MCQ Questions for Class 12 Commerce Maths Determinants Quiz 8

${x}^{a}{y}^{b}={e}^{m},{x}^{c}{y}^{d}={e}^{n}, { \triangle }_{ 1 }=\begin{vmatrix} m & b \\ n & d \end{vmatrix},{ \triangle }_{ 2 }=\begin{vmatrix} a & m \\ c & n \end{vmatrix}$ and

${ \triangle }_{ 3 }=\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ the value of

$x$ and

$y$ are respectively

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$\dfrac { { \triangle }_{ 1 } }{ { \triangle }_{ 3 } }$ and $\dfrac { { \triangle }_{ 2 } }{ { \triangle }_{ 3 } }$
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$\dfrac { { \triangle }_{ 2 } }{ { \triangle }_{ 1 } }$ and $\dfrac { { \triangle }_{ 3 } }{ { \triangle }_{ 1 } }$
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$\log { \left( \dfrac { { \triangle }_{ 1 } }{ { \triangle }_{ 3 } } \right) } and\log { \left( \dfrac { { \triangle }_{ 2 } }{ { \triangle }_{ 3 } } \right) }$
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${ e }^{ { \triangle }_{ 1 }/{ \triangle }_{ 3 } }\,\ and\,\ { e }^{ { \triangle }_{ 2 }/{ \triangle }_{ 3 } }$

Explanation

${ x }^{ a }{ y }^{ b }={ e }^{ m }$

Taking

${ \log }_{ e }$ on both sides,

$\log _{ e }{ \left( { x }^{ a }{ y }^{ b } \right) } =\log _{ e }{ { e }^{ m } }$

$a\log _{ e }{ x } +b\log _{ e }{ y } =m\rightarrow 1$

Similarly,

$c\log _{ e }{ x } +d\log _{ e }{ y } =n\rightarrow 2$

Solving

$1$ and

$2$ by Cramer's rule,

$\log _{ e }{ x } =\cfrac { \begin{vmatrix} m & b \\ n & d \end{vmatrix} }{ \begin{vmatrix} a & b \\ c & d \end{vmatrix} } =\cfrac { { \triangle }_{ 1 } }{ { \triangle }_{ 3 } }$

$\log _{ e }{ y } =\cfrac { \begin{vmatrix} a & m \\ c & n \end{vmatrix} }{ \begin{vmatrix} a & b \\ c & d \end{vmatrix} } =\cfrac { { \triangle }_{ 2 } }{ { \triangle }_{ 3 } }$

$\therefore x={ e }^{ \left( { { \triangle }_{ 1 } }/{ { \triangle }_{ 3 } } \right) },y={ e }^{ \left( { { \triangle }_{ 2 } }/{ { \triangle }_{ 3 } } \right) }$

Answer: D

The value of

$\dfrac { 1 }{ x-y } \begin{vmatrix} 1 & 0 & 0 \\ 3 & { x }^{ 3 } & 1 \\ 5 & { y }^{ 3 } & 1 \end{vmatrix}$ is-

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$x+y$
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${x}^{2}-xy+{y}^{2}$
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${x}^{2}+xy+{y}^{2}$
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${x}^{3}-{y}^{3}$

Explanation

Expanding along

$R_1$

$=\dfrac{1}{(x-y)}[1(x^3-y^3)]$

$=\dfrac{(x-y)(x^2+y^2+xy)}{x-y}=x^2+y^2+xy$

One of the roots of

$\begin{vmatrix}x + a & b & c\\ a & x + b & c\\ a & b & x + c\end{vmatrix} = 0$ is

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

Explanation

Given,

$\begin{vmatrix}x + a & b & c\\ a & x + b & c\\ a & b & x + c\end{vmatrix} = 0$

[

$C_1'=C_1+C_2+C_3$ ]

or,

$\begin{vmatrix}x + a+b+c & b & c\\ x+a+b+c & x + b & c\\ x+a+b+c & b & x + c\end{vmatrix} = 0$

or,

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

[

$R_2'=R_2-R_1$ and

$R_3'=R_3-R_1$ ]

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

or,

$x^2(x+a+b+c)=0$

So the roots are

$0.0,-(a+b+c)$ .

One root is

$-(a+b+c)$ .

Three straight lines

$2x+11y-5=0, 24x+7y-20=0$ and

$4x-3y-2=0$

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form a triangle
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are only congruent
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are concurrent with one line bisecting the angle between the other two
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None of above

$\Delta = \begin{vmatrix} x & 2y-z & -z \\ y & 2x-z & -z \\ y & 2y-z & \quad 2x-2y-z \end{vmatrix}$ ,then

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x-y is a factor of $\Delta$
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$(x-y)^2$ is a factor of $\Delta$
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(x-y)^3 is a factor of $\Delta$
0%
$\Delta$ is independent of z

Explanation

$\Delta = \begin{vmatrix} x & 2y-z & -z \\ y & 2x - z & - z \\y & 2y - z & 2x - 2y - z \end{vmatrix}$

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

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

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

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

So, we can say that

$\Delta$ is factor of

$(x - y)$

For distinct numbers

$a,b,c,x,y,z\ \epsilon R$ if

${ \Delta }_{ 1 }\left| \begin{matrix} \left( a-x \right) ^{ 2 } & \left( b-x \right) ^{ 2 } & \left( c-x \right) ^{ 2 } \\ \left( a-y \right) ^{ 2 } & \left( b-y \right) ^{ 2 } & \left( c-y \right) ^{ 2 } \\ \left( a-z \right) ^{ 2 } & \left( b-z \right) ^{ 2 } & \left( c-z \right) ^{ 2 } \end{matrix} \right| { \Delta }_{ 2 }\left| \begin{matrix} (ax+1)^{ 2 } & (bx+1)^{ 2 } & (cx+1)^{ 2 } \\ (ay+1)^{ 2 } & (by+1)^{ 2 } & (cy+1)^{ 2 } \\ (az+1)^{ 2 } & (bz+1)^{ 2 } & (cz+1)^{ 2 } \end{matrix} \right|$ then

$\frac { { \Delta }_{ 1 }^{ 2 } }{ { \Delta }_{ 2 }^{ 2 } } +\frac { { \Delta }_{ 2 }^{ 2 } }{ { \Delta }_{ 1 }^{ 2 } } =$

