MCQ Questions for Class 12 Commerce Maths Determinants Quiz 5

$A$ is any square matrix of order

$2$ , then

$adj \left ( adj A \right )$ =

0%
$A$
0%
$\det A$
0%
$A^{-1}$
0%
None of these

Explanation

We know that

$adj (adj A)=|A|^{n-2}A$

So, here

$adj (adj A)=|A|^{0}A$

$\Rightarrow adj (adj A)=A$

$\displaystyle A= \begin{bmatrix}0 &i-\sin x &i- \cos x\\sin x -i &0 &\sin x-i \\cos x-i &- \sin x+ i &0 \end{bmatrix}$ then

$\displaystyle \left | A \right |$ equals

0%
$0$
0%
$\displaystyle \sin x$
0%
$\displaystyle \cos x$
0%
$1$

Explanation

$\displaystyle A= \begin{bmatrix}0 &i-\sin x &i- \cos x\\sin x -i &0 &\sin x-i \\cos x-i &- \sin x+ i &0 \end{bmatrix}$

$\displaystyle \therefore \left | A\right | = \begin{vmatrix} 0 & i-\sin x & i-\cos x\\\sin x -i & 0 & \sin x-i \\\cos x-i & i-\sin x & 0\end{vmatrix}$

$\displaystyle =(i-\sin x)(\cos x-i)(\sin x-i) -(i-\cos x)(i-\sin x )^{2}$

$=(i-\sin x)^{2} (i-\cos x)-(i-\sin x)^{2} (i-\cos x)=0$

$\Delta =\begin{vmatrix} a & 5-i & 7+i\\ 5+i & b & 3+i\\ 7-i & 3-i & c \end{vmatrix}$ , then

$\Delta$ is always

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real
0%
imaginary
0%
$0$
0%
None of these

Explanation

$\Delta =\begin{vmatrix} a & 5-i & 7+i \\ 5+i & b & 3+i \\ 7-i & 3-i & c \end{vmatrix}$

$=abc-a(9-i^{ 2 })+(i-5)(5c+ic-(7-i)(3+i))+(7+i)[(5+i)(3-i)-b(7-i)]$

$=abc-10a+5ic-c-25c-5ic-(i-5)(21+4i+1)+(7+i)(16-2i)-50b$

$=abc-10a-26c-2i+114+114+2i-50b$

$\Delta=abc-10a-26c+228-50b$
Hence,

$\Delta$ is real.

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Both (A) & (R) are individually true & (R) is correct explanation of (A),
0%
Both (A)& (R) are individually true but (R) is not the correct (proper) explanation of (A).
0%
(A)is true but (R) is false,
0%
(A)is false but (R ) is true.

Explanation

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

$\displaystyle \therefore \left|A\right| =0$

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

$\displaystyle A^{T}=-A$

$\therefore$ Matrix A is, skew symmetric matrix, of odd order
Assertion (A) & Reason (R) both are individually true & Reason (R) is the correct explanation of Assertion (A)
Hence, option A.

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

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

$B$ is the adjoint of

$A$ then

$\displaystyle \alpha$ equals

0%
$2$
0%
$-1$
0%
$-2$
0%
$5$

Explanation

$\displaystyle A=\left [ \begin{matrix}1 &-1 &1 \\ 2 &1 &-3 \\ 1 &1 &1 \end{matrix} \right ]$
and

$\displaystyle B=\left [ \begin{matrix}4 &2 &2 \\ -5 &0 &\alpha \\ 1 &-2 &3 \end{matrix} \right ]$

Also given,

$adj A=B$

Now,

$adj A=C^{T}=\left[ \begin{matrix} 4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{matrix} \right] ^{T}$

$\Rightarrow adj A=\left[ \begin{matrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{matrix} \right] =\left [ \begin{matrix}4 &2 &2 \\ -5 &0 &\alpha \\ 1 &-2 &3 \end{matrix} \right ]$

On comparing ,we get

$\alpha=5$

$(-3, 11), (6, 2)$ and

$($

$k$

$, 4)$ are collinear points, then

$k$ is equal to

0%
8
0%
-8
0%
4
0%
-4

Explanation

$3$ points

$P\left( { x }_{ 1 },{ y }_{ 1 } \right) , Q\left( { x }_{ 2 },{ y }_{ 2 } \right)$ and

$R\left( { x }_{ 3 },{ y }_{ 3 } \right)$ are collinear, then slope

${ m }_{ PQ }=$ slope

${ m }_{ QR } =$ slope

${ m }_{ PR }$ .

The given points are

$P\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 3,11 \right) , Q\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 6,2 \right)$ and

$R\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( k,4 \right)$ .

Now if

$3$ points

$P\left( { x }_{ 1 },{ y }_{ 1 } \right) , Q\left( { x }_{ 2 },{ y }_{ 2 } \right)$ and

$R\left( { x }_{ 3 },{ y }_{ 3 } \right)$ are collinear, then

slope

${ m }_{ PQ }=$ slope

${ m }_{ QR } =$ slope

${ m }_{ PR }$

Slope

${ m }_{ PQ }=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\dfrac { 4-11 }{ k-3 }$ , slope

${ m }_{ PR }=\dfrac { { y }_{ 3 }-{ y }_{ 1 } }{ { x }_{ 3 }-{ x }_{ 1 } } =\dfrac { 4-2 }{ k-6 }$

$\therefore \dfrac { 4-11 }{ k-3 } =\dfrac { 4-2 }{ k-6 }$

$\Rightarrow 2k+6=-7k+42$

$\Rightarrow k=4$

The points

$X(-1, 3), Y (8, -3)$ and

$Z(2, 1)$

0%
are collinear
0%
not decided
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cannot be plotted
0%
are not collinear

