and

${R}_{3}$
If

$k=a(a+d)(a+2d)(a+d)(a+2d)(a+3d)(a+2d)(a+3d)(a+4d)$
or

$k=a(a+d)(a+2d)^{ 2 }(a+3d)^{ 2 }(a+4d)$
then

$\Delta =\dfrac { 1 }{ k } \left| \begin{matrix} (a+d)(a+2d) & a+2d & a \\ (a+2d)(a+3d) & a+3d & a+d \\ (a+3d)(a+4d) & a+4d & a+2d \end{matrix} \right|$
Apply

${C}_{2}-{C}_{3}$ and take

$2d$ common from

${C}_{2}$

$\Delta =\dfrac { 2d }{ k } \left| \begin{matrix} (a+d)(a+2d) & 1 & a \\ (a+2d)(a+3d) & 1 & a+d \\ (a+3d)(a+4d) & 1 & a+2d \end{matrix} \right|$
Now make two zeros by

${R}_{3}-{R}_{2}$ and

${R}_{2}-{R}_{1}$

$\Delta =\dfrac { 2d }{ k } \left| \begin{matrix} (a+d)(a+2d) & 1 & a \\ (a+2d).2d & 0 & d \\ (a+3d).2d & 0 & d \end{matrix} \right|$
Expanding with

$2nd$ column after taking

$2d$ and

$d$ common

$\Delta =\dfrac { 2d }{ k } \times 2{ d }^{ 2 }\left| \begin{matrix} a+2d & 1 \\ a+3d & 1 \end{matrix} \right| =\dfrac { 4{ d }^{ 4 } }{ k }$

$=\dfrac { { 4d }^{ 4 } }{ a\left( a+d \right) ^{ 2 }\left( a+2d \right) ^{ 2 }\left( a+3d \right) ^{ 2 }\left( a+4d \right) }$

Prove the following :
$\left| \begin{matrix} ax & by & cz \\ { x }^{ 2 } & { y }^{ 2 } & { z }^{ 2 } \\ 1 & 1 & 1 \end{matrix} \right| =\left| \begin{matrix} a & b & c \\ x & y & z \\ yz & zx & xy \end{matrix} \right|$

Multiply

${R}_{3}$ by

$xyz$ and divide by

$xyz$
$$\Delta =\cfrac { 1 }{ xyz } \begin{vmatrix} ax & by & cz \\ { x }^{ 2 } & { y }^{ 2 } & { z }^{ 2 } \\ xyz & xyz & xyz \end
{vmatrix}$$
Now take out

$x, y, z$ from

${ C }_{ 1 },{ C }_{ 2 }$ and

${ C }_{ 3 }$
$$\therefore \Delta =\cfrac { xyz }{ xyz }

$\begin{vmatrix} a & b & c \\ x & y & z \\ yz & zx & xy \end{vmatrix}$ =\begin{vmatrix} a & b & c \\ x
& y & z \\ yz & zx & xy \end{vmatrix}$$

Without expanding prove that the determinant $\left | \begin{array}{111} sinA & CosA & sin(A+\theta) \\ sin B & cos B & sin(B+\theta) \\ sin C & CosC & sin (c+\theta) \\ \end {array} \right |$ =0

Consider

$sin(A+\theta)$ in

$C_3$ .

$sin(A+\theta)sinA cos\theta +cos Asin\theta$
Hence apply

$C_3-cos\theta. -sin\theta.C_2$
Then each element in

$C_3$ is zero.

$\therefore \triangle =0$ .

A, B , C are the points (a, p), (b,q) and (c,r) respectively such that a, b, c are in A. P. and p, q ,r in G.P. If the points are collinear then prove that p = q = r

Since the points are collinear we must have

$\Delta \, = \, 0$

$\therefore \, \left |\begin{matrix}a &p &1 \\ b&q &1 \\ c &r &1\end{matrix} \right | \, = \, 0$

Since a, b, c are in A.P . Apply

$R_1 \, + \, R_3 \, - \, 2R_2$

$\therefore \, \left |\begin{matrix}0 &p+r-2q &0 \\ b&q &1 \\ c &r &1\end{matrix} \right | \, = \, 0$

$\therefore \, (p \, + \, r \, -2q) \, (b \, - \, c) \, = \, 0$

$\therefore \, 2q \, = \, p \, + \, r$

but

$q^2 \, = \, pr \, \, or\, \, \left ( \dfrac{P \, + \, r}{2} \right )^2 \, = \, pr$

$(p \, - \, r)^2 \, = \, 0$

$\therefore \, p \, = \, r$

$\therefore \, 2q \, = \, 2p \, \, or \, \, p\, =\, q \, \, \therefore \, p \, = \, q \, = \, r$

If $x,y,z \in R$ then find Determinant
$\left( {\matrix{ {{{({2^x} + {2^{ - x}})}^2}} & {{{({2^x} - {2^{ - x}})}^2}} & 1 \cr {{{({3^x} + {3^{ - x}})}^2}} & {{{({3^x} - {3^{ - x}})}^2}} & 1 \cr {{{({4^x} + {4^{ - x}})}^2}} & {{{({4^x} - {4^{ - x}})}^2}} & 1 \cr } } \right)$

$\det \begin{pmatrix}\left(2^x+2^{-x}\right)^2&\left(2^x-2^{-x}\right)^2&1\\ \left(3^x+3^{-x}\right)^2&\left(3^x-3^{-x}\right)^2&1\\ \left(4^x+4^{-x}\right)^2&\left(4^x-4^{-x}\right)^2&1\end{pmatrix}$

$=\left(2^x+2^{-x}\right)^2\begin{vmatrix}\left(3^x-3^{-x}\right)^2 & 1\\ \left(4^x-4^{-x}\right)^2 & 1\end{vmatrix}-\left(2^x-2^{-x}\right)^2\begin{vmatrix}\left(3^x+3^{-x}\right)^2 & 1\\ \left(4^x+4^{-x}\right)^2 & 1\end{vmatrix}+1\begin{vmatrix}\left(3^x+3^{-x}\right)^2 &\left(3^x-3^{-x}\right)^2 \\ \left(4^x+4^{-x}\right)^2 & \left(4^x-4^{-x}\right)^2\end{vmatrix}$

$\det \begin{pmatrix}\left(3^x-3^{-x}\right)^2&1\\ \left(4^x-4^{-x}\right)^2&1\end{pmatrix}=9^x+9^{-x}-16^x-16^{-x}$

$\det \begin{pmatrix}\left(3^x+3^{-x}\right)^2&1\\ \left(4^x+4^{-x}\right)^2&1\end{pmatrix}=9^x+9^{-x}-16^x-16^{-x}$

$\det \begin{pmatrix}\left(3^x+3^{-x}\right)^2&\left(3^x-3^{-x}\right)^2\\ \left(4^x+4^{-x}\right)^2&\left(4^x-4^{-x}\right)^2\end{pmatrix}=2^{-4x+2}+2^{4x+2}-4\cdot \:9^x-4\cdot \:9^{-x}$

$=(\left(2^x+2^{-x}\right)^2)(9^x+9^{-x}-16^x-16^{-x})-(-\left(2^x-2^{-x}\right)^2)(9^x+9^{-x}-16^x-16^{-x})+1(2^{-4x+2}+2^{4x+2}-4\cdot \:9^x-4\cdot \:9^{-x})$

$=0$

If A = diag (a, b, c) = $\begin{bmatrix} a & 0 & 0 \\[0.3em] 0 & b & 0 \\[0.3em] 0 & 0 & c \end{bmatrix}$ such that $abc \neq 0$
then $A^{-1} = diag (\frac{1}{a}, \frac{1}{b}, \frac{1}{c}) = \begin{bmatrix} \frac{1}{a} & 0 & 0 \\[0.3em] 0 & \frac{1}{b} & 0 \\[0.3em] 0 & 0 & \frac{1}{c} \end{bmatrix}$

|A| = abc

$\neq$ 0 and hence A is non-singular whose inverse will exit. Now

$A^{-1} = \frac{adj. (A)}{|A|}$
If C be the matrix formed by cofactors of the elements of A then C =

$\begin{bmatrix} bc & 0 & 0 \\[0.3em] 0 & ca & 0 \\[0.3em] 0 & 0 & ab \end{bmatrix}$

$\therefore$ adj. A = Transpose of C =

$\begin{bmatrix} bc & 0 & 0 \\[0.3em] 0 & ca & 0 \\[0.3em] 0 & 0 & ab \end{bmatrix}$

$Rows \rightarrow col \ and \ Col \rightarrow rows$

$\therefore \ \ \ A^{-1} = \frac{1}{|A|} Adj. A = \frac{1}{abc}\begin{bmatrix} bc & 0 & 0 \\[0.3em] 0 & ca & 0 \\[0.3em] 0 & 0 & ab \end{bmatrix}$
=

$\begin{bmatrix} \frac{1}{a} & 0 & 0 \\[0.3em] 0 & \frac{1}{b} & 0 \\[0.3em] 0 & 0 & \frac{1}{c} \end{bmatrix}$

If where r = 1,2,3 be the co-ordinates of the points A, B, C respectively, then prove the following:
The equation of internal bisector of angel A is
$b \, \begin{vmatrix} x & y & 1\\ x_1 & y_1 & 1\\ x_2 & y_2 & 1 \end{vmatrix} \, + \, c \, \begin{vmatrix} x & y & 1\\ x_1 & y_1 & 1\\ x_3 & y_3 & 1 \end{vmatrix} \, = \, 0$
where b = AC and c = AB.

In this case if AD be the internal bisector it will divide the opposite side BC in the ratio of the arms of the angel i.e. AB : AC or c : b

$\therefore \, \, \, \, D \, is \, \left(\dfrac{cx_3 \, + \, bx_2}{c \, + \, b}, \, \dfrac{cy_3 \, + \, by_2}{c \, + \, b}\right)$
Then, as above, on applying the condition of collinearly

$\begin{vmatrix}x & y & 1\\ x_1 & y_1 & 1\\ \dfrac{cx_3 \, + \, bx_2}{c \, + \, b} & \dfrac{cy_3 \, + \, by_2}{c \, + \, b} & 1\end{vmatrix}$
Multiply

$R_3$ by c + b and split into two determinants.

If $A$ is a matrix of order $3 \times 3$ then find $\left| adj\ A \right|$ where $\left| A \right| =2$

$| adj A | = |A|^{n-1}$ where n is order of matrix

hence

$| adj A| = |A|^{3-1}$ =

$2^2 = 4$

Find $a$ if $\begin{vmatrix} i & -2i & -1\\ 3i & i^{3} & -2\\ 1& -3 & -i\end{vmatrix}$ = $ai$ , where $i =\sqrt{-1}$

Given that

$\begin{vmatrix} i & -2i & -1\\ 3i & i^{3} & -2\\ 1& -3 & -i\end{vmatrix}=ai$

Expanding given Determinant

$i(-i^4-6)+2i(-3i^2+2)-1(-9i-i^3)$

$=-i^5-6i-6i^3+4i+9i+i^3$

$=-i-6i+6i+4i+9i-i$

$=11i$

Henece value of

$a=11$

Find the non-zero roots of the equation,
$\Delta = \begin{vmatrix} a & b & ax + b \\ b & c & bx + c \\ ax + b & bx + c & c\end{vmatrix} = 0$

Now,

$\begin{vmatrix} a & b & ax + b \\ b & c & bx + c \\ ax + b & bx + c & c\end{vmatrix} = 0$

[

$C_3'=C_3-(xC_1+C_2)$ ]

or,

$\begin{vmatrix} a & b & 0 \\ b & c & 0 \\ ax + b & bx + c & -ax^2-2bx\end{vmatrix} = 0$

or,

$(ac-b^2)(-ax^2-2bx)=0$

or,

$ax^2+2bx=0$ [Taking

$b^2-ac\ne 0$ ]

or,

$x=0,-\dfrac{2b}{a}$ .

So the non-zero root is

$-\dfrac{2b}{a}$ .

Solve: $\begin{vmatrix} 3x+4 & x+2 & 2x+3 \\ 4x+5 & 2x+3 & 3x+4 \\ 10x+17 & 3x+5 & 5x+8 \end{vmatrix}=0$ . The value of the determinant at $x=-1$ is:

$\begin{vmatrix} 3x+4 &x+2 &2x+3 \\ 4x+5 &2x+3 & 3x+4 \\ 10x+17&3x+5 & 5x+8 \end{vmatrix}=0$

$x=-1\begin{vmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\7 & 2 & 3\end{vmatrix}$

$R_2\rightarrow R_2-R_1=\begin{vmatrix}1&1&1\\0&0&0\\7&2&3\end{vmatrix}$

So,determinant is

$0$

By using properties of determinants prove that $\begin{vmatrix} (y+z)^{2}&xy& zx\\ xy&(x+z)^{2} & yz\\ xz & yz & (x+y)^{2}\end{vmatrix}$ is divisible by $(x+ y + z)^{n}$ . Find $n$ .

$\begin{vmatrix} { \left( y+z \right) }^{ 2 } & xy & zx \\ xy & { \left( x+z \right) }^{ 2 } & yz \\ xz & yz & { \left( x+y \right) }^{ 2 } \end{vmatrix}$

multiply

$R_1, R_2, R_3$ by

$x, y, z$

$\dfrac { 1 }{ xyz } \begin{vmatrix} x{ \left( y+z \right) }^{ 2 } & x^{ 2 }y & zx^{ 2 } \\ xy^{ 2 } & { y\left( x+z \right) }^{ 2 } & y^{ 2 }z \\ xz^{ 2 } & yz^{ 2 } & { z\left( x+y \right) }^{ 2 } \end{vmatrix}=\dfrac { xyz }{ xyz } \begin{vmatrix} { \left( y+z \right) }^{ 2 } & x^{ 2 } & x^{ 2 } \\ y^{ 2 } & { \left( x+z \right) }^{ 2 } & y^{ 2 } \\ z^{ 2 } & z^{ 2 } & { \left( x+y \right) }^{ 2 } \end{vmatrix}$

applying

$C_1\longrightarrow C_1-C_3$ and

$C_2\longrightarrow C_2-C_3$

$=\begin{vmatrix} { \left( y+z \right) }^{ 2 }-x^{ 2 } & 0 & x^{ 2 } \\ 0 & { \left( x+z \right) }^{ 2 }-y^{ 2 } & y^{ 2 } \\ z^{ 2 }-{ \left( x+y \right) }^{ 2 } & z^{ 2 }{ \left( x+y \right) }^{ 2 } & { \left( x+y \right) }^{ 2 } \end{vmatrix}$

$=\begin{vmatrix} { \left( y+z+x \right) \left( y+z-x \right) } & 0 & x^{ 2 } \\ 0 & { \left( x+z+y \right) }\left( x+z-y \right) & y^{ 2 } \\ { \left( z-x-y \right) }{ \left( z+x+y \right) } & { \left( z-x-y \right) }{ \left( z+x+y \right) } & { \left( x+y \right) }^{ 2 } \end{vmatrix}$

Taking

$(x+y+z)$ common from

$C_1$ and

$C_2$ .

$={ \left( x+y+z \right) }^{ 2 }\begin{vmatrix} { \left( y+z-x \right) } & 0 & x^{ 2 } \\ 0 & \left( z+x-y \right) & y^{ 2 } \\ { \left( z-x-y \right) } & { \left( z-x-y \right) } & { \left( x+y \right) }^{ 2 } \end{vmatrix}$

Applying

$R_3\longrightarrow R_3-(R_1+R_2)$

$={ \left( x+y+z \right) }^{ 2 }\begin{vmatrix} { \left( y+z-x \right) } & 0 & x^{ 2 } \\ 0 & \left( x+z-y \right) & y^{ 2 } \\ { -2y } & { -2x } & 2xy \end{vmatrix}$

Applying

$C_1\longrightarrow xC_1$ and

$C_2\longrightarrow yC_2$

$=\dfrac { { \left( x+y+z \right) }^{ 2 } }{ xy } \begin{vmatrix} { \left( y+z-x \right) x } & 0 & x^{ 2 } \\ 0 & \left( x+z-y \right) y & y^{ 2 } \\ { -2xy } & { -2xy } & 2xy \end{vmatrix}$

Applying

$C_1\longrightarrow C_1+C_3, C_2\longrightarrow C_2+C_3$

$=\dfrac { { \left( x+y+z \right) }^{ 2 } }{ xy } \begin{vmatrix} { xy+xz } & x^{ 2 } & x^{ 2 } \\ y^{ 2 } & xy+zy & y^{ 2 } \\ { 0 } & 0 & 2xy \end{vmatrix}$

Taking

$x, y$ and

$2xy$ common from

$R_1, R_2$ and

$R_3$

$=\dfrac { { \left( x+y+z \right) }^{ 2 } }{ xy } \times x\times y\times 2xy\begin{vmatrix} { y+z } & x & x \\ y & x+z & y \\ { 0 } & 0 & 1 \end{vmatrix}$

$=2xy{ \left( x+y+z \right) }^{ 2 }\begin{vmatrix} { y+z } & x & x \\ y & x+z & y \\ { 0 } & 0 & 1 \end{vmatrix}$

Expending along

$R_3\rightarrow$

$=2xy(x+y+z)^2[(y+z)(x+z)-xy)]$

$=2xy(x+y+z)^2[xy+yz+zx+z^2-xy]$

$=2xy(x+y+z)^2[z(x+y+z)]$

$=2xyz(x+y+z)^3$ So,

$n$ is

$3$

Find $n$ if: $\begin{vmatrix} 1+a^{ 2 }-b^{ 2 } & 2ab & -2b \\ 2ab & 1-a^{ 2 }+b^{ 2 } & 2a \\ 2b & -2a & 1-a^2-b^2 \end{vmatrix} = (1 + a^2 + b^2)^n$

find n ?

$\begin{vmatrix}1+a^2-b^2&2ab&-2b\\2ab&1-a^2+b^2&2a\\2b&-2a&1-a^2-b^2\end {vmatrix}=(1+a^2+b^2)^n$

Solve determinant

$C_2\rightarrow C_2+aC_3$

$\begin{vmatrix}1+a^2-b^2&0&-2b\\2ab&1+a^2+b^2&2a\\-2b&-a(1+a^2+b^2)&1-a^2-b^2\end {vmatrix}$

$(1+a^2+b^2)\begin{vmatrix}1+a^2-b^2&0&-2b\\2ab&1&2a\\2b&-a&1-a^2-b^2\end{vmatrix}$

$R_3\rightarrow R_3+aR_2$

$=(1+a^2+b^2)\begin{vmatrix}1+a^2-b^2&0-2b\\2ab&1&2a\\2b+2a^2b&0&1+a^2-b^2\end{vmatrix}$

Expand the determinant with respect to

$C_{22}$

$\Rightarrow (1+a^2+b^2)[(1+a^2-b^2)^2+2b(2b+2a^b)]$

$=(1+a^2+b^2)[(1+a^2)+(b^2)^2-2(1+a^2)b^2+4b^2(1+a^2)]$

$(1+a^2+b^2)[(1+a^2)^2+(b^2)^2+2b^2(1+a^)]$

$(1+a^2+b^2)(1+a^2+b^2)^2$

$(1+a^2+b^2)^3$

Compare both sides

$\boxed{n=3}$

$\begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix}=abc\left(1+ \dfrac { 1 }{ a } +\dfrac { 1 }{ b } +\dfrac { 1 }{ c } \right)$

$\left| { \begin{array} { *{ 20 }{ c } }{ 1+a } & 1 & 1 \\ 1 & { 1+b } & 1 \\ 1 & 1 & { 1+c } \end{array} } \right|$

Multiplying and dividing

$R_1,R_2,R_3$ by

$a,b,c$ respectively, and taking out common

$a,b,c$ from

$R_1,R_2,R_3$ respectively

$=abc\left| { \begin{array} { *{ 20 }{ c } }{ 1+\frac { 1 }{ a } } & { \frac { 1 }{ a } } & { \frac { 1 }{ a } } \\ { \frac { 1 }{ b } } & { 1+\frac { 1 }{ b } } & { \frac { 1 }{ b } } \\ { \frac { 1 }{ c } } & { \frac { 1 }{ c } } & { 1+\frac { 1 }{ c } } \end{array} } \right|$

$R_1 \to R_1+R_2+R_3$

$=abc\left| { \begin{array} { *{ 20 }{ c } }{ 1+\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } } & { 1+\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } } & { 1+\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } } \\ { \frac { 1 }{ b } } & { 1+\frac { 1 }{ b } } & { \frac { 1 }{ b } } \\ { \frac { 1 }{ c } } & { \frac { 1 }{ c } } & { 1+\frac { 1 }{ c } } \end{array} } \right|$

Taking out common

${1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$ from

$R_1$

$=abc\left( { 1+\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } } \right) \left| { \begin{array} { *{ 20 }{ c } }1 & 1 & 1 \\ { \frac { 1 }{ b } } & { 1+\frac { 1 }{ b } } & { \frac { 1 }{ b } } \\ { \frac { 1 }{ c } } & { \frac { 1 }{ c } } & { 1+\frac { 1 }{ c } } \end{array} } \right|$

$C_2 \to C_2-C_1$

$C_3 \to C_3-C_1$

$=abc\left( { 1+\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } } \right) \left| { \begin{array} { *{ 20 }{ c } }1 & 0 & 0 \\ { \frac { 1 }{ b } } & 1 & 0 \\ { \frac { 1 }{ c } } & 0 & 1 \end{array} } \right|$

Expanding along

$R_1$ , We get.

$= abc\left( {1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)$

Evaluate:
$\begin {vmatrix} x^{2} -x +1 & x-1\\ x+1 & x+1\end{vmatrix}$

$\begin {vmatrix} x^{2} -x +1 & x-1\\ x+1 & x+1\end{vmatrix}=(x^{2} -x +1)(x+1)-(x-1)(x+1)=x^3+1-(x^2-1)=x^3-x^2+2$

Prove that $\begin{vmatrix} \left( \beta +\gamma -\alpha -\delta \right) ^{ 4 } & \left( \beta +\gamma -\alpha -\delta \right) ^{ 2 } & 1 \\ \left( \gamma +\alpha -\beta -\delta \right) ^{ 4 } & \left( \gamma +\alpha -\beta -\delta \right) ^{ 2 } & 1 \\ \left( \alpha +\beta -\gamma -\delta \right) ^{ 4 } & \left( \alpha +\beta -\gamma -\delta \right) ^{ 2 } & 1 \end{vmatrix}=-64\left( \alpha -\beta \right) \left( \alpha -\gamma \right) \left( \alpha -\delta \right) \left( \beta -\gamma \right) \left( \beta -\delta \right) \left( \gamma -\delta \right)$

$\begin{vmatrix}(\beta +\gamma -\alpha -\delta )^4 & (\beta +\gamma -\alpha -\delta )^2 &1 \\ (\gamma +\alpha -\beta -\delta )^4 & (\gamma +\alpha -\beta -\delta )^2 &1 \\ (\alpha +\beta -\gamma -\delta )^4 & (\alpha +\beta -\gamma -\delta )^2 &1 \end{vmatrix}$

Let $\beta +\gamma -\alpha -\delta=X,\gamma +\alpha -\beta -\delta=Y,\alpha +\beta -\gamma -\delta =Z$

$\therefore$

$\begin{vmatrix}X^4 &X^2 &1 \\ Y^4 & Y^2& 1\\ Z^4& Z^2&1 \end{vmatrix}$

$=R_1\rightarrow R_1-R_2, R_2\rightarrow R_2-R_3$

$=\begin{vmatrix}X^4-Y^4 &X^2-Y^2 &0 \\ Y^4-Z^4 & Y^2-Z^2& 0\\ Z^4& Z^2&1 \end{vmatrix}$

$=(X^4-Y^4)(Y^2-Z^2)-(X^2-Y^2)(Y^4-Z^4)+0$

$=X^4Y^2-X^4Z^2-Y^6+Y^4Z^2-[X^2Y^4-X^2Z^4-Y^6+Y^2Z^4]$

$=X^4Y^2-X^4Z^2-Y^6+Y^4Z^2-X^2Y^4+X^2Z^4+Y^6-Y^2Z^4$

$=X^2Y^2(X^2-Y^2)+X^2Z^2(Z^2-X^2)+Y^2Z^2(Y^2-Z^2)$

$=(\beta +\gamma -\alpha -\delta)^2(\gamma +\alpha -\beta -\delta)^2[(\beta +\gamma -\alpha -\delta)^2-(\gamma +\alpha -\beta -\delta)^2]+(\beta +\gamma -\alpha -\delta)^2(\alpha +\beta -\gamma -\delta)^2[(\alpha +\beta -\gamma -\delta)^2-(\beta +\gamma -\alpha -\delta)^2]+(\gamma +\alpha -\beta -\delta)^2(\alpha +\beta -\gamma -\delta)^2[(\gamma +\alpha -\beta -\delta)^2-(\alpha +\beta -\gamma -\delta)^2]$

Solving the above equation, we get,

$=-64(\alpha -\beta )(\alpha -\gamma )(\alpha -\delta )(\beta -\gamma )(\beta -\delta )(\gamma -\alpha )$

Minor $m_{33}$ of the determinant $\begin{vmatrix} 2 & 3 & 5 \\ 2 & -1 & 8 \\ 1 & 2 & 4 \end{vmatrix}$ is

$m_{33}=$ minor of

$a_{33}=\begin{vmatrix}-1&8\\2&4\end{vmatrix}=-1*4-8*2=-4-16=-20$

Prove $\begin{vmatrix} x+4 & 2x & 2x\\ 2x & x+4 & 2x \\ 2x & 2x & x+4\end{vmatrix}=(5x+4)(4-x)^2$ .

Given:

$\left | \begin{matrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{matrix} \right |$

Use operation: $R_{1} = R_{1} + R_{2} + R_{3}$

$\left | \begin{matrix} 5x+4 & 5x+4 & 5x+4 \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{matrix} \right |$

Take common $5x + 4$ .

$\left (5x + 4 \right ) \left | \begin{matrix} 1 & 1 & 1 \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{matrix} \right |$

Use operation: $R_{2} = R_{2} - R_{3}$

$\left (5x + 4 \right ) \left | \begin{matrix} 1 & 1 & 1 \\ 0 & 4-x & x-4 \\ 2x & 2x & x+4 \end{matrix} \right |$

Take determinant with respect to first column.

$= 1\left ( 16-x^{2} - 2x^{2}+ 8x \right ) - 2x \left ( 4 -x - x + 4 \right )$

$= 1\left ( 16 - 3x^{2} + 8x \right ) - 2x \left ( 8 -2x \right )$

$= 16 - 3x^{2} + 8x - 16x + 4x^{2}$

$= 16 - 8x + x^{2}$

$= \left ( 4 - x \right )^{2}$

Thus, $\left | \begin{matrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{matrix} \right | = \left (5x + 4 \right ) \left ( 4 - x \right )^{2}$

Find the adjoint of matrix $A =\begin{bmatrix}2&0&1\\3&1&2\\-1&1&2\end{bmatrix}$

$A = \begin{bmatrix} 2 & 0 & 1\\3 & 1 & 2 \\-1 & 1 & 2 \end{bmatrix}$

Adjoint of a matrix is difined as the transpose of the co-factor matrix.

$C$ be the co-factor matrix then,

$C = \begin{bmatrix} C_{11} & C_{12} & C_{13}\\C_{21} & C_{22} & C_{23} \\C_{31} & C_{32} & C_{33} \end{bmatrix}$

$C_{11} = \begin{vmatrix} 1&2\\1&2\end{vmatrix} = 0$

$C_{12} = -\begin{vmatrix} 3&2\\-1&2\end{vmatrix} = -8$

$C_{13} = \begin{vmatrix} 3&1\\-1&1\end{vmatrix} = 4$

$C_{21} = -\begin{vmatrix} 0&1\\1&2\end{vmatrix} = 1$

$C_{22} = \begin{vmatrix} 2&1\\-1&2\end{vmatrix} = 5$

$C_{23} = -\begin{vmatrix} 2&0\\-1&1\end{vmatrix} = -2$

$C_{31} = \begin{vmatrix} 0&1\\1&2\end{vmatrix} = -1$

$C_{32} = -\begin{vmatrix} 2&1\\3&2\end{vmatrix} = -1$

$C_{33} = \begin{vmatrix} 2&0\\3&1\end{vmatrix} = 2$

Now,

$C = \begin{bmatrix} 0 & -8 & 4\\1 & 5 & -2 \\-1 & -1 & 2 \end{bmatrix}$

$C^{T} = \begin{bmatrix} 0 & 1 & -1\\-8 & 5 & -1\\4& -2 & 2 \end{bmatrix}$

$\therefore adjA = \begin{bmatrix} 0 & 1 & -1\\-8 & 5 & -1\\4& -2 & 2 \end{bmatrix}$

Find the adjoint of ${ \left( \dfrac { { 7 }^{ -4 } }{ { 4 }^{ -2 } } \right) }^{ \dfrac { 1 }{ 4 } }\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$

Let

$A={ \left( { \dfrac { { { 7^{ -4 } } } }{ { { 4^{ -2 } } } } } \right) ^{ \dfrac { 1 }{ 4 } } }\left[ { \begin{array} { *{ 20 }{ c } }1 & { -1 } & 2 \\ 3 & 0 & { -2 } \\ 1 & 0 & 3 \end{array} } \right]$

$==\left( { \dfrac { { \dfrac { 1 }{ 7 } } }{ { \dfrac { 1 }{ 2 } } } } \right) \left[ { \begin{array} { *{ 20 }{ c } }1 & { -1 } & 2 \\ 3 & 0 & { -2 } \\ 1 & 0 & 3 \end{array} } \right]$

$=\left( { \dfrac { 2 }{ 7 } } \right) \left[ { \begin{array} { *{ 20 }{ c } }1 & { -1 } & 2 \\ 3 & 0 & { -2 } \\ 1 & 0 & 3 \end{array} } \right]$

$adjA=\dfrac{2}{7}adj\left[ { \begin{array} { *{ 20 }{ c } }1 & { -1 } & 2 \\ 3 & 0 & { -2 } \\ 1 & 0 & 3 \end{array} } \right]$

$=\left( { \dfrac { 2 }{ 7 } } \right) { \left[ { \begin{array} { *{ 20 }{ c } }0 & { -11 } & 0 \\ 3 & 1 & { -1 } \\ 2 & 8 & 3 \end{array} } \right] ^{ \prime } }$

$=\left( { \dfrac { 2 }{ 7 } } \right) \left[ { \begin{array} { *{ 20 }{ c } }0 & 3 & 2 \\ { -11 } & 1 & 8 \\ 0 & { -1 } & 3 \end{array} } \right]$

$=\left[ { \begin{array} { *{ 20 }{ c } }0 & { \dfrac { 6 }{ 7 } } & { \dfrac { 1 }{ 7 } } \\ { \dfrac { { -22 } }{ 7 } } & { \dfrac { 2 }{ 7 } } & { \dfrac { { 16 } }{ 7 } } \\ 0 & { \dfrac { { -2 } }{ 7 } } & { \dfrac { 6 }{ 7 } } \end{array} } \right]$

If $x, y, z$ are non-zero real numbers and $\begin{vmatrix} 1+x & 1 & 1 \\ 1+y & 1+2y & 1 \\ 1+z & 1+z & 1+3z \end{vmatrix}=0$ then $-\left( \dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z } \right)$ is equal to________

$\left| \begin{matrix} 1+x & 1 & 1 \\ 1+y & 1+2y & 1 \\ 1+z & 1+z & 1+3z \end{matrix} \right| \\ dividing\quad { R }_{ 1 },{ R }_{ 2 }\& { R }_{ 3 }\quad by\quad { x }_{ 1 }y\& z\quad repectively\\ =xyz\left| \begin{matrix} 1+\dfrac { 1 }{ x } & \dfrac { 1 }{ x } & \dfrac { 1 }{ x } \\ 1+\dfrac { 1 }{ y } & 2+\dfrac { 1 }{ y } & \dfrac { 1 }{ y } \\ 1+\dfrac { 1 }{ z } & 1+\dfrac { 1 }{ x } & 3+\dfrac { 1 }{ z } \end{matrix} \right| =0\\ =xyz\left| \begin{matrix} 3+\dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z } & 3+\dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z } & 3+\dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z } \\ 1+\dfrac { 1 }{ y } & 2+\dfrac { 1 }{ y } & \dfrac { 1 }{ y } \\ 1+\dfrac { 1 }{ z } & 1+\dfrac { 1 }{ z } & 3+\dfrac { 1 }{ z } \end{matrix} \right| =0\\ Taking\quad out\quad common\quad \left( 3+\dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z } \right) from\quad { R }_{ 1 }\\ =\left( xyz \right) \left( \dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z } +3 \right) \left| \begin{matrix} 1 & 1 & 1 \\ 1+\dfrac { 1 }{ y } & 2+\dfrac { 1 }{ y } & \dfrac { 1 }{ y } \\ 1+\dfrac { 1 }{ z } & 1+\dfrac { 1 }{ z } & 3+\dfrac { 1 }{ z } \end{matrix} \right| =0\\ =\left( xyz \right) \left( \dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z } +3 \right) \left| \begin{matrix} 1 & 0 & 0 \\ 1+\dfrac { 1 }{ y } & 1 & -1 \\ 1+\dfrac { 1 }{ z } & 0 & 2 \end{matrix} \right| =0\\ Expanding\quad along\quad { R }_{ 1 }\\$

$\begin{array}{l} \Rightarrow xy=\left( { \dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z } +3 } \right) \left[ { 2-0 } \right] =0 \\ \Rightarrow 2xy=\left( { \dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z } +3 } \right) =0 \\ \Rightarrow \dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z } +3=0 \\ \Rightarrow -\left( { \dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z } } \right) =3 \end{array}$

Find $x$ if $\begin{bmatrix} 1 & 0 & 0 \\ 2 & 3 & 4 \\ 5 & -6 & x \end{bmatrix}=45$

We have the given matrix,

$\\ \left| \begin{matrix} 1 & 0 & 0 \\ 2 & 3 & 4 \\ 5 & -6 & x \end{matrix} \right| =45\\ For\quad the\quad value\quad of\quad x\quad we\quad can\quad solve\quad the\quad folowing\quad matrix\\ \Rightarrow 1\begin{vmatrix} 3 & 4 \\ -6 & x \end{vmatrix}=45\\ \Rightarrow 3x-24=45\\ \Rightarrow 3x=45+24\\ \Rightarrow 3x=69\\ \Rightarrow x=\dfrac { 69 }{ 3 } \\ \Rightarrow x=23$

Hence

$x=23$

Find the adjoint of $\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$

Let,

$A=\left| \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right|$

Let the minors,

$M_{11}=0,M_{12}=-11,M_{13}=0$

$M_{21}=3,M_{22}=1,M_{23}=-1$

$M_{31}=2,M_{32}=4,M_{33}=4$

Therefore,

$adjA=\left[ \begin{matrix} { M }_{ 11 } & { M }_{ 21 } & { M }_{ 31 } \\ { M }_{ 12 } & { M }_{ 22 } & { M }_{ 32 } \\ { M }_{ 13 } & { M }_{ 23 } & { M }_{ 33 } \end{matrix} \right]$

$=\left[ \begin{matrix} 0 & 3 & 2 \\ -11 & 1 & 4 \\ 0 & -1 & 4 \end{matrix} \right]$

Prove that $\begin{vmatrix}ax & by & cz\\x^{2}& y^{2}& z^{2}\\ 1&1&1 \end{vmatrix} = \begin{vmatrix}a&b&c\\x&y&z\\yz&xz&xy \end{vmatrix}$

$\begin{vmatrix} ax & by & cz\\x^2 & y^2 & z^2 \\1 & 1 & 1 \end{vmatrix}$

$= \dfrac{1}{xyz}\begin{vmatrix} ax & by & cz\\x^2 & y^2 & z^2 \\xyz & xyz & xyz \end{vmatrix}$

$= \dfrac{1}{xyz}.x.y.z\begin{vmatrix} a & b & c\\x & y & z \\yz & zx & xy \end{vmatrix}$

$= \begin{vmatrix} a & b & c\\x & y & z \\yz & zx & xy \end{vmatrix}$

Solve for $x$ :
$\begin{vmatrix}1&5&2\\2&6&4\\3&7&x\end{vmatrix}=0$

Now,

$\begin{vmatrix}1&5&2\\2&6&4\\3&7&x\end{vmatrix}=0$

[

$R_2'=R_2-R_1$ and

$R_3'=R_3-R_2$ ]

or,

$\begin{vmatrix}1&5&2\\1&1&2\\1&1&x-4\end{vmatrix}=0$

[

$R_3'=R_3-R_2$ ]

or,

$\begin{vmatrix}1&5&2\\1&1&2\\0&0&x-6\end{vmatrix}=0$

or,

$(x-6)(1-5)=0$

or,

$x=6$ .