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$\dfrac {5}{4}$
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$\dfrac {10}{3}$
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$\dfrac {1}{4}$
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$None\ of\ these$

Let

$\Delta =$

$\begin{vmatrix} sin\theta cos \phi & sin\theta sin\phi & cos\theta \\ cos\theta cos\phi & cos\theta sin\phi & -sin\theta \\ -sin\theta sin\phi & sin\theta cos\phi & 0\end{vmatrix}$ , then

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$\Delta$ is independent of $\theta$
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$\Delta$ is independent of $\phi$
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$\Delta$ is a constant
0%
none of these

Explanation

$\triangle = \begin{vmatrix} \sin \theta . \cos \phi & \sin \theta . \sin \phi & \cos \theta \\ \cos \theta . \cos \phi & \cos \theta . \sin \theta & -\sin \theta \\ -\sin \theta. \sin \theta & \sin \theta . \cos \phi & 0 \end{vmatrix}$

$= \sin \theta . \sin \phi (\sin^{2} \theta . \cos \theta) - \cos \theta (- \sin \theta . \cos \theta . \cos \theta)- \sin \theta . \sin \phi (- \sin^{2} \theta. \sin \phi - \cos^{2} \theta- \sin \theta)$

$=\sin^{3} \theta . \sin \phi. \cos \phi+ \cos^{2} \theta. \sin \theta. \cos^{2} \phi+ \sin^{3} \theta. \sin^{2} \phi + \sin \theta . \cos^{2} \theta . \sin^{2} \phi$

So, We can say that

$\triangle$ is depend

$d$ on both.

$\theta$ and

$\phi$

The determinant

$\Delta = \begin{vmatrix} a^ 2(1+x) & ab & ac \\ ab & b^ 2(1+x) & bc \\ ac & bc & c^ 2(1+x) \end{vmatrix}$ is divisible by

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$(x + 3)$
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$(1 + x)^2$
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$x^2$
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$(x^2 + 1)$

Explanation

b"b'It has multiple answers

$\Delta = \begin{vmatrix} a^2(1 + x) & ab & ac \\ ab & b^2(1 + n) & bc \\ ac & bc & c^2( 1+n) \end{vmatrix}$

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

$= (abc)^2 \begin{vmatrix} 1 + n & 1 & 1 \\ 1 & 1 + x & 1 \\ 1 & 1 & 1 + x \end{vmatrix}$

$= (ab c)^2 |(1 + x)| (1 + x)^2 - 1] - (1 + x - 1) + (1 - 1 - x)$

$= (abc)^2 [(1 + x)^3 - (1 + x) - x - x]$

$= (abc)^2 [ (1 + x)^3 - 1 - x - 2x]$

$= (abc)^2 [ 1 + x^3 + 3x (1 + x) - 1 - 3x]$

$= (abc)^2 [x^3 + 3x + 3x^2 - 3x]$

$= (abc)^2 [ x^2 (x + 3)]$

Divisible by

$x^2 \, b \, (x + 3)$

The determinant

$\Delta = \begin{vmatrix} b & c & b\alpha+c \\ c & d & c\alpha+d \\ b\alpha+c & c\alpha+d & a\alpha^3-c\alpha \end{vmatrix}$ is equal to zero if

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b,c,d are in A.P
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b,c,d are in G.P
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b,c,d are in H.P
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$\alpha$ is a root of $ax^3 - bx^2 - 3cx - d = 0$

Explanation

$\triangle = \begin{vmatrix} b & c & b\alpha + c \\ c & d & c\alpha+d \\ b\alpha+c & c\alpha+d & a\alpha^{3}-c\alpha \end{vmatrix}$

$\triangle = \begin{vmatrix} b\alpha & c\alpha & b\alpha^{2} + c\alpha \\ c & d & c\alpha+d \\ b\alpha+c & c\alpha+d & a\alpha^{3}-c\alpha \end{vmatrix}$

$\triangle = \begin{vmatrix} b\alpha+c & c\alpha+d & b\alpha^{2}+2c\alpha+d \\ c & d & c\alpha+d \\ b\alpha+c & c\alpha+d & a\alpha^{2}-c\alpha \end{vmatrix}$

$\triangle = \begin{vmatrix} 0 & 0 & b\alpha^{2} + 3c\alpha + d - a\alpha \\ c & d & c\alpha+d \\ b\alpha+c & c\alpha+d & a\alpha^{2}-c\alpha \end{vmatrix}$

Expand

$= 0 - 0 +\left ( b\alpha^{2} + 3c\alpha + d - q\alpha^{2} \right )$

$.\left ( c^{2}\alpha + cd - bd\alpha - cd \right )$

$=\left ( b\alpha^{2} + 3c\alpha + d - q\alpha^{2} \right )\left (c^{2}\alpha - bd\alpha \right )$

$b\alpha^{2} + 3c\alpha + d - a\alpha^{2} = 0$

$c^{2}\alpha - bd\alpha = 0$

$\alpha = 0 \,\,\,\,\, \boxed {c^{2} = bd }$

So we can Say that

b, c, d in G.P.

1071674_1071664_ans_d8ee999656054fb6908e97af2760cabc.jpg

Find the values of

$a$ and

$b$ so that the points

$( a, b, 3),( 2, 0, -1)$ and

$(1, -1, -3)$ are collinear.