Explanation

Three points

$A( { x }_{ 1 },{ y }_{ 1 }, B( { x }_{ 2 },{ y }_{ 2 })$ and

$C( { x }_{ 3 },{ y }_{ 3 })$ are said to be col-linear if

$slope\ { m }_{ AB }=slope\ { m }_{ BC } = slope { m }_{ AC }$

We know that, the slope of a line passing through the points

$(x_{1}, y_{1})\ \& \ (x_{2}, y_{2})$ is

$m=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } .$

Given ,

$X(-1 , 3), Y (8, -3), Z (2, 1)$

$m_{XY} = \dfrac{-3 - 3}{8 + 1}$

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

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

$m_{YZ} = \dfrac{1 + 3}{2 - 8}$

$= \dfrac{4}{-6}$

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

$m_{XZ} = \dfrac{1 - 3}{2 + 1}$

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

$\implies m_{XY} = m _{YZ} = m_{XZ}$
Hence, three points are collinear.

If the points

$\displaystyle \left(\frac{2}{5}, \frac{1}{3}\right), \left(\frac{1}{2} , k \right)$ and

$\displaystyle \left(\frac{4}{5}, 0 \right)$ are collinear then find the value of k

0%
$\displaystyle k = \frac{1}{2}$
0%
$\displaystyle k = \frac{1}{4}$
0%
$\displaystyle k = \frac{1}{5}$
0%
$\displaystyle k = -\frac{1}{2}$

Explanation

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 $( \dfrac {2}{5}, \dfrac {1}{3})$ and $( \dfrac {1}{2}, k) =$ Slope of line passing through $(\dfrac {1}{2}, k)$ and $( \dfrac {4}{5},0)$

$\dfrac { k- \dfrac{1}{3} }{ \dfrac {1}{2} - \dfrac {2}{5} } = \dfrac { 0 - k}{ \dfrac {4}{5} - \dfrac{1}{2} }$

On solving, $k = \dfrac {1}{4}$

The points

$A(7, 8), B(-5, 2)$ and

$C(3, 6)$

0%
are not collinear
0%
cannot be plotted
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are not defined
0%
are collinear

Explanation

The given points are

$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 7,8 \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( -5,2 \right)$ and

$C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 3,6 \right)$ .

Now if

$3$ points

$A\left( { x }_{ 1 },{ y }_{ 1 } \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right)$ and

$C\left( { x }_{ 3 },{ y }_{ 3 } \right)$ are collinear, then slope

${ m }_{ AB }=slope\quad { m }_{ BC } =$ slope

${ m }_{ AC }$

Slope

${ m }_{ AB }=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\dfrac { 2-8 }{ -5-7 } =\dfrac { 1 }{ 2 }$

Slope

${ m }_{ BC }=\dfrac { { y }_{ 3 }-{ y }_{ 2 } }{ { x }_{ 3 }-{ x }_{ 2 } } =\dfrac { 6-2 }{ 3+5 } =\dfrac { 1 }{ 2 }$ and slope

${ m }_{ AC }=\dfrac { { y }_{ 3 }-{ y }_{ 1 } }{ { x }_{ 3 }-{ x }_{ 1 } } =\dfrac { 6-8 }{ 3-7 } =\dfrac { 1 }{ 2 }$

Therefore, slope

${ m }_{ AB }=$ slope

${ m }_{ BC }$

$=$ slope

${ \quad m }_{ AC }$

Therefore, the points

$A, B$ and

$C$ are collinear.

If every element of third order determinant of

$\displaystyle \Delta$ is multiplied by 5 then value of new determinant equals to,

0%
$\displaystyle \Delta$
0%
$\displaystyle 5 \Delta$
0%
$\displaystyle 25 \Delta$
0%
$\displaystyle 125 \Delta$

Explanation

$|kA|=k^{n}|A|$ where n is order of matrix
Let

$\Delta =|A|$
So,

${\Delta}'=|5A|=5^{3}|A|$

${\Delta}'=125{\Delta}$

$\omega \neq 1$ is a complex cube root of unity, and

$x+iy=\begin{vmatrix} 1 & i & -\omega \\ -i & 1 & \omega ^{2}\\ \omega & -\omega ^{2} & 1 \end{vmatrix}$
then

0%
$x=-1, y=0$
0%
$x=1, y=-1$
0%
$x=1, y=1$
0%
none of these

Explanation

$x+iy=\begin{vmatrix} 1 & i & -\omega \\ -i & 1 & \omega ^{ 2 } \\ \omega & -\omega ^{ 2 } & 1 \end{vmatrix}\\$

$=1\left( 1+{ \omega }^{ 4 } \right) -i\left( -i-{ \omega }^{ 3 } \right) -\omega \left( -i-\omega \right) \\$

$=1+\omega -i\left( -i+\omega \right) +i\omega +{ \omega }^{ 2 }\\$

$=1+\omega -1-i\omega +i\omega +{ \omega }^{ 2 }=\omega +{ \omega }^{ 2 }=-1\\$

$\Rightarrow x=-1,y=0$

The points

$(a, b + c), (b, c + a)$ and

$(c, a + b)$ are

0%
vertices of an equilateral triangle
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concyclic
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vertices of a right angled triangle
0%
collinear

Explanation

${\textbf{Step1: Proving the three points are collinear }}$

${\text{Given points are A(a,b + c),B(b,c + a),C(c,a + b)}}$

${\text{ Lets the points of the triangle be }}\mathbf{{({x_1},{y_1}),({x_2},{y_2}),({x_3},{y_3})}}$

${\text{[}}{{\text{x}}_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})] = 0$

$\text{Comparing these with the given points}$

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

$=ac - ab + ab - bc + bc - ac$

$= 0$

${\textbf{Hence,The given points are collinear.}}$

The points

$(k, 2-2k)$ ,

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

$(-4-k, 6-2k)$ are collinear for

0%
all values of k
0%
$k=-1$
0%
$k=1/2$
0%
no value of k.