Prove that $\triangle =\begin{vmatrix} a & b & a\alpha +b \\ b & c & b\alpha +c \\ a\alpha +b & b\alpha +c & 0 \end{vmatrix}=0$ if a,b,c are in G.P.

Given

$\triangle =\begin{vmatrix} a & b & a\alpha +b \\ b & c & b\alpha +c \\ a\alpha +b & b\alpha +c & 0 \end{vmatrix}=0$

$\Rightarrow a\left( 0-{ \left( b\alpha +c \right) }^{ 2 } \right) -b\left( 0-\left( a\alpha +b \right) \left( b\alpha ++c \right) \right) +a\alpha +b\left( { b }^{ 2 }\alpha +bc-ac\alpha -bc \right) =0$

$\Rightarrow -a\left( { b }^{ 2 }{ \alpha }^{ 2 }+{ c }^{ 2 }+2bc\alpha \right) +b\left( ab{ \alpha }^{ 2 }+\left( ac+{ b }^{ 2 } \right) \alpha +bc \right) +a{ b }^{ 2 }{ \alpha }^{ 2 }-{ a }^{ 2 }c{ \alpha }^{ 2 }+{ b }^{ 3 }\alpha -abc\alpha =0$

$\Rightarrow \left( abc-{ a }^{ 2 }c \right) { \alpha }^{ 2 }+\left( 2{ b }^{ 3 }-2abc \right) \alpha +{ b }^{ 2 }c-a{ c }^{ 2 }=0$

$\Rightarrow ac\left( b-a \right) { \alpha }^{ 2 }+\left( 2{ b }^{ 3 }-2abc \right) \alpha +{ b }^{ 2 }c-a{ c }^{ 2 }=0$

By substituting

${ b }^{ 2 }=ac$ we get

$=0$

Show that:
$\left| \begin{matrix} 1 & { a }^{ 2 } & { a }^{ 3 } \\ 1 & { b }^{ 2 } & { b }^{ 3 } \\ 1 & { c }^{ 2 } & { c }^{ 3 } \end{matrix} \right| =(a-b)(b-c)(c-a)(ab+bc+ca)$

1333815_1064317_ans_389566e3dbc04713a6083e5057ccd995.jpg

show that $\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & b^{ 2 } \\ 1 & c & c^{ 2 } \end{matrix} \right| =\left( a-b \right) \left( b-c \right) \left( c-a \right)$

1333817_1064379_ans_8d00e59eaf8542a7b753ad96bd295b3e.jpg

Find the solution set of $\left| {\begin{array}{*{20}{c}}x&5&9\\{16}&{3x\, + \,8}&{36}\\3&1&7\end{array}} \right|\, = \,0$ .

$\left| \begin{matrix} x & 5 & 9 \\ 16 & 3x+8 & 36 \\ 3 & 1 & 7 \end{matrix} \right|=0$

$x\left(21x+56-36\right)-5\left(112-108\right)+9\left(16-9x-24\right)=0$

$x\left(21x+20\right)-5\left(4\right)+9\left(-9x-8\right)=0$

$\Rightarrow\,21{x}^{2}+20x-20-81x-72=0$

$\Rightarrow\,21{x}^{2}-61x-92=0$

$\Rightarrow\,21{x}^{2}-84x+23x-92=0$

$\Rightarrow\,21x\left(x-4\right)+23\left(x-4\right)=0$

$\Rightarrow\,\left(x-4\right)\left(21x+23\right)=0$

$\therefore\,x=4,\dfrac{-23}{21}$

Solve:
$\left| {\begin{array}{*{20}{c}}1&1&1\\1&{1 + \sin \theta }&1\\{1 + \cos \theta }&1&1\end{array}} \right|$

We have,

$D=\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 1+\sin\theta & 1 \\ 1+\cos\theta & 1 & 1 \end{matrix} \right|$

${{R}_{1}}\to {{R}_{1}}-{{R}_{3}}$

$D=\left| \begin{matrix} -\cos \theta & 0 & 0 \\ 1 & 1+\sin\theta & 1 \\ 1+\cos\theta & 1 & 1 \end{matrix} \right|$

On expanding with respect to

$R_1$ , we get

$D=-\cos \theta (1 + \sin \theta -1)$

$D=-\cos \theta (\sin \theta )$

$D=- \sin \theta\cos \theta$

Hence, this is the answer.

Solve:
$\begin{vmatrix} 1 & 1 & 2 \\ 1 & -2 & 3 \\ 1 & (x+1) & x \end{vmatrix}=0$

$1(-2x-3x-3)-(4x-2x-2)+(3+4)=0$

$-2-5x-3+x+2+7=0$

$-4x=-6$

$x=\dfrac{3}{2}$

Find the area of quadrilateral ABCD, the co ordinates of whose vertices are $A(-3,2), B(5,4), C(7,-6), D(-5,-4)$

Using the shoelace formula

$A\left( \Box ABCD \right) =\cfrac { 1 }{ 2 } \begin{vmatrix} -3 & 2 \\ 5 & 4 \\ 7 & -6 \\ -5 & -4 \end{vmatrix}$

$=\cfrac { 1 }{ 2 } \left| \left[ -3\times 4+5\times -6+7\times -4+2\times -5-5\times 2-\left( -3 \right) \times \left( -4 \right) -7\times 4-\left( -5 \right) \times -6 \right] \right|$

$=\cfrac { 1 }{ 2 } \left| \left[ -12-30-28-10-10-12-28-30 \right] \right|$

$=\cfrac { 1 }{ 2 } \times 160$

$A\left( \Box ABCD \right) =80$ sq. units.

Prove that, $\left| \begin{matrix} x+\lambda & 2x & 2x \\ 2x & x+\lambda & 2x \\ 2x & 2x & x+\lambda \end{matrix} \right| =(5x+\lambda)( \lambda -x)^{2}$ .

1328686_1083895_ans_d05297da2cbf4b98842022e81015d602.jpg

Show that $\begin{vmatrix} b+c & c+a & a+b \\ a+b & b+c & c+a \\ a & b & c \end{vmatrix} = a^3 + b^3 +c^3 - 3abc$

$\begin{vmatrix} b+c & c + a & a + b \\ a + b & b + c & c + a\\a & b & c \end{vmatrix} = a^3 + b^3 + c^3 - 3abc$

Taking L.H.S

$= (b + c)[(bc + c^2 - bc - ab)] - (a + b) (c^2 + ac - ab - b^2) + a (c^2 + a^2 + 2a - ab - b^2 - ac - bc)$

$= (b + c) (c^2 - ab) - (a + b) (c^2 + ac - ab - b^2) + a (c^2 + a^2 - b^2 + ac - ab - bc)$

$= bc^2 - ab^2 + c^3 - abc - [ac^2 + a^2c - a^2b - ab^2 + bc^2 + abc - ab^2 + b^3 ] + ac^2 + a^3 - ab^2 + a^2c - a^2b - abc$

$= bc^2 - ab^2 + c^3 - abc - ac^2 + a^2c + a^2b + ab^2 - bc^2 - abc + ab^2 + b^3 + ac^2 + a^3 - ab^2 + a^2 c + a^2b - abc$

$= a^3 + b^3 + c^3 - 3abc$

If $a \ne p, b \ne q$ and $c \neq r$ $\begin{vmatrix} p & b & c \\ a & q & c \\ a & b & r \end{vmatrix}= 0$ then, find the value of :-
$\dfrac{p}{p-a} + \dfrac{q}{q-b} + \dfrac{r}{r-c} = ?$

It should have some corrections in the answer
1100649_1081069_ans_284c4486a8084bb7b65e4a9d8176c21a.jpg

Find the value of following determinant.
$\begin{vmatrix} -1 & 7\\ 2 & 4\end{vmatrix}$ .

$\begin{vmatrix} -1 & 7\\ 2 & 4\end{vmatrix}=-1*4-7*2=-18$

Find the value of the following determinant.
$\begin{vmatrix} 2i & -3i \\ { i }^{ 3 } & -2{ i }^{ 3 } \end{vmatrix}$ , where $i=\sqrt {-1}$ .

Given,

$\begin{vmatrix} 2i & -3i \\ { i }^{ 3 } & -2{ i }^{ 3 } \end{vmatrix}$ , where

$i=\sqrt {-1}$ .

Expanding the determinant we get

$=2i\times(-2i^3)-(-3i)\times i^3$

$=-4i^4+3i^4$

$=-i^4$

$=-1$

If $f\left( x \right) = a + bx + c{x^2}$ and $\alpha ,\beta ,\gamma$ are the roots of the equation ${x^3} = 1$ , then $\left| {\begin{array}{*{20}{c}}a & b & c\\b & c & a\\c & a & b\end{array}} \right|$ is equal to

Given,

$f(x)=a+bx+cx^2$

and

$\alpha ,\beta , \gamma$ are roots

$x^3=1$

which means

$\alpha ,\beta , \gamma$ are

$1, w, w^2$

and we know ,

$w^3=1$ and

$1+w+w^2=0$

so,

$f(\alpha)=f(1)=a+b+c$ ,

$f(\beta)=f(w)=a+bw+cw^2$

and

$f(\gamma)=f(w^2)=a+bw^2+cw$

now,

$=\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$

$=a(bc-a^2)-b(b^2-ac)+c(ab-c^2)$

$=abc-a^3+abc-b^3+abc-c^3$

$=3abc-a^3-b^3-c^3$

$=-(a^3+b^3+c^3-3abc)$

$=-(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$=-(a+b+c)(a^2+b^2+c^2+(-1)ab+(-1)bc+(-1)ca)$

$=-(a+b+c)(a^2+b^2+c^2+(w+w^2)ab+(w+w^2)bc+(w+w^2)ca)$

$=-(a+b+c)(a+bw+cw^2)(a+bw^2+cw)$

$=-f(1)f(w)f(w^2)$

$=-f(\alpha)f(\beta)f(\gamma)$

If $A$ is a skew-symmetric matrix of order $3$ then find $\left| A \right|$

Let

$A=\begin{bmatrix} o & a & b \\ -a & o & c \\ -b & -c & o \end{bmatrix}$ be a skew symmetric matrix

$|A|=\begin{bmatrix} o & a & b \\ -a & o & c \\ -b & -c & o \end{bmatrix}$

$|A|=-a(bc)+b(ac)$

$=-abc+abc$

$=0$

Evaluate $\left| \begin{matrix} 1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b \end{matrix} \right|$ without direct expansion.

1381170_1130488_ans_3280335b3fc1404987d756535d583d81.jpg

Find the value of the determinant: $\begin{vmatrix} \sqrt{2} & 2\sqrt{3}\\ 3\sqrt{2} & \sqrt{3}\end{vmatrix}$ .

$\begin{array}{l} \left( { \sqrt { 2 } .\sqrt { 3 } +3\sqrt { 2. } \, 2\sqrt { 3 } } \right) \\ \sqrt { 6 } +6\sqrt { 6 } \\ 7\sqrt { 6 } \end{array}$

$\begin{vmatrix}1 & 2 & 3\\ 0 & 2 & 4\\ 0 & 0 & 5\end{vmatrix}$ .

$\begin{vmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{vmatrix}$

solving across Row

$3$ and

$C_{3}$

$0+0+5(2)=10$

$\therefore\ = 10$

If $\vec { a } =\tilde { i } +\tilde { j } +\tilde { k } ,\vec { b } =\tilde { i } -\tilde { j } +\tilde { k } ,\vec { c } =\tilde { i } +2\tilde { j } -\tilde { k }$ , then the value of $\left| \begin{matrix} \overrightarrow { a } .\overrightarrow { a } & \overrightarrow { a } .\overrightarrow { b } & \overrightarrow { a } .\overrightarrow { c } \\ \overrightarrow { b } .\overrightarrow { a } & \overrightarrow { b } .\overrightarrow { b } & \overrightarrow { b } .\overrightarrow { c } \\ \overrightarrow { c } .\overrightarrow { a } & \overrightarrow { c } .\overrightarrow { b } & \overrightarrow { c } .\overrightarrow { c } \end{matrix} \right| =$

$\vec{a}= \hat{i}+ \hat{j}+ \hat{k}$

$b= \hat{i}- \hat{j}+ \hat{k}$

$c= \hat{i}+ 2\hat{j}- \hat{k}$

$\vec{a}.\vec{a}= 1\times 1 + 1\times 1 + 1\times 1=\left(3\right)$

$\vec{a}.\vec{b}= 1\times 1 + 1\times -1 + 1\times 1=\left(1\right)$

$\vec{a}.\vec{c}= 1 + 1\times 2 + 1\times -1=\left(2\right)$

$\vec{b}.\vec{a}= 1 + -1\times 1 + 1\times 1=\left(1\right)$

$\vec{b}.\vec{b}= 1 + -1\times -1 + 1\times 1=\left(3\right)$

$\vec{b}.\vec{c}= 1 + -1\times 2 + 1\times -1=\left(-2\right)$

$\vec{c}.\vec{a}= 1\times 1 + 2\times -1 - 1\times 1=\left(2\right)$

$\vec{c}.\vec{b}= 1\times 1 + 2\times -1 - 1\times 1=\left(-2\right)$

$\vec{c}.\vec{c}= 1\times 1 + 2\times 2 + -1\times -1=\left(6\right)$

$\therefore \begin{vmatrix} \vec { a }. \vec { a } & \vec { a }. \vec { b } & \vec { a }. \vec { c } \\ \vec { b }. \vec { a } & \vec { b }. \vec { b } & \vec { b }. \vec { c } \\ \vec { c }. \vec { a } & \vec { c }. \vec { b } & \vec { c }. \vec { c } \end{vmatrix}= \begin{vmatrix} 3 & 1 & 2 \\ 1 & 3 & -2 \\ 2 & -2 & 6 \end{vmatrix}$

$R_{3} \rightarrow R_{3}/2$

$= \begin{vmatrix} 3 & 1 & 2 \\ 1 & 3 & 2 \\ 1 & -1 & 3 \end{vmatrix}$

$R_{1}\rightarrow R_{1}-3R_{3}$

$R_{2}\rightarrow R_{2}-R_{1}$

$\Rightarrow \begin{vmatrix} 0 & 4 & -7 \\ 0 & 4 & -1 \\ 1 & -1 & 3 \end{vmatrix}$

$R_{2} \rightarrow R_{2}-R_{1}$

$\Rightarrow \begin{vmatrix} 0 & 0 & -6 \\ 0 & 4 & -1 \\ 1 & -1 & 3 \end{vmatrix}$

$= 1\left(-1\times 0-\left(-6\right)\times 4\right)=1 (0-\left(-24\right)$

$=\boxed{24}$

Prove that $\left| \begin{matrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{matrix} \right| =(a+b+c)^3$

1349008_1134313_ans_177784654783444f9343cf5da8db0b55.jpg

If $\Delta=\left| \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right|$ and $A_{IJ}$ is cofactors of $a_{ij}$ then the value of $\Delta$ is given by

1347478_1134574_ans_640679f3e9584dc9899ac4c648fcad4a.jpg

$\left| \begin{matrix} 1 & w & { w }^{ 2 } \\ w & { w }^{ 2 } & 1 \\ { w }^{ 2 } & w & 1 \end{matrix} \right|$ Where w is a complex cube root of unity.

We know that

$w^2+w+1=0\\w^3=1$

Now, Expanding the given Determinant

$1(w^2-w)-w(w-w^2)+w^2(w^2-w^4)=w^2-w-w^2+w^3+w^4-w^6=-w+1+w^4-1^2=-w+w^4\\=-w+w.w^3=-w+w.1=-w+w=0$

Find the value of $k$ if points $\left (k,3\right),\left (6,-2\right)\ and \left (-3,4\right)$ are collinear.

(k,3),(6,-2),(-3,4) are

collinear

here

$x_1 = k, y_1 = 3$

$x_2 = 6, y_2 = -2$

$x_3 = -3,y_3 = 4$

So, as given that the points are collinear, then,

$x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) = 0$

$k(-2-4)+6(4-3)+3(3+2) = 0$

$k(-6)+6(1)+(-3)\times 5 = 0$

$-6k+6-15 = 0$

$-6k -9 = 0$

$k = \dfrac{9}{6} = \dfrac{3}{2}$

1135655_1144588_ans_3993e8d420854e8faa0045f4d7848f02.jpg

Find the value of $\begin{vmatrix} 53 & 106 & 159 \\ 52 & 65 & 91 \\ 102 & 153 & 221 \end{vmatrix}$

$\begin{bmatrix} 53& 106 & 159\\ 52 &65 & 91\\ 102 & 153 & 221\end{bmatrix}$

$=53\times \begin{bmatrix} 65& 91\\ 153 &221 \end{bmatrix}-106\times \begin{bmatrix} 52& 91\\ 102 & 221\end{bmatrix}+159\times \begin{bmatrix}52 &65 \\ 102 &153 \end{bmatrix}$

$=0$

Solve $\left| \begin{matrix} x & -4 & -4 \\ 3 & 2 & 1 \\ -2 & 4 & 1 \end{matrix} \right|$ =0

$\begin{vmatrix} x & -4 & -4 \\ 3 & 2 & 1 \\ -2 & 4 & 1 \end{vmatrix}=0$

$x(2-4)+4(3+2)-4(12+4)=0$

$-2x+20-64=0$

$-2x=44$

$x=-22$

Find the value of K if the point A $(2,3)$ ,B $(4,K)$ and C $( 6,-3)$ are collinear?

Given that the points

$A \left( 2, 3 \right), B \left( 4, k \right)$ and

$C \left( 6, -3 \right)$ are collinear.

As we know that if three points are collinear then they will lie of a same plane and thus will not form a triangle.

Therefore,

Area of triangle formed by these points will be

$0$

Therefore,

Now,

Area of

$\triangle$ formed by these points

$= \cfrac{1}{2} \times \begin{vmatrix} 2 & 3 & 1 \\ 4 & k & 1 \\ 6 & -3 & 1 \end{vmatrix}$

Therefore,

$\begin{vmatrix} 2 & 3 & 1 \\ 4 & k & 1 \\ 6 & -3 & 1 \end{vmatrix} = 0$

$\left[ 2 \left( k - \left( -3 \right) \right) - 3 \left( 4 - 6 \right) + 1 \left( \left( -12 \right) - 6k \right) \right] = 0$

$2k + 6 + 6 - 12 - 6k = 0$

$-4k = 0$

$\Rightarrow k = 0$

Thus the value of

$k$ is

$0$ .

Hence the correct answer is

$0$ .

Find the values of the following determinants
$\begin{vmatrix} 1+3i & i-2 \\ { -i-2 } & 1-3{ i } \end{vmatrix}$
where $i=\sqrt{-1}$ .

$\begin{vmatrix} 1+3i & i-2\\ -i-2 & 1-3i\end{vmatrix}$

$=(1+3i)(1-3i)-(2-2)(-i-2)$

$=(1-9i^2)-(-1^2-2i+2i+4)$

$=(1+9)-(1+4)$

$=10-5$

$=5$ .

1072128_1160083_ans_b19beead150148a887253b4ad2049c30.png

If $\left| \begin{matrix} a & b & a\alpha +b \\ b & c & b\alpha +c \\ a\alpha +b & b\alpha +c & 0 \end{matrix} \right| =0$ . Prove that $a,b,c$ are in G.P. or $\alpha$ is a root of $ax^{2}+2bx+c=0$

$\begin{vmatrix} a & b & a\alpha +b \\ b & c & b\alpha +c \\ a\alpha +b & b\alpha +c & 0 \end{vmatrix}=\begin{matrix} 0{ b }^{ 2 }=ac \\ { ax }^{ 2 }+2bx+c=0 \end{matrix}$

$\begin{vmatrix} a-b & b & a\alpha +b \\ 0 & 0 & -\left( b\alpha +c \right) \\ \left( a-b \right) \alpha +\left( b-c \right) & b\alpha +c & 0 \end{vmatrix}$

$a(b\alpha +c)-(-(b\alpha +c))(a-b)\alpha =0$

$ab\alpha +ca +(a-b)\alpha =0$

$ab\alpha +ca +a\alpha -b\alpha =0$

$\alpha (ab++a-b)=-ca$

Without expanding , find the value of $\begin{vmatrix} a+b & 2a+b & 3a+b \\ 2a+b & 3a+b & 4a+b \\ 4a+b & 5a+b & 6a+b \end{vmatrix}$

$\begin{vmatrix} a+b & 2a + b & 3a + b \\ 2a + b & 3a + b & 4a + b \\ 4a + b & 5a + b & 6a + b \end{vmatrix}$

$R_3 \rightarrow R_3 - R_2 \,\,\, R_2 \rightarrow R_2 - R_1$

$\begin{vmatrix} a+ b & a & a \\ 2a + b & a & a \\ 4a + b & a & a \end{vmatrix}$

$0$ .

Find the roots of equation $\left| {\begin{array}{*{20}{c}}1 & 4 & {20}\\1 & { - 2} & 5\\1 & {2x} & {5{x^2}}\end{array}} \right| = 0$ .

$\begin{vmatrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & 5{ x }^{ 2 } \end{vmatrix}=0$

$\Rightarrow \ 1(-10x^2-10x)-4(5x^2-5)+20(2x+2)=0$

$\Rightarrow \ -10x^2-10x-20x^2+20+40x+40=0$

$\Rightarrow \ -30x^2+30x+60=0$

$\Rightarrow \ -30(x^2-x-2)=0$

$\Rightarrow \ x^2-x-2=0$

$\Rightarrow \ x=\dfrac {1\pm \sqrt {1+8}}{2}=\dfrac {1+3}{2}$

$\therefore \ x=2$ and

$x=-1$

Solve
$\left| {\begin{array}{*{20}{c}}1&1&{ - 1}\\6&4&{ - 5}\\{ - 4}&{ - 2}&3\end{array}} \right|$

$\left| \begin{matrix} 1 & 1 & -1 \\ 6 & 4 & -5 \\ -4 & -2 & 3 \end{matrix} \right|$

$=1\left(12-10\right)-1\left(18-20\right)-1\left(-12+16\right)$

$=2+2-4=0$

Verify whether the points $(1,5)$ , $(2,3)$ ,and $(-2,-1)$ are collinear or not.

Given points

$(1,5), (2,3)(-2,-1)$

Lets find a equation of line passing through

$( 1,5)(2,3)$

$\rightarrow \dfrac{y-5}{x-1}=\dfrac{2-1}{3-5}$

$\rightarrow -2(y-5)=x-1$

$\rightarrow -2y+10=x-1$

$\rightarrow x+2y=11$

Point

$(-2,-1)\Rightarrow -2+2(-1)=-4\neq 11$

So, point

$(-2,-1)$ do uot passes through line all thsee points are not collinear.

Solve $D = \left| \begin{array} { c c c } { 1 } & { - 2 } & { 1 } \\ { 2 } & { 1 } & { - 1 } \\ { 1 } & { 3 } & { 1 } \end{array} \right]$

$D = \left| \begin{array} { c c c } { 1 } & { - 2 } & { 1 } \\ { 2 } & { 1 } & { - 1 } \\ { 1 } & { 3 } & { 1 } \end{array} \right]$

$D= 1(1+3)+2(2+1)+1(6-1)$

$D = 4+6+5 = 15$

$D = 15$

$\left| \begin{matrix} 0 \\ -a \\ b \end{matrix}\begin{matrix} a \\ 0 \\ c \end{matrix}\begin{matrix} -b \\ -c \\ 0 \end{matrix} \right| =0$

$\begin {vmatrix} 1 & a & -b \\ -a & 0 & -c \\ b & c & 0\end {vmatrix}$

$= 0(0 + c^2) - a (0 + bc) - b (-ac - 0)$

$= - abc + abc$

$= 0$

Prove that points $(2,-2),(-3,8),(-1,4)$ are collinear.

Slope of the line joining by the points

$(2, -2), (-3, 8) = \dfrac{8+2}{-3-2} = \dfrac{10}{-5} = - 2$

Slope of the line joining by the points

$(-3, 8), (-1, 4) = \dfrac{4 - 8}{-1 + 3} = \dfrac{-4}{2} = - 2$

Sloap of both the line is same.

Hence,

$(2, -2), (-3, 8), (-1, 4)$ are collinear.

If the points $A\ (x,2),B\ (-3,-4),C\ (7,-5)$ are collinear, then find the value of $x$ .

$A(X,2)$

$B(-3,-4)$

$C(7,-5)$

$x= ?$

slope AB = slope BC = slope AC

$\frac{-3-x}{-4-2} = \frac{7+3}{-5+4} = \frac{7-x}{-5-2}$

$\frac{x+3}{6} = -10 = \frac{x-7}{7}$

$x = -63$

1113289_1204151_ans_cd6354e75da64d1b86fa58d149f53e35.jpg

Evaluate :
$\left| { \begin{array} { *{ 20 }{ c } }3 & { -1 } & { -2 } \\ 0 & 0 & { -1 } \\ 3 & { -5 } & 0 \end{array} } \right|$

$\begin{array}{l} \left| { \begin{array} { *{ 20 }{ c } }3 & { -1 } & { -2 } \\ 0 & 0 & { -1 } \\ 3 & { -5 } & 0 \end{array} } \right| =3\left( { 0-\left( { -1 } \right) \left( { -5 } \right) } \right) -\left( { -1 } \right) \left( { 0-3\left( { -1 } \right) } \right) +\left( { -2 } \right) \left( { 0\left( { -5 } \right) -0\left( 3 \right) } \right) \\ =3\left( { -5 } \right) -\left( { -3 } \right) +0 \\ =-15+3 \\ =-12 \end{array}$

Find minor & cofactors of elements '6', '5', '0' & '4' of the determinant $\begin{vmatrix} 2 & 1 & 3 \\ 6 & 5 & 7 \\ 3 & 0 & 4 \end{vmatrix}$

Minor of

$6 = \begin{vmatrix} 1 & 3 \\ 0 & 4 \end{vmatrix} = 1 \times 4 - 3 \times 0 = -4$
Cofactor =

$(-1)^{2 + 1} (4) = -4$
Minor of

$5 = \begin{vmatrix} 2 & 3 \\ 3 & 4 \end{vmatrix} = 2 \times 4 - 3 \times 3 = -1$
Cofactor =

$(-1)^{2 + 2} (-1) = -1$
minor of

$0 = \begin{vmatrix} 2 & 3 \\ 6 & 7 \end{vmatrix} = 2 \times 7 - 3 \times 6 = -4$
Cofactor =

$(-1)^{3 + 2} (-4) = 4$

Solve:
$\left| \begin{matrix} 1 & xy & xy(x+y) \\ 1 & yz & yz(y+z) \\ 1 & zx & zx(2+x) \end{matrix} \right|$

1150807_1188482_ans_8723583ae4b646d5bcc366260576b632.jpg

$\left| {\begin{array}{*{20}{c}}2&1&1\\1&{ - 2}&{ - 3}\\3&2&4\end{array}} \right| = ?$

$\begin{vmatrix} 2 & 1 & 1 \\ 1 & -2 & -3 \\ 3 & 2 & 4 \end{vmatrix}$

$=2\left( -2\times 4-2\times -3 \right) -1\left( 1\times 4-3\times -3 \right) +1\left( 1\times 2-3\times -2 \right)$

$=2\left( -8+6 \right) -1\left( 4+9 \right) +1\left( 2+6 \right)$

$=2\times -2-13+8=-4-13+8$

$=-17+8=-9$

Find $x$ if: $\left| \begin{matrix} 2 & 1 & x+1 \\ -1 & 3 & -4 \\ 0 & -5 & 3 \end{matrix} \right|$ $= 1$

$\begin{vmatrix} 2 & 1 & x+1 \\ -1 & 3 & -4 \\ 0 & -5 & 3 \end{vmatrix}=1$

$\Rightarrow \ 2[3\times 3-(-4)(-5)]-1[(-1)(3)-(0)(-4)]+(x+1)[(-1)(-1)-(0)3]$

$\Rightarrow \ 2[9-20]-1[3]+(x+1)[5]=1$

$\Rightarrow \ -22+3+(x+1)(5)=1$

$\Rightarrow \ -19+(x+1)(5)=1$

$\Rightarrow \ (x+1)(5)=20$

$\Rightarrow \ (x+1)=4$

$\Rightarrow \ x=3$

Show that:

$\left| \begin{array} { c c c } { a ^ { 2 } } & { b ^ { 2 } } & { c ^ { 2 } } \\ { a } & { b } & { c } \\ { 1 } & { 1 } & { 1 } \end{array} \right| = - ( a - b ) ( b - c ) ( c - a )$

Given

$\left| \begin{array} { c c c } { a ^ { 2 } } & { b ^ { 2 } } & { c ^ { 2 } } \\ { a } & { b } & { c } \\ { 1 } & { 1 } & { 1 } \end{array} \right|$

$c_1 \rightarrow c_1 - c_2$

$c_2 \rightarrow c_2 - c_3$

$\begin{vmatrix} a^2-b^2 & b^2 - c^2 & c^2 \\ a-b & b-c & c \\ 0 & 0 & 1 \end{vmatrix}$

$=1[(a^2 - b^2) (b - c) - (a - b) (b^2 - c^2)]$

$=( a- b) (b - c) [a + b - b - c]$

$=(-1) ( a - b) ( b - c) ( c- a)$ .