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$a=4,b=2$
0%
$a=0,b=2$
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$a=4,b=-2$
0%
$a=-4,b=-2$

Explanation

$A(a,b,3)\\B(2,0,1)\\C(1,-1,-3)\\if\>A,Band\>C\>are\>collinear\>,then\\\>\vec{AB}\>\>=\lambda\>\vec{BC}\\\vec{AB}\>(2-a)\hat{i}-b\hat{j}-4\hat{k}\\\vec{BC}=-\hat{i}-\hat{j}-2\hat{k}\\now\>2-a=\lambda\>(-1)\\or\>2-a=-\lambda\>\rightarrow\>(1.)\\\>and\>-b=\>\lambda\>(-1)\lambda\>\\or\>-b=-\lambda\>\rightarrow\>(2.)\\\therefore\>from\>eq^n\>(1.)and(2.)\\2-a=-b\\or\>a-b=\>2\\now\>-4=\lambda\>(-2)\\\therefore\>\lambda\>=2\\then\>b=\lambda\>=2\\and\>a=2+\lambda\>=2+2=4$

$\Delta =\left| \begin{matrix} 0 & i-100 & i-500 \\ 100-i & 0 & 1000-i \\ 500-i & i-1000 & 0 \end{matrix} \right|$ is equal to

0%
$100$
0%
$500$
0%
$1000$
0%
$0$

Let

$A =\begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix}$ and suppose that det.(A) =2 then the det.(B) equals, where

$B =\begin{vmatrix} 4x & 2a & -p \\ 4y & 2b & -q \\ 4z & 2c & -t \end{vmatrix}$

0%
$det(B) = -2$
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$det(B) = -8$
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$det(B) = -16$
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$det(B) = 8$

Explanation

$A=\begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix}$

Let

$(A)=2$ , Let

$(B)=?$

$B=\begin{vmatrix} 4x & 2a & -p \\ 4y & 2b & -q \\ 4z & 2c & -r \end{vmatrix}$

$=-8\begin{vmatrix} x & a & p \\ y & b & q \\ z & c & r \end{vmatrix}$

$=-8\begin{vmatrix} x & y & z \\ a & b & c \\ p & q & r \end{vmatrix}$

$=+8\begin{vmatrix} a & b & c \\ x & y & z \\ p & q & r \end{vmatrix}$

$=-8\begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix}$

So Let

$|B|=-8\times 2=-16$

The value of determinant

$\begin{vmatrix} a^ 2 & a & 1 \\ cos(nx) & cos(n+1)x & cos(n+2)x \\ sin(nx) & sin(n+1)x & sin(n+2)x \end{vmatrix}$ is independent of

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n
0%
a
0%
x
0%
a , n and x

$\begin{vmatrix} 1 & 2 & 7 \\ 3 & 7 & -5 \\ -1 & 4 & 3 \end{vmatrix}$ , cofactor of

$2=___________$ and cofactor of

$-1=___________.$

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$-4,-59$
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$4,-59$
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$-4,59$
0%
$59,4$

$f(x) = \left| \begin{array}{111} x-3 & 2x^2 -18 & 3x^3 -81\\ x-5 & 2x^2-50 & 4x^3-500\\ 1 & 2 & 3 \\ \end {array} \right|$ then

$f(1). f(3) + f(3) .f(5) + f(5) .f(1)$ is equal to-

0%
$f(1)$
0%
$f(3)$
0%
$f(1) + f(3)$
0%
$f(1) + f(5)$

Explanation

$f(x)=\begin{vmatrix} x-3 & \quad \quad \quad { 2x }^{ 2 }-18 & \quad \quad \quad { 3x }^{ 3 }-81 \\ x-5 & { \quad \quad \quad 2x }^{ 2 }-50 & \quad \quad \quad { 4x }^{ 2 }-500 \end{vmatrix}$

Now

$f(3)=\begin{vmatrix} 3-3 & \quad \quad { 2x3 }^{ 2 }-18 & \quad \quad { 3x3 }^{ 2 }-81 \\ 3-5 & \quad \quad { 2x3 }^{ 2 }-50 & \quad \ { 3 }^{ 3 }-500 \\ 1 & 2 & 3 \end{vmatrix}$

$=0$

Also ,

$f(5)=0$

Hence,

$f(1), f(3) + f(3), f(3)+f(3)+f(5), f(1)$

$=0$

$f(3)$

The points

$({X}_{1},{Y}_{1})$ ,

$({X}_{2},{Y}_{2})$ ,

$({X}_{1},{Y}_{2})$ and

$({X}_{2},{Y}_{1})$ are always

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Collinear
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Concyclic
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Vertices of a square
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Vertices of rectangle

Explanation

As all the points lie on the same line this implies that area of the triangle formed by the points is zero.

So,using area formula in determinant form

${ X }_{ 1 }∗({ Y }_{ 2 }−{ Y }_{ 3 })−{ X }_{ 2 }∗({ Y }_{ 1 }−{ Y }_{ 3 })+{ X }_{ 3 }∗({ Y }_{ 1 }−{ Y }_{ 2 })=0$

${ X }_{ 1 }∗({ Y }_{ 2 }−{ Y }_{ 3 })+{ X }_{ 2 }∗({ Y }_{ 3 }−{ Y }_{ 1 })+{ X }_{ 3 }∗({ Y }_{ 1 }−{ Y }_{ 2 })=0$

Now,divide both sides by ${ X }_{ 1 }∗{ X }_{ 2 }∗{ X }_{ 3 }$

$({ Y }_{ 2 }−{ Y }_{ 3 })+{ X }_{ 2 }∗({ Y }_{ 3 }−{ Y }_{ 1 })+{ X }_{ 3 }∗({ Y }_{ 1 }−{ Y }_{ 2 })=0$

$\dfrac { { Y }_{ 2 }−{ Y }_{ 3 } }{ { X }_{ 2 }−{ X }_{ 3 } } +\dfrac { { Y }_{ 3 }−{ Y }_{ 1 } }{ { X }_{ 3 }−{ X }_{ 1 } } +\dfrac { { Y }_{ 1 }−{ Y }_{ 2 } }{ { X }_{ 1 }−{ X }_{ 2 } } =0$

Its collinear.