Explanation

The given points are collinear if

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

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

$\left [ R_{2}\rightarrow R_{2}-R_{1}, R_{3}\rightarrow R_{3}-R_{1} \right ]$

$\Rightarrow 4(-2k+1)-(-4-2k)(4k-2)=0$

$\Rightarrow (1-2k)(4-8-4k)=0$

$\Rightarrow (1-2k)(k+1)=0$

$\Rightarrow k=-1 \ or \ k=1/2$

$\triangle ABC$ is not a right triangle, then value of

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

0%
$-1$
0%
$2$
0%
$3$
0%
$0$

Explanation

$\Delta ABC$
We can use the below stated standard for the triangle

$ABC$ , the result can be proved easily by using sum and difference angles formula for

$\tan$

$A+B+C=\pi \Rightarrow \tan { A } +\tan { B } +\tan { C } =\tan { A } \tan { B } \tan { C }$
Thus

$\Delta =\begin{vmatrix} \tan { A } & 1 & 1 \\ 1 & \tan { B } & 1 \\ 1 & 1 & \tan { C } \end{vmatrix}\\ =\tan { A } \left( \tan { B } \tan { C } -1 \right) -1\left( \tan { C } -1 \right) +\left( 1-\tan { B } \right) \\ =\tan { A } \tan { B } \tan { C } -\tan { A } -\tan { C } +1+1-\tan { B } =2$

$\displaystyle \Delta =\begin{vmatrix} 0 &b-a &c-a \\ a-b &0 &c-b \\ a-c &b-c &0 \end{vmatrix}$ then

$\displaystyle \Delta$ is equal to

0%
$\displaystyle a+b+c$
0%
$\displaystyle -(a+b+c)$
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$\displaystyle abc$
0%
$\displaystyle 0$

Explanation

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

$\displaystyle \Delta' =\begin{vmatrix} 0 &a-b &a-c \\ b-a &0 &b-c \\ c-a &c-b &0 \end{vmatrix}=-\Delta$

Thus

$\Delta$ is an skew-symmetric matrix

And we know determinant of a skew-symmetric matrix of odd order is zero

Hence option 'D' is the correct choice.

If points

$(x,0)$ ,

$(0,y)$ and

$(1,1)$ are collinear then the relation is-

0%
$x+y=1$
0%
$x+y=xy$
0%
$x+y+1=0$
0%
$x+y+xy=0$

Explanation

Area of a triangle with vertices

$({ x }_{ 1 },{ y }_{ 1 })$ ;

$({ x }_{ 2 },{ y }_{ 2 })$ and

$({ x }_{ 3 },{ y }_{ 3 })$ is

$\left| \dfrac { { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right|$

Since the given points are collinear, they do not form a triangle, which means area of the triangle is zero.

Hence, substituting the points

$({ x }_{ 1 },{ y }_{ 1 }) = (x,0)$ ;

$({ x }_{ 2 },{ y }_{ 2 }) = (0,y)$ and

$({ x }_{ 3 },{ y }_{ 3 }) = (1,1)$ in the area formula, we get

$\left| \dfrac { { x }({ y }-1)+0(1-0)+1(0-{ y }) }{ 2 } \right| = 0$

$=> xy - x -y = 0$

$=> x + y = xy$
Hence, option 'B' is correct.

$a, b, c> 1$ ,

$\Delta =\begin{vmatrix} \log _{a}\left ( abc \right ) & \log _{a}b & \log _{a}c\\ \log _{b}\left ( abc \right ) & 1 & \log _{b}c\\ \log _{c}\left ( abc \right ) & \log _{c}b & 1 \end{vmatrix}$ is

0%
$0$
0%
$\log _{a}b+\log _{b}c+\log _{c}a$
0%
$\log _{abc}\left ( a+b+c \right )$
0%
none of these

Explanation

$\Delta =\begin{vmatrix} \log _{ a } \left( abc \right) & \log _{ a } b & \log _{ a } c \\ \log _{ b } \left( abc \right) & 1 & \log _{ b } c \\ \log _{ c } \left( abc \right) & \log _{ c } b & 1 \end{vmatrix}\\$

$=\log _{ a } \left( abc \right) \left( 1-\log _{ c } b\log _{ b } c \right) -\log _{ b } \left( abc \right) \left( \log _{ a } b-\log _{ c } b\log _{ a } c \right) +\log _{ c } \left( abc \right) \left( \log _{ a } b\log _{ b } c-\log _{ a } c \right) \\$

$=\cfrac { logabc }{ logb } \left( \cfrac { logb }{ loga } -\cfrac { logb }{ logc } \cfrac { logc }{ loga } \right) +\cfrac { logabc }{ logc } \left( \cfrac { logb }{ loga } \cfrac { logc }{ logb } -\cfrac { logc }{ loga } \right) =0$

Hence, option 'A' is correct.

For which value of '

$k$ ' the points

$(7, -2), (5, 1), (3, k)$ are collinear?