$[hence \quad proved]$

If $\Delta=\begin{vmatrix} 1 & \sin{\theta} & 1\\ -\sin{\theta} & 1 & \sin{\theta} \\ -1 & -\sin{\theta} & 1 \end{vmatrix}$ then prove that $2\le \Delta\le 4$

1353780_1241862_ans_594ce9022133490fb4aa489e09c38d9d.PNG

Expand: $\begin{vmatrix} 3 & 2 & 5\\ 9 & -1 & 4 \\ 2 & 3 & -5 \end{vmatrix}$

1353781_1241947_ans_0bc41f4a0b9d44dea7091b2872c93f69.PNG

Expand: $\begin{vmatrix} 1 & 2 & 3\\ 4 & 6 & 2 \\ 5 & 9 & 4 \end{vmatrix}$

1353785_1245144_ans_db480640fa154438b946145abf67ac3d.PNG

Prove that
$\begin{vmatrix} bc & a & { a }^{ 2 } \\ ca & b & { b }^{ 2 } \\ ab & c & { c }^{ 2 } \end{vmatrix}=\begin{vmatrix} 1 & { a }^{ 2 } & { a }^{ 3 } \\ 1 & { b }^{ 2 } & { b }^{ 3 } \\ 1 & { c }^{ 2 } & { c }^{ 3 } \end{vmatrix}$

$\begin{vmatrix} bc & a & a^2\\ ca & b & b^2\\ ab & c & c^2\end{vmatrix}$

on multiply and divide abc

$\dfrac{abc}{abc}\begin{vmatrix} bc & a & a^2\\ ca & b & b^2\\ ab & c & c^2\end{vmatrix}$

$R_1, R_2, R_3$

$\dfrac{1}{abc}\begin{vmatrix} abc & a^2 & a^3\\ abc & b^2 & b^3\\ abc & c^2 & c^3\end{vmatrix}$

Common from

$C_1$ to abc

$\dfrac{abc}{abc}\begin{vmatrix} 1 & a^2 & a^3\\ 1 & b^2 & b^3\\ 1 & c^2 & c^3\end{vmatrix}$

$\begin{vmatrix} 1 & a^2 & a^3\\ 1 & b^2 & b^3\\ 1 & c^2 & c^3\end{vmatrix}$ .

R.H.S.

1200226_1242814_ans_12bcc6f63504481cb16573bb1b9e1ee0.png

Find the value of k so that point colinear(7,-2)(5,1)(3,k).

Points are collinear

$(7, -2) (5, 1)(3, k)$

Area of

$\Delta =0$

$=\dfrac{1}{2}\begin{vmatrix} 7 & -2 & 1\\ 5 & 1 & 1\\ 3 & k & 1\end{vmatrix}=0$

$=\dfrac{1}{2}[7(1-k)+2(5-3)+1(5k-3)]=0$

$=7(1-k)+4+5k-3=0$

$=7-7k+1+5k=0$

$=8-2k=0$

$k=4$ .

Solve
$\left| \begin{matrix} 4 \\ 3 \end{matrix}\begin{matrix} x+1 \\ x \end{matrix} \right| =5$

$\begin{vmatrix} 4 & x+1\\ 3 & x\end{vmatrix}=5$

$ad-bc$

$4(x)-(x+1)3=5$

$4x-3x-3=5$

$4x-3x=8$

$x=8$ .

If for any 2 x 2 square matrix A, A(adj A) = $\left[ \begin{array}{l}8\,\,0\\0\,\,8\end{array} \right]$ ,then write the value of $\left| A \right|$ .

Given :

$A (adj \ A ) = \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix}$

$\because A (adj \ A) = |A| I_n$

$\Rightarrow |A (adj \ A)| = |A|^n \ ;$ where

$n \rightarrow$ order of matrix

$\because n = 2$

$\Rightarrow |A (adj \ A) | = |A|^2 \Rightarrow |A|^2 = 64$

$\Rightarrow |A| = 8$

Find the value of the determinant $\begin{vmatrix} -1 & 2 & 1\\ 3+2\sqrt{2} & 2+2\sqrt{2} & 1 \\ 3-2\sqrt{2} & 2-2\sqrt{2} & 1 \end{vmatrix}$

1353787_1245728_ans_d2f7ea05a1dd4ab3861f700c7346f2a8.PNG

Show that $\begin{vmatrix} a & \quad a+b & \quad a+b+c\quad \\ 2a & 3a+2b & 4a+3b+2c \\ 3a\quad & 6a+3b & 10a+6b+3c \end{vmatrix}={ a }^{ 3 }$

L.H.S =

$\begin{vmatrix} a & \quad a+b & \quad a+b+c\quad \\ 2a & 3a+2b & 4a+3b+2c \\ 3a\quad & 6a+3b & 10a+6b+3c \end{vmatrix}$

$R_{2}\rightarrow R_{2}-2R_{1}, R_{3}\rightarrow R_{3}-2R_{1}$

$\begin{vmatrix} a & \quad a+b & \quad a+b+c\quad \\ 0 & a & 2a+b \\ 0\quad & 3a & 7a+3b \end{vmatrix}$

Expanding along first column

$= a[a(7a+3b)-3a(2a+b)]-0+0$

$= a[7a^{2}+3ab-6a^{2}-3ab]$

$= a(a)^{2} = a^{3} = R.H.S.$

1215654_1250067_ans_3befe8399e1f4f9a9ae059a168a237ba.jpg

Prove that $\begin{vmatrix} {a}^{2}+1 & ab & ac\\ ab & {b}^{2}+1 & bc \\ ac & bc & {c}^{2}+1 \end{vmatrix}=1+{a}^{2}+{b}^{2}+{c}^{2}$

1353802_1245804_ans_41c5d8603ebf40db830ad2d89083a0f9.PNG

Solve for $x$
$\begin{vmatrix} 4x & 6x+2 & 8x+1\\ 6x+2 & 9x+3 & 12x \\ 8x+1 & 12x & 16x+2 \end{vmatrix}=0$

1353800_1245861_ans_9eb76de56e014457ad6d0d53c24d9f23.PNG

Given that matrix $A=\left[ \begin{matrix} x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z \end{matrix} \right]$ . If $xyz=2013$ and $8x+4y+3z=2012$ then A(adj A) is equal to

1210238_1248669_ans_bb01f4ac8de44de482d1d743e56c6405.jpg

If $f(x)=\left| \begin{matrix} 2x & { x }^{ 2 } & { x }^{ 3 } \\ { x }^{ 2 }+2x & 1 & 3x+1 \\ 2x & 1-3{ x }^{ 2 } & 5x \end{matrix} \right|$ , then find $f'(1)$

Que.

$f(x)=\left| \begin{matrix} 2x & { x }^{ 2 } & { x }^{ 3 } \\ { x }^{ 2 }+2x & 1 & 3x+1 \\ 2x & 1-3{ x }^{ 2 } & 5x \end{matrix} \right|$ , then

$f'(1)$

$f(x) = 2x(5x-(3x+1)(1-3x^{2}))-x^{2}(5x(x^{2}+2x)-2x(3x+1))+x^{3}((x^{2}+2x)(1-3x^{2})-2x)$

$= 2x(5x-(3x-9x^{3}+1-3x^{2}))-x^{2}(5x^{3}+10x^{2}-6x^{2}-2x)+x^{3}(x^{2}-3x^{4}+2x-6x^{3}-2x)$

$= 2x(2x+9x^{3}+3x^{2}-1)-x^{2}(5x^{3}+10x^{2}-6^{2}-2x)+x^{3}(-3x^{4}-6x^{3}+x^{2})$

$= 4x^{2}+18x^{4}+6x^{3}-2x-5x^{5}-10x^{4}+6x^{4}+2x^{3}-3x^{7}-6x^{6}+x^{5}$

$f(x) = -3x^{7}-6x^{6}-4x^{5}+14x^{4}+8x^{3}+4x^{2}-2x$

$f^{1}(x) = -3\times 7x^{6}-6\times 6x^{5}-4\times 5x^{4}+14\times 4x^{3}+8\times 3x^{2}+4\times 2x-2$

$f^{1}(x) = -21\times x^{6}-36x^{5}-20x^{4}+56x^{3}+24x^{2}+8x-2$

$f^{1}(1) = -21-36-20+56+24+8-2$

$= -79+88$

$f^{1}(1) = 9$

1131948_1250036_ans_827288fc404e4155aacc5c60bfda9c11.jpg

Show that the points $A(1,2,7),\ B(2,6,3)$ and $C(3,10,-1)$ are collinear.

We have,

Points are

$A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 1,2,7 \right)$

$B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)=\left( 2,6,3 \right)$

$C\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)=\left( 3,10,-1 \right)$

Now,

Using distance formula

$AB=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}+{{\left( {{z}_{1}}-{{z}_{2}} \right)}^{2}}}$

$AB=\sqrt{{{\left( 1-2 \right)}^{2}}+{{\left( 2-6 \right)}^{2}}+{{\left( 7-3 \right)}^{2}}}$

$AB=\sqrt{1+16+16}$

$AB=\sqrt{33}$

$BC=\sqrt{{{\left( {{x}_{2}}-{{x}_{3}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{3}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{3}} \right)}^{2}}}$

$BC=\sqrt{{{\left( 2-3 \right)}^{2}}+{{\left( 6-10 \right)}^{2}}+{{\left( 3+1 \right)}^{2}}}$

$BC=\sqrt{1+16+16}$

$BC=\sqrt{33}$

$CA=\sqrt{{{\left( {{x}_{3}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{3}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{3}}-{{z}_{1}} \right)}^{2}}}$

$CA=\sqrt{{{\left( 3-1 \right)}^{2}}+{{\left( 10-2 \right)}^{2}}+{{\left( -1-7 \right)}^{2}}}$

$CA=\sqrt{4+64+64}$

$CA=\sqrt{132}$

$CA=\sqrt{2\times 2\times 33}$

$CA=2\sqrt{33}$

Then,

$CA=AB+BC$

$2\sqrt{33}=\sqrt{33}+\sqrt{33}$

$2\sqrt{33}=2\sqrt{33}$

Then,

It's points are collinear.

Hence, this is the answer.

$(k,k ), (2,3)$ and $(4,-1)$ are collimear. So find the value of $k$ .

We know that,

Area of

$\Delta ABC = 0$

$= \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] = 0$

Here,

${x_1}=k,{{y_1}}=k$

$\begin{matrix} { x_{ 2 } }=2,{ y_{ 2 } }=3 \\ { x_{ 3 } }=4,{ y_{ 3 } }=-1 \\ \end{matrix}$

Putting values,

$= \dfrac{1}{2}\left[ {k\left( {3 - \left( { - 1} \right)} \right) + 2\left( {\left( { - 1} \right) - k} \right) + 4\left( {k - 3} \right)} \right] = 0$

$= \dfrac{1}{2}\left[ {3k + k - 2 - 2k + 4k - 12} \right] = 0$

$= \dfrac{1}{2}\left[ {6k - 14} \right] = 0$

$\begin{matrix} \dfrac { { 6k-14 } }{ 2 } =0 \\ 6k-14=0 \\ 6k=14 \\ k=\dfrac { { 14 } }{ 6 } =\dfrac { 7 }{ 3 } \\ \end{matrix}$

$k = \dfrac{7}{3}$

Hence, this is the answer.

Solve the matrix
$\left| {\begin{array}{*{20}{c}}{\frac{1}{b}{b^2}} & {ca}\\{\frac{1}{c}{c^2}} & {ab}\end{array}} \right|$

1352015_1270785_ans_b4cda735632e49cda1446830e1f502d2.PNG

Prove that point $(1,1),(-2,7)$ and $(3,-3)$ are collinear.

A (1, 1)

B (-2, 7)

C (3, -3)

Slope of AB

$=\dfrac { { 7-1 } }{ { -2-1 } } =\dfrac { 6 }{ { -3 } } =-2$

Slope of BC

$=\dfrac { { 7-\left( { -3 } \right) } }{ { -2-3 } } =\dfrac { { 10 } }{ { -5 } } =-2$

Slope of AC

$=\dfrac { { 1-\left( { -3 } \right) } }{ { 1-3 } } =\dfrac { 4 }{ { -2 } } =-2$

Since slope of AB = Slope of BC = Slope of AC

Hence, points A,B,C are collinear.

If the points ( a,0),(0,b)and (3,2) are col linear, prove that $\dfrac { 2 }{ b } +\dfrac { 3 }{ a } =1$

We have,

$A\left( {{x}_{1}},{{y}_{1}} \right)=\left( a,0 \right)$

$B\left( {{x}_{2}},{{y}_{2}} \right)=\left( 0,b \right)$

$C\left( {{x}_{3}},{{y}_{3}} \right)=\left( 3,2 \right)$

Given that,

Area of triangle $=0$

$\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]=0$

$\Rightarrow \left[ a\left( b-2 \right)+0\left( 2-0 \right)+3\left( 0-b \right) \right]=0$

$\Rightarrow ab-2a-3b=0$

$\Rightarrow 2a+3b=ab$

$\Rightarrow \dfrac{2a+3b}{ab}=1$

$\Rightarrow \dfrac{2a}{ab}+\dfrac{3b}{ab}=1$

$\Rightarrow \dfrac{2}{b}+\dfrac{3}{a}=1$

$\Rightarrow \dfrac{3}{a}+\dfrac{2}{b}=1$

Hence proved.

Hence, this is the answer.

Evaluate $\left| \begin{matrix} x & x^{ 2 } & x^{ 2 } \\ y & y^{ 2 } & y^{ 2 } \\ z & z^{ 2 } & z^{ 3 } \end{matrix} \right|$

$\begin{pmatrix}x&x^2&x^2\\ y&y^2&y^2\\ z&z^2&z^3\end{pmatrix}$

$=x \begin{pmatrix}y^2&y^2\\ z^2&z^3\end{pmatrix}-x^2 \begin{pmatrix}y&y^2\\ z&z^3\end{pmatrix}+x^2 \begin{pmatrix}y&y^2\\ z&z^2\end{pmatrix}$

$=x\left(y^2z^3-y^2z^2\right)-x^2\left(yz^3-y^2z\right)+x^2\left(yz^2-y^2z\right)$

$=-x^2yz^3+x^2yz^2+xy^2z^3-xy^2z^2$

Prove that. $\left| \begin{matrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{matrix} \right| (a-b)(b-c)(c-a)$

1189511_1354814_ans_6d23c71f602346bb80d43e5373e80474.jpg

Prove that :
$\begin{vmatrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{vmatrix}=0$ .

LHS

$=$

$\begin {vmatrix}0&a&-b\\-a&0&-c\\b&c&0 \end {vmatrix}$

Expanding the determinant along

$R_1$

$=0\begin{vmatrix}0&c\\c&0\end{vmatrix}$

$-a\begin{vmatrix}0&c\\c&0\end{vmatrix}$

$-b\begin{vmatrix}0&c\\c&0\end{vmatrix}$

$=0-a(bc)-b(-ac)\\=abc-abc$

$=0=$ RHS

Hence proved.

Find the value of the following determinants.
$\begin{vmatrix} \dfrac { 7 }{ 3 } & \dfrac { 5 }{ 3 } \\ \dfrac { 3 }{ 2 } & \dfrac { 1 }{ 2 } \end{vmatrix}$

$\begin{vmatrix} \dfrac { 7 }{ 3 } & \dfrac { 5 }{ 3 } \\ \dfrac { 3 }{ 2 } & \dfrac { 1 }{ 2 } \end{vmatrix}$

$= (\frac{7}{3})(\frac{1}{2}) - (\frac{5}{3}) (\frac{3}{2})$

$\Rightarrow \frac{7}{6} - \frac{15}{6} = \frac{-8}{6} = \frac{-4}{3}$

1185860_1348700_ans_da93bb2abb654d1a84b323326c40bc74.jpg

Given $\begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} X=\begin{bmatrix} 7 \\ 6 \end{bmatrix}$ . Write :
the order of the matrix $X$
the matrix $X$ .

$\begin{bmatrix}2&1\\-3&4 \end{bmatrix}X$

$=\begin{bmatrix}7\\6 \end{bmatrix}$

As per matrix multiplication rule

$X$ matrix must be in orderof

$(2\times 1)$ .

Let

$X=\begin{bmatrix}m\\n \end{bmatrix}$

Then

$\begin{bmatrix}2&1\\-3&4 \end{bmatrix}$

$\begin{bmatrix}m\\n \end{bmatrix}$

$=\begin{bmatrix}7\\6 \end{bmatrix}$

$\implies \begin{bmatrix}2m+n\\-3m+4n \end{bmatrix}$

$=\begin{bmatrix}7\\6 \end{bmatrix}$

Equating each element we get two linear equations

$2m+n=7$ and

$-3m+4n=6$

Solving these two equations we get

$m=2$ and

$n=3$

Hence

$X=\begin{bmatrix}2\\3 \end{bmatrix}$

If $A=\begin{bmatrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{bmatrix}$ the $A^{n}=$

1188298_1344091_ans_606fcbd7e55241d187b43fd4d3cdf598.jpg

Find the value of $\lambda$ for which the points $(6,-1,2),(8,-7,\lambda)$ and $(5,2,4)$ are collinear.

Set B cuts AC in the ratio

$l:1$

${ x }_{ 2 }=\cfrac { l{ x }_{ 3 }+{ x }_{ 1 } }{ l+1 }$

$8=\cfrac { l\left( 5 \right) +1\left( 6 \right) }{ l+1 }$

$5l+6=8l+8$

$3l+2=0$

$l=\cfrac { -2 }{ 3 }$

$\therefore$ B cuts Ac in the ratio

$\cfrac { -2 }{ 3 } :1$

${ z }_{ 2 }=\cfrac { l{ z }_{ 3 }+{ z }_{ 1 } }{ l+1 }$

$\lambda =\cfrac { \left( \cfrac { -2 }{ 3 } \right) \left( 4 \right) +\left( 2 \right) }{ \cfrac { -2 }{ 3 } +1 }$

$=\cfrac { \cfrac { -8 }{ 3 } +2 }{ \cfrac { 1 }{ 3 } } =\cfrac { \cfrac { -2 }{ 3 } }{ \cfrac { 1 }{ 3 } } =-2$

$\therefore \lambda =-2$

1173335_1341205_ans_cb9664d847714a9d944d122aa925055a.png

Find the value of the following determinants.
$\begin{vmatrix} -1 & 7 \\ 2 & 4 \end{vmatrix}$

$\begin{vmatrix} -1 & 7 \\ 2 & 4 \end{vmatrix}$

$\Rightarrow (-1) (4) - (2) (7)$

$\Rightarrow -4 - 14$

$\Rightarrow -18$

1185820_1348685_ans_d6332a882db44da3807802a533a00ba5.jpg

Find the value of the following determinants.
$\begin{vmatrix} 5 & 3 \\ -7 & 0 \end{vmatrix}$

$\begin{vmatrix} 5 & 3 \\ -7 & 0 \end{vmatrix}$

$(5)(0) - (-7) (3)$

= 0-(-21) = 21

1185857_1348694_ans_3d453e793ba345edb6b8431b3a0967fd.jpg

If $A=[aij]$ is a matrix of order 2x2 such that $|A|=15$ and cij represents the co factor of aij then find ${ a }_{ 21 }{ c }_{ 21 }+{ a }_{ 22 }{ c }_{ 22 }$ .

Determinant of a

$2\times 2$ matrix is given by

$|A|=a_{21}c_{21}+a_{22}c_{22}$

Given,

$|A|=15$

Hence,

$a_{21}c_{21}+a_{22}c_{22}=15$

A square matrix $\mathrm { B }$ of order $3 ,$ has $| B | = 7 ,$ find $| B$ adjB $|$

Given

$B$ is a matrix of order

$3$

$\implies |B|=7$

$\implies |adj B|=|B|^{(3-1)}=|B|^2=7^2=49$

$|B\ \mathrm{adj} B|=|B|\times |\mathrm{adj}B|=7\times 49=343$

A. square matrix $A$ of order $3 ,$ has $| A | = 5 ,$ find $| A$ adja|

Given

$A$ is a matrix of order

$3$

$\implies |A|=5$

$\implies |adj A|=|A|^{(3-1)}=|A|^2=5^2=25$

$|A\ \mathrm{adj} A|=|A|\times |\mathrm{adj}A|=5\times 25=125$

Find equation of line joining (1,2) and (3,6) using determinants.

$\textbf{Step 1: Find the equation using suitable condition.}$

$\text{Let, $$P(x, y)$$ be the general point on the line joining $$A(1,2)$$ and $$B(3,6)$$}$

$\text{We know, area of a triangle with vertices }(x_1,y_1),(x_2,y_2)\text{ and }(x_3,y_3)\text{ is given by,}$

$\Rightarrow\text{Area }=\dfrac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\x_2 & y_2 & 1 \\x_3 & y_3 & 1\end{array}\right|$

$\text{From the figure, it is obvious that the area of }\triangle \text{APB } \text{is zero.}$

$\Rightarrow\dfrac{1}{2}\left|\begin{array}{lll}x & y & 1 \\1 & 2 & 1 \\3 & 6 & 1\end{array}\right|=0$

$\Rightarrow\left|\begin{array}{lll}x & y & 1 \\1 & 2 & 1 \\3 & 6 & 1\end{array}\right|=0$

$\Rightarrow x(2-6)-y(1-3)+1(6-6)=0$

$\Rightarrow-4 x+2 y=0$

$\Rightarrow4 x-2 y=0$

$\Rightarrow2 x- y=0$

$\textbf{Hence, the required equation of the line is : }\mathbf{2x-y=0.}$

2178009_1361967_ans_80e026c5d8f64b6cab6cb3b6a52a85fc.jpg

Prove that $\begin {vmatrix} a^2 & bc & ac + c^2 \\ a^2 + ab & b^2 & ac \\ ab & b^2 + bc & c^2 \end {vmatrix}$ = 4 $a^2b^2c^2$

L.H.S.=

$\begin{vmatrix} a^{2} & bc & ac+c^{2} \\ a^{2} + ab & b^{2} & ac \\ ab & b^{2}+bc & c^{2} \end{vmatrix}$

$=abc\begin{vmatrix} a & c & a+c \\ a+b & b & a \\ b & b+c & c \end{vmatrix}$

$=abc\begin{vmatrix} 2a+2b & 2b+2c & 2c+2a \\ a+b & b & a \\ b & b+c & c \end{vmatrix}$

$(R_{1} \rightarrow R_{1} + R_{2} + R_{3})$

$=2abc\begin{vmatrix} a+b & b+c & c+a \\ a+b & b & a \\ b & b+c & c \end{vmatrix}$

$=2abc\begin{vmatrix} 0 & c & c \\ a+b & b & a \\ b & b+c & c \end{vmatrix}$

$(R_{1} \rightarrow R_{1}-R_{2})$

$=2abc\begin{vmatrix} 0 & c & 0 \\ a+b & b & a-b \\ b & b+c & -b \end{vmatrix}$

$(C_{3} \rightarrow C_{3}-C_{2})$

$=2abc[-c(-ab-b^{2}-ba+b^{2})]$

$=2abc(2abc) = 4a^{2}b^{2}c^{2}$ = R.H.S.

Prove that $\begin{vmatrix} a & a & a \\ a & b & b \\ a & b & c \end{vmatrix}=a(b-c)(a-b)$ , Hence find the value of $\begin{vmatrix} 3 & 3 & 3 \\ 3 & 5 & 5 \\ 3 & 5 & 7 \end{vmatrix}.$

$$\begin{vmatrix} a & a & a\\ a & b & b\\ a & b & c \end{vmatrix} = a(b-c)(a-b)$$

Solving LHS we get

$$\begin{vmatrix} a & a & a\\ a & b & b\\ a & b & c \end{vmatrix}$$

$\Rightarrow a(bc-b)^2-a(ac-ab)+a(ab-ab)$

$\Rightarrow ab[c-b]-a^2[c-b]+0$

$\Rightarrow a(c-b)(b-a)$

$\Rightarrow a (b-c)(a-b)=RHS$

now

$$\begin{vmatrix} 3 & 3 & 3 \\ 3 & 5 & 5\\ 3 & 5 & 7 \end{vmatrix}$$

$= 3 (5-7) (3-5) = 3(-2) (-2)$

$=12$

1214649_1366866_ans_60a1cfd711b148a59d29d500a7fe105a.JPG

A point $P\left(2,-1\right)$ is equidistant from points $\left(a,7\right)$ and $\left(-3,a\right)$ . Find $a$ .

$\sqrt{(2-(-3))^2+(-1-a)^2}=\sqrt{(2-a)^2+(-1-7)^2}$

Squaring both sides

$(2+3)^2+(1+a)^2=(2-a)^2+(-8)^2$

$\Rightarrow 25+1+a^2+2a=4+a^2-4a+64$

$\Rightarrow 6a=42$

$\Rightarrow a=7$ .

1189562_1356134_ans_0ec19f7c8b5849e98b6c44f1a262b609.jpg

Prove that : $\begin {vmatrix} a & c & a + c \\ a + b & b & a \\ b & b + c & c \end {vmatrix}$ = $4abc$

L.H.S.

$\begin{vmatrix} a & c & a+c \\ a+b & b & a \\ b & b+c & c \end{vmatrix}$

$= \begin{vmatrix} 2a+2b & 2b+2c & 2a+2c \\ a+b & b & a \\ b & b+c & c \end{vmatrix}$

$(R_{1} \rightarrow R_{1}+R_{2}+R_{3})$

$=2\begin{vmatrix} a+b & b+c & a+c \\ a+b & b & a \\ b & b+c & c \end{vmatrix}$

$=2\begin{vmatrix} 2a+2b+2c & b+c & a+c \\ 2a+2b & b & a \\ 2b+2c & b+c & c \end{vmatrix}$

$(C_{1} \rightarrow C_{1}+C_{2}+C_{3})$

$=4\begin{vmatrix} a+b+c & b+c & a+c \\ a+b & b & a \\ b+c & b+c & c \end{vmatrix}$

$=4\begin{vmatrix} a & 0 & a \\ a+b & b & a \\ b+c & b+c & c \end{vmatrix}$

$(R_{1} \rightarrow R_{1}-R_{3})$

$= 4\begin{vmatrix} a & 0 & a \\ a & b & a \\ 0 & b+c & c \end{vmatrix}$

$(C_{1} \rightarrow C_{1} - C_{2})$

$=4\begin{vmatrix} 0 & -b & 0 \\ a & b & a \\ 0 & b+c & c \end{vmatrix}$

$(R_{1} \rightarrow R_{1}-R_{2})$

$=4b(ac) = 4abc$ = R.H.S.

Prove that:
$\begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix}=2{ \left( a+b+c \right) }^{ 3 }$

To prove:-

$\begin{vmatrix} a + b + 2c & a & b \\ c & b + c + 2a & b \\ c & a & c + a + 2b \end{vmatrix} = 2 {\left( a + b + c \right)}^{3}$

Proof:-

Taking L.H.S., we have

$\begin{vmatrix} a + b + 2c & a & b \\ c & b + c + 2a & b \\ c & a & c + a + 2b \end{vmatrix}$

Applying

${C}_{1} \rightarrow {C}_{1} + {C}_{2} + {C}_{3}$ , we get

$= \begin{vmatrix} 2a + 2b + 2c & a & b \\ 2a + 2b + 2c & b + c + 2a & b \\ 2a + 2b + 2c & a & c + a + 2b \end{vmatrix}$

$= 2 \left( a + b + c \right) \begin{vmatrix} 1 & a & b \\ 1 & b + c + 2a & b \\ 1 & a & c + a + 2b \end{vmatrix}$

$= 2 \left( a + b + c \right) \left[ \left( b + c + 2a \right) \left( c + a + 2b \right) - ab - a \left( c + a + 2b \right) - ab + ab - b \left( b + c + 2a \right) \right]$

$= 2 \left( a + b + c \right) \left[ {\left( a + b + c \right)}^{2} + ab + a \left( c + a + 2b \right) + b \left( b + c + 2a \right) - a \left( c + a + 2b \right) - ab - b \left( b + c + 2a \right) \right]$

$= 2 \left( a + b + c \right) {\left( a + b + c \right)}^{2}$

$= 2 {\left( a + b + c \right)}^{3}$

$=$ R.H.S.

Hence proved.

Fill in the blanks with correct number
$\left| \begin{matrix} 3 \\ 4 \end{matrix}\begin{matrix} 2 \\ 5 \end{matrix} \right| =3\times \Box -\Box \times 4=\Box -8=\Box$

1894623_1491647_ans_6da89423211d450094816b366e97506f.jpeg

Using properties of determinant solve :-
$\left| \begin{matrix} 1 & a & { a }^{ 2 }-bc \\ 1 & b & { b }^{ 2 }-ac \\ 1 & c & { c }^{ 2 }-ab \end{matrix} \right| =$

$\begin{vmatrix} 1 & a & a^2-bc\\ 1 & b & b^2-ac\\ 1 & c & c^2-ab\end{vmatrix}$

$R_1\rightarrow R_1-R_2;$

$R_2\rightarrow R_2-R_3$

$\begin{vmatrix} 0 & a-b & (a^2-b^2)+c(a-b)\\ 0 & b-c & (b^2-c^2)+a(b-c)\\ 1 & c & c^2-ab\end{vmatrix}$

Expanding the determinant

$1[(a-b)[(b-c)(a+b+c)]-(b-c)[(a-b)(a+b+c)]]$

$\Rightarrow 0$ .

1203594_1447688_ans_835c6ca582914423b7ccd2dff27f50d8.jpg

Find the values of x, if
$\begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix}=\begin{vmatrix} 6 & 5 \\ 8 & 3 \end{vmatrix}$

$\begin{vmatrix}2x&5\\8&x\end{vmatrix}=\begin{vmatrix}6&5\\8&3\end{vmatrix}$

$\Rightarrow$

$(2x\times x)-(5\times 8)=(6\times 3)-(5\times 8)$

$\Rightarrow$

$2x^2-40=18-40$

$\Rightarrow$

$2x^2=18$

$\Rightarrow$

$x^2=\dfrac{18}{2}$

$\Rightarrow$

$x^2=9$

$\Rightarrow$

$x=\pm 3$

Prove that $\left| \begin{matrix} cos\theta & cos\phi \\ sin\theta & sin\phi \end{matrix} \right| =-sin\left( \theta -\phi \right)$

1237288_1505870_ans_16ad6846b5934206b65cde2a79251672.PNG

Find the cofactors of the elements of the following matrices :
(i) $\begin{bmatrix} -1 & 2 \\ -3 & 4 \end{bmatrix}$

(ii) $\left[ \begin{matrix} 1 & -1 & 2 \\ -2 & 3 & 5 \\ -2 & 0 & -1 \end{matrix} \right]$

1237324_1502488_ans_403759ff98bc4b1b81d6122ec400aefe.jpg

Find the minor of $\begin{bmatrix} 2 & 7 & 3 \\ -4 & 3 & -1 \\ 0 & -3 & 7 \end{bmatrix}$ .