$a,b,c$ are non-zeros, then the system of equation :

$(\alpha+a)x+\alpha+\alpha z=0$ ;

$\alpha x+(a+b)y+\alpha z=0$ ;

$\alpha x+\alpha y+(\alpha +c)z=0$ has a non-trivial solution if

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$\dfrac{1}{\alpha}=-(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})$
0%
${\alpha}^{-1}=a+b+c$
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$\alpha+a+b+c=1$
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None of the above

Explanation

Given that

$\Rightarrow \begin{vmatrix} \alpha+a&\alpha&\alpha\\\alpha & \alpha+b&\alpha\\\alpha &\alpha &\alpha +c\end {vmatrix}0$

$\Rightarrow R_1\rightarrow R_1-R_3$

$\Rightarrow \begin{vmatrix} a&0&-c\\0 & b&-c\\\alpha &\alpha &\alpha +c\end {vmatrix}=0$

$\Rightarrow a(b\alpha+bc+c\alpha)+\alpha bc=0$

$\Rightarrow (ab+bc+ca)\alpha=-abc$

$\Rightarrow \dfrac{1}{\alpha}=-\dfrac{ab+bc+ca}{abc}$

$\Rightarrow \dfrac{1}{\alpha}=-(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})$

$p{\lambda ^4} + p{\lambda ^3} + p{\lambda ^2} + s\lambda + t =$

$\left| {\begin{array}{*{20}{c}}{{\lambda ^2} + 3\lambda } & {\lambda + 1} & {\lambda + 3}\\{\lambda + 1} & {2 - \lambda } & {\lambda - 4}\\{\lambda - 3} & {\lambda + 4} & {3\lambda }\end{array}} \right|$ , then value of t is

0%
$16$
0%
$18$
0%
$17$
0%
$19$

Explanation

first we solve RHS then compare with LHS

$=>(\lambda^2+3\lambda)[(2-\lambda)(3\lambda)-(\lambda+4)(\lambda-4)]-(\lambda+1)[(\lambda+1)(3\lambda)-(\lambda-3)(\lambda-4)]$

$+(\lambda+3)$

$[(\lambda+1)(\lambda+4-(\lambda-3)(2-\lambda))]$

$=>[-4\lambda^2+6\lambda^3+16\lambda^2-12\lambda^3+18\lambda^2+48\lambda-2\lambda^3-10\lambda^2+12\lambda-2\lambda^2-10\lambda+12+2\lambda^3+10\lambda+6\lambda^2+30]$

$=>[-4\lambda^4-6\lambda^3+28\lambda^2+60\lambda+18]$

Now compare both side

$=>[p\lambda^4+p\lambda^3+p\lambda^2+s\lambda+t=-4\lambda^4-6\lambda^3+28\lambda^2+60\lambda+18]$

hence,

$t=18$

State whether following statement is true or false.

$A$ is a square matrix of order

$n$ , then

$|Adj (Adj \,A)|$ is of order

$(n^2)$ .

0%
True
0%
False

$A=\left[ \begin{matrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{matrix} \right]$ ; If

$\left| { A }^{ 2 } \right| =25$ , then

$\left| \alpha \right| =$

0%
$5$
0%
$5^{2}$
0%
$1$
0%
$\dfrac{1}{5}$

Explanation

$A=\begin{bmatrix} 5 & 5\alpha & \alpha\\ 0 & \alpha & 5\alpha\\ 0 & 0 & 5\end{bmatrix}$

$A^2=\begin{bmatrix} 5 & 5\alpha & \alpha\\ 0 & \alpha & 5\alpha\\ 0 & 0 & 5\end{bmatrix}\begin{bmatrix} 5 & 5\alpha & \alpha\\ 0 & \alpha & 5\alpha\\ 0 & 0 & 5\end{bmatrix}=\begin{bmatrix} 25 & 30\alpha & 25\alpha^2+10\alpha\\ 0 & \alpha^2 & 5\alpha^2+25\alpha\\ 0 & \alpha^2 & 5\alpha^2 +25\alpha\\ 0 & 0 & 25\end{bmatrix}$

$|A^2|=25$

$25\begin{vmatrix} \alpha^2 & 5\alpha^2+25\alpha\\ 0 & 25\end{vmatrix} -30\alpha \begin{vmatrix} 0 & 5\alpha^2+25\alpha\\ 0 & 25\end{vmatrix} +25\alpha^2+10\alpha\begin{vmatrix} 0 & \alpha^2\\ 0 & 0\end{vmatrix}=25$

$25\times 25\alpha^2=25$

$25\alpha^2=1$

$\alpha =\dfrac{1}{5}$ .

the following relation is

$\begin{vmatrix} a & a + b & a + b + c \\ 2a & 3a + 2b & 4a + 3b + 2c = a^3 \\ 3a & 6a + 2b & 10a + 6b + 3c \end{vmatrix} = a ^3$

0%
True
0%
False

$\begin{vmatrix} \lambda^2 + 3 \lambda & \lambda -1 & \lambda +3 \\ \lambda + 1 & 2- \lambda & \lambda - 4 \\ \lambda-3 & \lambda + 4 & 3 \lambda \end{vmatrix} = p \lambda^4 + q \lambda^3 + r \lambda^2 + s \lambda + t$ then

$t =$

0%
$16$
0%
$17$
0%
$18$
0%
$19$

Explanation

To get

$t$ ,

Put

$\lambda=0$

We get, value of determinant as,

$[0(0+16)+1(0-12)+3(4+6)]=t$

$t=18$

$\begin{vmatrix}\dfrac{1}{a} &bc &a^2 \\ \dfrac{1}{b} &ca & b^2\\ \dfrac{1}{c} & ab & c^2\end{vmatrix}$ is equal to -

0%
$(a - b)(b-c) ( c-a)$
0%
$abc(a-b)(b-c)(c-a)$
0%
$0$
0%
1

$\begin{vmatrix} log e & log e^2 & log e^3\\ log e^2 & log e^3 & log e^4\\ log e^3 & log e^4 & log e^5\end{vmatrix}=$ ?

0%
$0$
0%
$1$
0%
$4$ log e
0%
$5$ log e

Explanation

$\begin{vmatrix} \log { e } & \log { { e }^{ 2 } } & \log { e^{ 3 } } \\ \log { { e }^{ 2 } } & \log { { e }^{ 3 } } & \log { e^{ 4 } } \\ \log { e^{ 3 } } & \log { { e }^{ 4 } } & \log { { e }^{ 3 } } \end{vmatrix}$

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

${ \left( \log { e } \right) }^{ 3 }\begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}\xrightarrow [ { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 3 } ]{ { R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 } }$

${ \left( \log { e } \right) }^{ 3 }\begin{vmatrix} 1- & -1 & -1 \\ -1 & -1 & -1 \\ 3 & 4 & 5 \end{vmatrix}=0$ (

$\because$ two rowa are same)

$f(x)=\begin{vmatrix} 1 & x & x+1\\ 2x & x(x-1) & (x+1)x\\ 3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1)\end{vmatrix}$ then

$f(100)$ is equal to?