0%
$-4$
0%
$4$
0%
$8$
0%
$-8$

Explanation

Since points are collinear,

so,

$x_1(y_2 - y_3)+x_2(y_3 - y_1) + x_3(y_1 - y_2)= 0$

Take (

$7, -2$ ) as

$(x_1 , y_1)$ , (

$5, 1$ ) as

$(x_2 , y_2)$ and (

$3 , k$ ) as

$(x_3 , y_3)$ , we have

$7(1 -k ) + 5(k + 2) + 3(-2 -1) = 0$

$\Rightarrow$

$7- 7k + 5k + 10 -9 =0$

$\Rightarrow$

$2k =8$

$\Rightarrow$

$k = 4$

$(3,2)$ ,

$(4,k)$ and

$(5,3)$ are collinear, then

$k$ is equal to:

0%
$\dfrac { 3 }{ 2 }$
0%
$\dfrac { 2 }{ 5 }$
0%
$\dfrac { 5 }{ 2 }$
0%
$\dfrac { 3 }{ 5 }$

Explanation

As the points are collinear, the slope of the line joining any two points, should be same as the slope of the line joining two other points.

Slope of the line passing through points

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

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

$\dfrac { { y }_{ 2 }-{y }_{ 1 } }{ { x }_{ 2 }-x_{ 1 } }$
So, Slope of the line joining

$(3,2) , (4,k) =$ Slope of the line joining

$(3,2)$ and

$(5,3)$

$\dfrac { k - 2 }{ 4 - 3 } = \dfrac { 3- 2 }{ 5 - 3 }$

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

$k = \dfrac { 5 }{ 2 }$

Hence, option C.

Let

$A$ be a non-singular matrix. Then

$\left| adjA \right|$ is equal to

0%
${ \left| A \right| }^{ n }$
0%
${ \left| A \right| }^{ n-1 }$
0%
${ \left| A \right| }^{ n-2 }$
0%
None of these

Explanation

Since A is non-singular

$\therefore { A }^{ -1 }$ exists

Now ,

$A$

$\left( adjA \right)$ =

$\displaystyle\left| A \right| I=$

$(adjA)A$

$\displaystyle \Rightarrow \left| A \right| \left| adjA \right| ={ \left| A \right| }^{ n }=\left| adjA \right| \left| A \right|$

$\displaystyle \therefore \left| adjA \right| ={ \left| A \right| }^{ n-1 }$

The value of the determinant

$\begin{vmatrix} -a& b & c\\ a & -b &c \\ a & b &-c \end{vmatrix}$ is equal to

0%
$0$
0%
$(a-b)(b-c)(c-a)$
0%
$(a+b)(b+c)(c+a)4abc$
0%
$4abc$

Explanation

$\begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix}=\left( -a \right) \left( bc-bc \right) -b\left( -ac-ac \right) +c\left( ab+ab \right) \\ =2abc+2abc=4abc$

$A=\begin{bmatrix} 0&\sin \alpha & \sin \alpha\sin \beta \\-\sin \alpha &0 &\cos \alpha \sin \beta \\-\sin \alpha \sin \beta &-\cos \alpha \cos \beta &0 \end{bmatrix}$ , then which of the following is true?

0%
$|A|$ is independent of $\alpha$ and $\beta$ .
0%
$A^{-1}$ depends only on $\alpha$ .
0%
$A^{-1}$ depends only on $\beta$ .
0%
None of these

Explanation

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

A is a skew symmetric matrix

$\Rightarrow \left| A \right| =0$

$\Rightarrow { A }^{ -1 }$ does not exist and

$\left| A \right|$ is independent of

$\alpha \& \beta$

Option A

$D_p=\begin{vmatrix} p& 15 & 8\\ p^2 & 35 & 9\\ p^3 & 25 & 10\end{vmatrix}$ , then

$D_1+D_2+D_3+D_4+D_5$ is equal to

0%
0
0%
25
0%
625
0%
none of these

Explanation

$D_{ p }=\begin{vmatrix} p & 15 & 8 \\ p^{ 2 } & 35 & 9 \\ p^{ 3 } & 25 & 10 \end{vmatrix}=5\begin{vmatrix} p & 3 & 8 \\ p^{ 2 } & 7 & 9 \\ p^{ 3 } & 5 & 10 \end{vmatrix}\\ =5\left( p\left( 70-45 \right) -{ p }^{ 2 }\left( 30-40 \right) +{ p }^{ 3 }\left( 27-56 \right) \right) \\ =5\left( 25p+10{ p }^{ 2 }-29{ p }^{ 3 } \right)$

$D_{ 1 }+D_{ 2 }+D_{ 3 }+D_{ 4 }+D_{ 5 }\\ =5\left( 25+10-29 \right) +5\left( 25.2+10.2^{ 2 }-29.{ 2 }^{ 3 } \right) +5\left( 25.3+10{ .3 }^{ 2 }-29{ .3 }^{ 3 } \right) \\ +5\left( 25.4+10{ .4 }^{ 2 }-29{ .4 }^{ 3 } \right) +5\left( 25.5+10{ .5 }^{ 2 }-29{ .5 }^{ 3 } \right) \\ =28000$

$\displaystyle a_{0}$ equals

0%
0
0%
1
0%
2
0%
3

Explanation

$\triangle \left( x \right) =\begin{vmatrix} 2{ x }^{ 3 }-3{ x }^{ 2 } & 5x+7 & 2 \\ 4{ x }^{ 3 }-7x & 3x+2 & 1 \\ 7{ x }^{ 3 }-8{ x }^{ 2 } & x-1 & 3 \end{vmatrix}={ a }_{ 0 }+{ a }_{ 1 }x+....{ a }_{ 4 }{ x }^{ 4 }$