$\begin{bmatrix} 2 & 7 & 3 \\ -4 & 3 & -1 \\ 0 & -3 & 7 \end{bmatrix}$ .

Minor of

$2$ is

$\begin{bmatrix} 3 & -1 \\ -3 & 7 \end{bmatrix}=21-3=18$

Minor of

$7$ is

$\begin{bmatrix} -4 & -1 \\ 0 & 7 \end{bmatrix}=-28-0=-28$

Minor of

$3$ is

$\begin{bmatrix} -4 & 3 \\ 0 & -3 \end{bmatrix}=12-0=12$

Minor of

$-4$ is

$\begin{bmatrix} 7 & 3 \\ -3 & 7 \end{bmatrix}=49+9=58$

Minor of

$3$ is

$\begin{bmatrix} 2 & 3 \\ 0 & 7 \end{bmatrix}=14-0=14$

Minor of

$-1$ is

$\begin{bmatrix} 2 & 7 \\ 0 & -3 \end{bmatrix}=-6-0=-6$

Minor of

$0$ is

$\begin{bmatrix} 7 & 3 \\3 & -1 \end{bmatrix}=-7-9=-16$

Minor of

$-3$ is

$\begin{bmatrix} 2 & 3 \\ -4 & -1 \end{bmatrix}=-2+12=10$

Minor of

$7$ is

$\begin{bmatrix} 2 & 7 \\ -4 & 3 \end{bmatrix}=6+28=34$

Prove that $\left| \begin{matrix} \sin\alpha & \cos\alpha & \cos(a+\delta ) \\ \sin\beta & \cos\beta & cos(\beta +\delta ) \\ \sin\gamma & \cos\gamma & \cos(\gamma +\delta ) \end{matrix} \right| =0.$

1233036_1502281_ans_225594f529b0480d9f354d42f8dae15e.jpeg

For the matrix $A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10 \end{bmatrix}$ , show that $A (adj A) = 0$ .

1416602_1661739_ans_12c66d0c60504da09b90a1349a25162f.jpg

Find the values of x, if
$\begin{vmatrix} 3 & x \\ x & 1 \end{vmatrix}$ = $\begin{vmatrix} 3 & 2 \\ 4 & 1 \end{vmatrix}$

$\begin{vmatrix} 3 & x \\ x & 1 \end{vmatrix}$ =

$\begin{vmatrix} 3 & 2 \\ 4 & 1 \end{vmatrix}$

$\Rightarrow$

$(3\times 1)-x\times x=3\times 1-4\times 2$

$\Rightarrow$

$3-x^2=3-8$

$\Rightarrow$

$3-x^2=-5$

$\Rightarrow$

$x^2=8$

$\Rightarrow$

$x=\pm2\sqrt{2}$

Solve:
$\begin{bmatrix} 1 & 5 & -2 \\ 3x & 2 & 4 \\ 5 & -1 & 0 \end{bmatrix}=0$

$\begin{bmatrix} 1 & 5 & -2 \\ 3x & 2 & 4 \\ 5 & -1 & 0 \end{bmatrix}=0$

$\begin{vmatrix} 1 & 5 & -2 \\ 3x & 2 & 4 \\ 5 & -1 & 0 \end{vmatrix}=0$

$\Rightarrow\,1\left(0+4\right)-5\left(0-20\right)-2\left(-3x-10\right)=0$

$\Rightarrow\,4+100+6x+60=0$

$\Rightarrow\,6x=-164$

$\Rightarrow\,x=\dfrac{-164}{6}=-\dfrac{82}{3}$

$\therefore\,x=-\dfrac{82}{3}$

If $A = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}$ , show that adj $A = A$ .

1416599_1661742_ans_f7b0638909ad4c7da9eba4e1aa3f7606.jpg

Find the values of x, if
$\begin{vmatrix} 4 & x+1 \\ x+2 & 2 \end{vmatrix}$ = $\begin{vmatrix} 4 & 4 \\ 5 & 2 \end{vmatrix}$

Given,

$\begin{vmatrix} 4 & x+1 \\ x+2 & 2 \end{vmatrix}$ =

$\begin{vmatrix} 4 &4 \\ 5& 2 \end{vmatrix}$

or,

$8-(x+1)(x+2)=8-20$

or,

$x^2+3x+2=20$

or,

$x^2+3x-18=0$

or,

$(x+6)(x-3)=0$

or,

$x=3,-6$ .

Find the adjoint of the following matrice:
$\begin{bmatrix} \cos\,\alpha & \sin\,\alpha \\ \sin\,\alpha & \cos\,\alpha \end{bmatrix}$
Verify that $(adj A ) A = |A| I = A (adj A)$ for the above matrice.

1416196_1661660_ans_b05d41181cab4ba1a7061e6e2d3c7335.jpg

If $A = \begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}$ ,show that adj $A = 3A^{T}.$

1416598_1661746_ans_825f90c7ecf04989addcc7847b6469dd.jpg

Solve:
$\begin{vmatrix} x+2 & 1 \\ -1 & x\end{vmatrix}=0$

$\begin{vmatrix} x+2 & 1 \\ -1 & x\end{vmatrix}=0$

$\Rightarrow\,x\left(x+2\right)+1=0$

$\Rightarrow\,{x}^{2}+2x+1=0$

$\Rightarrow\,{\left(x+1\right)}^{2}=0$

$\Rightarrow\,x+1=0$

$\therefore\,x=-1$

If $A$ is a square matrix of order $n$ , prove that $\left | A\: adj \: A \right |= \left | A \right |^{n}$ .

$A^{-1} = \dfrac{Adj \, A}{|A|}$

$Adj\, A = A^{-1} . |A|$

$|A . A^{-1} |A||$

$|I. |A||$

$[|A.k| = k^n.|A|]$

$= |A|^n. |I|$

$= |A|^n . I$

$= |A|^n$ .

Find the values of x, if
$\begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix}=\begin{vmatrix} 6 & 5 \\ 8 & 3 \end{vmatrix}$

$\begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix}=\begin{vmatrix} 6 & 5 \\ 8 & 3 \end{vmatrix}$

$\Rightarrow$

$(2x)(x)-(5)(8)=(6)(3)-(5)(8)$

$\Rightarrow$

$2x^2-40=18-40$

$\Rightarrow$

$2x^2=18$

$\Rightarrow$

$x^2=\dfrac{18}{2}$

$\Rightarrow$

$x^2=9$

$\therefore$

$x=\pm 3$

Evaluate the following determinant :

$\begin{vmatrix} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \end{vmatrix}$

Given

$\begin{vmatrix} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \end{vmatrix}$

$=1(-2-10)+3(8-6)+2(20+3)$

$=-12+6+46$

$=40$

Evaluate the following determinant : $=\begin{vmatrix} 67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{vmatrix}$

Given

$\begin{vmatrix} 67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{vmatrix}$

Expanding the determinant

$=67(338-336)-19(1014-1134)+21(936-1053)$

$=134+2280-2457$

$=-43$

Find the integral value of x, if $\begin{vmatrix} x^{2} & x & 1\\ 0 & 2 & 1 \\ 3 & 1 & 4 \end{vmatrix}= 28$

$\begin{vmatrix} x^{2} & x & 1\\ 0 & 2 & 1 \\ 3 & 1 & 4 \end{vmatrix}= 28$

Expanding det on both sides.

$x^2(8-1)-x(0-3)+1(0-6)=28$

$7x^2+3x-6-28=0$

$7x^2+3x-34=0$

$x=2$

Evaluate the following determinant :

$\begin{vmatrix} 1 & 3 & 5 \\ 2 & 6 & 10 \\ 31 & 11 & 38 \end{vmatrix}$

Given

$\begin{vmatrix} 1 & 3 & 5 \\ 2 & 6 & 10 \\ 31 & 11 & 38 \end{vmatrix}$

$=1(228-110)-3(76-310)+5(22-186)$

$=118+702-820$

$=0$

Evaluate the following :

$\begin{vmatrix} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{vmatrix}$

Let

$\Delta = \begin{vmatrix} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{vmatrix}$ . Applying

$C_{1}\rightarrow C_{1}+C_{2}+C_{3},$

we get

$\begin{vmatrix} 3x+\lambda & x & x \\ 3x+\lambda & x+\lambda & x \\ 3x+\lambda & x & x+\lambda \end{vmatrix} = (3x+\lambda ) \begin{vmatrix} 1 & x & x \\ 1 & x+\lambda & x \\ 1 & x & x+\lambda \end{vmatrix}$ [Taking

$(3x+\lambda )$ common from

$C_{1}$ ]

$\Delta = (3x+\lambda ) \begin{vmatrix} 1 & x & x \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{vmatrix}$ [Applying

$R_{2}\rightarrow R_{2}-R_{1},R_{3}\rightarrow R_{3}-R_{1}]$

$\Rightarrow \Delta = (3x+\lambda ) \begin{vmatrix} \lambda & 0 \\ 0 & \lambda \end{vmatrix}$ [Expanding along

$C_{1}$ ]

$\Rightarrow \Delta = \lambda ^{2}(3x+\lambda )$

Evaluate the following :
$\begin{vmatrix} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{vmatrix}$

Let

$\Delta = \begin{vmatrix} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{vmatrix}$ .

Applying

$C_{1}\rightarrow C_{1}+C_{2}+C_{3},$ we get

$= \begin{vmatrix} 3x+\lambda & x & x \\ 3x+\lambda & x+\lambda & x \\ 3x+\lambda & x & x+\lambda \end{vmatrix}$

[Taking

$(3x+\lambda )$ common from

$C_{1}$ ]

$= (3x+\lambda ) \begin{vmatrix} 1 & x & x \\ 1 & x+\lambda & x \\ 1 & x & x+\lambda \end{vmatrix}$

[Applying

$R_{2}\rightarrow R_{2}-R_{1},R_{3}\rightarrow R_{3}-R_{1}]$

$\Delta = (3x+\lambda ) \begin{vmatrix} 1 & x & x \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{vmatrix}$

$\Rightarrow \Delta = (3x+\lambda ) \begin{vmatrix} \lambda & 0 \\ 0 & \lambda & \end{vmatrix}$

[Expanding along

$C_{1}$ ]

$\Rightarrow \Delta = \lambda ^{2}(3x+\lambda )$

Evaluate the following determinant :

$\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}$

Given

$\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}$

Expanding determinant

$=a(bc-f^2)-h(hc-gf)+g(hf-gb)$

$=abc-af^2-ch^2+gfh+ghf-bg^2$

$=abc-af^2-bg^2-ch^2$

If a,b,c are real numbers such that $\begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}= 0$ , then show that either $a+b+c=0$ or, $a=b=c$ .

We have,

$\begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}= 0$

$\Rightarrow \begin{vmatrix} 2(a+b+c) & c+a & a+b \\ 2(a+b+c) & a+b & b+c \\ 2(a+b+c) & b+c & c+a \end{vmatrix}= 0$ [Applying

$C_{1}\rightarrow C_{1}+C_{2}+C_{3}]$

$\Rightarrow 2(a+b+c)\begin{vmatrix} 1 & c+a & a+b \\ 1 & a+b & b+c \\ 1 & b+c & c+a \end{vmatrix}= 0$

$\Rightarrow 2(a+b+c)\begin{vmatrix} 1 & c+a & a+b \\ 0 & b-c & c-a \\ 0 & b-a & c-b \end{vmatrix}= 0$ Applying

$R_{2}\rightarrow R_{2}-R_{1}, R_{3}\rightarrow R_{3}-R_{1}$

$\Rightarrow 2(a+b+c)\left \{ (b-c)(c-b)-(c-a)(b-a) \right \} =0$ [Expanding along

$C_{1}]$

$\Rightarrow 2(a+b+c)\left \{ -(b^{2}-2bc+c^{2})-(bc-ca-ab+a^{2}) \right \} =0$

$\Rightarrow 2(a+b+c)(-a^{2}-b^{2}-c^{2}+bc+ca+ab)=0$

$\Rightarrow 2(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)=0$

$\Rightarrow (a+b+c)(2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca)=0$

$\Rightarrow (a+b+c)\left \{ (a-b)^{2}+(b-c)^{2}+(c-a)^{2} \right \}=0$

$\Rightarrow a+b+c=0$ or,

$(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$

$\Rightarrow a+b+c=0$ or

$a=b=c$

Evaluate the following :

$\begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix}$

Let

$\Delta = \begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix}$ .

Applying

$R_{2}\rightarrow R_{2}-R_{1},R_{3}\rightarrow R_{3}-R_{1},$ we get

$\Delta = \begin{vmatrix} 1 & a & bc \\ 0 & b-a & ca-bc \\ 0 & c-a & ab-bc \end{vmatrix}$

$= \begin{vmatrix} 1 & a & bc \\ 0 & b-a & c(a-b) \\ 0 & c-a & b(a-c) \end{vmatrix}$

$\Rightarrow \Delta = (a-b)(a-c) \begin{vmatrix} 1 & a & bc \\ 0 & -1 & c \\ 0 & -1 & b \end{vmatrix}$

$[Taking \ (a-b) \ and \ (a-c) \ common \ from \$

$R_{1}$ and

$R_{3}$

$respectively]$

$\Rightarrow \Delta = (a-b)(a-c)(-b+c)$

$[Expanding \ along \ first \ column]$

$\Rightarrow \Delta = (a-b)(b-c)(c-a)$

Evaluate the following determinant :

$\begin{vmatrix} 15 & 11 & 7 \\ 11 & 17 & 14 \\ 10 & 16 & 13 \end{vmatrix}$

Given

$\begin{vmatrix} 15 & 11 & 7 \\ 11 & 17 & 14 \\ 10 & 16 & 13 \end{vmatrix}$

Expanding determinant on both sides

$=15(17(13)-16(14))-11(11(13)-14(10))+7(11(16)-10(17))$

$=15(221-224)-11(143-140)+7(176-170)$

$=-45-33+42$

$=-36$

Find the adjoint of the given matrix and verify in each case that $A \cdot (adj A) = (adj A)\cdot A = |A| \cdot I$ .
$\begin{bmatrix} 3& -1 & 1\\ -15 & 6 & -5 \\ 5 & -2 & 2\end{bmatrix}$

1536582_1705135_ans_31f3d4a74226466e960c89013ece71f9.jpg

If $a, b, c$ are $p^{th}, q^{th}$ and $r^{th}$ terms respectively of a G.P, then prove that
$\begin{vmatrix}\log a& p & 1 \\ \log b& q & 1\\ \log c &r & 1 \end{vmatrix} =0$

It is given that in a G. P., $p^{th}$ term $=a$ , $q^{th}$ term $=b$ and $r^{th}$ term $=c$ .

Let $A$ and $R$ be the first term and the common ratio of the G.P.

Then,

$a = AR^{p-1} \Rightarrow \log a = \log (AR^{p-1}) =\log A + \log R^{p-1} = \log A+(p-1)\log R$

i.e., $\log a = \log A + (p – 1) \log R$

Similarly,

$b = AR^{q-1} \Rightarrow \log b = \log A + (q-1) \log R$

$c = AR^{r-1} \Rightarrow \log c = \log A + (r - 1) \log R$

Now, $\Delta = \begin{vmatrix}\log a & p & 1 \\ \log b&q&1 \\\log c&r&1 \end{vmatrix}$

$\Rightarrow \Delta = \begin{vmatrix} \log A + (p-1) \log R&p&1 \\\log A + (q-1)\log R &q & 1\\ \log A + (r-1)\log R & r&1 \end{vmatrix}$

$C_2 \to C_2 - C_3$

$\Rightarrow \Delta = \begin{vmatrix} \log A + (p-1) \log R&p-1&1 \\\log A + (q-1)\log R &q-1 & 1\\ \log A + (r-1)\log R & r-1&1 \end{vmatrix}$

$C_1 \to C_1 - (\log A)C_3 - (\log R) C_2$

$\Rightarrow \Delta = \begin{vmatrix} 0&p-1&1 \\0 &q-1 & 1\\ 0 & r-1&1 \end{vmatrix}$

Hence proved.

Find the cofactors of all the elements of $\begin{bmatrix}1&-2 \\ 4 & 3 \end{bmatrix}$

$M_{11}=3$

$M_{12}=4$

$M_{21}=-2$

$M_{22}=1$

$C_{11}=(-1)^{1+1}M_{11}$

$=3$

$C_{12}=(-1)^{1+2}M_{12}$

$=-(4)$

$=-4$

$C_{21}=(-1)^{2+1}M_{21}$

$=-(-2)$

$=2$

$C_{22}=(-1)^{2+2}M_{22}$

$=1$

$\therefore$ Co-factor of

$\begin{bmatrix} 1& -2\\ 4 & 3\end{bmatrix}$ is

$\begin{bmatrix} 3 & -4 \\ 2 & 1 \end{bmatrix}$

Find the adjoint of the given matrix and verify in each case that $A \cdot (adj A) = (adj A)\cdot A = |A| \cdot I$ .
$\begin{bmatrix} 0& 1 & 2\\ 1 & 2 & 3 \\ 3 & 1 & 1\end{bmatrix}$

1536539_1705136_ans_b024173ad2c64d4baf21c7febcdb8561.jpg

Find the adjoint of the given matrix and verify in each case that $A \cdot (adj A) = (adj A)\cdot A = |A| \cdot I$ .
$\begin{bmatrix} 2& 3\\ 5 & 9\end{bmatrix}$

1535660_1705131_ans_d7bf25a27d614e8fbd5be8450452d45a.jpg

Find the adjoint of the given matrix and verify in each case that $A \cdot (adj A) = (adj A)\cdot A = |A| \cdot I$ .
$\begin{bmatrix} \cos \alpha& \sin \alpha\\ \sin \alpha & \cos \alpha\end{bmatrix}$

1535721_1705133_ans_cbd4d65dd64341a4ac5b0437dd50531e.jpg

Find the adjoint of the given matrix and verify in each case that $A \cdot (adj A) = (adj A)\cdot A = |A| \cdot I$ .
$\begin{bmatrix} 9& 7 & 3\\ 5 & -1 & 4 \\ 6 & 8 & 2\end{bmatrix}$

1536433_1705137_ans_715faca6a98045c2b18979e072d4c334.jpg

If $A$ and $B$ are square matrices each of order $3$ and $|A|= 5, \, |B| = 3$ , then the value of $|3AB|$ is ______

As we know

$|KA| = K^n|A|$

where

$n$ is the order of the matrix.

$\therefore |3AB| = 3^3|A| |B|$

(

$\because$ order given is

$3$ )

$\therefore |3AB| = 3^3 \times |A| |B|$

$= 27 \times 5 \times 3$

$= 27 \times 15$

$= 405$

Find the adjoint of the given matrix and verify in each case that $A \cdot (adj A) = (adj A)\cdot A = |A| \cdot I$ .
$\begin{bmatrix} 1& -1 & 2\\ 3 & 1 & -2 \\ 1 & 0 & 3\end{bmatrix}$

1536615_1705134_ans_366f0783c4504b8c86a12f29d01a494a.jpg

Find the adjoint of the given matrix and verify in each case that $A \cdot (adj A) = (adj A)\cdot A = |A| \cdot I$ .
$\begin{bmatrix} 3& -5\\ -1 & 2\end{bmatrix}$

1535674_1705132_ans_486aa06e177d4b98b3fefb5c366b46fb.jpg

Let $A$ be a square matrix of order $3$ , write the value of $|2A|$ , where $|A|=4$ .

Given

$A$ is a square matrix of order

$3$

then we know

$|2A|=2^3|A|$

$=8|A|$

given

$|A|=4$

then

$|2A|=8|A|$

$=8\times 4$

$=32$

Hence

$|2A|=32$ .

Evaluate $\begin{vmatrix} \sqrt { 6 } & \sqrt { 5 } \\ \sqrt { 20 } & \sqrt { 24 } \end{vmatrix}$ .

Consider

$\Delta =\begin{vmatrix} \sqrt { 6 } & \sqrt { 5 } \\ \sqrt { 20 } & \sqrt { 24 } \end{vmatrix}$ .

$\Delta =\sqrt{6}.\sqrt{24}-\sqrt{20}.\sqrt{5}$

$=\sqrt{144}-\sqrt{100}$

$=12-10$

$=2$

If $A=\begin{vmatrix} 3 & 4 \\ 1 & 2 \end{vmatrix}$ , find the value of $3|A|$ .

Given

$|A|$

$=\begin{vmatrix} 3 & 4 \\ 1 & 2 \end{vmatrix}$ then

$3.|A|$

$|A|=6-4$

$=2$

Hence

$3.|A|=3.2$

$=6.$

Find the adjoint of the given matrix and verify in each case that $A \cdot (adj A) = (adj A)\cdot A = |A| \cdot I$ .
$\begin{bmatrix} 4& 5 & 3\\ 1 & 0 & 6 \\ 2 & 7 & 9\end{bmatrix}$

1536687_1705138_ans_433c0db737744cb6a4927aca89fc8c6b.jpg

If $A$ is a $2\times 2$ matrix such that $|A|\neq 0$ and $|A|=5$ , write the value of $|4A|$ .

We know if

$A$ is matrix of

$n\times n$ then

$|KA|=k^n|A|$

here order of

$A$ is

$2$

$|4A|=4^2|A|$

$=16\times 5$

$=80$

If $A$ is a $3\times 3$ matrix such that $|A|\neq 0$ and $|3A|=k|A|$ , then write the value of $k$ .

We know if

$A$ is a matrix of order

$n\times n$ then

$|KA|=k^n|A|$

here

$|3A|=3^3|A|$

$=27|A|$

Hence

$K=27$ .

Evaluate $\begin{vmatrix} \sin { { 60 }^{ o } } & \cos { { 60 }^{ o } } \\ -\sin { { 30 }^{ o } } & \cos { { 30 }^{ o } } \end{vmatrix}$ .

Consider

$\Delta =\begin{vmatrix} \sin { { 60 }^{ o } } & \cos { { 60 }^{ o } } \\ -\sin { { 30 }^{ o } } & \cos { { 30 }^{ o } } \end{vmatrix}$ .

By expanding we get

$\Delta =\sin 60^0\cos 30^0+\sin 30^0\cos 60^0$

$=\sin (60^0+30^0)$

$=\sin 90^0$

$\Delta =1$

Hence

$\begin{vmatrix} \sin { { 60 }^{ o } } & \cos { { 60 }^{ o } } \\ -\sin { { 30 }^{ o } } & \cos { { 30 }^{ o } } \end{vmatrix}=1$ .

Find the adjoint of the given matrix and verify in each case that $A \cdot (adj A) = (adj A)\cdot A = |A| \cdot I$ .
$\begin{bmatrix} \cos \alpha& -\sin \alpha & 0\\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{bmatrix}$

1683708_1705139_ans_5b5c49247a3d4ce6ac9f7fb200daa0a4.jpeg

Evaluate $\begin{vmatrix} 2\cos { \theta } & -2\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{vmatrix}$ .

Consider

$\Delta=\begin{vmatrix} 2\cos { \theta } & -2\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{vmatrix}$

$=2\cos^2\theta+2\sin^2\theta$

$=2(\cos^2\theta+\sin^2\theta)$

$=2$

Hence

$\Delta=2$

Evaluate $\begin{vmatrix} \cos { \alpha } & -\sin { \alpha } \\ \sin { \alpha } & \cos { \alpha } \end{vmatrix}$ .

Consider

$\Delta=\begin{vmatrix} \cos { \alpha } & -\sin { \alpha } \\ \sin { \alpha } & \cos { \alpha } \end{vmatrix}$ .

$=\cos^2\alpha+\sin^2\alpha$

$=1$

Hence

$\Delta=1$

Evaluate $\begin{vmatrix} \cos { { 65 }^{ o } } & \sin { { 65 }^{ o } } \\ \sin { 25^{ o } } & \cos { 25^{ o } } \end{vmatrix}$ .

Consider

$\Delta=\begin{vmatrix} \cos { { 65 }^{ o } } & \sin { { 65 }^{ o } } \\ \sin { 25^{ o } } & \cos { 25^{ o } } \end{vmatrix}$

By expanding we get

$\Delta=\cos 65^0 \cdot \cos 25^0-\sin 25^0 \cdot \sin 65^0$

$=\cos(65^0+25^0)$ .................

$\cos (a+b)=\cos a \cdot \cos b-\sin a \cdot \sin b$

$=\cos 90^0$

Hence

$\Delta=0$

Evaluate $\begin{vmatrix} \cos { { 15 }^{ o } } & \sin { { 15 }^{ o } } \\ \sin { 75^{ o } } & \cos { 75^{ o } } \end{vmatrix}$ .

Consider

$\Delta =\begin{vmatrix} \cos { { 15 }^{ o } } & \sin { { 15 }^{ o } } \\ \sin { 75^{ o } } & \cos { 75^{ o } } \end{vmatrix}$

By expanding we get

$\Delta =\cos 15^0\cos 75^0-\sin 15^0\sin 75^0$

$=\cos(15^0+75^0)$ .................

$\cos (a+b)=\cos a \cdot \cos b-\sin a \cdot \sin b$

$=\cos 90^0$

Hence

$\Delta =0$

Evaluate $\begin{bmatrix} \sqrt { 3 } & \sqrt { 5 } \\ -\sqrt { 5 } & 3\sqrt { 3 } \end{bmatrix}$ .

Let us consider

$\begin{bmatrix} \sqrt { 3 } & \sqrt { 5 } \\ -\sqrt { 5 } & 3\sqrt { 3 } \end{bmatrix}$

$\sqrt{3}.3\sqrt{3}+\sqrt{5}.\sqrt{5}$

$=9+5$

$=14$

Prove that the three straight lines whose equations are
$15x - 18y + 1 = 0, 12x + 10y -3 = 0$ , and $6x + 66y - 11 = 0$
all meet in a point.
Show also that the third line bisects the angle between the other two.

1686966_1740718_ans_47880aad60bd4aa4bf9c2631dc9180e7.jpg

Prove that the following sets of three lines meet in a point.
$2x - 3y = 7, 3x - 4y = 13$ , and $8x - 11y = 33$ .

1766791_1740717_ans_47ac5e80ea904feb8df87a6db6b13bb3.jpg

If $\begin{vmatrix} { \left( \beta +\gamma -\alpha -\delta \right) }^{ 4 } & { \left( \beta +\gamma -\alpha -\delta \right) }^{ 2 } & 1 \\ { \left( \gamma +\alpha -\beta -\delta \right) }^{ 4 } & { \left( \gamma +\alpha -\beta -\delta \right) }^{ 2 } & 1 \\ { \left( \alpha +\beta -\gamma -\delta \right) }^{ 4 } & { \left( \alpha +\beta -\gamma -\delta \right) }^{ 2 } & 1 \end{vmatrix}=k\left( \alpha -\beta \right) \left( \alpha -\gamma \right) \left( \alpha -\delta \right) \left( \beta -\gamma \right) \left( \beta -\delta \right) \left( \gamma -\delta \right) \quad$ then the value of ${(k)}^{1/2}$ is

let

$D=\begin{vmatrix} { \left( \beta +\gamma -\alpha -\delta \right) }^{ 4 } & { \left( \beta +\gamma -\alpha -\delta \right) }^{ 2 } & 1 \\ { \left( \gamma +\alpha -\beta -\delta \right) }^{ 4 } & { \left( \gamma +\alpha -\beta -\delta \right) }^{ 2 } & 1 \\ { \left( \alpha +\beta -\gamma -\delta \right) }^{ 4 } & { \left( \alpha +\beta -\gamma -\delta \right) }^{ 2 } & 1 \end{vmatrix}$

applying

${ R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 3 };{ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 3 }$

$=\begin{vmatrix} { \left( \beta +\gamma -\alpha -\delta \right) }^{ 4 } & { \left( \beta +\gamma -\alpha -\delta \right) }^{ 2 }-{ \left( \alpha +\beta -\gamma -\delta \right) }^{ 2 } & 0 \\ { \left( \gamma +\alpha -\beta -\delta \right) }^{ 4 } & { \left( \gamma +\alpha -\beta -\delta \right) }^{ 2 }-{ \left( \alpha +\beta -\gamma -\delta \right) }^{ 2 } & 0 \\ { \left( \alpha +\beta -\gamma -\delta \right) }^{ 4 } & { \left( \alpha +\beta -\gamma -\delta \right) }^{ 2 } & 1 \end{vmatrix}\\=4\left( \beta -\delta \right) \left( \gamma -\alpha \right) .4\left( \alpha -\delta \right) \left( \gamma -\beta \right) \times \begin{vmatrix} { \left( \beta +\gamma -\alpha -\delta \right) }^{ 2 }-{ \left( \alpha +\beta -\gamma -\delta \right) }^{ 2 } & 1 & 0 \\ { \left( \gamma +\alpha -\beta -\delta \right) }^{ 2 }-{ \left( \alpha +\beta -\gamma -\delta \right) }^{ 2 } & 1 & 0 \\ { \left( \alpha +\beta -\gamma -\delta \right) }^{ 4 } & { \left( \alpha +\beta -\gamma -\delta \right) }^{ 2 } & 1 \end{vmatrix}\quad$

applying

${ R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 }$

$=16\left( \beta -\delta \right) \left( \gamma -\alpha \right) .4\left( \alpha -\delta \right) \left( \gamma -\beta \right) \begin{vmatrix} 1 & 1 & 0 \\ { \left( \gamma +\alpha -\beta -\delta \right) }^{ 2 }-{ \left( \alpha +\beta -\gamma -\delta \right) }^{ 2 } & 1 & 0 \\ { \left( \alpha +\beta -\gamma -\delta \right) }^{ 4 } & { \left( \alpha +\beta -\gamma -\delta \right) }^{ 2 } & 1 \end{vmatrix}$

$=64\left( \alpha -\beta \right) \left( \alpha -\gamma \right) \left( \alpha -\delta \right) \left( \beta -\gamma \right) \left( \beta -\delta \right) \left( \gamma -\delta \right)$

$\therefore \,k=64\Rightarrow\,k^{\frac12}=8$

Given $A=\begin{vmatrix} a & b & 2c \\ d & e & 2f \\ l & m & 2n \end{vmatrix};B=\begin{vmatrix} f & 2d & e \\ 2n & 4l & 2m \\ c & 2a & b \end{vmatrix}$ , then the value of $\dfrac{B}{A}$ is ________.

Given

$B=\left| \begin{matrix} f & 2d & e \\ 2n & 4l & 2m \\ c & 2a & b \end{matrix} \right|$

Take 2 as a common from

$R_{2}$ and 2 as a common from

${ C }_{ 2 }$

$\therefore B=2\times 2\left| \begin{matrix} f & d & e \\ n & l & m \\ c & a & b \end{matrix} \right|$

$\therefore B=2\left| \begin{matrix} 2f & d & e \\ 2n & l & m \\ 2c & a & b \end{matrix} \right|$

${ R }_{ 3 }\leftrightarrow { R }_{ 2 }$

$\therefore B=2\left| \begin{matrix} 2f & d & e \\ 2c & a & b \\ 2n & l & m \end{matrix} \right|$

${ R }_{ 2 }\leftrightarrow { R }_{ 1 }$

$\therefore B=2\left| \begin{matrix} 2c & a & b \\ 2f & d & e \\ 2n & l & m \end{matrix} \right|$

${ C }_{ 1 }\leftrightarrow { C }_{ 2 }$

$\therefore B=2\left| \begin{matrix} a & 2c & b \\ d & 2f & e \\ l & 2n & m \end{matrix} \right|$

${ C }_{ 2 }\leftrightarrow { C }_{ 3 }$

$\\ \therefore B=2\left| \begin{matrix} a & b & 2c \\ d & e & 2f \\ l & m & 2n \end{matrix} \right|$ (1)

Now, Given

$\\ A=\left| \begin{matrix} a & b & 2c \\ d & e & 2f \\ l & m & 2n \end{matrix} \right|$ (2)

From equations (1) and (2), we get,

$B=2A$

$\therefore \frac { B }{ A } =2$

Find the minor of the element of second row and third column ( $a 23$ ) in the following determinant:
$\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$

We have,

$\left|\begin{array}{ccc}2 & -3 & 5 \\6 & 0 & 4 \\1 & 5 & -7\end{array}\right|\\$
Minor of an element

$\quad a_{23}=M_{23}=\left|\begin{array}{cc}2 & -3 \\ 1 & 5\end{array}\right|=10+3=13$

If $A$ is an invertible metrix of order $3$ and $|A| = 5$ , then find $|adj \ A|$ .