0%
$0$
0%
$1$
0%
$100$
0%
$-100$

Explanation

$f\left( x \right) =\begin{vmatrix} 1 & x & x+1 \\ 2x & x\left( n-1 \right) & \left( n+1 \right) x \\ 3z\left( n-1 \right) & x\left( n-1 \right) \left( n-2 \right) & \left( n+1 \right) x\left( n-1 \right) \end{vmatrix}$

$\left( n-1 \right) \left( n \right) \begin{vmatrix} 1 & x & n+1 \\ 2 & x-1 & n+1 \\ 3 & n-2 & n+1 \end{vmatrix}\begin{matrix} { R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 } \\ \longrightarrow \\ { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 3 } \end{matrix}$

$\left( n-1 \right) \left( n \right) \begin{vmatrix} -1 & 1 & 0 \\ -1 & 1 & 0 \\ 3 & n-2 & n+1 \end{vmatrix}$

$(n-1)(n) (0)=0$

so,

$t(100)=0$

the value of the determinant of order

$3$ remains unchanged if its rows and columns are interchanged. that statement is ___

0%
True
0%
False

If none of

$a, b, c$ is zero, Whether the given equation

$\begin{vmatrix} -bc & { b }^{ 2 }+bc & { c }^{ 2 }+bc \\ { a }^{ 2 }+ac & -ac & { c }^{ 2 }+ac \\ { a }^{ 2 }+ab & { b }^{ 2 }+ab & -ab \end{vmatrix}={ \left( bc+ca+ab \right) }^{ 3 }$ is ?

0%
True
0%
False

Explanation

Let

$\begin{vmatrix} -bc & b^2 + bc & c^2 + bc \\ a^2 + ac & -ac & c^2 + ac \\ a^2+ab & b^2 + ab & -ab \end{vmatrix}$

$\therefore D = \dfrac{1}{abc} \begin{vmatrix} -abc & ab^2 + abc & ac^2 + abc \\ a^2b + abc & -abc & bc^2 + abc \\ ca^2+ab & cb^2 + abc & -abc \end{vmatrix} [R_1^{'} = aR_1 \, , \, R_2^{'} = bR_2 \, , \, R_3^{'} = cR_3 \\ = \begin{vmatrix} -bc & ab+ac & ac+ab \\ ab+bc & -ac & bc+ab \\ ca + bc & bc+ac & -ab \end{vmatrix} [\textrm{taking} \, a, b, c \, \textrm{common from} \, c_1 , c_2 \, \text{ and }\, c_3 \, \textrm{respectively}] \\ = \begin{vmatrix} ab+bc+ca & ab + bc + ca & ab+bc+ca \\ ab+bc & -ca & bc+ab \\ ca+bc & bc+ca & -ab \end{vmatrix} \, [R_1^{'} = R_1 + R_2 + R_3] \\ = (ab+bc+ca) \begin{vmatrix} 1 & 1 & 1 \\ ab+bc & -ca & bc+ab \\ ca+bc & bc+ca & -ab \end{vmatrix} \, [\textrm{taking common from 1st row}] \\ = (ab+bc + ca) \begin{vmatrix} 0 & 0 & 0 \\ ab+bc+ca & -(ab+bc+ca) & bc+ab \\ 0 & ab+bc + ca & -ab \end{vmatrix} \, [c_1^{'} = c_1 - c_2 \, \textrm{and} \, c_2^{'} = c_2- c_3]$

Now expanding w.r.t.

$R_1$ we get,

$= (ab+bc+ca) [(ab+bc+ca)^2 - 0] \\ = (ab+bc+ca)^3$

The value of the determinant

$\left| \begin{matrix} { b }^{ 2 }-ab & b-c & bc-ac \\ ab-{ b }^{ 2 } & a-b & { b }^{ 2 }-ab \\ bc-ac & c-a & ab-{ b }^{ 2 } \end{matrix} \right|$

0%
$abc$
0%
$a+b+c$
0%
$0$
0%
$ab+bc+ca$

$a>0$ and discriminant of

${ax}^{2}+{2bx}+{c}$ is

$-ve$ , then

$\left| \begin{matrix} a & b & ax+b \\ b & c & bx+c \\ ax+b & bx+c & 0 \end{matrix} \right|$ is

0%
$+ve$
0%
${(ac-b^2)(ax^2+2bx+c)}$
0%
$-ve$
0%
$0$

$\left| \begin{matrix} \sqrt { 13 } +\sqrt { 3 } & 2\sqrt { 5 } & \sqrt { 5 } \\ \sqrt { 15 } +\sqrt { 26 } & 5 & \sqrt { 10 } \\ 3+\sqrt { 65 } & \sqrt { 15 } & 5 \end{matrix} \right| =$

0%
$15\sqrt{2}-25\sqrt{3}$
0%
$15\sqrt{5}-25\sqrt{6}$
0%
$25\sqrt{2}-15\sqrt{3}$
0%
$0$

Explanation

$R_{3}\rightarrow R_{3}-\sqrt{5}R_{1}$

$R_{2}\rightarrow R_{2}-\sqrt{2}R_{1}$

$\begin{vmatrix} \sqrt { 13 } +\sqrt { 3 } & 2\sqrt { 5 } & \sqrt { 5 } \\ \sqrt { 15 } -\sqrt { 6 } & 5-2\sqrt { 10 } & 0 \\ 3-\sqrt { 15 } & \sqrt { 15 } -10 & 0 \end{vmatrix}$

$\Rightarrow \sqrt{5}(\sqrt{15}-\sqrt{6})(\sqrt{15}-10)(5-2\sqrt{10})(3-\sqrt{15})$

$\Rightarrow \sqrt{5}(15-10\sqrt{15}-3\sqrt{10}+10\sqrt{6})-(15-5\sqrt{15}-6\sqrt{10}+2.5.\sqrt{16})$

$\Rightarrow \sqrt{5}(15-10\sqrt{15}-3\sqrt{10}+10\sqrt{6}-15+5\sqrt{15}+6\sqrt{10}-10\sqrt{6})$