Putting

$x=0$ we get

${ a }_{ 0 }=?$

$\triangle \left( 0 \right) ={ a }_{ 0 }=\begin{vmatrix} 0 & 7 & 2 \\ 0 & 2 & 1 \\ 0 & -1 & 3 \end{vmatrix}=0$

$\therefore { a }_{ 0 }=0$

Option A

The minors and cofactors of -4 and 9 in determinant

$\begin{vmatrix} -1 & -2 & 3 \\ -4 & -5 & -6 \\ -7 & 8 & 9\end{vmatrix}$ are respectively

0%
$42, 42 ; 3, 3$
0%
$-42, 42 ; -3, -3$
0%
$42, -42 ; 3, -3$
0%
$42, 3 ; 42, 3$

Explanation

Minor of -4 is

$\begin{vmatrix} -2\quad & 3 \\ 8 & 9 \end{vmatrix}=-42$
Cofactor of -4 is

${ \left( -1 \right) }^{ 1+2 }\left( 42 \right) =42$

Minor of 9 is

$\begin{vmatrix} -1\quad & -2 \\ -4 & -5 \end{vmatrix}=-3$
Cofactor of 9 is

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

$A=\begin{bmatrix} 4 & 2 \\ 3 & 3 \end{bmatrix}$ , then adj (adj

$A$ ) is equal to

0%
$\begin{bmatrix} 3 & -2 \\ -3 & 4 \end{bmatrix}$
0%
$\begin{bmatrix} 4 & 2 \\ 3 & 3 \end{bmatrix}$
0%
$6\begin{bmatrix} 4 & 2 \\ 3 & 3 \end{bmatrix}$
0%
None of these

Explanation

For

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

$adj\left( A \right) =\begin{bmatrix} 3 & -3 \\ -2 & 4 \end{bmatrix}$

$adj\left( adj\left( A \right) \right) =\begin{bmatrix} 4 & 2 \\ 3 & 3 \end{bmatrix}$

A set of points which lie on same line are called as

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collinear
0%
non-collinear
0%
concurrent
0%
none of these

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

0%
AdjA is zero matrix
0%
$AdjA=\begin{bmatrix} 0& 0 & -1\\ 0& -1 & 0\\ -1 &0 & 0\end{bmatrix}$
0%
$A^{-1} = A$
0%
$A^2 = I$

Explanation

$A=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}\\ AdjA=\begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}\\ { A }^{ -1 }=\cfrac { 1 }{ -1 } \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$

The point with the co-ordinates

$(2a,\ 3a),\ (3b,\ 2b)$ &

$(c,\ c)$ are collinear _____________ .

0%
for no value of $a,\ b,\ c$
0%
for all values of $a,\ b,\ c$
0%
if $a,\ \dfrac {c}{5},\ b$ are in $H.P$ .
0%
if $a,\ \dfrac {2}{5}c,\ b$ are in $H.P$ .

Explanation

$\Delta=\dfrac {1}{2}\begin{vmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\x_3 & y_3 & 1\end{vmatrix}=0$

$\Delta=\dfrac {1}{2}\begin{vmatrix}2a & 3a & 1\\ 3b & 2b & 1\\c & c & 1\end{vmatrix}=0$

$\Rightarrow (2a-c)(2b-c)-(3a-c)(3b-c)=0$

$\Rightarrow 4ab-2ac-2bc+c^2=(9ab-3ac-3bc+c^2)=0$

$\Rightarrow ac+bc-5ab=0$

$\dfrac {1}{a}+\dfrac {1}{b}=\dfrac {5}{c}\Rightarrow \dfrac {1}{a}+\dfrac {1}{b}=2\left (\dfrac {5}{2c}\right )$

$\because$

$\dfrac {1}{a},\dfrac {5}{2c}\, and\, \dfrac{1}{b}$ are in

$A.P.$

$\therefore a, \dfrac {2c}{5}, b$ are in H.P.

If lines AB, AC, AD and AE are parallel to a line then_______

0%
A, B, C, D, E are collinear points
0%
A, B, C, D, E are noncolinear points
0%
AB & AC are parallel and AD & AE are perpendicular
0%
None of these

Explanation

Gievn Line

$AB,AC,AD,AE$ are parallel so they are only and only they are collinear

Because if we draw line AB To point C we find line AC similarly in moving forward we Get line AC.AD and AE

Hence

$A,B,C,D,E$ are collinear points

Three lines

$px + qy + r = 0, qx + ry + p = 0$ and

$rx + py + q = 0$ are concurrent if

0%
$p + q + r = 0$
0%
$p^2 + q^2 + r^2 = pr + qr + pq$
0%
$p^3 + q^3 + r^3 = 3pqr$
0%
none of these

Explanation

These lines are concurrent when area formed from these lines is zero

$\begin{vmatrix} p & q & r \\ q & r & p \\ r & p & q \end{vmatrix}=0$
Applying

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

$\Rightarrow \begin{vmatrix} p+q+r & p+q+r & p+q+r \\ q & r & p \\ r & p & q \end{vmatrix}=0\\ \Rightarrow \left( p+q+r \right) \begin{vmatrix} 1 & 1 & 1 \\ q & r & p \\ r & p & q \end{vmatrix}=0$
Again applying

${ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 1 },{ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }$

$\Rightarrow \left( p+q+r \right) \begin{vmatrix} 1 & 0 & 0 \\ q & r-q & p-q \\ r & p-r & q-r \end{vmatrix}\\ \Rightarrow \left( p+q+r \right) \left( \left( r-q \right) \left( q-r \right) -\left( p-r \right) \left( p-q \right) \right) =0$ \

$\Rightarrow p^3+q^3+r^3-3pqr =0$
Hence, options A, B and C are correct.