$|adj \ A| = |A|^{3 - 1} = 5^2 = 25$ [

$\because |adj \ A| = |A|^{n - 1}$ , where

$n$ is order of A]

If $\Delta=\left|\begin{array}{lll}5 & 3 & 8 \\2 & 0 & 1 \\1 & 2 & 3\end{array}\right| \text { , then write the minor of the element } a_{23}$

$\textbf{Step-1: Apply the concept of determinant.}$

$\text{Given, }\Delta=\left|\begin{array}{lll}5 & 3 & 8 \\2 & 0 & 1 \\1 & 2 & 3\end{array}\right|$

$\text {Minor of } a_{23}=M_{23}=\left|\begin{array}{cc}5 & 3 \\1 & 2\end{array}\right|=10-3=7$

$\textbf{Hence, the answer is 7}$

What positive value of $x$ makes the following pair of determinants equal?
$\left|\begin{array}{cc}2 x & 3 \\ 5 & x\end{array}\right|,\left|\begin{array}{cc}16 & 3 \\ 5 & 2\end{array}\right|$

$\begin{aligned}&\begin{array}{l}\because \quad\left|\begin{array}{cc}2 x & 3 \\5 & x\end{array}\right|=\left|\begin{array}{cc}16 & 3 \\5 & 2\end{array}\right| \\\Rightarrow \quad 2 x^{2}-15=32-15 \quad \Rightarrow \quad 2 x^{2}=32 \\\Rightarrow \quad x^{2}=16 \quad \Rightarrow \quad x=\pm 4\end{array}\\&\Rightarrow \quad x=4 \quad(\text { +ve value })\end{aligned}$

Find the cofactor of $a_{12}$ in the following:
$\left|\begin{array}{ccc}1 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$

$\begin{array}{l}a_{12}=(-1)^{1+2} \cdot M_{12} \\\quad=-\left|\begin{array}{cc}6 & 4 \\1 & -7\end{array}\right| \\=-(-42-4)=46\end{array}$

$\begin{aligned}& \\&\text { Evaluate: }\left|\begin{array}{ll}\cos 15^{\circ} & \sin 15^{\circ} \\\sin 75^{\circ} & \cos 75^{\circ}\end{array}\right|\end{aligned}$

Expanding the determinant, we get

$\begin{array}{l}\cos 15^{\circ} \cdot \cos 75^{\circ}-\sin 15^{\circ} \cdot \sin 75^{\circ}=\cos \left(15^{\circ}+75^{\circ}\right)=\cos 90^{\circ}=0 \\ \text{Note : } \cos (A+B)=\cos A \cdot \cos B-\sin A \cdot \sin B\end{array}\\$

If $A$ is idempotent matrix satisfying $( I - 0.4 A)^{-1} = I - \alpha A$ where $I$ is the unit matrix of the same order as that of $A$ then find the value of $| 9 \alpha |$ .

1691764_1750143_ans_679c1e550ef246ec87baf459fc37a59b.PNG

Find the equation of line joining (1,2) and (3,6) using determinants.

$\textbf{Step 1: Find the equation using suitable condition.}$

$\text{Let, $$P(x, y)$$ be the general point on the line joining $$A(1,2)$$ and $$B(3,6)$$}$

$\text{We know, area of a triangle with vertices }(x_1,y_1),(x_2,y_2)\text{ and }(x_3,y_3)\text{ is given by,}$

$\Rightarrow\text{Area }=\dfrac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\x_2 & y_2 & 1 \\x_3 & y_3 & 1\end{array}\right|$

$\text{From the figure, it is obvious that the area of }\triangle \text{APB } \text{is zero.}$

$\Rightarrow\dfrac{1}{2}\left|\begin{array}{lll}x & y & 1 \\1 & 2 & 1 \\3 & 6 & 1\end{array}\right|=0$

$\Rightarrow\left|\begin{array}{lll}x & y & 1 \\1 & 2 & 1 \\3 & 6 & 1\end{array}\right|=0$

$\Rightarrow x(2-6)-y(1-3)+1(6-6)=0$

$\Rightarrow-4 x+2 y=0$

$\Rightarrow4 x-2 y=0$

$\Rightarrow2 x- y=0$

$\textbf{Hence, the required equation of the line is : }\mathbf{2x-y=0.}$

2177866_1781588_ans_e349bd4cce68429287a7137c4571a3c7.jpg

Find the value of $\left|\begin{array}{cccc}\sin A & -\sin B \\\cos A & \cos B\end{array}\right|$ $\text { where } A=53^{\circ}, B=37^{\circ}$

$\begin{array}{l}\text { Let } \Delta=\left|\begin{array}{lll}\sin A & -\sin B \\\cos A & \cos B\end{array}\right| \\\Delta=\sin A \cos B+\sin B \cos A \quad \Rightarrow \quad \Delta=\sin (A+B) \\\Delta=\sin (53+37)=\sin 90^{\circ}=1\end{array}$

If $A = \begin{bmatrix} \alpha & 0 & 0 \\ 0 & \alpha & 0 \\ 0 & 0 & \alpha \end{bmatrix}$ , find the value of $|adj \ A|$ .

$|adj \ A| = |A|^2$

$[\because |adj \ A|= |A|^{n-1}$ , where n is the order of matrix]

$$=\begin{bmatrix}
\alpha & 0 & 0 \\
0 & \alpha & 0 \\
0 & 0 & \alpha

\end{bmatrix}^2$$

$(\alpha^3)^2 = \alpha^6$

$[\because$ Determinant of a triangular matrix is equal to the product of its diagonal elements]

If $\Delta= \left| \begin{matrix} 0 & b-a & c-a \\ a-b & 0 & c-b \\ a-c & b-c & 0 \end{matrix} \right|$ then show that $\Delta$ is equal to zero.

We have

$\Delta_1 = \left| \begin{matrix} 1 & 1 & 1 \\ yz & zx & xy \\ x & y & z \end{matrix} \right|$
Intercjanging rows and columns, we get

$\Delta_1= \left| \begin{matrix} 1 & yz & x \\ 1 & zx & y \\ 1 & xy & z \end{matrix} \right| =\frac { 1 }{ xyz } \left| \begin{matrix} x & xyz & { x }^{ 2 } \\ y & xyz & { y }^{ 2 } \\ z & xyz & { z }^{ 2 } \end{matrix} \right|$

$= \frac{xyz}{xyz} \left| \begin{matrix} x & 1 & { x }^{ 2 } \\ y & 1 & { y }^{ 2 } \\ z & 1 & { z }^{ 2 } \end{matrix} \right|$
Interchanging

$C_1$ and

$C_2$

$= (-1) \left| \begin{matrix} 1 & x & { x }^{ 2 } \\ 1 & y & y^{ 2 } \\ 1 & z & { z }^{ 2 } \end{matrix} \right| = - \Delta$

$\Rightarrow \Delta_1 + \Delta=0$

Write the adjoint of the following matrix:
$\Big[\begin{matrix} 2 & -1 \\ 4 & 3 \end{matrix}\Big]$

Given,

$\Big[\begin{matrix}2 & -1 \\4 & 3\end{matrix}\Big]$

$C_{11} = 3, C_{12} = -4; C_{21} = 1, C_{22} = 2$

$\therefore$ adj A =

$\Big[\begin{matrix}3 & 1 \\-4 & 2\end{matrix}\Big]$

If $\left|\begin{array}{ll}x & x \\1 & x\end{array}\right|=\left|\begin{array}{ll}3 & 4 \\1 & 2\end{array}\right|$ , then write the positive value of $x$

$\begin{array}{l}\text { We have }\left|\begin{array}{cc}x & x \\1 & x\end{array}\right|=\left|\begin{array}{cc}3 & 4 \\1 & 2\end{array}\right| \\\Rightarrow \quad x^{2}-x=6-4 \quad \Rightarrow \quad x^{2}-x-2=0\end{array}\\$

$\Rightarrow x^{2}-2 x+x-2=0 \quad \Rightarrow \quad x(x-2)+1(x-2)=0\\$

$\Rightarrow (x-2)(x+1)=0 \quad \Rightarrow \quad x=2 \quad$ or

$\quad x=-1 \quad(\text { Not accepted })$

$\Rightarrow x=2\\$

Find the equation of the line joining $A(1,3)$ and $B(0,0)$ using determinants and find $k$ if $D(k, 0)$ is a point such that the area of $\Delta A B D$ is 3 sq units.

Let

$P(x, y)$ be any point on the line

$A B$ . Then

$\operatorname{area}(\Delta A B P)=0$

$\Rightarrow \quad \cfrac{1}{2}\left|\begin{array}{lll}1 & 3 & 1 \\0 & 0 & 1 \\x & y & 1\end{array}\right|=0 \quad$

$\Rightarrow \quad \cfrac{1}{2}\{1(0-y)-3(0-x)+1(0-0)\}=0\\$

$\Rightarrow \quad 3 x-y=0,$ which is the required equation of line

$A B$

Now, area

$(\Delta A B D)=3$ sq units

$\Rightarrow \quad \cfrac{1}{2}\left|\begin{array}{lll}1 & 3 & 1 \\0 & 0 & 1 \\k & 0 & 1\end{array}\right|=\pm 3 \quad$

$\Rightarrow \quad\left|\begin{array}{lll}1 & 3 & 1 \\0 & 0 & 1 \\k & 0 & 1\end{array}\right|=\pm 6\\$

$\Rightarrow \quad 1(0-0)-3(0-k)+1(0-0)=\pm 6$

$\Rightarrow 3 k=\pm 6$

$\Rightarrow k=\pm 2$

The determinant $\Delta = \left| \begin{matrix} \sqrt {23} + \sqrt {3} & \sqrt {5} & \sqrt {5} \\ \sqrt {15} + \sqrt {46} & 5 & \sqrt {10} \\ 3 +\sqrt {115} & \sqrt {15} & 5 \end{matrix} \right|$ is equal to ...............

Answer is

$0$ , Apply

$R_2 \rightarrow R_2-R_1, R_3 \rightarrow R_3-R_1$ .

Show that if the determinant $\Delta = \left| \begin{matrix} 3 & -2 & sin\quad 3\quad \theta \\ -7 & 8& cos \quad 2 \theta \\ -11 & 14 & 2 \end{matrix} \right| = 0$ then $Sin \theta = 0$ or $\frac {1}{2}$

Applying

$R_2 \rightarrow R_2+ 4R_1$ and

$R_3 \rightarrow R_3 + 7R_1$ , we get

$\left| \begin{matrix} 3 & -2 & sin\quad 3\theta \\ 5 & 0 & cos2\theta +4sin3\theta \\ 10 & 0 & 2+7sin3\theta \end{matrix} \right| =0$
or

$2 [5(2+7 sin 3 \theta)-10(cos2 \theta+4sin3 \theta)]=0$
or

$2+7 sin 3 \theta-2 cos2 \theta - 8 sin 3 \theta =0$
or

$2-2cos 2 \theta -sin 3 \theta = 0$

$sin \theta (4sin^2 \theta+ 4 sin \theta - 3)=0$
or

$sin \theta =0 or (2 sin\theta -1) = 0 or (2sin \theta + 3)=0$
or

$sin \theta = 0 or sin \theta = \frac{1}{2}$ .

Write the value of $\left|\begin{array}{cc}\sin 20^{\circ} & -\cos 20^{\circ} \\ \sin 70^{\circ} & \cos 70^{\circ}\end{array}\right|$

$\left|\begin{array}{cc}\sin 20^{\circ} & -\cos 20^{\circ} \\\sin 70^{\circ} & \cos 70^{\circ}\end{array}\right|=\sin 20^{\circ} \cdot \cos 70^{\circ}+\cos 20^{\circ} \cdot \sin 70^{\circ}\\$

$=\sin \left(20^{\circ}+70^{\circ}\right)=\sin 90^{\circ}=1$

Evaluate :
$\left| \begin{matrix} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{matrix} \right| =\quad 4xyz$

$\left| \begin{matrix} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{matrix} \right| =\quad 4xyz$

$\therefore LHS = \left| \begin{matrix} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{matrix} \right|$

$\left| \begin{matrix} y+z+z+y & z & y \\ z+z+x+x & z+x & x \\ y+x+x+y & x & x+y \end{matrix} \right| [ \because C_1 \rightarrow C_1+C_2+C_3]$

$= 2 \left| \begin{matrix} (y+z) & z & y \\ (z+x) & z+x & x \\ (x+y) & x & x+y \end{matrix} \right| [Taking\quad 2\quad common\quad from\quad C_1]$

$= 2\left| \begin{matrix} y & z & y \\ 0 & z+x & x \\ y & x & x+y \end{matrix} \right| [ \because R_1 \rightarrow R_1-R_3]$

$= 2 [ y (xz-x^2 + xz +x^2) ]$

$= 4xyz =RHS$ hence proved.

If $A$ is a matrix or order $3 \times 3$ then number of minors in determinant of $A$ are__________

$A$ is a matrix or order

$3 \times 3$ then number of minors in determinant of

$A$ are

$9$ .

[Since in a

$3 \times 3$ matrix , these are

$9$ elements ]

Fill in the blank.
If $A$ is matrix of order $3 \times 3$ then $|3A|$ is equal to_________

${\textbf{Step -1: Formulate the determinant using the given matrix value}}{\textbf{.}}$

${\text{Given: A is a 3}} \times {\text{3 matrix}}{\text{.}}$

${\text{For any }}n \times n,{\text{ here 3}} \times {\text{3 matrix also called square matrix}}{\text{.}}$

${\text{For square matrix of order n its determinant will be }}\left| A \right|.$

${\textbf{Step -2: Finding the value of }}$

${\mathbf{\left| {3A} \right|.}}$

${\text{So, for any scalar k, }}\left| {kA} \right| = {k^n}\left| A \right|$

$\left[ {{\textbf{Property of matrix}}} \right]$

$\;{\text{Hence, for n = 3, }}\left| {3A} \right| = {3^3}\left| A \right|$

$= 27\left| A \right|$

${\textbf{Hence, value of }}$

${\mathbf{\left| {3A} \right| = {\text{27}}\left| A \right|}}$

${\textbf{.}}$

If $x+y+z = 0$ prove that $\left| \begin{matrix} xa & yb & zc \\ yc & za & xb \\ zb & xc & ya \end{matrix} \right| \quad =\quad xyz\quad \left| \begin{matrix} a & b & c \\ c & a & b \\ b & c & a \end{matrix} \right|$

Since

$x+y+z = 0$ also we have to prove

$\left| \begin{matrix} xa & yb & zc \\ yc & za & xb \\ zb & xc & ya \end{matrix} \right| \quad =\quad xyz\quad \left| \begin{matrix} a & b & c \\ c & a & b \\ b & c & a \end{matrix} \right|$

$\therefore LHS = \left| \begin{matrix} xa & yb & zc \\ yc & za & xb \\ zb & xc & ya \end{matrix} \right|$

$= xa(za.ya-xbxc) -yb(yc.ya-xbzb) +zc(yc.xc-za.zb)$

$=xa(a^2yz -x^2bc)-yb(y^2ac -b^2xz) +zc (c^2xy -z^2 ab)$

$= xyza^3 -x^3 abc -y63 abc +b^3 xyz +c^3 xyz -z^3 abc$

$=xyz (a^3 +b^3 +c^3 )-abc (x^3 +y^3 +z^3)$

$=xyz (a^3 +b^3 +c^3)-abc(3xyz)$

$[ \because x+y+z = 0 \Rightarrow x^3 +y^3 +z^3 - 3xyz ]$

$=xyz (a^3 +b^3 +c^3 - 3abc)......(i)$

Now

$RHS = xyz \left| \begin{matrix} a & b & c \\ c & a & b \\ b & c & a \end{matrix} \right| =xyz\left| \begin{matrix} a+b+c & b & c \\ a+b+c & a & b \\ a+b+c & c & a \end{matrix} \right| [ \because C_1 \rightarrow C_1+C_2+C_3 ]$

$= xyz ( a+b+c) \left| \begin{matrix} 1 & b & c \\ 1 & a & b \\ 1 & c & a \end{matrix} \right|$

[taking

$(a+b+c)$ common from

$c_1$ ]

$= xyz (a+b+c) \left| \begin{matrix} 0 & b-c & c-a \\ 0 & c-a & b-a \\ 1 & c & a \end{matrix} \right|$

$[ \because R_1 \rightarrow R_1 -R_3$ and

$R_2 \rightarrow R_2 -R_3 ]$

expanding along

$C_1$

$=xyz(a+b+c) [ 1(b-c)(b-a) -(a-c)(c-a) ]$

$= xyz ( a+b+c)(b^2 -ab -bc +ac +a^2 +c^2 -2ac)$

$= xyz ( a+b+c)(a^2 +b^2 +c^2 -ab-bc-ca)$

$= xyz (a^3 +b^3 +c^3 -3abc) .....(ii)$

From eqs (i) and (ii)

$\left| \begin{matrix} xa & yb & zc \\ yc & za & xb \\ zb & xc & ya \end{matrix} \right| \quad =\quad xyz\quad \left| \begin{matrix} a & b & c \\ c & a & b \\ b & c & a \end{matrix} \right|$

hence proved.

Evaluate :
$\left| \begin{matrix} 3x & -x+y & -x+z \\ x-y & 3y & z-y \\ x-z & y-z & 3z \end{matrix} \right|$

We have

$\left| \begin{matrix} 3x & -x+y & -x+z \\ x-y & 3y & z-y \\ x-z & y-z & 3z \end{matrix} \right|$
Applying

$C_1 \rightarrow C_1 +C_2 + C_3$

$= \left| \begin{matrix} x+y+z & -x+y & -x+z \\ x+y+z & 3y & z-y \\ x+y+z & y-z & 3z \end{matrix} \right|$

$=(x+y+z) \left| \begin{matrix} 1 & -x+y & -x+z \\ 1 & 3y & z-y \\ 1 & y-z & 3z \end{matrix} \right|$
[ Taking

$x+y+z$ common from column

$C_1$ ]

$=(x+y+z) \left| \begin{matrix} 1 & -x+y & -x+z \\ 0 & 2y+x & x-y \\ 0 & x-z & 2z+x \end{matrix} \right|$

$[ \because R_2 \rightarrow R_2 -R_1 \quad and \quad R_3 \rightarrow R_3-R_1 ]$
Now , expanding along first column , we get

$( x +y+ z) . 1 [(2y+x)(2z+x) -(x-y)(x-z) ]$

$= (x +y+ z)(4yz +2yx +2xz + x^2 -x^2 +xz+yx -yz)$

$=( x+y+z+) (3yz +3yx + 3xz )$

$= 3 (x +y+z)(yz +yx +xz)$

If $A +B +C = 0$ then prove that $\left| \begin{matrix} 1 & cos\quad C & cos\quad B \\ cos\quad C & 1 & cos\quad A \\ cos\quad B\quad & cos\quad A\quad & 1 \end{matrix} \right| =0$

we have

$\left| \begin{matrix} 1 & cos\quad C & cos\quad B \\ cos\quad C & 1 & cos\quad A \\ cos\quad B\quad & cos\quad A\quad & 1 \end{matrix} \right| =0$

$\therefore \left| \begin{matrix} 1 & cos\quad C & cos\quad B \\ cos\quad C & 1 & cos\quad A \\ cos\quad B\quad & cos\quad A\quad & 1 \end{matrix} \right| =0$

$= 1 (1-cos^2A) - cos C ( cos C - cos A . cos B ) +cos B ( cos C .cos A - cos B )$

$= sin^2 A -cos^2 C+ cos A.cos B .cos C + cos A. cos B.cos C - cos^2 B$

$=sin^2 A - cos^2 B + 2cos A .cos B.cos C - cos^2 C$

$= -cos^(A+B).cos (A-B) + 2cos A.cosB.cos C -cos^2 C$

$[ \because cos^2 B -sin^2 A =cos (A+B) . cos (A-B) ]$

$= -cos (-C). cos (A-B) + cos C ( 2cos A.cos B - cos C ) [ \because cos (-\theta) - cos \theta ]$

$= - cos C (cosA.cos B + sin A . sin B -2cos A.cos B +cos C )$

$= cos C ( cos A. cos B - sin A .sin B -cos C )$

$= cos C [ cos (A+B) - cos C ]$

$= cos C (cos C -cos C) = 0 = RHS$ hence proved.

If $Cos 2 \theta = 0$ then $\left| \begin{matrix} 0 & cos \theta &sin \theta \\ cos \theta & sin \theta & 0 \\ sin \theta & 0 & cos \theta \end{matrix} \right|^2 =$ _________

Since

$cos 2 \theta = 0$

$\Rightarrow cos 2 \theta = cos \frac { \pi}{2} \Rightarrow 2 \theta = \frac { \pi}{2}$

$\Rightarrow \theta = \frac { \pi }{4}$

$\therefore sin \frac { \pi}{4} = \frac {1}{ \sqrt {2} }$ and

$cos \frac { \pi}{4} = \frac {1}{\sqrt {2}}$

$\therefore \left| \begin{matrix} 0 & \frac {1}{\sqrt {2}} & \frac {1}{\sqrt {2}} \\ \frac {1}{\sqrt {2}} & \frac {1}{\sqrt {2}} & 0 \\ \frac {1}{\sqrt {2}} & 0 & \frac {1}{\sqrt {2}} \end{matrix} \right|$

Expanding along

$R_1$

$[ - \frac {1}{ \sqrt {2} } ( \frac {1}{2}) + \frac { 1}{ \sqrt {2} } ( - \frac {1}{2}) ]^2$ =

$[ \frac {-2}{2 \sqrt {2} } ]^2$ =

$( \frac {-1}{ \sqrt {2}})^2 = \frac {1}{2}$

If $a+ b +c \neq 0$ and $\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right| = 0$ then prove that $a = b = c$ .

We have

$\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right|$

$[ \because R_1 \rightarrow R_1 +R_2 +R_3 ]$

$= \left| \begin{matrix} a+b+c & a+b+c & a+b+c \\ b & c & a \\ c & a & b \end{matrix} \right|$

$=(a + b+ c) \left| \begin{matrix} 1 & 1 & 1 \\ b & c & a \\ c & a & b \end{matrix} \right|$

$[ \because C_1 \rightarrow C_1 -C_3$ and

$C_2 \rightarrow C_2 -C_3$ ]

$=( a +b +c) \left| \begin{matrix} 0 & 0 & 1 \\ b-a & c-a & a \\ c-b & a-b & b \end{matrix} \right|$
Expanding along

$R_1$

$= ( a+ b+c )[ 1(b-a)(a-b)-(c-a)(c-b) ]$

$= (a+b+c)(ba-b^2 -a^2 +ab -c^2 +cb +ac -ab)$

$= - \frac {1}{2} ( a+ b +c) \times (-2)(-a^2 -b^2 -c^2 +ab +bc+ca )$

$= \frac {-1}{2} ( a+b+c) [a^2 +b^2 +c^2 -2ab -2bc-2ca +a^2 +b^2 +c^2 ]$

$= - \frac {1}{2} ( a+b+c) [a^2+b^2 -2ab +b^2 +c^2 - 2bc +c^2 +a^2 -2ac ]$

$= \frac {-1}{2} (a +b+ c) [ (a-b)^2 + (b-c)^1 + (c-a)^2 ]$
Also

$A = 0$

$= \frac {-1}{2} (a +b+c) [ (a-b)^2 + (b-c)^2 +(c-a)^2 ] = 0$

$(a-b)^2 +(b-c)^2 + (c-a)^2 = 0$ [

$\because a+b+c \neq 0$ given ]

$\Rightarrow a-b = b-c =c-a = 0$

$a = b = c$ hence proved.

Find the value of the following determinant
$\begin{vmatrix} 5 & {-2}\\{-3} & 1\end{vmatrix}$ .

$\begin{vmatrix} 5 & {-2}\\{-3} & 1\end{vmatrix} = 5 \times 1 - (-2) \times (-3) = 5 - 6 = -1$ .

Find the value of the following determinant
$\begin{vmatrix} 3 & {-1}\\{1} & 4 \end{vmatrix}$ .

$\begin{vmatrix} 5 & {-2}\\{-3} & 1\end{vmatrix} = 3 \times 4 - (- 1) \times 1 = 12 + 1 = 13$ .

Find the adjoint of the matrix $A = \begin{bmatrix} 1 & 2\\ 3 & -5\end{bmatrix}$ and verify the result $A(adj \,A) = (adj \,A) = A |A| \cdot I$ .

Given

$A=\begin{bmatrix}1&2\\3&-5\end{bmatrix}$

Here

$a_{11}=1,a_{12}=2,a_{21}=3,a_{22}=-5$

Co-factor of

$a_{11}=(-1)^{1+1}\times [-5]=-5$

Co-factor of

$a_{12}=(-1)^{1+2}\times [3]=-3$

Co-factor of

$a_{21}=(-1)^{2+1}\times [2]=-2$

Co-factor of

$a_{22}=(-1)^{2+2}\times [1]=1$

Co-factor matrix of

$A=\begin{bmatrix}-5&-3\\-2&1\end{bmatrix}$

We know that

$adj(A)$ is the transpose of co-factor matrix

$adj(A)=\begin{bmatrix}-5&-3\\-2&1\end{bmatrix}^{T}=$

$\begin{bmatrix}-5&-2\\-3&1\end{bmatrix}$

$A=\begin{vmatrix}1&2\\3&-5\end{vmatrix}=1(-5)-(2)(3)=-5-6=-11$

$A(adj A)=$

$\begin{bmatrix}1&2\\3&-5\end{bmatrix}$

$\begin{bmatrix}-5&-2\\-3&1\end{bmatrix}=$

$\begin{bmatrix}1(-5)+2(-3)&1(-2)+2(1)\\3(-5)+(-5)(-3)&3(-2)+(-5)(1)\end{bmatrix}$

$=\begin{bmatrix}-11&0\\0&-11\end{bmatrix}=-11 I=|A| I$

$(adj A)A=$

$\begin{bmatrix}-5&-2\\-3&1\end{bmatrix}$

$\begin{bmatrix}1&2\\3&-5\end{bmatrix}=$

$\begin{bmatrix}1(-5)+(-2)(3)&(-5)(2)+(-2)(-5)\\(-3)(1)+(1)(3)&(-3)(2)+(-5)(1)\end{bmatrix}$

$=\begin{bmatrix}-11&0\\0&-11\end{bmatrix}=-11 I=|A| I$

$\therefore A(adj A)=(adj A)A=|A|I$

Hence Verified

Find the adjoint of the following matrix.
$\begin{bmatrix} 1& -1 & 2 \\ -2 & 3 & 5 \\ -2 & 0 & -1 \end{bmatrix}$

Let

$A = \begin{bmatrix} 1& -1 & 2 \\ -2 & 3 & 5 \\ -2 & 0 & -1 \end{bmatrix}$

Now,

$M_{11} = \begin{vmatrix} 3 & 5 \\ 0 & -1 \end{vmatrix} = -3 - 0 = -3$

$\therefore \, A_{11} = (-1)^{1 + 1} (-3) = -3$

$M_{12} = \begin{vmatrix}-2 & 5 \\ -2 & -1 \end{vmatrix} = 2 + 10 = 12$

$\therefore \,A_{12} = (-1)^{1 + 2} (12) = -12$

$M_{13} = \begin{vmatrix}-2 & 3 \\ -2 & 0 \end{vmatrix} = 0 + 6 = 6$

$\therefore \,A_{13} = (-1)^{1 + 3} (6) = 6$

$M_{21} = \begin{vmatrix}-1 & 2 \\ 2 & -1 \end{vmatrix} = 1 - 0 = 1$

$\therefore \, A_{21} = (-1)^{2 + 1} (1) = -1$

$M_{22} \begin{vmatrix} 1 & 2 \\ -2 & -1 \end{vmatrix} = -1 + 4 = 3$

$\therefore \, A_{22} = (-1)^{2 + 1} (3) = 3$

$M_{23} = \begin{vmatrix}1 & -1 \\ -2 & 0 \end{vmatrix} = 0 - 2 = -2$

$\therefore \, A_{23} = (-1)^{2 + 3} (-2) = 2$

$M_{31} = \begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix} = -5 - 6 = -11$

$\therefore \, A_{31} = (-1)^{3 + 1}(-11) = -11$

$M_{32} = \begin{vmatrix}1 & 2 \\ -2 & 5 \end{vmatrix} = 5 + 4 = 9$

$\therefore \, A_{32} = (-1)^{3 + 2} (9) = -9$

$M_{33} = \begin{vmatrix}1 & -1 \\ -2 & 3 \end{vmatrix} = 3 - 2 = 1$

$\therefore \, A_{33} = (-1)^{3 + 3} (1) = 1$

the cofactor matrix

$= \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31}& A_{32} & A_{33} \end{bmatrix}$

$= \begin{bmatrix}-3 & -12 & 6 \\ -1 & 3 & 2 \\ -11 & -9 & 1 \end{bmatrix}$

$\therefore \, adj \, A = \begin{bmatrix}-3 & -1 & -11 \\ -12 & 3 & -9 \\ 6 & 2 & 1 \end{bmatrix}$

The sum if the products of element of any row with the co-factors of corresponding elements is equal to __________

The sum if the products of element of any row with the co-factors of corresponding elements is equal to value of the determinant

let

$\Delta = \left| \begin{matrix} a_{ 11 } & a_{ 12 } & a_{ 13 } \\ a_{ 21 } & a_{ 22 } & a_{ 23 } \\ a_{ 31 } & a_{ 32 } & a_{ 33 } \end{matrix} \right|$

Expanding along

$R_1$

$\Delta = a_{11}A_{11} +a_{12}A_{12} +a_{13}A_{13}$

=Sum of products of elements of

$R_1$ with their corresponding cofactors.

Find the adjoint of the following matrix.
$\begin{bmatrix} 2 & -3 \\ 3 & 5 \end{bmatrix}$

Let

$A = \begin{bmatrix} 2 & -3 \\ 3 & 5 \end{bmatrix}$

Here,

$a_{11} = 2 , M_{11} = 5$

$\therefore \, A_{11} = (-)^{1 + 1} (5) = 5$

$a_{12} = -3 , M_{12} = 3$

$\therefore\, A_{12} = (-)^{1 + 2} (3) = -3$

$a_{21} = 3 , M_{21} = -3$

$\therefore \, A_{21} = (-)^{2 + 1}(-3) = 3$

$a_{22} = 5, M_{22} = 2$

$\therefore \, A_{22} = (-)^{2 + 2} = 2$

$\therefore$ the cofactor matrix

$= \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix}$

$= \begin{bmatrix} 5 & -3 \\ 3 & 2 \end{bmatrix}$

$\therefore \, adj \, A = \begin{bmatrix}5 & 3 \\ -3 & 2 \end{bmatrix}$

Evaluate the determinations:
$\begin{vmatrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{vmatrix}$

$\cos { \theta } \times \cos { \theta } +\sin { \theta } \times \sin { \theta }$

$={\cos}^{2}{\theta}+{\sin}^{2}{\theta} =1$

If $A=\begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix}$ find $\left| A \right|$

Given,

$\left| A \right| =\begin{vmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{vmatrix}$

$=\begin{vmatrix} 1 & -3 \\ 4 & -9 \end{vmatrix}-1\begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix}-2\begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix}$

$=-9-(-12)-1(-18-(-15)-2(8-5)$

$=-9+12+18-15-16+10=40-40-0$

since

$\left| A \right| =0$ the matrix is called singular matrix

Find the values of determinant.
$\begin{vmatrix} -1&7 \\ 2&4 \end{vmatrix}$

$\begin{vmatrix} -1&7 \\ 2&4 \end{vmatrix} = -1(4) - 7(2) = -4 - 14 = -18$

Evaluate the determinations:
$\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$

$2\times (-1)-4\times (-5)$

$=-2+20=18$

If $A=\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$ , then show that $\left| 2A \right| =4\left| A \right|$

$\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$

$\left| A \right| =\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$

$=1\times 2-4\times 2=2-8=-6$

$4\left| A \right| =4\times -6=-24$ .....(1)

$\left| 2A \right| =\begin{bmatrix} 2 & 4 \\ 8 & 2 \end{bmatrix}$

$=2\times 2-4\times 8$

$\left| 2A \right| =8-32=-24$ .....(2)

from (1) and (2)

$\left| 2A \right| =4\left| A \right|$

Find the values of following determinant.
$\begin{vmatrix} 5&3 \\ -7&0 \end{vmatrix}$

$\begin{vmatrix} 5&3 \\ -7&0 \end{vmatrix} = 5 \times 0 - 3 \times (-7) = 0 + 21 = 21$

Evaluate the determinations:
$\begin{vmatrix} { x }^{ 2 }-x+1 & x-1 \\ x+1 & x+1 \end{vmatrix}$

$({ x }^{ 2 }-x+1)(x+1)-(x-1)(x+1)$

$={x}^{3}+1-({x}^{2}-1)$

$={x}^{3}-{x}^{2}+2$

there are n points in a plane of which m points are collinear. how many triangles will be formed by joining three points?