$\sqrt{5}(-5\sqrt{15}+3\sqrt{10})$

$\Rightarrow -5.5\sqrt{3}+3.6\sqrt{2}$

$\Rightarrow 15\sqrt{2}-25\sqrt{3}$

$\Rightarrow 5(3\sqrt{2}-5\sqrt{3})$

The value of

$\left| \begin{matrix} 1+w & { w }^{ 2 } & -w \\ 1+{ w }^{ 2 } & w & -{ w }^{ 2 } \\ { w }^{ 2 }+w & w & -{ w }^{ 2 } \end{matrix} \right|$ is equal to

0%
$0$
0%
$2 \omega$
0%
$2 {\omega}^{2}$
0%
$-3 {\omega}^{2}$

Explanation

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

Multiplying

$C_{2}$ by

$w$ and adding to

$C_{3}$

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

$\begin{bmatrix} 1+w & { w }^{ 2 } & { w }^{ 2 }-w \\ 1+{ w }^{ 2 } & w & 0 \\ { w }^{ 2 }+w & w & 0 \end{bmatrix}$

Now finding determinant

$(-w+w^{3})\left[w(1+w^{2})-w(w^{2}+w)\right]$

$(-w+w)(w+w^{3}-w^{3}-w^{2})$

$\because w^{3}=1$ , then

$(-w+1)(w-w^{2})$

$-w^{2}+w^{3}+w-w^{2}$

$-w^{2}+1+w-w^{2}$

$-2w^{2}+1+w+w^{2}-w^{2}$

$(1+w+w^{2})-3w^{2}$

As we know ,

$1+w+w^{2}=0$

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

Option

$(d)$ is correct

If the points

$A(at^{2}_{1},2at_{1}), B(at^{2}_{2},2at_{2})$ and

$C(\alpha,0)$ are collinear, then

$t_{1} t_{2}$ equals

0%
$2$
0%
$-1$
0%
$1$
0%
$None\ of\ these$

Let

$F(x)=$

$\left | \left| \right| \begin{matrix}1 &1+sin\ x &1+sin\ x+cos\ x \\ 2 &3+2\ sin\ x &4+3\ sin\ x+2\ cos\ x \\ 3&6+3\ sin\ x&10+6\ sin\ x+3\ cos\ x \end{matrix} \right |$ then

$F'\ \left ( \dfrac{\pi}{2} \right )$ is equal to

0%
$-1$
0%
$0$
0%
$1$
0%
$2$

Explanation

$f(x) = \begin{vmatrix} 1 & 1+\sin x & 1+ \sin x +\cos x \\ 2 & 3+2 \sin x & 4+3 \sin x +2 \cos x \\ 3 & 6+3 \sin x & 10+6 \sin x +3 \cos x \end{vmatrix}$

$f'(x)= \begin{vmatrix} 0 & \cos x & \cos x- \sin x \\ 2 & 3+2 \sin x & 4+3 \sin x + 2 \cos x \\ 3 & 6+3 \sin x & 10+6 \sin x+ \cos x \end{vmatrix}+ \begin{vmatrix} (1) & (1+ \sin x) & 1+\sin x+ \cos x \\ 0 & 2 \cos x & 3 \cos x-2 \sin x \\ 3 & 6+3 \sin x & 10+6 \sin x+3 \cos \end{vmatrix}+ \begin{vmatrix} 1 & 1+ \sin x & 1+ \sin x + \cos x \\ 2 & 3+2 \sin x & 4+ 3 \sin x + 2 \cos x \\ 0 & 3 \cos x & 6 \cos -3 \sin x \end{vmatrix}$ (differentiating row

$mse$ )

$f' \left( \dfrac{\pi}{2} \right) = \begin{vmatrix} 0 & 0 & -1 \\ 2 & 5 & 7 \\ 3 & 9 & 16 \end{vmatrix}+ \begin{vmatrix} 1 & 2 & 2 \\ 0 & 0 & -2 \\ 3 & 9 & 16 \end{vmatrix}+ \begin{vmatrix} 1 & 2 & 2 \\ 2 & 5 & 7 \\ 0 & 0 & -3 \end{vmatrix}$

$- (18-15)+ 2 (9-6)-3 (5-4)$

$=-3+6-3=0$

$\therefore$ Option

$B$ is correct.

$A = \left[ {\begin{array}{*{20}{c}}a&0&0\\0&a&0\\0&0&a\end{array}} \right]$ then find the value of

$\left| A \right|\left| {adjA} \right|$

0%
${a^3}$
0%
${a^6}$
0%
${a^9}$
0%
a

Explanation

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

$\left( A \right) =\begin{bmatrix} { a }^{ 2 } & 0 & 0 \\ 0 & { a }^{ 2 } & 0 \\ 0 & 0 & { a }^{ 2 } \end{bmatrix}Adj\left( A \right) =\begin{bmatrix} { a }^{ 2 } & 0 & 0 \\ 0 & { a }^{ 2 } & 0 \\ 0 & 0 & { a }^{ 2 } \end{bmatrix}$

$|A|=a^3$

$|Adj (A)|=a^6$

$|A|.|Adj (A)|=a^3a^6=a^9$

$C$ is correct

$\begin{vmatrix} 1+x & 2 & 3 \\ 1 & 2+x & 3 \\ 1 & 2 & 3+x \end{vmatrix}=0$ then

$x=$

0%
$1$
0%
$-1$
0%
$-6$
0%
$6$

Explanation

We have

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

it gives

$6x^2+x^3=0$

$x^2(6+x)=0\implies x=0,0,-6$

option c gives correct answer

$A = {\left[ {\begin{array}{*{20}{c}}a\\b\\c\end{array}\begin{array}{*{20}{c}}p\\q\\r\end{array}} \right]_{3 \times 2}}$ then determinant

$\left( {A{A^T}} \right)$ is equal to

0%
$0$
0%
${a^2} + {b^2} + {c^2}$
0%
${p^2} + {q^2} + {r^2}$
0%
${p^2} + {q^2}$

Explanation

$A=\begin{bmatrix} a & b \\ c & p \\ q & r \end{bmatrix} \Rightarrow A^{T}= \begin{bmatrix} a & b & c \\ p & q & r \end{bmatrix}$