The adjoint of the matrix

$\begin{bmatrix} 1 & -2 & 3 \\ 0 & 2 & -1 \\ -4 & 5 & 2 \end{bmatrix}$ is:

0%
$\begin{bmatrix} 9 & 19 & -4 \\ 4 & 14 & 1 \\ 8 & 3 & 2 \end{bmatrix}$
0%
$\begin{bmatrix} 9 & 4 & 8 \\ 19 & 14 & 3 \\ -4 & 1 & 2 \end{bmatrix}$
0%
$\begin{bmatrix} 9 & -19 & -4 \\ -4 & 14 & -1 \\ 8 & -3 & 2 \end{bmatrix}$
0%
none of these

Explanation

Cofactor of the matrix is given by

$\begin{bmatrix} M11 & -M12 & M13 \\ -M21 & M22 & -M23 \\ M31 & -M32 & M33 \end{bmatrix}$

where

$M11$ =

$\begin{bmatrix} 2 & -1\\ 5 & 2 \end{bmatrix} = 4+5=9$

$M12$ =

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

$M13$ =

$\begin{bmatrix} 0 & 2 \\ -4 & 5 \end{bmatrix} = 0+8=8$

$M21$ =

$\begin{bmatrix} -2 & 3 \\ 5 & 2\end{bmatrix} = -4-15=-19$

$M22$ =

$\begin{bmatrix} 1 & 3 \\ -4& 2 \end{bmatrix} = 2+12=14$

$M23$ =

$\begin{bmatrix} 1 & -2 \\ -4 & 5 \end{bmatrix} = 5-8=-3$

$M31$ =

$\begin{bmatrix} -2& 3 \\ 2 & -1 \end{bmatrix} = 2-6=-4$

$M32$ =

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

$M33$ =

$\begin{bmatrix} 1 & -2\\ 0 & 2 \end{bmatrix} = 2-0=2$

substituting all the values in the Cofactor of the matrix formula we get,

$\begin{bmatrix} 9 & 4 & 8 \\ 19 & 14 & 3\\ -4 & 1 & 2\end{bmatrix}$

the adjoint of the matrix is transpose of cofactor matrix

$\therefore$ Adjoint of the matrix is

$\begin{bmatrix} 9 & 19 & -4 \\ 4 & 14& 1 \\ 8 & 3 & 2 \end{bmatrix}$

$A$ and

$B$ are square matrices of same orders then adj.

$(AB)$ equals

0%
adj $A$ . adj $B$
0%
adj $B$ /adj $A$
0%
adj $A+$ adj $B$
0%
adj $A-$ adj $B$

Explanation

$(AB)^{-1}=\displaystyle\frac{adj(AB)}{|AB|}$

$\Rightarrow B^{-1}A^{-1}=\displaystyle\frac{adj(AB)}{|AB|}$

$\Rightarrow \displaystyle\frac{adj B}{|B|}\frac{adj A}{|A|}=\frac{adj(AB)}{|A||B|}$

$\therefore adj(AB)=(adj B)(adj A)$
Hence, option B.

If cofactor of 2x in the determinant

$\begin{vmatrix}x & 1 & -2 \\ 1 & 2x & x-1 \\ x-1 & x & 0\end{vmatrix}$ is zero, then

$x$ equals to

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

Explanation

Cofactor of

$2x=0$

$\displaystyle \Rightarrow { C }_{ 22 }=0$

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

$\displaystyle \Rightarrow x\times 0-\left( -2 \right) \left( x-1 \right) =0$

$\displaystyle \Rightarrow 2\left( x-1 \right) =0\Rightarrow x=1$

$A = (a_{ij})$ is a

$4\times 4$ matrix and

$C_{ij}$ is the co-factor of the element

$a_{ij}$ in Det (A), then the expression

$a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} + a_{14}C_{14}$ equals

0%
$0$
0%
$-1$
0%
$1$
0%
$Det. (A)$

Consider the determinant

$\Delta=\begin{vmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix}$

$M_{ij} =$ Minor of the element of

$i^{th}$ row &

$j^{th}$ column.

$C_{ij} =$ Cofactor of element of

$i^{th}$ row &

$j^{th}$ column.

$a_3M_{13} - b_3M_{23} + c_3M_{33}$ is equal to

0%
$0$
0%
$4\Delta$
0%
$2\Delta$
0%
$\Delta$

Explanation

${ a }_{ 3 }{ M }_{ 13 }-{ b }_{ 2 }{ M }_{ 23 }+{ c }_{ 3 }{ M }_{ 33 }$

$={ a }_{ 3 }\begin{vmatrix} { b }_{ 1 }\quad & { b }_{ 2 } \\ { c }_{ 1 } & { c }_{ 2 } \end{vmatrix}-{ b }_{ 3 }\begin{vmatrix} { a }_{ 1 }\quad & { a }_{ 2 } \\ { c }_{ 1 } & { c }_{ 2 } \end{vmatrix}+{ c }_{ 3 }\begin{vmatrix} { a }_{ 1 }\quad & { a }_{ 2 } \\ { b }_{ 1 } & { b }_{ 2 } \end{vmatrix}$

Is equal to the expansion of

$\triangle$ along

${ C }_{ 3 }$

Let

$A = [a_{ij}]_{n\times n}$ be a square matirx and let

$c_{ij}$ be cofactor of

$a_{ij}$ in A. If

$C = [c_{ij}]$ , then

0%
$|C|=|A|$
0%
$|C|=|A|^{n-1}$
0%
$|C|=|A|^{n-2}$
0%
none of these

Explanation

$C\rightarrow$ Cofactor matrix

$AdjA= \left (C \right )^{^{T}}$

But det of

$AdjA= Det \quad of \quad C$

Because they are transpore of each other .