For making a triangle we need three points

thus if

$3$ points out of n points are not in line then

$^nC_3$ triangle can be formed by n points

But m point is a line thus

$^mC_3$ triangle forms less

Hence required number of triangles =

$^nC_3 -^mC_3$

If matrix $A=\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$ then find $adj.A$ also prove that $A(adj A)= |A| I_3= (adj A)A$

Given matrix

$A=\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$
Cofactors of matrix

$A$

${ F }_{ 11 }=+\begin{vmatrix} 0 & -2 \\ 0 & 3 \end{vmatrix}=0\\ { F }_{ 12 }=-\begin{vmatrix} 3 & -2 \\ 1 & 3 \end{vmatrix}=-11\\ { F }_{ 13 }=+\begin{vmatrix} 3 & 0 \\ 1 & 0 \end{vmatrix}=0\\ { F }_{ 21 }=-\begin{vmatrix} -1 & 2 \\ 0 & 3 \end{vmatrix}=3\\ { F }_{ 22 }=+\begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix}=1\\ { F }_{ 23 }=1\begin{vmatrix} 1 & -1 \\ 1 & 0 \end{vmatrix}=-1\\ { F }_{ 31 }=+\begin{vmatrix} -1 & 2 \\ 0 & -2 \end{vmatrix}=2\\ { F }_{ 32 }=-\begin{vmatrix} 1 & 2 \\ 3 & -2 \end{vmatrix}=8\\ { F }_{ 33 }=+\begin{vmatrix} 1 & -1 \\ 3 & 0 \end{vmatrix}=3\\$
Matrix made from adjoint of matrix

$A$ ,

$B=\begin{bmatrix} 0 & -11 & 0 \\ 3 & 1 & -1 \\ 2 & 8 & 3 \end{bmatrix}\\ Now,\quad adj.A={ B }^{ T }={ \begin{bmatrix} 0 & -11 & 0 \\ 3 & 1 & -1 \\ 2 & 8 & 3 \end{bmatrix} }^{ T\\ }\\ So,\quad adj.A=\begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix}\\ \because \quad =\begin{vmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{vmatrix}$

$=1(0-0)+1(9+2)+2(0-0)$

$|A|=11$
Now,

$\\ A.(adj.A)=\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}\begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix}\\ =\begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}\\ A.(adj.A)=11\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
So,

$A.(adj.A)=|A|I_3 ...(i)$

$\\ (adj.A).A=\begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix}\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}\\ =\begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}\\ A.(adj.A)=11\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
So,

$(adj.A).A=|A|I_3$
From

$(i)$ and

$(ii)$ ,

$(adj.A).A=|A|I_3=(adj.A).A$
Hence proved.

Find the minors of elements of second row of determinant $\begin{vmatrix} 2 & 3 & 4\\ 3 & 6 & 5\\ 1 & 8 & 9 \end{vmatrix}$ .

$A=\begin{vmatrix} 2 & 3 & 4\\ 3 & 6 & 5\\ 1 & 8 & 9 \end{vmatrix}=0$

Minor of

$a_{21}(=3)$

$A_{21}=\begin{vmatrix} 3 & 4\\ 8 & 9\end{vmatrix}=27-32=5$

Minor of

$a_{22}(=6)$

$A_{22}=\begin{vmatrix} 2 & 4\\ 1 & 9\end{vmatrix}=18-4=14$

Minor of

$a_{23}(=5)$

$A_{23}=\begin{vmatrix} 2 & 3\\ 1 & 8\end{vmatrix}=16-3=13$

If $A$ be a square matrix of order $3\times 3$ then find $\left| kA \right|$
$\left| KA \right| ={ k }^{ n }\left| A \right|$ when $n$ is the order

$Using \space property \space of \space Determinant$

$For \space a \space matrix \space of \space order \space "n"$

$|kA|=k^{n}|A|$

$Since \space it \space is \space given \space order \space is \space 3$

$n=3$

$So\space,\left| kA \right| ={ k }^{ 3 }\left| A \right|$

For which value of k, det $\begin{vmatrix} k & 2\\ 4 & -3 \end{vmatrix}$ will be zero?

$$\begin{vmatrix} k & 2\\
4 & -3 \end{vmatrix}=0$$

$\Rightarrow -3k-8=0 \Rightarrow -3k=8 \Rightarrow k=\dfrac{-8}{3}$

Evaluate
$\begin{vmatrix} \sin { 30 } & \cos { 30 } \\ -\sin { 60 } & \cos { 60 } \end{vmatrix}$

$\begin{vmatrix} 1/2 & \sqrt { 3 } /2 \\ -\sqrt { 3 } /2 & 1/2 \end{vmatrix}$

$=\cfrac{1}{2}\times \cfrac{1}{2}+\cfrac{\sqrt 3}{2}\times (\cfrac{-\sqrt 3}{2})$

$=\cfrac{1}{4}+\cfrac{3}{4}=1$

$\sin 30 \cos 60 + \cos 30 \sin 60$

$=\sin (30+60)=\sin 90=1$

Find adjoint of matrix :
$\begin{vmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{vmatrix}$

${M}_{11}=3$

${M}_{12}$

${M}_{13}=6$

${M}_{21}=-1$

${M}_{22}=5$

${M}_{23}-2$

${M}_{31}=-11$

${M}_{32}=1$

${M}_{33}=5$

${C}_{11}=3$

${C}_{12}=-12$

${C}_{13}=6$

${C}_{21}=1$

${C}_{22}=5$

${C}_{23}=2$

${C}_{31}=-11$

${C}_{32}=-1$

${C}_{33}=5$
co factor matrix

$\begin{bmatrix} 3 & -12 & 6 \\ 1 & 5 & 2 \\ -11 & -1 & 5 \end{bmatrix}$
Adj A = Transpose of co-factor matrix

$\begin{bmatrix} 3 & -12 & 6 \\ 1 & 5 & 2 \\ -11 & -1 & 5 \end{bmatrix}$
Adj A

$=\begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}$

$A = \begin{bmatrix}1 & \tan x\\ -\tan x & 1\end{bmatrix}$ and $f(x)$ is defined as $f(x) = det. (A^T A^{-1})$ then find the value of $\underset{\text{n times}}{\underbrace{f(f(f(f.......f(x))))}}$ is $(n \geq 2)$ .

Given

$A = \begin{bmatrix}1 & \tan x\\ -\tan x & 1\end{bmatrix}$

$|A|=1+\tan ^{ 2 }{ x }$

${ A }^{ T }=\begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}$
Here,

${ c }_{ 11 }=1,{ c }_{ 12 }=\tan x\\ { c }_{ 21 }=-\tan x,{ c }_{ 22 }=1$

$adj A=\begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}^{ T }=\begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}$

${ A }^{ -1 }=\displaystyle \frac { 1 }{ 1+\tan ^{ 2 }{ x } } \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}$

Now,

$\displaystyle { A }^{ T }{ A }^{ -1 }=\frac { 1 }{ 1+\tan ^{ 2 }{ x } } \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} =\begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}$

$\displaystyle { A }^{ T }{ A }^{ -1 }=\frac { 1 }{ 1+\tan ^{ 2 }{ x } } \begin{bmatrix} 1-\tan ^{ 2 }{ x } & -2\tan x \\ 2\tan x & 1-\tan ^{ 2 }{ x } \end{bmatrix}$

$\Rightarrow \displaystyle { A }^{ T }{ A }^{ -1 }=\begin{bmatrix} \cos { 2x } & -\sin { 2x } \\ \sin { 2x } & \cos { 2x } \end{bmatrix}$

$|A^TA^{-1}|=\sin^{2x}+\cos^{2x}=1$
Given ,

$f(x) = det. (A^T A^{-1})$

$\Rightarrow f(x)=1$

$\Rightarrow \underset{\text{n times}}{\underbrace{f(f(f(f.......f(x))))}}=1$

If A is a square matrix of order 3, then find $|(A - A^T)^{2011}|$ .

$Given\quad A\quad is\quad a\quad square\quad matrix\quad of\quad order\quad 3\\ \left( A-A^{ T } \right) ^{ T }\quad =\quad A^{ T }-\left( A^{ T } \right) ^{ T }\quad =\quad -\left( A-A^{ T } \right) \\ i.e\quad A-A^{ T }\quad is\quad a\quad skew-symmetric\quad matrix\\ \Rightarrow \quad \left| A-A^{ T } \right| \quad =\quad 0\\ \left| \left( A-A^{ T } \right) ^{ 2011 } \right| \quad =\quad \left( \left| A-A^{ T } \right| \right) ^{ 2011 }\quad =\quad 0$

Match the statements in Column I with statements in column II

$f(x)=\begin{vmatrix} x & { \left( 1+\sin { x } \right) }^{ 3 } & \cos { x } \\ 1 & \log { \left( 1+x \right) } & 2 \\ { x }^{ 2 } & 1+{ x }^{ 2 } & 0 \end{vmatrix}$

$=x\left( 0-2\left( 1+{ x }^{ 2 } \right) \right) -{ \left( 1+\sin { x } \right) }^{ 3 }\left( 0-2{ x }^{ 2 } \right) +\cos { x } \left( \left( 1+{ x }^{ 2 } \right) -\left( { x }^{ 2 }\log { \left( 1+x \right) } \right) \right)$
Coefficient of

$x$ is

$-2$

B)

$\begin{vmatrix} 1 & 3\cos { \theta } & 1 \\ \sin { \theta } & 1 & 3\cos { \theta } \\ 1 & \sin { \theta } & 1 \end{vmatrix}$

$=1\left( 1-3\sin { \theta } \cos { \theta } \right) -3\cos { \theta } \left( \sin { \theta } -3\cos { \theta } \right) +1\left( \sin ^{ 2 }{ \theta } -1 \right) \\ =9\cos ^{ 2 }{ \theta } -6\sin { \theta } \cos { \theta } +\sin ^{ 2 }{ \theta } ={ \left( 3\cos { \theta } -\sin { \theta } \right) }^{ 2 }\ge { \left( \sqrt { 10 } \right) }^{ 2 }=10$

C) As

$a,b,c$ is in A.P then

$a+c=2b$

$f(x)=\begin{vmatrix} x+a & { x }^{ 2 }+1 & 1 \\ x+b & 2{ x }^{ 2 }-1 & 1 \\ x+c & 3{ x }^{ 2 }-2 & 1 \end{vmatrix}\\ =\left( x+a \right) \left( \left( { 2x }^{ 2 }-1 \right) -\left( 3{ x }^{ 2 }-2 \right) \right) -\left( x+b \right) \left( \left( { x }^{ 2 }+1 \right) -\left( 3{ x }^{ 2 }-2 \right) \right) +\left( x+c \right) \left( \left( { x }^{ 2 }+1 \right) -\left( { 2x }^{ 2 }-1 \right) \right) \\ =\left( x+a \right) \left( 1-{ x }^{ 2 } \right) -\left( x+b \right) \left( 3-2{ x }^{ 2 } \right) +\left( x+c \right) \left( 2-{ x }^{ 2 } \right) \\ =a-3b+2c-{ x }^{ 2 }\left( a+c-2b \right) =a-3b+2c\\ f'\left( x \right) =0\Rightarrow f'\left( 0 \right) =0$

D)

$\begin{vmatrix} x & 2 & x \\ 1 & x & 6 \\ x & x & x+1 \end{vmatrix}=x\left( x\left( x+1 \right) -6x \right) -2\left( x+1-6x \right) +x\left( x-{ x }^{ 2 } \right)$

$=x\left( { x }^{ 2 }-5x \right) -2\left( 1-5x \right) +x\left( x-{ x }^{ 2 } \right) ={ x }^{ 3 }-5{ x }^{ 2 }-2+10x+{ x }^{ 2 }-{ x }^{ 3 }=-4{ x }^{ 2 }+10x-2\\ \Rightarrow { a }_{ 0 }=-2$

Prove that $\begin{vmatrix} bc & bc'+b'c & b'c'\\ ca & ca'+c'a & c'a'\\ ab & ab'+a'b & a'b' \end{vmatrix}=(ab'-a'b)(bc-b'c)(ca'-c'a)$

Let

$\Delta =\begin{vmatrix} bc & bc'+b'c & b'c' \\ ca & ca'+c'a & c'a' \\ ab & ab'+a'b & a'b' \end{vmatrix}$
Taking

$\left( b'c' \right) ,\left( c'a' \right)$ and

$\left( a'b' \right)$ common from

${ R }_{ 1 },{ R }_{ 2 }$ and

${ R }_{ 3 }$ respectively, we get

$\Delta =\left( b'c' \right) \left( c'a' \right) \left( a'b' \right) \begin{vmatrix} \cfrac { bc }{ b'c' } & \cfrac { b }{ b' } +\cfrac { c }{ c' } & 1 \\ \cfrac { ca }{ c'a' } & \cfrac { c }{ c' } +\cfrac { a }{ a' } & 1 \\ \cfrac { ab }{ a'b' } & \cfrac { a }{ a' } +\cfrac { b }{ b' } & 1 \end{vmatrix}$
Applying

${ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 },{ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }$

$\Delta ={ \left( a'b'c' \right) }^{ 2 }\begin{vmatrix} \cfrac { bc }{ b'c' } & \cfrac { b }{ b' } +\cfrac { c }{ c' } & 1 \\ \cfrac { c }{ c' } \left( \cfrac { a }{ a' } -\cfrac { b }{ b' } \right) & \left( \cfrac { a }{ a' } -\cfrac { b }{ b' } \right) & 0 \\ \cfrac { b }{ b' } \left( \cfrac { a }{ a' } -\cfrac { c }{ c' } \right) & \cfrac { a }{ a' } -\cfrac { c }{ c' } & 0 \end{vmatrix}\\ \Delta ={ \left( a'b'c' \right) }^{ 2 }\left( \cfrac { a }{ a' } -\cfrac { b }{ b' } \right) \left( \cfrac { a }{ a' } -\cfrac { c }{ c' } \right) \begin{vmatrix} \cfrac { bc }{ b'c' } & \cfrac { b }{ b' } +\cfrac { c }{ c' } & 1 \\ \cfrac { c }{ c' } & 1 & 0 \\ \cfrac { b }{ b' } & 1 & 0 \end{vmatrix}\\ \Delta ={ \left( a'b'c' \right) }^{ 2 }\cfrac { ab'-a'b }{ a'b' } \cfrac { ac'-a'c }{ a'c' } \left( \cfrac { c }{ c' } -\cfrac { b }{ b' } \right) \\ =(ab'-a'b)(bc-b'c)(ca'-c'a)$

Let

$\Delta _{1}=\begin{vmatrix} 1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1 \end{vmatrix}$

and $\Delta _{2}=\begin{vmatrix} 0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0 \end{vmatrix}$ .

If $\Delta _{1}=\Delta _{2}$ , find $\sin ^{2}\alpha +\sin ^{2}\beta +\sin ^{2}\gamma$

$Given \quad \Delta _{ 1 }=\begin{vmatrix} 1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1 \end{vmatrix}$

$expanding\quad the\quad determinant\quad gives$

$\Delta _{ 1 }\quad =\quad \sin ^{ 2 }{ \gamma } -\cos ^{ 2 }{ \alpha } -\cos ^{ 2 }{ \beta } +2\cos { \alpha } \cos { \beta } \cos { \gamma }$

$\Delta _{ 2 }=\begin{vmatrix} 0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0 \end{vmatrix}$

$\Delta _{ 2 }\quad =\quad 2\cos { \alpha } \cos { \beta } \cos { \gamma }$

$but\quad \Delta _{ 1 }\quad =\quad \Delta _{ 2 }$

$\Rightarrow \quad \sin ^{ 2 }{ \gamma } -\cos ^{ 2 }{ \alpha } -\cos ^{ 2 }{ \beta } \quad =\quad 0$

$\therefore \quad \sin ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \beta } +\sin ^{ 2 }{ \gamma } \quad =\quad 2$ [

$\because \cos ^{ 2 }{ \alpha }=1-\sin ^{ 2 }{ \alpha }$ ]

Prove that
$\begin{vmatrix} bc-a^2 & ca-b^2 & ab-c^2\\ -bc+ca+ab & bc-ca+ab & bc+ca-ab\\ (a+b)(a+c) & (b+c)(b+a) & (c+a)(c+b) \end{vmatrix}=3.(b-c)(c-a)(a-b)(a+b+c)(ab+bc+ca)$

Consider,

$\begin{vmatrix}bc-a^2 & ca-b^2 & ab-c^2\\ -bc+ca+ab & bc-ca+ab & bc+ca-ab\\ (a+b)(a+c) & (b+c)(b+a) & (c+a)(c+b)\end{vmatrix}$

$=\begin{vmatrix} bc-a^{ 2 } & ca-b^{ 2 } & ab-c^{ 2 } \\ -bc+ca+ab & bc-ca+ab & bc+ca-ab \\ a^{ 2 }+ab+bc+ac & b^{ 2 }+ab+bc+ac & c^{ 2 }+ab+bc+ac \end{vmatrix}$

$R_3\rightarrow R_3+R_1$

$=\begin{vmatrix} bc-a^{ 2 } & ca-b^{ 2 } & ab-c^{ 2 } \\ -bc+ca+ab & bc-ca+ab & bc+ca-ab \\ ab+2bc+ac & ab+bc+2ac & 2ab+bc+ac \end{vmatrix}$

$C_{ 1 }\rightarrow { C }_{ 1 }-{ C }_{ 2 },C_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 3 }$

$=\begin{vmatrix} (b-a)(a+b+c) & (c-b)(a+b+c) & ab-c^{ 2 } \\ 2c(a-b) & 2a(b-c) & bc+ca-ab \\ c(b-a) & a(c-b) & 2ab+bc+ac \end{vmatrix}$

$=(a-b)(b-c)\begin{vmatrix} -(a+b+c) & -(a+b+c) & ab-c^{ 2 } \\ 2c & 2a & bc+ca-ab \\ -c & -a & 2ab+bc+ac \end{vmatrix}$

$R_{ 2 }\rightarrow { R }_{ 2 }+{ 2R }_{ 3 }$

$=(a-b)(b-c)\begin{vmatrix} -(a+b+c) & -(a+b+c) & ab-c^{ 2 } \\ 0 & 0 & 3bc+3ca+3ab \\ -c & -a & 2ab+bc+ac \end{vmatrix}$

Expanding along second row, we get

$=3(a-b)(b-c)(ab+bc+ca)(c-a)(a+b+c)$

If $ax_1^2+by_1^2+cz_1^2=ax_2^2+by_2^2+cz_2^2=ax_3^2+by_3^2+cz_3^2=d$ and $ax_2x_3+by_2y_3+cz_2z_3=ax_3x_1+by_3y_1+cz_3z_1=ax_1x_2+by_1y_2+cz_1z_2=f$ , then prove that $\begin{vmatrix} x_1 & y_1 & z_1\\ x_2 & y_2 & z_2\\ x_3 & y_3 & z_3 \end{vmatrix}=(d-f)\left [\frac {d+2f}{abc}\right ]^{1/2}(a, b, c\neq 0)$

Given ,

$ax_1^2+by_1^2+cz_1^2=ax_2^2+by_2^2+cz_2^2=ax_3^2+by_3^2+cz_3^2=d$ .....(1)

$ax_2x_3+by_2y_3+cz_2z_3=ax_3x_1+by_3y_1+cz_3z_1=ax_1x_2+by_1y_2+cz_1z_2=f$ ....(2)

Let

$\Delta=\begin{vmatrix}x_1 & y_1 & z_1\\ x_2 & y_2 & z_2\\ x_3 & y_3 & z_3 \end{vmatrix}$

$\Delta^2=\frac {1}{abc}\begin{vmatrix}x_1 & y_1 & z_1\\ x_2 & y_2 & z_2\\ x_3 & y_3 & z_3\end{vmatrix}\times \begin{vmatrix}ax_1 & by_1 & cz_1\\ ax_2 & by_2 & cz_2\\ ax_3 & by_3 & cz_3\end{vmatrix}$

$=\frac {1}{abc}\begin{vmatrix} d & f & f\\ f & d & f\\ f & f & d\end{vmatrix}$ (by (1) and (2))

$C_1\rightarrow C_1+C_2+C_3$

$=\dfrac {1}{abc}\begin{vmatrix}d+2f & f & f\\ d+2f & d & f\\ d+2f & f & d \end{vmatrix}=\frac {(d+2f)}{abc}\begin{vmatrix}1 & f & f\\ 1 & d & f\\ 1 & f & d \end{vmatrix}$
On solving it we get

$\Delta^2=\frac {(d+2f)}{abc}(d-f)^2$

$\Delta=(d-f)\left [\frac {d+2f}{abc}\right ]^{\frac {1}{2}}$

Are the three points $A(2,3),B(5,6)$ and $C(0,2)$ collinear?

Three points

$A$ ,

$B$ and

$C$ are collinear if

$AB+BC=AC$ .

Therefore, we calculate:

$AB=\sqrt { \left( 5-2 \right) ^{ 2 }+\left( 6-3 \right) ^{ 2 } } =\sqrt { 3^{ 2 }+3^{ 2 } } =\sqrt { 9+9 } =\sqrt { 18 } \\$

$BC=\sqrt { \left( 0-5 \right) ^{ 2 }+\left( 2-6 \right) ^{ 2 } } =\sqrt { (-5)^{ 2 }+(-4)^{ 2 } } =\sqrt { 25+16 } =\sqrt { 41 } \\$

$AC=\sqrt { \left( 0-2 \right) ^{ 2 }+\left( 2-3 \right) ^{ 2 } } =\sqrt { (-2)^{ 2 }+(-1)^{ 2 } } =\sqrt { 4+1 } =\sqrt { 5 }$

Since,

$\sqrt { 18 } +\sqrt { 41 } \neq \sqrt { 5 }$ that is

$AB+BC\neq AC$ .

Hence, the given points are not collinear.

Find the maximum value of $\begin{vmatrix} 1& 1 & 1\\ 1 & 1 +\sin \theta & 1\\ 1 & 1 & 1 + \cos \theta\end{vmatrix}$

$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 + \sin{\theta} & 1 \\ 1 & 1 & 1 + \cos{\theta} \end{vmatrix}$

$= 1[(1 + \sin{\theta}) \times (1 + \cos{\theta}) - 1] - 1[1 + \cos{\theta} - 1] + 1[1 - 1 - \sin{\theta}]$

$=1 + \sin{\theta} + \cos{\theta} + \sin{\theta}\cos{\theta} - 1 - \cos{\theta} - \sin{\theta}$

$= \sin{\theta} \times \cos{\theta}$

Since we need the maximum value of the above expression, we differentiate it w.r.t

$\theta$

We have

$\cos^2{\theta} - \sin^2{\theta} = 0$

$\Rightarrow \cos{2\theta} = 0$ or

$\theta = \dfrac{\pi}{4}$

Maximum value therefore becomes

$\sqrt{2}$

If $a^2+b^2+c^2=1$ then is the value of the determinant $\begin{vmatrix} a^2+(b^2+c^2)cos\theta & ba(1-cos\theta) & ca(1-cos\theta)\\ ab(1-cos\theta) & b^2(c^2+a^2)cos\theta & cb(1-cos\theta)\\ ac(1-cos\theta) & bc(1-cos\theta) & c^2+(a^2+b^2)cos\theta \end{vmatrix}$ independent of a,b,c?
If yes enter 1 else enter 0.

Let

$\Delta =\begin{vmatrix} a^{ 2 }+\left( b^{ 2 }+c^{ 2 } \right) \cos \phi & ab\left( 1-\cos \phi \right) & ac\left( 1-\cos \phi \right) \\ ba\left( 1-\cos \phi \right) & b^{ 2 }+\left( c^{ 2 }+a^{ 2 } \right) \cos \phi & bc\left( 1-\cos \phi \right) \\ ca\left( 1-\cos \phi \right) & cb\left( 1-\cos \phi \right) & c^{ 2 }+\left( a^{ 2 }+b^{ 2 } \right) \cos \phi \end{vmatrix}$
Multiplying

${ C }_{ 1 }$ by

$a$ ,

${ C }_{ 2 }$ by

$b$ and

${ C }_{ 3 }$ by

$c$ we get

$\Delta =\cfrac { 1 }{ abc } \begin{vmatrix} a^{ 3 }+a\left( b^{ 2 }+c^{ 2 } \right) \cos \phi & ab^{ 2 }\left( 1-\cos \phi \right) & ac^{ 2 }\left( 1-\cos \phi \right) \\ ba^{ 2 }\left( 1-\cos \phi \right) & b^{ 3 }+b\left( c^{ 2 }+a^{ 2 } \right) \cos \phi & bc^{ 2 }\left( 1-\cos \phi \right) \\ ca^{ 2 }\left( 1-\cos \phi \right) & cb^{ 2 }\left( 1-\cos \phi \right) & c^{ 3 }+c\left( a^{ 2 }+b^{ 2 } \right) \cos \phi \end{vmatrix}$
Taking

$a$ common from

${ R }_{ 1 }$ ,

$b$ from

${ R }_{ 2 }$ and

$c$ from

${ R }_{ 3 }$ , we get

$\Delta =\cfrac { abc }{ abc } \begin{vmatrix} a^{ 2 }+\left( b^{ 2 }+c^{ 2 } \right) \cos \phi & b^{ 2 }\left( 1-\cos \phi \right) & c^{ 2 }\left( 1-\cos \phi \right) \\ a^{ 2 }\left( 1-\cos \phi \right) & b^{ 2 }+\left( c^{ 2 }+a^{ 2 } \right) \cos \phi & c^{ 2 }\left( 1-\cos \phi \right) \\ a^{ 2 }\left( 1-\cos \phi \right) & b^{ 2 }\left( 1-\cos \phi \right) & c^{ 2 }+\left( a^{ 2 }+b^{ 2 } \right) \cos \phi \end{vmatrix}$
Applying

${ C }_{ 1 }\rightarrow { C }_{ 1 }+{ C }_{ 2 }+{ C }_{ 3 }$

$\Delta =\begin{vmatrix} 1 & b^{ 2 }\left( 1-\cos \phi \right) & c^{ 2 }\left( 1-\cos \phi \right) \\ 1 & b^{ 2 }+\left( c^{ 2 }+a^{ 2 } \right) \cos \phi & c^{ 2 }\left( 1-\cos \phi \right) \\ 1 & b^{ 2 }\left( 1-\cos \phi \right) & c^{ 2 }+\left( a^{ 2 }+b^{ 2 } \right) \cos \phi \end{vmatrix}$
As

$a^{ 2 }+b^{ 2 }+c^{ 2 }=1$
Applying

${ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 },{ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }$

$\Delta =\begin{vmatrix} 1 & b^{ 2 }\left( 1-\cos \phi \right) & c^{ 2 }\left( 1-\cos \phi \right) \\ 0 & \left( b^{ 2 }+c^{ 2 }+a^{ 2 } \right) \cos \phi & 0 \\ 1 & 0 & \left( c^{ 2 }+a^{ 2 }+b^{ 2 } \right) \cos \phi \end{vmatrix}$
Expanding along

${ C }_{ 1 }$
We get

$\Delta =\cos ^{ 2 }{ \phi }$

A line of slope 2 passes through the point A(1,3).
a) Check whether B(3,7) is a point on this line.
b) Write down the equation of this line.
c) Find the coordinates of a point C on the line such that BC = 2AB.

Slope of the line is 2 and passes through A(1, 3)

$\dfrac{y-3}{x-1}=2$

$y-3=2x-2$

$y=2x+1$

(a) For point B(3, 7)
2(3) + 1 = 7
Hence the point B(3, 7)lies on the line y = 2x + 1

(b) Equation of the line is y = 2x+1.

(c) Let

$(x_1, y_1)$ be the coordinates of point C on line.
Therefore it satisfies the equation of line a

$y = 2x + 1$

$\therefore y_1=2x_1+1$

$BC=2AB$

$\Rightarrow \sqrt{(x_1-3)^2+(y_1-7)^2}=2\sqrt{(3-1)^2+(7-3)^2}$
Squaring both sides, we get

$\Rightarrow (x_1-3)^2+(y_1-7)^2=4[(3-1)^2+(7-3)^2]$

$(x_1-3)^2+(2x_1+1-7)^2=4[(2)^2+(4)^2]$

$[\therefore y_1=2x_1+1]$

$5(x_1-3)^2=80$

$(x_1-3)^2=16$

$x_1-3=\pm 4$

$x_1=7$ or

$-1$

$y_1=2x_1+1=2(7)+1=15$

$y_1=2x_1+1=2(-1)+1=-1$
The coordinates of point C are

$(7, 15)$ or

$(-1, -1)$ .

If $\begin{vmatrix}x+1&x-1\\x-3&x+2\end{vmatrix}=\begin{vmatrix}4&-1\\1&3\end{vmatrix}$ , then write the value of $x$ .

Compute the determinants as shown below:

$\begin{vmatrix} x+1 & x-1 \\ x-3 & x+2 \end{vmatrix}=\begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix}\\$

$\left( x+1 \right) \left( x+2 \right) -\left( x-3 \right) \left( x-1 \right) =4\times 3-1\times -1\\$

${ \Rightarrow x }^{ 2 }+2x+x+2-\left( { x }^{ 2 }-x-3x+3 \right) =12+1\\$

${ \Rightarrow x }^{ 2 }+3x+2-{ x }^{ 2 }+4x-3=13\\$

$\Rightarrow 7x-1=13\\$

$\Rightarrow 7x=14\\$

$\Rightarrow x=2\\$

Find determinent
$\left( {\matrix{ 7 & 1 & { - 6} \cr { - 6} & { - 4} & {13} \cr 2 & 5 & { - 8} \cr } } \right)$

$=7(32-65)-1(48-26)-6(-30+8)\\ =7(-33)-1(22)-6(22)\\ =7(-33)-7(22)\\ =7(-33-22)\\ -7\times55\\ =-385$

Find the value of the determinant:
$\begin{vmatrix}\cos(\theta+\phi)&-\sin (\theta+\phi)&\cos 2\phi\\ \sin \theta& \cos \theta &\sin \phi\\ -\cos \theta &\sin \theta &\cos \phi\end{vmatrix}$

$\begin{vmatrix} \cos (\theta +\phi ) & -\sin (\theta +\phi ) & \cos 2\phi \\ \sin \theta & \cos \theta & \sin \phi \\ -\cos \theta & \sin \theta & \cos \phi \end{vmatrix}$

Take

$\theta =\dfrac { \pi }{ 2 } ;\phi =0$ (Assume)

$=\begin{vmatrix} \cos (\dfrac { \pi }{ 2 } ) & -\sin (\dfrac { \pi }{ 2 } ) & \cos 0 \\ \sin \dfrac { \pi }{ 2 } & \cos \dfrac { \pi }{ 2 } & \sin 0 \\ -\cos \dfrac { \pi }{ 2 } & \sin \dfrac { \pi }{ 2 } & \cos 0 \end{vmatrix}\\ =\begin{vmatrix} 0 & -1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{vmatrix}\\ =0+1(1-0)-1(1-0)\\ =1-1=0$

For what values of p and q, the system of equations
$2x + py + 6z = 8, x + 2y + qz = 5, x + y + 3z = 4$
has (i) no solution (ii) a unique solution (iii) infinitely many solutions.