$AA^{T}= \begin{bmatrix} a & b & c \\ p & q & r \end{bmatrix} \begin{bmatrix} a & b & c \\ p & q & r \end{bmatrix}= \begin{bmatrix} a^{2}+ p^{2} & ab+pq & ac+pr \\ ab+pq & b^{2}+q^{2} & bc+qr \\ ac+pr & bc+qr & c^{2}+r^{2} \end{bmatrix}$

$det (AA^{T}) = (a^{2}+ p^{2}) {(b^{2}+q^{2})(c^{2}-r^{2})-(bc+qr)^{2}}$

$-(ab+pq){(ab+pq)(c^{2}-r^{2})-(ac+pr)(bc+qr)}$

$+(ac+pr) {(ab+pq)(bc+qr)- (ac+pr)(b^{2}+q^{2})}$

$=0$

${t_{1,}}{t_2}\,$ and

${t_3}$ distinct. and the points

$\left( {{t_1}.2a{t_1} + a{t_1}^3} \right).\left( {{t_2}.2a{t_2} + a{t_2}^3} \right),\left( {{t_3}.2a{t_3} + a{t_3}^3} \right)$ are collinear, then

${t_1} + {t_2} + {t_3} =$

0%
$t_1 t_2 t_3 =-1$
0%
$t_1 +t_2 +t_3 =t_1 t_2 t_3$
0%
$t_1 +t_2 +t_3 =0$
0%
$t_1 +t_2 +t_3 =-1$

Explanation

Let points are

$P(t_1. 2at_1 + at_1^3)$

$,Q(t_1, 2at_2+at_1^3)$ and

$R(t_3, 2at_3+at_3^3)$

$P,Q$ and

$R$ are collinear

$\Rightarrow$ area (

$\Delta PQR$ )

$= 0$

$\Rightarrow \begin{vmatrix} 1&t_1&2at_1 + at_1^3 \\1&t_2& 2at_2+at_2^3\\1& t_2&2at_3+at_3^3 \end{vmatrix}=0$

$\Rightarrow \begin{vmatrix} 1&t_1 &2at_1 \\ 1&t_2 &2at_2 \\1&t_3&2at_3 \end{vmatrix} + \begin{vmatrix}1&t_1 &at_1^3 \\ 1&t_2&at_2^3\\1&t_3&at_3^3 \end{vmatrix}=0$

$\Rightarrow 2a\begin{vmatrix}1 &t_1&t_1 \\ 1&t_2&t_2\\1&t_3&t_3 \end{vmatrix}+a\begin{vmatrix}1 &t_1&t_1^3\\1&t_2&t_2^3\\1 & t_3&t_3^3 \end{vmatrix}=0$

Determinant is zero if two rows of a determinant are same

$\Rightarrow 0+a \begin{vmatrix} 1&t_1&t_1^3 \\1&t_2 &t_2^3\\ 1&t_3&t_3^3 \end{vmatrix}=0$

$R_2 \to R_2 - R_1$

$R_3\to R_3-R_1$

$\Rightarrow \begin{vmatrix} 1&t_1&t_1^3 \\ 1&t_2-t_1&t_2^3-t_1^3\\0&t_2-t_1&t_3^3-t_1^3 \end{vmatrix}=0$

$\Rightarrow (t_2-t_1)(t_3-t_1) \begin{vmatrix} 1&t_1&t_1^3 \\0&1&t_1^2+t_1t_2+t_2^2\\0&1&t_1^2+t_1t_3+t_3^2\end{vmatrix}=0$

$\Rightarrow (t_2-t_1)(t_2-t_1)[t_1^2 + t_!t_2+t_3^2 - t_1^2 - t_1t_2-t_2^2]=0$

$\Rightarrow (t_2-t_1)(t_3-t_1)[t_2^3 + t_1t_3-t_1t_2-t_2^2] = 0$

$\Rightarrow (t_2-t_1)(t_3-t_1)(t_3-t_2)(t_1+t_2+t_3)=0$

$t_1\neq t_2 \neq t_3$

$\Rightarrow t_2-t_1$ or

$t_3-t_1$ or

$t_3-t_2$ can't be zero

$\Rightarrow t_1+t_2+t_3 = 0$

Using properties of determinants it can be proved

$\begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix}=4abc$

0%
True
0%
False

Explanation

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

$= \begin{vmatrix} 0 & -2c & -2b \\ b & c+a & b \\ c & c & a+b \end{vmatrix} [R_1^{'} = R_1 - (R_2 + R_3)]$

expanding w.r.t. row 1, we get,

$= 0 + 2c (b(a+b)-bc)) - 2b (bc- c(c+a)) \\ = 2abc + 2b^2c - 2bc^2 - 2b^2c + 2bc^2 + 2abc \\ = 4abc$

$\Delta = \left| {\begin{array}{*{20}{c}}1&1&1\\1&{1 + x}&1\\1&1&{1 + y}\end{array}} \right|$ for

$x \ne 0,\,y \ne 0$ then

$\Delta$ is

0%
Divisible by neither $x$ nor $y$
0%
Divisible by both $x$ and $y$
0%
Divisible by $x$ but not $y$
0%
Divisible by $y$ but not $x$

Explanation

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

$\Delta =\left( 1+x \right) \left( 1+y \right) -1-1\left( 1+y-1 \right) +1\left( 1-1-x \right)$

$\Delta =\left( 1+x \right) \left( 1+y \right) -1-y-x$

$\Delta =xy+1+x+y-1-y-x$

$\Delta =xy$

It is divisible by both

$x$ and

$y$

$B$ is correct.