$\Rightarrow\left | AdjA \right | = \left | C \right |= \left | A \right |^{n-1}$

Option-B

If in the determinant

$\Delta=\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}, A_i, B_i, C_i$ etc. be the co-factors of

$a_i, b_i, c_i$ etc., then which of the following relations is incorrect?

0%
$a_1A_1 + b_1B_1 + c_1C_1=\Delta$
0%
$a_2A_2 + b_2B_2 + c_2C_2=\Delta$
0%
$a_3A_3 + b_3B_3 + c_3C_3=\Delta$
0%
$a_1A_2 + b_1B_2 + c_1C_2=\Delta$

Explanation

$\Delta=\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}, A_1, B_1, C_1$ etc. be the co-factors of

$a_1, b_1, c_1$ etc.
Sum of product of any row with its corresponding cofactors is Determinant.

$\Rightarrow a_1A_1+b_1B_1+c_1C_1=a_2A_2+b_2B_2+c_2C_2=a_3A_3+b_3B_3+c_3C_3=\Delta$
Sum of product of any row with another row's cofactors is

$0$ .

$\Rightarrow a_1A_2+b_1B_2+c_1C_2=0$
Hence, option D is incorrect.

Consider the determinant

$\Delta=\begin{vmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix}$

$M_{ij} =$ Minor of the element of

$i^{th}$ row &

$j^{th}$ column.

$C_{ij} =$ Cofactor of element of

$i^{th}$ row &

$j^{th}$ column.

Value of

$b_1.C_{31} + b_2.C_{32} + b_3.C_{33}$ is

0%
$0$
0%
$\Delta$
0%
$2\Delta$
0%
$\Delta^2$

Explanation

Value of

${ b }_{ 1 }.{ C }_{ 31 }+{ b }_{ 2 }.C_{ 32 }+b_{ 3 }.{ C }_{ 33 }$

$={ b }_{ 1 }.{ \left( -1 \right) }^{ 3+1 }\begin{vmatrix} { a }_{ 2 }\quad & { a }_{ 3 } \\ { b }_{ 2 } & { b }_{ 3 } \end{vmatrix}+{ b }_{ 3 }.{ \left( -1 \right) }^{ 3+2 }\begin{vmatrix} { a }_{ 1 }\quad & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 3 } \end{vmatrix}+{ b }_{ 3 }.{ \left( -1 \right) }^{ 3+3 }\begin{vmatrix} { a }_{ 1 }\quad & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } \end{vmatrix}$

$={ b }_{ 1 }\left( { b }_{ 2 }{ a }_{ 3 }-{ a }_{ 2 }{ b }_{ 3 } \right) -{ b }_{ 2 }\left( { b }_{ 1 }{ a }_{ 3 }-{ a }_{ 1 }{ b }_{ 3 } \right) +{ b }_{ 3 }\left( { b }_{ 1 }{ a }_{ 3 }-{ a }_{ 1 }{ b }_{ 2 } \right)$

$=0$

$A+B+C=\pi$ , then

$\begin{vmatrix}sin(A+B+C) & sin B & cos C\\ -sinB & 0 & tan A\\ cos(A+B) & -tanA & 0\end{vmatrix}$ equals

0%
0
0%
2 sin B tan A cos C
0%
1
0%
none of these

Explanation

$\Delta =\begin{vmatrix} \sin { \left( A+B+C \right) } & \sin { B } & \cos { C } \\ -\sin { B } & 0 & \tan { A } \\ \cos { \left( A+B \right) } & -\tan { A } & 0 \end{vmatrix}$

$=\begin{vmatrix} \sin { \left( \pi \right) } & \sin { B } & \cos { C } \\ -\sin { B } & 0 & \tan { A } \\ \cos { \left( \pi -C \right) } & -\tan { A } & 0 \end{vmatrix}$

$=\begin{vmatrix} 0 & \sin { B } & \cos { C } \\ -\sin { B } & 0 & \tan { A } \\ -\cos { \left( C \right) } & -\tan { A } & 0 \end{vmatrix}$

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

$f(x)=\begin{vmatrix} cos x & 1 & 0 \\ 1 & cos x & 1 \\ 0 & 1 & cos x\end{vmatrix}$ the

$f'(\dfrac \pi 3)$ equals

0%
$\dfrac {11\sqrt 3}{8}$
0%
$\dfrac {5\sqrt 3}{8}$
0%
$-\dfrac {5\sqrt 3}{8}$
0%
none of these

Explanation

Given,

$f\left( x \right) =\begin{vmatrix} \cos { x } & 1 & 0 \\ 1 & \cos { x } & 1 \\ 0 & 1 & \cos { x } \end{vmatrix}\\ =\cos { x } \left( \cos ^{ 2 }{ x-1 } \right) -1\cos { x } \\ =\cos ^{ 3 }{ x } -\cos { x } -\cos { x } \\ =\cos ^{ 3 }{ x } -2\cos { x }$
Now

$f^{ ' }\left( x \right) =-3\cos ^{ 2 }{ x } \sin { x } +2\sin { x }$

$\displaystyle { f }^{ ' }\left( \frac { \pi }{ 3 } \right) =\frac { 5\sqrt { 3 } }{ 8 }$