1312047_1003872_ans_281d9245d53049bd9c48bff03b8ce9d4.jpg

Solve:
$\begin{vmatrix} 1+\sin ^2 \theta & \cos ^2\theta & 4\sin 4\theta \\ \sin ^2 \theta &1+\cos ^2 \theta & 4\sin 4\theta \\ \sin ^2 \theta & \cos ^2 \theta & 1+4\sin 4\theta \end{vmatrix}$ $=0$ ; where, $0<\theta <\large{\cfrac{\pi}{2}}$ .

Given

$\begin{vmatrix} 1+\sin ^2 \theta & \cos ^2\theta & 4\sin 4\theta \\ \sin ^2 \theta &1+\cos ^2 \theta & 4\sin 4\theta \\ \sin ^2 \theta & \cos ^2 \theta & 1+4\sin 4\theta \end{vmatrix}$

$=0$ ; where,

$0<\theta <\large{\cfrac{\pi}{2}}$ .

$\left(C_1 \rightarrow C_1+C2 \right)$

$\Rightarrow \ \begin{vmatrix} 2 & \cos ^2\theta & 4\sin 4\theta \\ 2 &1+\cos ^2 \theta & 4\sin 4\theta \\ 1 & \cos ^2 \theta & 1+4\sin 4\theta \end{vmatrix}$

$=0$

$\left(R_2 \rightarrow R_2-R_1\right.$ &

$\left.R_3 \rightarrow R_3-R_1\right)$

$\Rightarrow \begin{vmatrix} 2 & \cos ^2\theta & 4\sin 4\theta \\ 0 &1 & 0 \\ -1 & 0 & 1 \end{vmatrix}$

$=0$

$\Rightarrow 2+4 \sin 4\theta=0$

$\Rightarrow \sin 4\theta =-\cfrac{1}{2}$

$\Rightarrow 4\theta= \cfrac{7\pi}{6}$ ...[Since

$0<\theta <\cfrac{\pi}{2}$ ]

$\Rightarrow \theta=\cfrac{7\pi}{24}$

The straight lines $\displaystyle \imath_1, \imath_2 \, and \, \imath_3$ are parallel and lie in the same plane. A total of m points are taken on the line $\displaystyle \imath_1,$ n points on $\displaystyle \imath_2$ , and k points on $\displaystyle \imath_3$ . How many triangles are there whose vertices are at these points?

There are total

$n+m+k$ points.

And a triangle can't be formed if all the three points are on the same line.

Total number of ways to select three points

$=^{n+m+k}C_3$

And, number of ways to select three points on the same line

$=^mC_3+^nC_3+^kC_3$

Hence, required number of triangles formed

$=^{n+m+k}C_3-(^nC_3+^mC_3+^kC_3)$

Write minors and cofactors of the elements of the following determinants:
$(i) \begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}$
$(ii) \begin{vmatrix} a & c \\ b & d \end{vmatrix}$
$(iii) \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{vmatrix}$
$(iv) \begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}$

$(i)\quad \begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}\\ Minor\Rightarrow (6-0)=6\\ cofactors\begin{vmatrix} 3 & 0 \\ -4 & 2 \end{vmatrix}\\ (ii)\quad \begin{vmatrix} a & c \\ b & d \end{vmatrix}\\ Minor\Rightarrow (ad-bc)=6\\ cofactors\quad \begin{vmatrix} d & c \\ c & a \end{vmatrix}\\ (iii)\quad \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}\\ Minor=\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}\Rightarrow 1\\ cofactors=\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}\\ (iv)\quad \begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}\\ Minor=\begin{vmatrix} 5 & -1 \\ 1 & 2 \end{vmatrix}=10+1=11\\ cofactors=\begin{vmatrix} 11 & -6 & 3 \\ 4 & 2 & -1 \\ -20 & 13 & 5 \end{vmatrix}$

Evaluate the determinants: $\left | \begin{array}{111} 21 & 17 & 7 & 10 \\ 24 & 22 & 6 & 10 \\ 6 & 8 & 2 & 3 \\ 5 & 7 & 1 & 2 \\ \end {array} \right |$

1970621_1004402_ans_1ced46f0d956422caee1f4d6743455eb.jpeg

$ax + by + cz = k$
$a^2x + b^2y + c^2z = k^2$
$a^3x + b^3y + c^3z = k^3$
Solve by Crammer's rule

$ax+by+cz = k$

$a^2x + b^2y+c^z = k^2$

$a^3x + b^3y+c^z = k^3$

using cramer's rule,

$x = \dfrac{D_1}{D}, \quad y = \dfrac{D_2}{D}, \quad Z=\dfrac{D_3}{D}$

where,

$D = \begin{bmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{bmatrix} = abc \begin{bmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2\end{bmatrix}$

$=abc \begin{bmatrix} 0 & 1 & 0 \\ a-b & b & c-b \\ a^2-b^2 & b^2 & c^2 - b^2 \end{bmatrix} = abc(a-b)(c-b) \begin{bmatrix} 0 & 1 & 0 \\ 1 & b & 1 \\ a+b & b^2 & c+b \end{bmatrix}$

$=(a-b)(c-b)(a+b-b-c)abc$

$=(a-b)(b-c)(c-a)abc$

$D_1 = \begin{bmatrix} k & b & c \\ k^2 & b^2 & c^2 \\ k^3 & b^3 & c^3 \end{bmatrix} = kbc \begin{bmatrix} 1& 1 & 1\\ k & b & c\\ k^2& b^2 & c^2 \end{bmatrix} = kbc \begin{bmatrix} 0 & 1 & 0 \\ k-b & b & c-b\\ k^2 - b^2 & b^2 & c^2 - b^2 \end{bmatrix}$

$- kbc (k-b)(c-b)\begin{bmatrix} 0 & 1 & 0\\ 1 & b & 1\\ k+b & b^2 & c+b \end{bmatrix}$

$= kbc(k-b)(c-b)(k+b-c-b)$

$=kbc(k-b)(b-c)(c-k)$

$\therefore x = \dfrac{D_1}{D} = \dfrac{kbc(k-b)(b-c)(c-k)}{kbc(a-b)(b-c)(c-a)}$

$= \dfrac{k(k-b)(b-c)(c-k)}{a(a-b)(b-c)(c-a)}$

$D_2 = \begin{bmatrix} a & k & c \\ a^2 & k^2 & c^2 \\ a^3 & k^3 & c^3 \end{bmatrix} = akc \begin{bmatrix} 1 & 1 & 1\\ a & k & c\\ a^2& k^2 & c^2 \end{bmatrix} = akc \begin{bmatrix} 0 & 1 & 0 \\ a-b & k & c-k\\ a^2-k^2 & k^2 & c^2 - k^2 \end{bmatrix}$

$- akc (a-k)(c-k)\begin{bmatrix} 0 & 1 & 0\\ 1 & k & 1\\ a+k & k^2 & c+k \end{bmatrix}$

$= akc(a-k)(c-k)(a+k-c-k)$

$=akc(a-k)(k-c)(c-a)$

$\therefore y = \dfrac{D_2}{D} = \dfrac{akc(a-k)(k-c)(c-a)}{abc(a-b)(b-c)(c-a)}$

$= \dfrac{k(a-k)(k-c)(c-a)}{b(a-b)(b-c)(c-a)}$

$D_3 = \begin{bmatrix} a & b & k \\ a^2 & b^2 & k^2 \\ a^3 & b^3 & k^3 \end{bmatrix} = abk \begin{bmatrix} 1 & 1 & 1\\ a & b & k\\ a^2& b^2 & k^2 \end{bmatrix} = abk \begin{bmatrix} 0 & 1 & 0 \\ a-b & b & k-b\\ a^2-b^2 & b^2 & k^2 - b^2 \end{bmatrix}$

$- abk (a-b)(k-b)\begin{bmatrix} 0 & 1 & 0\\ 1 & b & 1\\ a+b & b^2 & k+b \end{bmatrix}$

$= abk(a-b)(k-b)(a+b-k-b)$

$=abk(a-b)(b-k)(k-a)$

$\therefore z = \dfrac{D_3}{D} = \dfrac{abk(a-b)(b-k)(k-a)}{abc(a-b)(b-c)(c-a)}$

$= \dfrac{k(a-b)(b-k)(k-a)}{c(a-b)(b-c)(c-a)}$

Using properties of determinants, find the value of $n$ in the following.
$\left | \begin{array}{111} 3x & -x+y & -x+z \\ x-y & 3y & z-y \\ x-z & y-z & 3z \\ \end {array} \right |$ = $n( x+y + z)(xy + yz + zx)$

2018734_1028141_ans_61e9fa2bf70a409894ece72c11d7aed0.jpg

The value of determinant $\begin{vmatrix} 2 & 7 & 65\\ 3 & 8 & 75 \\ 5 & 9 & 86\end{vmatrix} equals$ .

We know that to solve a determinant, we apply the following formula:

$\left| \begin{matrix} { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| ={ a }_{ 1 }\left[ \left( { b }_{ 2 }\times { c }_{ 3 } \right) -\left( { c }_{ 2 }\times { b }_{ 3 } \right) \right] -{ a }_{ 2 }\left[ \left( { b }_{ 1 }\times { c }_{ 3 } \right) -\left( { c }_{ 1 }\times { b }_{ 3 } \right) \right] +{ a }_{ 3 }\left[ \left( { b }_{ 1 }\times { c }_{ 2 } \right) -\left( { c }_{ 1 }\times { b }_{ 2 } \right) \right]$

Now, consider the given determinant and solve as follows:

$\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right| \\ =2\left[ \left( 8\times 86 \right) -\left( 9\times 75 \right) \right] -7\left[ \left( 3\times 86 \right) -\left( 5\times 75 \right) \right] +65\left[ \left( 3\times 9 \right) -\left( 5\times 8 \right) \right] \\ =2\left( 688-675 \right) -7\left( 258-375 \right) +65\left( 27-40 \right) \\ =(2\times 13)-(7\times -117)+(65\times -13)\\ =26+819-845\\ =845-845\\ =0$

Hence,

$\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right| =0$

$\left| \begin{matrix} { a }^{ 2 }+1 & ab & ac \\ ab & { b }^{ 2 }+1 & bc \\ ca & cb & { c }^{ 2 }+1 \end{matrix} \right| =1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }$

$\begin{vmatrix} { a }^{ 2 }+1 & ab & ac \\ ab & { b }^{ 2 }+1 & bc \\ ca & cb & { c }^{ 2 }+1 \end{vmatrix}=\left( { a }^{ 2 }+1 \right) \left( \left( { b }^{ 2 }+1 \right) \left( { c }^{ 2 }+1 \right) -{ b }^{ 2 }{ c }^{ 2 } \right) -ab\left( ab{ c }^{ 2 }+ab-ab{ c }^{ 2 } \right) +ac\left( a{ b }^{ 2 }c-a{ b }^{ 2 }c-ac \right)$

$=\left( { a }^{ 2 }+1 \right) \left( { b }^{ 2 }{ c }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+1-{ b }^{ 2 }{ c }^{ 2 } \right) -ab\left( ab \right) +ac\left( -ac \right)$

$={ a }^{ 2 }{ b }^{ 2 }+{ a }^{ 2 }{ c }^{ 2 }+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+1-{ a }^{ 2 }{ b }^{ 2 }-{ a }^{ 2 }{ c }^{ 2 }$

$={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+1$

= R.H.S

Using cofactor elements of $2$ nd row, find the value of determinant, $\Delta =\begin{vmatrix} -3 & 0 & 2\\ 4 & -1 & 3\\ 5 & 0 & -2\end{vmatrix}$ .

1321004_1084384_ans_77f59d53e41a4c88ac982d4a82c80514.jpg

Show that the points are collinear $(-5, 1) (5, 5) (10, 7)$

Let the three collinear points be

$({ x }_{ 1 },{ y }_{ 1 })$ ,

$({ x }_{ 2 },{ y }_{ 2 })$ and

$({ x }_{ 3 },{ y }_{ 3 })$

The condition of collinearity is

${ x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 })=0$

Here, the given points are

$(-5,1)$ ,

$(5,5)$ and

$(10,7)$

$-5(5-7)+5(7-1)+10(1-5)\\ =-5(-2)+5(6)+10(-4)\\ =10+30-40\\ =0$

Hence, points

$(-5,1)$ ,

$(5,5)$ and

$(10,7)$ are collinear

Find the value of determinant $\begin{vmatrix} 5 & -3 \\ -7 & -4 \end{vmatrix}$

$\left| A \right| =\left| \begin{pmatrix} { a }_{ 11 } & { a }_{ 12 } \\ { a }_{ 21 } & { a }_{ 22 } \end{pmatrix} \right| ={ a }_{ 11 }{ a }_{ 22 }-{ a }_{ 21 }{ a }_{ 12 }$

$\left| \begin{matrix} 5 & -3 \\ -7 & 4 \end{matrix}- \right| =\left( 5\ast -4 \right) -\left( -7\ast -3 \right) =-20-14=-34$

Find the relation between $x$ and $y$ , if the points $x,y$ i.e $(1,2)$ and $(7,0)$ are collinears.

$\\ \\ \\ \\ (x,y),(1,2)and(7,0)are\quad collinear\quad points.$

$So\quad slope\quad of\quad \quad any\quad two\quad points\quad must\quad be\quad same.$

$\frac { (y-2) }{ (x-1) } =\frac { (0-2) }{ (7-1) }$

$\frac { (y-2) }{ (x-1) } =\frac { (-1) }{ (3) }$

$3(y-2) = -(x-1)$

$3y - 6 = -x + 1$

$So, x + 3y = 7$

$Thus,for\quad given\quad points\quad to\quad be\quad collinear,x\quad and\quad y\quad must\quad be\quad related\quad as:$

$x + 3y = 7$

If $A =\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$ , then verify that $A(adjA)=(adjA)A= I \left| A \right|$ .

1193787_1223777_ans_47ad188e02c44d3b9df87c02a0b2e62d.jpg

Find the value of $a$ if the points $P ( 1,5 ) , Q ( a , 1 )$ and $R ( 4,11 )$ are collinear .

1086072_1176354_ans_6dedf5d5038d426fab4e084ae6d8bd36.png

Find the largest value of a third order determinant whose elements are $1$ or $-1$

$\textbf{Step -1: Find the largest value of third order determinant whose elements are 1 or -1.}$

$\text{Let }\triangle=\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}\text{ be a determinant of order }3.$

$\text{Then, }\triangle =a_1b_2c_3+a_3b_1c_2+a_2b_3c_1-a_1b_3c_2-a_2b_1c_3-a_3b_2c_1$

$=(a_1b_2c_3+a_3b_1c_2+a_2b_3c_1)-(a_1b_3c_2+a_2b_1c_3+a_3b_2c_1)$

$\text{Since, each element of }\triangle \text{ is either }1\text{ or }-1.$

$\therefore \text{The value of the determinant cannot exceed }6.$

$\text{Clearly, the value of }\triangle \text{ is }6\text{ when the value of each term in the first bracket is }1.$

$\text{And the value of each term in the second bracket is }-1.$

$\text{But }a_1b_2c_3=a_3b_1c_2=a_2b_3c_1=1,$

$\text{implies that the product of the nine elements of the determinant }\triangle\text{ is }1$

$\text{And }a_1b_3c_2=a_2b_1c_3=a_3b_2c_1=-1,$

$\text{implies that the product of the nine elements of the determinant }\triangle\text{ is }-1$

$\text{So, this is not possible which implies that the maximum value of }\triangle\neq6$

$\therefore\text{Take each element of two terms of first bracket as }1\text{ and each element of the remaining term as }-1.$

$\text{Say, }a_1=b_2=c_3=a_3=b_1=c_2=1\text{ and }a_2=b_3=c_1=-1$

$\Rightarrow \triangle=\begin{vmatrix}1&1&-1\\-1&1&1\\1&-1&1\end{vmatrix}$

$\Rightarrow \triangle=4$

$\textbf{Hence, the largest value of a third order determinant whose elements are 1 or -1 is 4. }$

Find the cofactor matrix of the determinant $\begin{vmatrix} 1 & 2 & 3\\ -4 & 3 & 6 \\ 2 & -7 & 9 \end{vmatrix}$

1353794_1246069_ans_712ccef0d3c2414e8401113e218e40ca.PNG

If the points $(x_{1},\ y_{1}),\ (x_{2},\ y_{2})$ and $(x_{3},\ y_{3})$ are collinear, show that $\displaystyle \sum { \left( \frac { { y }_{ 1 }-{ y }_{ 2 } }{ { x }_{ 1 }{ x }_{ 2 } } \right) } =0$ , i.e
$\dfrac {y_{1}-y_{2}}{x_{1}x_{2}}+\dfrac {y_{2}-y_{3}}{x_{2}x_{3}}+\dfrac {y_{3}-y_{1}}{x_{3}x_{1}}=0$

Given

$(x_{1},y_{1})(x_{2},y_{2})(x_{3},y_{3})$ are collinear

$\begin{vmatrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{vmatrix}=0$

$x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})=0$ ____ (1)

By solving first column

Divide (1) by

$x_{1}x_{2}x_{3}$

$\dfrac{x_{1}(y_{2}-y_{3})}{x_{1}x_{2}x_{3}}+\dfrac{x_{2}(y_{3}-y_{1})}{x_{1}x_{2}x_{3}}+\dfrac{x_{3}(y_{1}-y_{2})}{x_{1}x_{2}x_{3}}=0$

$\dfrac{y_{2}-y_{3}}{x_{2}x_{3}}+\dfrac{y_{3}-y_{1}}{x_{1}x_{3}}+\dfrac{y_{1}-y_{2}}{x_{1}x_{2}}=0$

$\therefore \displaystyle\sum (\dfrac{y_{1}-y_{2}}{x_{1}x_{2}})=0$

$\therefore$ Hence proved.

If $A = \left[ {\begin{array}{*{20}{c}}1 & { - 2} & 3\\4 & 0 & { - 1}\\{ - 3} & 1 & 5\end{array}} \right]$ , then ${\left( {adj\,A} \right)}$ is equal to

1339866_1244369_ans_020eac6059884b879b3ae23c2a3a8a74.PNG

Find the largest value of a third order determinant whose elements are $0$ or $1$

$\textbf{Step -1: Find the largest value of third order determinant whose elements are 0 or 1.}$

$\text{Let }\triangle=\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}\text{ be a determinant of order }3.$

$\text{Then, }\triangle =a_1b_2c_3+a_3b_1c_2+a_2b_3c_1-a_1b_3c_2-a_2b_1c_3-a_3b_2c_1$

$=(a_1b_2c_3+a_3b_1c_2+a_2b_3c_1)-(a_1b_3c_2+a_2b_1c_3+a_3b_2c_1)$

$\text{Since, each element of }\triangle \text{ is either }0\text{ or }1.$

$\therefore \text{The value of the determinant cannot exceed }3.$

$\text{Clearly, the value of }\triangle \text{ is }3\text{ when the value of each term in the first bracket is }1.$

$\text{And the value of each term in the second bracket is zero.}$

$\text{But }a_1b_2c_3=a_3b_1c_2=a_2b_3c_1=1\text{ implies that every element of the determinant }\triangle\text{ is }1$

$\text{and in this case }\triangle=0\text{ so, the maximum value of }\triangle \neq3$

$\therefore\text{Take two of the three term as }1\text{ and each element of the remaining term as }0.$

$\text{Say, }a_1b_2c_3=a_3b_1c_2=1\text{ and }a_2=b_3=c_1=0$

$\Rightarrow \triangle=\begin{vmatrix}1&1&0\\0&1&1\\1&0&1\end{vmatrix}$

$\Rightarrow \triangle=2$

$\textbf{Hence, the largest value of a third order determinant whose elements are 0 or 1 is 2. }$

$adj (adj A) = |A|^{n - 2} A$

${\textbf {Step 1: Using the Inverse formula to prove the given relation.}}$

$\text{We know that}$

$A^{-1}=\dfrac{adjA}{|A|}$

$\therefore A.adjA=|A|I\;\;...........(1)$

$\text{Substituting }A\text{ with }adjA$

$adjA.adj(adjA)=|adjA|I$

$\text{From (1)}$

$adjA=\dfrac{|A|I}{A}\;\;.............(2)$

$\text{Substituting (2) in (1)}$

$\dfrac{|A|I}{A}adj(adjA)=|adjA|I\;\;...............(3)$

$\text{Taking determinant on both sides of (1)}$

$|A.adjA|=||A|I|$

$\implies |A||adjA|=|A|^n$

$\implies |adjA|=|A|^{n-1}\;\;...........(4)$

$\text{Substituting (4) in (3)}$

$\dfrac{|A|I}{A}adj(adjA)=|A|^{n-1}I$

$\implies adj(adjA)=|A|^{n-2}A$

$\textbf{Hence, we proved that }\mathbf{adj(adjA)=|A|^{n-2}A}$ .

Prove that $\begin{vmatrix} { yz-x }^{ 2 } & { zx-y }^{ 2 } & xy-z^{ 2 } \\ { zx-y }^{ 2 } & { xy }-z^{ 2 } & { yz-x }^{ 2 } \\ { xy-z }^{ 2 } & { yz-x }^{ 2 } & { zx-y }^{ 2 } \end{vmatrix}$ is divisible by $(x+y+z)$ and hence find the quotient

Let

$\Delta=\left|\begin{matrix} yz-{x}^{2} & zx-{y}^{2} & xy-{2}^{2} \\ zx-{y}^{2} & xy-{2}^{2} & yz-{x}^{2}\\ xy-{2}^{2} & yz-{x}^{2} & zx-{y}^{2} \end{matrix}\right|$

Adding

${C}_{1}\rightarrow {C}_{1}+{C}_{2}+{C}_{3}$ , we get

$\Delta=\left|\begin{matrix} xy+yz+zx-{x}^{2}-{y}^{2}-{z}^{2} & zx-{y}^{2} & xy-{2}^{2} \\ xy+yz+zx-{x}^{2}-{y}^{2}-{z}^{2} & xy-{2}^{2} & yz-{x}^{2}\\ xy+yz+zx-{x}^{2}-{y}^{2}-{z}^{2} & yz-{x}^{2} & zx-{y}^{2} \end{matrix}\right|$

$\Delta=\left(xy+yz+zx-{x}^{2}-{y}^{2}-{z}^{2}\right)\left|\begin{matrix} 1 & zx-{y}^{2} & xy-{2}^{2} \\ 1 & xy-{2}^{2} & yz-{x}^{2}\\1 & yz-{x}^{2} & zx-{y}^{2} \end{matrix}\right|$

Applying

${R}_{2}\rightarrow{R}_{2}-{R}_{1}$ and

${R}_{3}\rightarrow{R}_{3}-{R}_{1}$ we get

$\Delta=\left(xy+yz+zx-{x}^{2}-{y}^{2}-{z}^{2}\right)\left|\begin{matrix} 1 & zx-{y}^{2} & xy-{2}^{2} \\ 0 & \left(x+y+z\right)\left(y-z\right) & \left(x+y+z\right)\left(z-x\right)\\0 & \left(x+y+z\right)\left(y-x\right) & \left(x+y+z\right)\left(z-y\right) \end{matrix}\right|$

$\Delta=\left(xy+yz+zx-{x}^{2}-{y}^{2}-{z}^{2}\right){\left(x+y+z\right)}^{2}\left|\begin{matrix} 1 & zx-{y}^{2} & xy-{2}^{2} \\ 0 & \left(y-z\right) & \left(z-x\right)\\0 & \left(y-x\right) & \left(z-y\right) \end{matrix}\right|$

$\Delta=\left(xy+yz+zx-{x}^{2}-{y}^{2}-{z}^{2}\right){\left(x+y+z\right)}^{2}\left[\left(y-z\right)\left(z-y\right)-\left(z-x\right)\left(y-x\right)\right]$

$\Delta=\left({x}^{2}+{y}^{2}+{z}^{2}-xy-yz-zx\right){\left(x+y+z\right)}^{2}$

$\Delta$ is divisible by

$\left(x+y+z\right)$

The quotient when

$\Delta$ is divisible by

$\left(x+y+z\right)$ is

$\left({x}^{2}+{y}^{2}+{z}^{2}-xy-yz-zx\right)\left(x+y+z\right)$

If $A=\begin{pmatrix} 2 & 3 & 4 \\ 0 & -2 & 1 \\ 3 & -1 & 2 \end{pmatrix},B=\begin{pmatrix} 2 & 0 & -3 \\ 4 & 0 & -1 \\ 3 & 4 & 5 \end{pmatrix}$ and $C=\begin{pmatrix} 5 & 6 & 7 \\ -1 & 2 & 3 \\ 4 & -5 & 4 \end{pmatrix}$ Prove that $A(BC)$ and $(AB)C$ .

$A=\left( \overset { 2 }{ \underset { 3 }{ 0 } } \quad \overset { 3 }{ \underset { -1 }{ -2 } } \quad \overset { 4 }{ \underset { 2 }{ 1 } } \right) ;\quad B=\left( \overset { 2 }{ \underset { 3 }{ 4 } } \quad \overset { 0 }{ \underset { 4 }{ 0 } } \quad \overset { -3 }{ \underset { 5 }{ -1 } } \right) ;\quad C=\left( \overset { 5 }{ \underset { 4 }{ -1 } } \quad \overset { 6 }{ \underset { -5 }{ 2 } } \quad \overset { 7 }{ \underset { 4 }{ 3 } } \right)$

$BC=\left( \overset { -2 }{ \underset { 31 }{ 16 } } \quad \overset { 17 }{ \underset { 1 }{ 29 } } \quad \overset { 2 }{ \underset { 53 }{ 24 } } \right)$

$A\left( BC \right) =\left( \overset { 168 }{ \underset { 40 }{ -1 } } \quad \overset { 145 }{ \underset { 24 }{ -57 } } \quad \overset { 288 }{ \underset { 88 }{ 5 } } \right)$

Now,

$AB=\left( \overset { 28 }{ \underset { 8 }{ -5 } } \quad \overset { 16 }{ \underset { 8 }{ 4 } } \quad \overset { 11 }{ \underset { 2 }{ 7 } } \right)$

$\left( AB \right) C=\left( \overset { 168 }{ \underset { 40 }{ -1 } } \quad \overset { 145 }{ \underset { 24 }{ -57 } } \quad \overset { 288 }{ \underset { 88 }{ 5 } } \right)$

$\therefore$

$A\left( BC \right) =\left( AB \right) C$ Proved.

Using properties of determinant, prove that $\begin{vmatrix} b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{vmatrix}=3abc-{ a }^{ 3 }-{ b }^{ 3 }-{ c }^{ 3 }$

To prove:-

$\begin{vmatrix} b+c & a - b & a \\ c + a & b - c & b \\ a + b & c - a & c \end{vmatrix} = 3abc - {a}^{3} - {b}^{3} - {c}^{3}$

Proof:-

Taking L.H.S.-

$\begin{vmatrix} b+c & a - b & a \\ c + a & b - c & b \\ a + b & c - a & c \end{vmatrix}$

Applying

${R}_{1} \rightarrow {R}_{1} + {R}_{2} + {R}_{3}$

$= \begin{vmatrix} 2 \left( a + b + c \right) & 0 & a + b + c \\ c + a & b - c & b \\ a + b & c - a & c \end{vmatrix}$

$= \left( a + b + c \right) \begin{vmatrix} 2 & 0 & 1 \\ c + a & b - c & b \\ a + b & c - a & c \end{vmatrix}$

Now, expanding along

${R}_{1}$ , we have

$= \left( a + b + c \right) \left[ 2 \left( bc - {c}^{2} - bc + ab \right) - 0 + 1 \left( {c}^{2} - {a}^{2} - ab - {b}^{2} + ac + bc \right) \right]$

$= \left( a + b + c \right) \left( 2ab - 2 {c}^{2} + {c}^{2} - {a}^{2} - {b}^{2} - ab + ac + bc \right)$

$= \left( a + b + c \right) \left( ab + bc + ca - {a}^{2} - {b}^{2} - {c}^{2} \right)$

$= - \left( a + b + c \right) \left( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca \right)$

$= - \left( {a}^{3} + {b}^{3} + {c}^{3} - 3abc \right)$

$= 3abc - {a}^{3} - {b}^{3} - {c}^{3}$

$=$ R.H.S.

Hence proved

Compute the adjoint of the following matrice:
$\begin{bmatrix} 2 & -1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & -1 \end{bmatrix}$
Verify that $(adj A)A = |A|I = A(adj A)$ for the above matrice.

1622637_1661701_ans_38edb8054fc24e299e0f0a732b4fca15.jpg

Compute the adjoint of the following matrice:
$\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$

Verify that $(adj A)A = |A|I = A(adj A)$ for the above matrice.

$\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$

${C}_{11}={\left(-1\right)}^{1+1}\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}=1-4=-3$

${C}_{12}={\left(-1\right)}^{1+2}\begin{vmatrix} 2 & 2 \\ 2 & 1 \end{vmatrix}=-\left(2-4\right)=2$

${C}_{13}={\left(-1\right)}^{1+3}\begin{vmatrix} 2 & 1 \\ 2 & 2 \end{vmatrix}=4-2=2$

${C}_{21}={\left(-1\right)}^{2+1}\begin{vmatrix} 2 & 2 \\ 2 & 1 \end{vmatrix}=-\left(2-4\right)=2$

${C}_{22}={\left(-1\right)}^{2+2}\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}=1-4=-3$

${C}_{23}={\left(-1\right)}^{2+3}\begin{vmatrix} 1 & 2 \\ 2 & 2 \end{vmatrix}=-\left(2-4\right)=2$

${C}_{31}={\left(-1\right)}^{3+1}\begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix}=4-2=2$

${C}_{32}={\left(-1\right)}^{3+2}\begin{vmatrix} 1 & 2 \\ 2 & 2 \end{vmatrix}=-\left(2-4\right)=2$

${C}_{33}={\left(-1\right)}^{3+3}\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}=1-4=-3$

$adj{A}={\begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}}^{T}$

$adj{A}=\begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}$

$(adj A)A=\begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$

$(adj A)A=\begin{bmatrix} -3+4+4 & -6+2+4 & -6+4+2 \\ 2-6+4 & 4-3+4 & 4-6+2 \\2+4-6 & 4+2-6 & 4+4-3 \end{bmatrix}$

$(adj A)A=\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\0 & 0 & 5 \end{bmatrix}=5\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix}=5I$ ...........

$(1)$

$\left|A\right|=\begin{vmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{vmatrix}$

$=1\left(1-4\right)-2\left(2-4\right)+2\left(4-2\right)$

$=-3+4+4=5$

$\left|A\right|I=5I$ ..........

$(2)$

$A\left(adj{A}\right)=\begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$

$A\left(adj{A}\right)=\begin{bmatrix} -3+4+4 & -6+2+4 & -6+4+2 \\ 2-6+4 & 4-3+4 & 4-6+2 \\ 2+4-6 & 4+2-6 & 4+4-3 \end{bmatrix}$

$A\left(adj{A}\right)=\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$

$=5\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=5I$ ..........

$(3)$

From

$(1),(2)$ and

$(3)$ we have

$(adj A)A = |A|I = A(adj A)$

Hence verified.

Compute the adjoint of the following matrice:
$\begin{bmatrix} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1 \end{bmatrix}$
Verify that $(adj A)A = |A|I = A(adj A)$ for the above matrice.

1668989_1661694_ans_90c05a7139774cc09375c1ca9d67b0e4.jpg

Find the adjoint of the following matrice:
$\begin{bmatrix} -3 & 5 \\ 2 & 4 \end{bmatrix}$
Verify that $(adj A ) A = |A| I = A (adj A)$ for the above matrice.

$Adj \ A = \begin{bmatrix} +4 & -5 \\ -2& -3 \end{bmatrix} = \begin{bmatrix} 4 &-5 \\ -2 & -3 \end{bmatrix}$

$A = \begin{bmatrix} -3 &5 \\2 & 4 \end{bmatrix} \Rightarrow det \ A = -12 - 10 = -22$

$(Adj . A)A = \begin{bmatrix} +4 & -5 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} -3& 5 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} -22 &0 \\ 0 & -22 \end{bmatrix}$

$= -22 \begin{bmatrix} 1 & 0 \\ 0&1 \end{bmatrix}$

$= |A|. I.$

$A. (adj \, A) = \begin{bmatrix} -3 & 5 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 4& -5\\ -2& -3 \end{bmatrix} = \begin{bmatrix} -22 & 0 \\ 0& -22 \end{bmatrix}$

$= -22 \begin{bmatrix} 1 &0 \\0 & 1 \end{bmatrix}$

$= |A|. I$

Find the adjoint of the following matrice:

$\begin{bmatrix} 1 & \tan\,\alpha /2 \\ -\tan\,\alpha /2 & 1 \end{bmatrix}$

Verify that $(adj A ) A = |A| I = A (adj A)$ for the above matrice.