$A=\quad \begin{bmatrix} 1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 0 \end{bmatrix}, B=(adj\quad A)$ and

$C=5A$ , then

$\cfrac { \left| adj\quad B \right| }{ \left| C \right| }$ is equal to

0%
$5$
0%
$25$
0%
$-1$
0%
$1$

Explanation

We have

$\begin{matrix} A=\left[ { \begin{array} { *{ 20 }{ c } }1 & { -1 } & 1 \\ 0 & 2 & { -3 } \\ 2 & 1 & 0 \end{array} } \right] \\ B=\left( { adj\, A } \right) \\ And, \\ C=5A \\ Now \\ \frac { { \left| { adj\, B } \right| } }{ { \left| C \right| } } =? \\ \left| A \right| =1\left( { 0+3 } \right) +1\left( 6 \right) +1\left( { -4 } \right) \\ \left| A \right| =5 \\ \frac { { \left| { adj\, \left( { adj\, A } \right) } \right| } }{ { \left| { 5A } \right| } } =\frac { { { { \left| A \right| }^{ { { \left( { 3-1 } \right) }^{ 2 } } } } } }{ { 125\left| A \right| } } \\ =\frac { { { { \left| A \right| }^{ 4 } } } }{ { 125|A| } } \\ =\frac { { |A| } }{ { 125 } } =1 \\ \end{matrix}$

Hence,

The option

$(A)$ is the correct answer.

Find the value of the determinant

$\begin{vmatrix} 1 & 0 & 0 \\ 2 & \cos { x } & \sin { x } \\ 3 & \sin { x } & \cos { x } \end{vmatrix}$ .

0%
$\cos{2x}$
0%
$1$
0%
$0$
0%
$\sin{2x}$

Explanation

Given

$\begin{vmatrix} 1 & 0 & 0 \\ 2 & cosx & sinx \\ 3 & sinx & cosx \end{vmatrix}$

$=1(\cos x\times\cos x-\sin x\times\sin x)$

$=\cos^x-\sin^2x$

We know that

$\cos 2x=\cos^2x-\sin^2x$

$=\cos 2x$

If the points

$(k, 2-2k)$ ,

$(1-k, 2k)$ and

$(-k-4, 6-2k)$ be collinear, the number of possible values of

$k$ are

0%
$4$
0%
$2$
0%
$1$
0%
$3$

Explanation

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

$(-k-4, 6-2k)$

if collinear the

$m_1=m_2$

$\dfrac{2k-2+2k}{1-k-k}=\dfrac{6-2k-2k}{-k-4-1+k}$

$\dfrac{4k-2}{1-2k}=\dfrac{6-4k}{-5}$

$\dfrac{2k-1}{1-2k}=\dfrac{3-2k}{-5}$

$5=3-2k$

$2k=-2$

$k=-1$

Only

$1$ value possible.

1108128_1196912_ans_4544716b664b484f8db29bdf7752b252.jpg

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

${A}^{-1}$ =

0%
$A$
0%
$\dfrac{1}{9} {A}^{T}$
0%
$\dfrac{1}{9}A$
0%
$\dfrac{1}{9}{A}^{-1}$

The determinant

$\begin{bmatrix} b_{1}+c_{1} & c_{1}+a_{1} & a_{1}+b_{1} \\ b_{2}+c_{2} & c_{2}+a_{2} & a_{2}+b_{2} \\ b_{3}+c_{3} & c_{3}+a_{3} & a_{3}+b_{3}\end{bmatrix}=$ _____

0%
$\begin{bmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{bmatrix}$
0%
$2\begin{bmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{bmatrix}$
0%
$3\begin{bmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{bmatrix}$
0%
$4\begin{bmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{bmatrix}$

In a triangle ABC, with usual notations, if

$\begin{vmatrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{vmatrix}=0,$ then

$4sin^2A+24sin^2B+36sin^2C$ is equal to

0%
$48$
0%
$50$
0%
$44$
0%
$34$

Explanation

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

using

$R_2 \rightarrow R_2 - R_1$ and

$R_3 \rightarrow R_3 - R_1$ property.

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

Expanding along first column.

$\Rightarrow 1[(c -a ) (c - b) - (b - a)(a - b)] + 0 = 0$

$\Rightarrow c^2 - ac - bc + ab - (ab - a^2 - b^2 + ab) = 0$

$\Rightarrow c^2 - ac - bc + ab - ab + a^2 + b^2 - ab = 0$

$\Rightarrow a^2 + b^2 + c^2 - ab - bc - ca = 0$

using identify

$a^2 + b^2 + c^2 - ab - bc - ca = \dfrac{1}{2}\left [(a - b)^2 + (b - c)^2 + (c- a)^2\right]$

$\Rightarrow \dfrac{1}{2} [(a - b)^2 + (b - c)^2 + (c - a)^2] = 0$

It us only possible if

$a + b = c$

$\Rightarrow$ Triangle is equilateral

$\Rightarrow \angle A = \angle B = \angle C = 60^{\circ}$

$\Rightarrow 4 \sin^2 A + 24 \sin^2 B + 36 \sin^2 c$

$= 4 \sin^2 60^{\circ} + 24 \sin^2 60^{\circ} + 36 \sin^2 60^{\circ}$

$= 64 \sin^2 60^{\circ} = 64 \times \left(\dfrac{\sqrt{3}}{2} \right)^2$

$= 64 \times \dfrac{3}{4}$

$= 48$

If adj

$A=\begin{bmatrix}20 & -20 \\ 10 & 10 \end{bmatrix}$ , then

$|A|=$ .....

0%
400
0%
200
0%
$\pm 20$
0%
0

29 If

$z=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ where 0, I are 2x2 null and identity matrix then det

$\left( \left[ z \right] \right)$ is _______________.

0%
1
0%
-1
0%
0
0%
None of these

Sum of the real roots of the equation

$\begin{vmatrix} 1 & 4 & 20\\ 1 & -2 & 5 \\ 1 & 2x & 5{x}^{2} \end{vmatrix}=0$ is

0%
$-2$
0%
$-1$
0%
$0$
0%
$1$

The cofactor of the element

$4$ in the determinant

$\begin{vmatrix} 1 & 3 & 5 & 1\\ 2 & 3 & 4 & 2\\ 8 & 0 & 1 & 1\\ 0 & 2 & 1 & 1\end{vmatrix}$ is?

0%
$4$
0%
$10$
0%
$-10$
0%
$-4$

Find the values of

$x$ if,

$\left| \begin{matrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & 5x^{ 2 } \end{matrix} \right| =0$

0%
$-1, 2$
0%
$-1, -2$
0%
$1, -2$
0%
$1, 2$

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 8 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 8 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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