We have

$A= \begin{bmatrix} 1 & \tan\,\alpha /2 \\ -\tan\,\alpha /2 & 1 \end{bmatrix}$

${C}_{11}={\left(-1\right)}^{1+1}{M}_{11}=1$

${C}_{12}={\left(-1\right)}^{1+2}{M}_{12}=-\tan{\dfrac{\alpha}{2}}$

${C}_{21}={\left(-1\right)}^{2+1}{M}_{21}=\tan{\dfrac{\alpha}{2}}$

${C}_{22}={\left(-1\right)}^{2+2}{M}_{22}=1$

$adj{A}={\begin{bmatrix} 1 & -\tan{\dfrac{\alpha}{2}} \\ \tan{\dfrac{\alpha}{2}} & 1 \end{bmatrix}}^{T}$

$=\begin{bmatrix} 1 & \tan{\dfrac{\alpha}{2}} \\ -\tan{\dfrac{\alpha}{2}} & 1 \end{bmatrix}$

Now

$\left(adj{A}\right)A=\begin{bmatrix} 1 & \tan{\dfrac{\alpha}{2}} \\ -\tan{\dfrac{\alpha}{2}} & 1 \end{bmatrix}\begin{bmatrix} 1 & -\tan{\dfrac{\alpha}{2}} \\ \tan{\dfrac{\alpha}{2}} & 1 \end{bmatrix}$

$=\begin{bmatrix} 1+{\tan}^{2}{\dfrac{\alpha}{2}} & -\tan{\dfrac{\alpha}{2}}+\tan{\dfrac{\alpha}{2}} \\ -\tan{\dfrac{\alpha}{2}}+\tan{\dfrac{\alpha}{2}} & {\tan}^{2}{\dfrac{\alpha}{2}}+1 \end{bmatrix}$

$=\begin{bmatrix} {\sec}^{2}{\dfrac{\alpha}{2}} & 0 \\ 0 & {\sec}^{2}{\dfrac{\alpha}{2}} \end{bmatrix}$ ..........

$(1)$

$\left|A\right|=\begin{vmatrix} 1 & \tan{\dfrac{\alpha}{2}} \\ -\tan{\dfrac{\alpha}{2}} & 1 \end{vmatrix}$

$=1+{\tan}^{2}{\dfrac{\alpha}{2}}={\sec}^{2}{\dfrac{\alpha}{2}}$

$\left|A\right|I={\sec}^{2}{\dfrac{\alpha}{2}}\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$

$=\begin{bmatrix} {\sec}^{2}{\dfrac{\alpha}{2}} & 0\\ 0 & {\sec}^{2}{\dfrac{\alpha}{2}} \end{bmatrix}$ .........

$(2)$

$A\left(adj{A}\right)=\begin{bmatrix} 1 & \tan\,\alpha /2 \\ -\tan\,\alpha /2 & 1 \end{bmatrix} \begin{bmatrix} 1 & \tan{\dfrac{\alpha}{2}} \\ -\tan{\dfrac{\alpha}{2}} & 1 \end{bmatrix}$

$=\begin{bmatrix} {\sec}^{2}{\dfrac{\alpha}{2}} & 0 \\ 0 & {\sec}^{2}{\dfrac{\alpha}{2}} \end{bmatrix}$ ..........

$(3)$

$\therefore\,$ from

$(1),(2)$ and

$(3)$ it is verified that

$(adj A ) A = |A| I = A (adj A)$

Find A (adj A) for the matrix $A = \begin{bmatrix} 1 & -2 & 3 \\ 0 & 2 & 1 \\ -4 & 5 & 2 \end{bmatrix}$ .

1416594_1661754_ans_ea7d5757779f4be6b1d22dda21e97dad.jpg

Find the adjoint of the matrix $A= \begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}$ and hence show that $A(adj\: A)= \left | A \right |I_{3}$ .

2039699_1661810_ans_6186456754454077b17d09dabc1cda2f.png

Compute the adjoint of the following matrice:
$\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 1 & 1 & 3 \end{bmatrix}$
Verify that $(adj A)A = |A|I = A(adj A)$ for the above matrice.

1622633_1661707_ans_1ed60b5d3be740d98235ba4a23645e1e.jpg

$\left| \begin{matrix} 0 & xyz & x-z \\ y-x & 0 & y-z \\ z-x & z-y & 0 \end{matrix} \right|$ =_________

Let

$D=\left| \begin{matrix} 0 & xyz & x-z \\ y-x & 0 & y-z \\ z-x & z-y & 0 \end{matrix} \right|$

${ R }_{ 2 }-{ R }_{ 3 }\rightarrow { R }_{ 2 }$

$\therefore D=\left| \begin{matrix} 0 & xyz & x-z \\ y-z & y-z & y-z \\ z-x & z-y & 0 \end{matrix} \right|$

$\therefore D=\left( y-z \right) \left| \begin{matrix} 0 & xyz & x-z \\ 1 & 1 & 1 \\ z-x & z-y & 0 \end{matrix} \right|$

$\therefore D=\left( y-z \right) \left\{ -xyz\left[ 0-\left( z-x \right) \right] +\left( x-z \right) \left[ \left( z-y \right) -\left( z-x \right) \right] \right\}$

$\therefore D=\left( y-z \right) \left\{ -xyz\left[ x-z \right] +\left( x-z \right) \left[ z-y-z+x \right] \right\}$

$\therefore D=\left( y-z \right) \left\{ -xyz\left[ x-z \right] +\left( x-z \right) \left[ x-y \right] \right\}$

$\therefore D=\left( y-z \right) \left( x-z \right) \left\{ -xyz+x-y \right\}$

Determinants - Class 12 Commerce Maths - Extra Questions

If A is 2×22\times 2 matrix, detA=4det A=4 then find the product of det(3A)det(3A) and det(A−1)det(A^{-1})

Write the value of the determinant [pp+1p−1p]\begin{bmatrix} p & p+1 \\ p-1 & p\end{bmatrix}when p=1342p=1342

Using the properties of determinants, find the value of |0a−b−a0−cbc0|\begin{vmatrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{vmatrix}

Evaluate the determinants|24−5−1|\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}

If A=[1002345−6x]A=\begin{bmatrix} 1 & 0 & 0 \\ 2 & 3 & 4 \\ 5 & -6 & x \end{bmatrix} and detA=45\det A = 45; then find xx.

Find adjoint of matrix A=[1234]\ \ \ A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}

Show that the points A(2,3),B(4,0)A(2,3),B(4,0) and C(6,−3)C(6,-3) are collinear.

Write the value of △=|x+yy+zz+xzxy−3−3−3|\triangle = \begin{vmatrix}x+y&y+z&z+x\\z&x&y\\-3&-3&-3\end{vmatrix}.

The area of a triangle is zero, then the three points are said to be ______ points.

Find the value of the determinant:|5−2−31|\begin{vmatrix}5 & -2\\ -3 & 1\end{vmatrix}

If for any 2×22 \times 2 square matrix A, A(adjA)=[8008]A (adj A) = \begin{bmatrix} 8& 0\\ 0 & 8\end{bmatrix}, them write the value of det[A].

Find a value of xx if |x218x|=|62186|\begin{vmatrix}x& 2 \\18&x\end{vmatrix} = \begin{vmatrix}6& 2 \\18&6\end{vmatrix}

|111abca2b3c3|\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ { a }^{ 2 } & { b }^{ 3 } & { c }^{ 3 } \end{vmatrix}Prove that : = (a−b)(b−c)(c−a)(a+b+c).(a - b) (b - c) (c - a) (a + b + c).

Find the values of xx for which |3xx1|=|3241|\begin{vmatrix}3&x\\x&1\end{vmatrix}=\begin{vmatrix}3&2\\4&1\end{vmatrix}.

Evaluate the following deteminants :i)|x−7x5x+1|\begin{vmatrix} x& -7\\ x &5x+1 \end{vmatrix}ii)|cos15∘sin15∘sin75∘cos75∘|\begin{vmatrix} \cos 15^{\circ} & \sin 15^{\circ}\\ \sin 75^{\circ} & \cos 75^{\circ} \end{vmatrix}

Find the determinant in following case:A=[52002]A=\left [ \begin{array}{c}5\hspace{0.5cm}{20}\\0\hspace{0.5cm}{2}\end{array} \right ]

Find if the points (0, 8/3), (1/3) , and (82, 30) are collinear.

Prove the following :|y+zzyzz+xxyxx+y|=4xyz\left| \begin{matrix} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{matrix} \right| =4xyz

Prove that the points (a , b), (c , d) and (a - c, b - d) are collinear if ad = bc. Also show that the straight line passing through these points passes through origin.

Show that|sin10∘−cos10∘sin80∘cos80∘|=1\begin{vmatrix} \sin 10^{\circ}& -\cos 10^{\circ}\\ \sin 80^{\circ}&\cos 80^{\circ} \end{vmatrix}=1

Find the value of KK, for which the given points are collinear:A(7,−2)B(5,1)C(3,K)\begin{matrix} A \\ \left( 7,-2 \right) \end{matrix}\begin{matrix} B \\ \left( 5,1 \right) \end{matrix}\begin{matrix} C \\ \left( 3,K \right) \end{matrix}

In each of the following find the value of kk, for which the points are collinear..(ii) (8,1),(k,−4),(2,−5)(8, 1), (k, -4), (2, -5)

Let A=[58813]A=\begin{bmatrix}5&8\\8&13\end{bmatrix} then show that AA satisfies the equation x2−18x+1=0x^2-18x+1=0.

In each of the following find the value of kk, for which the points are collinear.(i) (7,−2),(5,1),(3,k)(7, -2), (5, 1), (3, k)

Find the value of xx if the points (x,5),(2,3)(x,5),(2,3) and (−2,−11)(-2,-11) are collinear.

Show that △\triangleABC is an isosceles triangle, if the determinant|1111+cosA1+cosB1+cosCcos2A+cosAcos2B+cosBcos2C+cosC|=0\begin{vmatrix} 1 & 1 & 1 \\ 1+\cos A & 1+\cos B & 1+\cos C \\ \cos^2A +\cos A & \cos^2B+\cos B & \cos^2C+\cos C\end{vmatrix}=0

Using properties of determinants, prove the following|1+a2−b22ab−2b2ab1−a2+b22a2b−2a1−a2−b2|=(1+a2+b2)3\begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{vmatrix}=(1+a^2+b^2)^3.

Solve:|0−3xx+131415|\begin{vmatrix} 0& -3 &x \\ x+1& 3 &1 \\ 4& 1 & 5\end{vmatrix} =00

If A=|1−1−11|A = \left| {\begin{array}{*{20}{c}} 1&{ - 1} \\ { - 1}&1 \end{array}} \right|, then show that A2=2A&A3=4A{A^2} = 2A\,\,\& \,\,{A^3} = 4A

Find |462309715|\begin{vmatrix}4&6&2\\3&0&9\\7&1&5\end{vmatrix}

Show that the point P(a,b+c)P (a,b+c), Q(b,c+a)Q(b,c+a) and R(c,a+b)R(c,a+b) are collinear.

Find the adjoint of |1−1230−2103|\begin{vmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{vmatrix}

Find the value of following determinant.|73533212|\begin{vmatrix} \dfrac{7}{3} & \dfrac{5}{3}\\ \dfrac{3}{2} & \dfrac{1}{2}\end{vmatrix}.

If A=[1121],A=\left[ \begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix} \right], find |A||A|

If A is a square matrix of order 33 such |A|=5|A|=5, then find the value of |adjA||adjA| ?

If three points (x1,y1),(x2,y2)(x_1 , y_1) , (x_2 , y_2 ) and (x3,y3)(x_3 , y_3 ) lie on the same line, prove that y2−y3x2x3+y3−y1x3x1+y1−y2x1x2=0 \dfrac {y_2 - y_3}{x_2 x_3} + \dfrac {y_3 - y_1}{x_3 x_1}+ \dfrac {y_1 - y_2}{x_1 x_2} = 0

Evaluate |abc−11−11−11|\begin{vmatrix} a & b & c \\ -1 & 1 & -1 \\ 1 & -1 & 1 \end{vmatrix}.

Find the values of the following determinants |2i−3ii3−2i5|\begin{vmatrix} 2i & -3i \\ { i }^{ 3 } & -2{ i }^{ 5 } \end{vmatrix} where i=√−1i=\sqrt{-1}.

A=[2468]A = \begin{bmatrix} 2&4 \\ 6&8 \end{bmatrix} find det(A)

Compute the following determinant :|a2ababb2|\begin{vmatrix} { a }^{ 2 } & ab \\ ab & { b }^{ 2 } \end{vmatrix}

Show that |111abca2b2c2|=(a−b)(b−c)(c−a)\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ { a }^{ 2 } & b^{ 2 } & c^{ 2 } \end{matrix} \right| =(a-b)(b-c)(c-a).

Compute the following determinant :|35−26|\begin{vmatrix} 3 & 5 \\ -2 & 6 \end{vmatrix}

Find value of the determinantA=|3257|A = \begin{vmatrix}3 & 2\\ 5 & 7\end{vmatrix}

Prove that |1ωω2ωω21ω21ω|=0\left| \begin{matrix} 1 & \omega & { \omega }^{ 2 } \\ \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & { \omega } \end{matrix} \right| =0

For what value of pp are the points (2,1),(p,−1)(2,1),(p,-1) and (−1,3)(-1,3) collinear.

Using determinants show that points A(a,b+c),B(b,c+a)A(a,b+c),B(b,c+a) and C(c,a+b)C(c,a+b) are col-linear.

Prove that the adjoint of a symmetric matrix is also a symmetric matrix.

Find deteminant [1002]\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}

Prove that: 2|8−5−26|=|14−2−46|2\left| \begin{matrix} 8 & -5 \\ -2 & 6 \end{matrix} \right|=\left| \begin{matrix} 14 & -2 \\ -4 & 6 \end{matrix} \right|

In the diagram on a lunar eilpse, if the positions od sun,Earth and moon are shown by (−4,6),(k,−2)(-4,6), (k,-2) and(5,−6) (5,-6) respectively, then find the value of k.

If the points (3,−2),(x,2)(3,-2), (x,2) and (8,8)(8,8) are collinear find 10x10x using determinant.

If the value of determinant |m6−54|\left| \begin{matrix} m & 6 \\ -5 & 4 \end{matrix} \right| is 5858, then find the value of mm.

If A(x1,y1),B(x2,y2),C(x3,y3)A(x_{1},y_{1}),B(x_{2},y_{2}),C(x_{3},y_{3}) are vertices of an equilateral triangle whose each side is equal to a, then prove that |x1y12x2y22x3y32|=3a4\begin{vmatrix} x_{1} & y_{1} & 2 \\ x_{2} & y_{2} & 2 \\ x_{3} & y_{3} & 2 \end{vmatrix}=3a^4.

Find the value of kk if det |K+3123−21−4−33|=0\begin{vmatrix} K+3 & 1 & 2 \\ 3 & -2 & 1 \\ -4 & -3 & 3 \end{vmatrix}=0

Using the method of slope, show that the following points are collinear : (i)A(4,8),B(5,12),C(9,28)( i ) A \left( 4,8\right ) , B \left( 5,12 \right) , C \left( 9,28 \right) (ii)A(16,−18),B(3,−6),C(−10,6)(ii) A \left( 16 , - 18 \right) , B \left( 3 , - 6 \right) , C \left( - 10,6\right )

Show that the points P(−2,3,5),Q(1,2,3)P(-2,3,5), Q(1,2,3) and R(7,0,−1)R(7,0,-1) are collinear.

Show that the points P(-2, 3, 5), Q(1, 2, 3)P(-2, 3, 5), Q(1, 2, 3) and R(7, 0, -1)R(7, 0, -1) are collinear.

Solve:\left[ \begin{matrix} 1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & 6 \end{matrix} \right] \left[ \begin{matrix} 1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & 6 \end{matrix} \right]

If A= \left[ {\matrix{10 & 0 \cr 4 & 10\cr } } \right]A= \left[ {\matrix{10 & 0 \cr 4 & 10\cr } } \right], then find \left| A \right|\left| A \right|.

Construct a 3\times 23\times 2 matrix whose elements are given by a_{1}=\dfrac{1}{2}|i-3j|a_{1}=\dfrac{1}{2}|i-3j|.

Write the value of the determinant \left| \begin{array}{cc}{3} & {-1} \\ {2} & {1}\end{array}\right| \left| \begin{array}{cc}{3} & {-1} \\ {2} & {1}\end{array}\right|

\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { b }^{ 2 } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| =(a-b)(b-c)(c-a)\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { b }^{ 2 } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| =(a-b)(b-c)(c-a)

Solve:\left| \begin{matrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{matrix} \right| =0\left| \begin{matrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{matrix} \right| =0

Find the value of x for which |_{ x }^{ 3 }\quad _{ 1 }^{ x }|=|_{ 8 }^{ 3 }\quad _{ 1 }^{ 2 }||_{ x }^{ 3 }\quad _{ 1 }^{ x }|=|_{ 8 }^{ 3 }\quad _{ 1 }^{ 2 }|.

If the points (-3,6), (-9,a) and (0,15) are collinear, then find a.

Convert \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} into an identify matrix by suitable row transformations.

If A = \left| \begin{array} { c c } { 2 } & { 3 } \\ { - 1 } & { 2 } \end{array} \right|A = \left| \begin{array} { c c } { 2 } & { 3 } \\ { - 1 } & { 2 } \end{array} \right| Find AA.

Evaluate \Delta = \begin{vmatrix} \cos 2\alpha & \sin 2\alpha \\-\sin 3\alpha & \cos 3\alpha \end{vmatrix}\Delta = \begin{vmatrix} \cos 2\alpha & \sin 2\alpha \\-\sin 3\alpha & \cos 3\alpha \end{vmatrix}

The value of \begin{vmatrix} \cos15^{\circ} & \sin15^{\circ} \\ \sin75^{\circ} & \cos75^{\circ} \end{vmatrix}\begin{vmatrix} \cos15^{\circ} & \sin15^{\circ} \\ \sin75^{\circ} & \cos75^{\circ} \end{vmatrix} is

Find the value of x, if \begin{vmatrix} x+2 & x \\ x-4 & x+3 \end{vmatrix}=\begin{vmatrix} 4 & 2 \\ -2 & 5 \end{vmatrix}\begin{vmatrix} x+2 & x \\ x-4 & x+3 \end{vmatrix}=\begin{vmatrix} 4 & 2 \\ -2 & 5 \end{vmatrix}

Evaluate:\begin{vmatrix} \cos30^{\circ} & \sin30^{\circ} \\ \sin105^{\circ} & \cos105^{\circ} \end{vmatrix}\begin{vmatrix} \cos30^{\circ} & \sin30^{\circ} \\ \sin105^{\circ} & \cos105^{\circ} \end{vmatrix}

Evaluate the determinant :\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}

Find the determinant of the matrix AAA=\begin{bmatrix} -1 & 4 \\ 2 & 3 \end{bmatrix}A=\begin{bmatrix} -1 & 4 \\ 2 & 3 \end{bmatrix}

Show that the points A(-7, 4, -2), B(-2, 1, 0)A(-7, 4, -2), B(-2, 1, 0) and C(3, -2, 2)C(3, -2, 2) are collinear.

Find the coordinates of a point AA, where ABAB is diameter of a circle whose centre is (2, -3)(2, -3) and BB is the point (1, 4)(1, 4).

Evaluate the following determinants :\begin{vmatrix} \cos15^{\circ} & \sin15^{\circ} \\ \sin75^{\circ} & \cos75^{\circ} \end{vmatrix}\begin{vmatrix} \cos15^{\circ} & \sin15^{\circ} \\ \sin75^{\circ} & \cos75^{\circ} \end{vmatrix}

Find the values of xx, if \begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}= \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}

Are the following pairs of sets equal? Give reasons.A=\{x : xA=\{x : x is a letter of the word "WOLF"\}\},B=\{x : xB=\{x : x is a letter of the word "FOLLOW"\}\}.

The value of the determinant \begin{vmatrix} \cos30^{\circ} & \sin60^{\circ} \\ \sin30^{\circ} & \cos60^{\circ} \end{vmatrix}\begin{vmatrix} \cos30^{\circ} & \sin60^{\circ} \\ \sin30^{\circ} & \cos60^{\circ} \end{vmatrix} is

If A is $2\times 2$ matrix, $det A=4$ then find the product of $det(3A)$ and $det(A^{-1})$

Write the value of the determinant $\begin{bmatrix} p & p+1 \\ p-1 & p\end{bmatrix}$
when $p=1342$

Using the properties of determinants, find the value of

$\begin{vmatrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{vmatrix}$

Evaluate the determinants
$\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$

If $A=\begin{bmatrix} 1 & 0 & 0 \\ 2 & 3 & 4 \\ 5 & -6 & x \end{bmatrix}$ and $\det A = 45$ ; then find $x$ .

Find adjoint of matrix $\ \ \ A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

Show that the points $A(2,3),B(4,0)$ and $C(6,-3)$ are collinear.

Write the value of $\triangle = \begin{vmatrix}x+y&y+z&z+x\\z&x&y\\-3&-3&-3\end{vmatrix}$ .

Find the value of the determinant:
$\begin{vmatrix}5 & -2\\ -3 & 1\end{vmatrix}$

If for any $2 \times 2$ square matrix A, $A (adj A) = \begin{bmatrix} 8& 0\\ 0 & 8\end{bmatrix}$ , them write the value of det[A].

Find a value of $x$ if $\begin{vmatrix}x& 2 \\18&x\end{vmatrix} = \begin{vmatrix}6& 2 \\18&6\end{vmatrix}$

$\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ { a }^{ 2 } & { b }^{ 3 } & { c }^{ 3 } \end{vmatrix}$
Prove that : = $(a - b) (b - c) (c - a) (a + b + c).$

Find the values of $x$ for which $\begin{vmatrix}3&x\\x&1\end{vmatrix}=\begin{vmatrix}3&2\\4&1\end{vmatrix}$ .

Evaluate the following deteminants :
i) $\begin{vmatrix} x& -7\\ x &5x+1 \end{vmatrix}$

ii) $\begin{vmatrix} \cos 15^{\circ} & \sin 15^{\circ}\\ \sin 75^{\circ} & \cos 75^{\circ} \end{vmatrix}$

Find the determinant in following case:
$A=\left [ \begin{array}{c}5\hspace{0.5cm}{20}\\0\hspace{0.5cm}{2}\end{array} \right ]$

Prove the following :
$\left| \begin{matrix} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{matrix} \right| =4xyz$

Show that
$\begin{vmatrix} \sin 10^{\circ}& -\cos 10^{\circ}\\ \sin 80^{\circ}&\cos 80^{\circ} \end{vmatrix}=1$

Find the value of $K$ , for which the given points are collinear:
$\begin{matrix} A \\ \left( 7,-2 \right) \end{matrix}\begin{matrix} B \\ \left( 5,1 \right) \end{matrix}\begin{matrix} C \\ \left( 3,K \right) \end{matrix}$

In each of the following find the value of $k$ , for which the points are collinear..
(ii) $(8, 1), (k, -4), (2, -5)$

Let $A=\begin{bmatrix}5&8\\8&13\end{bmatrix}$ then show that $A$ satisfies the equation $x^2-18x+1=0$ .

In each of the following find the value of $k$ , for which the points are collinear.(i) $(7, -2), (5, 1), (3, k)$

Find the value of $x$ if the points $(x,5),(2,3)$ and $(-2,-11)$ are collinear.

Show that $\triangle$ ABC is an isosceles triangle, if the determinant
$\begin{vmatrix} 1 & 1 & 1 \\ 1+\cos A & 1+\cos B & 1+\cos C \\ \cos^2A +\cos A & \cos^2B+\cos B & \cos^2C+\cos C\end{vmatrix}=0$

Using properties of determinants, prove the following
$\begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{vmatrix}=(1+a^2+b^2)^3$ .

Solve:
$\begin{vmatrix} 0& -3 &x \\ x+1& 3 &1 \\ 4& 1 & 5\end{vmatrix}$ = $0$

If $A = \left| {\begin{array}{*{20}{c}} 1&{ - 1} \\ { - 1}&1 \end{array}} \right|$ , then show that ${A^2} = 2A\,\,\& \,\,{A^3} = 4A$

Find $\begin{vmatrix}4&6&2\\3&0&9\\7&1&5\end{vmatrix}$

Show that the point $P (a,b+c)$ , $Q(b,c+a)$ and $R(c,a+b)$ are collinear.

Find the adjoint of $\begin{vmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{vmatrix}$

Find the value of following determinant.
$\begin{vmatrix} \dfrac{7}{3} & \dfrac{5}{3}\\ \dfrac{3}{2} & \dfrac{1}{2}\end{vmatrix}$ .

If $A=\left[ \begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix} \right],$ find $|A|$

If A is a square matrix of order $3$ such $|A|=5$ , then find the value of $|adjA|$ ?

If three points $(x_1 , y_1) , (x_2 , y_2 )$ and $(x_3 , y_3 )$ lie on the same line, prove that
$\dfrac {y_2 - y_3}{x_2 x_3} + \dfrac {y_3 - y_1}{x_3 x_1}+ \dfrac {y_1 - y_2}{x_1 x_2} = 0$

Evaluate $\begin{vmatrix} a & b & c \\ -1 & 1 & -1 \\ 1 & -1 & 1 \end{vmatrix}$ .

Find the values of the following determinants
$\begin{vmatrix} 2i & -3i \\ { i }^{ 3 } & -2{ i }^{ 5 } \end{vmatrix}$ where $i=\sqrt{-1}$ .

$A = \begin{bmatrix} 2&4 \\ 6&8 \end{bmatrix}$ find det(A)

Compute the following determinant :

$\begin{vmatrix} { a }^{ 2 } & ab \\ ab & { b }^{ 2 } \end{vmatrix}$

Show that $\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ { a }^{ 2 } & b^{ 2 } & c^{ 2 } \end{matrix} \right| =(a-b)(b-c)(c-a)$ .

Compute the following determinant :

$\begin{vmatrix} 3 & 5 \\ -2 & 6 \end{vmatrix}$

Find value of the determinant
$A = \begin{vmatrix}3 & 2\\ 5 & 7\end{vmatrix}$

Prove that $\left| \begin{matrix} 1 & \omega & { \omega }^{ 2 } \\ \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & { \omega } \end{matrix} \right| =0$

For what value of $p$ are the points $(2,1),(p,-1)$ and $(-1,3)$ collinear.

Using determinants show that points $A(a,b+c),B(b,c+a)$ and $C(c,a+b)$ are col-linear.

Find deteminant $\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$

Prove that: $2\left| \begin{matrix} 8 & -5 \\ -2 & 6 \end{matrix} \right|=\left| \begin{matrix} 14 & -2 \\ -4 & 6 \end{matrix} \right|$

In the diagram on a lunar eilpse, if the positions od sun,Earth and moon are shown by $(-4,6), (k,-2)$ and $(5,-6)$ respectively, then find the value of k.

If the points $(3,-2), (x,2)$ and $(8,8)$ are collinear find $10x$ using determinant.

If the value of determinant $\left| \begin{matrix} m & 6 \\ -5 & 4 \end{matrix} \right|$ is $58$ , then find the value of $m$ .

If $A(x_{1},y_{1}),B(x_{2},y_{2}),C(x_{3},y_{3})$ are vertices of an equilateral triangle whose each side is equal to a, then prove that $\begin{vmatrix} x_{1} & y_{1} & 2 \\ x_{2} & y_{2} & 2 \\ x_{3} & y_{3} & 2 \end{vmatrix}=3a^4$ .

Find the value of $k$ if det $\begin{vmatrix} K+3 & 1 & 2 \\ 3 & -2 & 1 \\ -4 & -3 & 3 \end{vmatrix}=0$

Using the method of slope, show that the following points are collinear :
$( i ) A \left( 4,8\right ) , B \left( 5,12 \right) , C \left( 9,28 \right)$
$(ii) A \left( 16 , - 18 \right) , B \left( 3 , - 6 \right) , C \left( - 10,6\right )$

Show that the points $P(-2,3,5), Q(1,2,3)$ and $R(7,0,-1)$ are collinear.

Show that the points $P(-2, 3, 5), Q(1, 2, 3)$ and $R(7, 0, -1)$ are collinear.

Solve:
$\left[ \begin{matrix} 1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & 6 \end{matrix} \right]$

If $A= \left[ {\matrix{10 & 0 \cr 4 & 10\cr } } \right]$ , then find $\left| A \right|$ .

Construct a $3\times 2$ matrix whose elements are given by $a_{1}=\dfrac{1}{2}|i-3j|$ .

Write the value of the determinant $\left| \begin{array}{cc}{3} & {-1} \\ {2} & {1}\end{array}\right|$

$\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { b }^{ 2 } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| =(a-b)(b-c)(c-a)$

Solve:
$\left| \begin{matrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{matrix} \right| =0$

Find the value of x for which $|_{ x }^{ 3 }\quad _{ 1 }^{ x }|=|_{ 8 }^{ 3 }\quad _{ 1 }^{ 2 }|$ .

Convert $\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$ into an identify matrix by suitable row transformations.

If $A = \left| \begin{array} { c c } { 2 } & { 3 } \\ { - 1 } & { 2 } \end{array} \right|$ Find $A$ .

Evaluate $\Delta = \begin{vmatrix} \cos 2\alpha & \sin 2\alpha \\-\sin 3\alpha & \cos 3\alpha \end{vmatrix}$

The value of $\begin{vmatrix} \cos15^{\circ} & \sin15^{\circ} \\ \sin75^{\circ} & \cos75^{\circ} \end{vmatrix}$ is

Find the value of x, if
$\begin{vmatrix} x+2 & x \\ x-4 & x+3 \end{vmatrix}=\begin{vmatrix} 4 & 2 \\ -2 & 5 \end{vmatrix}$

Evaluate: $\begin{vmatrix} \cos30^{\circ} & \sin30^{\circ} \\ \sin105^{\circ} & \cos105^{\circ} \end{vmatrix}$

Evaluate the determinant :
$\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}$

Find the determinant of the matrix $A$
$A=\begin{bmatrix} -1 & 4 \\ 2 & 3 \end{bmatrix}$

Show that the points $A(-7, 4, -2), B(-2, 1, 0)$ and $C(3, -2, 2)$ are collinear.

Find the coordinates of a point $A$ , where $AB$ is diameter of a circle whose centre is $(2, -3)$ and $B$ is the point $(1, 4)$ .

Evaluate the following determinants :
$\begin{vmatrix} \cos15^{\circ} & \sin15^{\circ} \\ \sin75^{\circ} & \cos75^{\circ} \end{vmatrix}$

Find the values of $x$ , if
$\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}$ = $\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$

Are the following pairs of sets equal? Give reasons.
$A=\{x : x$ is a letter of the word "WOLF" $\}$ ,
$B=\{x : x$ is a letter of the word "FOLLOW" $\}$ .

The value of the determinant $\begin{vmatrix} \cos30^{\circ} & \sin60^{\circ} \\ \sin30^{\circ} & \cos60^{\circ} \end{vmatrix}$ is

Evaluate $\begin{bmatrix} 14 & 9 \\ -8 & -7 \end{bmatrix}$ .

Find the adjoint of the following matrice:
$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$

The value of the determinant $\begin{vmatrix} \cos15^{\circ} & \sin15^{\circ} \\ \sin75^{\circ} & \cos75^{\circ} \end{vmatrix}$ is______

If $A,B$ and $C$ are the angles of non-right angles triangle $ABC$ , then find the value of $\begin{vmatrix} \tan { A } & 1 & 1 \\ 1 & \tan { B } & 1 \\ 1 & 1 & \tan { C } \end{vmatrix}$

If the value of $\begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{vmatrix}$ is $k$ , then find $\dfrac{-k}{13}$ .

If $\begin{vmatrix} x & x+y & x+y+z \\ 2x & 3x+2y & 4x+3y+2z \\ 3x & 6x=3y & 10x+6y+3z \end{vmatrix}=64$ , then the real value of $x$ is .........