$x, y, z$ are positive numbers, then value of the determinant

$\begin{vmatrix}1 & log_xy & log_xz \\ log_yx & 1 & log_yz\\ log_zx & log_zy & 1\end{vmatrix}$ is equal to

0%
$0$
0%
$3$
0%
$log xyz$
0%
none

Explanation

The value of the determinant is

$\begin{vmatrix} 1 & \log _{ x }{ y } & \log _{ x }{ z } \\ \log _{ y }{ x } & 1 & \log _{ y }{ z } \\ \log _{ z }{ x } & \log _{ z }{ y } & 1 \end{vmatrix}$

$=\left( 1-\log _{ y }{ z\log _{ z }{ y } } \right) -\log _{ x }{ y } \left( \log _{ y }{ x } -\log _{ y }{ z\log _{ z }{ x } } \right) +\log _{ x }{ z } \left( \log _{ y }{ x\log _{ z }{ y } -\log _{ z }{ x } } \right)$

$[\because log{_n}^m=\dfrac{logm}{logn}]$

$=0-1+1+1-1=0$

$\Delta =\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{vmatrix}$ and

$A_2, B_2, C_2$ are respectively cofactors of

$a_2, b_2, c_2$ then

$a_1A_2 + b_1B_2 + c_1C_2$ is equal to

0%
$-\Delta$
0%
0
0%
$\Delta$
0%
none of these

Explanation

$\Delta =\begin{vmatrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \end{vmatrix}\\ { A }_{ 2 }=-\begin{vmatrix} { b }_{ 1 } & { c }_{ 1 } \\ { b }_{ 3 } & { c }_{ 3 } \end{vmatrix}={ b }_{ 3 }{ c }_{ 1 }-{ b }_{ 1 }{ c }_{ 3 }\\ B_{ 2 }=\begin{vmatrix} { a }_{ 1 } & { c }_{ 1 } \\ { a }_{ 3 } & { c }_{ 3 } \end{vmatrix}={ c }_{ 3 }{ a }_{ 1 }-{ c }_{ 1 }{ a }_{ 3 }\\ C_{ 3 }=-\begin{vmatrix} { a }_{ 1 } & { b }_{ 1 } \\ { a }_{ 3 } & { b }_{ 3 } \end{vmatrix}={ a }_{ 3 }{ b }_{ 1 }-{ a }_{ 1 }{ b }_{ 3 }\\ \therefore { a }_{ 1 }{ A }_{ 2 }+{ b }_{ 1 }B_{ 2 }+{ c }_{ 1 }C_{ 3 }={ a }_{ 1 }{ b }_{ 3 }{ c }_{ 1 }-{ a }_{ 1 }{ b }_{ 1 }{ c }_{ 3 }+{ a }_{ 1 }{ b }_{ 1 }{ c }_{ 3 }-{ a }_{ 3 }{ b }_{ 1 }{ c }_{ 1 }+{ a }_{ 3 }{ b }_{ 1 }{ c }_{ 1 }-{ a }_{ 1 }{ b }_{ 3 }{ c }_{ 1 }=0$

If [a] denotes the greatest integer less than or equal to a and

$-1 \leq x < 0, 0 \leq y < 1, 1 \leq z < 2$ , then

$\begin{vmatrix}[x]+1 & [y] & [z] \\ [x] & [y]+1 & [z] \\ [x] & [y] & [z]+1\end{vmatrix}$ is equal to

0%
$[x]$
0%
$[y]$
0%
$[z]$
0%
None of these

Explanation

$- 1 \leq x < 0\rightarrow [x] =- 1$

$0 \leq y < 1 \rightarrow [y] = 0$

$1 \leq z < 2 \rightarrow [z] = 1$

$\begin{vmatrix}0 & 0 & 1 \\ -1 & 1 & 1 \\ -1 & 0 & 2\end{vmatrix}$

$=1=[z]$

$A =$

$\displaystyle \begin{bmatrix}\alpha & 2\\ 2 & \alpha \end{bmatrix}$ and

$\displaystyle \left | A \right |^{3}=125$ then the value of

$\displaystyle \alpha$ is

0%
$\displaystyle \pm 1$
0%
$\displaystyle \pm 2$
0%
$\displaystyle \pm 3$
0%
$\displaystyle \pm 5$

Explanation

We know

$\left| { A }^{ n } \right| ={ \left| A \right| }^{ n }$
Since,

$\left| { A }^{ 3 } \right| =125\Rightarrow { \left| A \right| }^{ 3 }=125$

$\Rightarrow \begin{vmatrix} \alpha & 2 \\ 2 & \alpha \end{vmatrix}=5$

$\Rightarrow { \alpha }^{ 2 }-4=5$

$\Rightarrow \alpha =\pm 3$

If the points

$A(1, 2) ,\ O (0, 0)$ and

$C (a, b)$ are collinear then

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

Explanation

$\textbf{Step -1: Area of triangle in co-ordinate geometry.}$

$\text{Given points are }A(1, 2), O(0, 0), C(a, b)$

$\text{The points are said to be collinear if the area of }\triangle AOC = 0$

$\text{Area of }\triangle AOC =$

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

$\textbf{Step -2: Substituting the given points in the formula.}$

$\text{Area of }\triangle AOC =$

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

$\Rightarrow - b +2a = 0$

$\Rightarrow 2a = b$

$\textbf{Hence, the points A(1, 2) , O(0, 0) and C(a, b) are collinear when }$ $\textbf{2a = b.}$

The points (-a ,-b), (0, 0) (a, b) and

$\displaystyle (a^{2},ab)$ are

0%
Collnear
0%
Vertices of parallelogram
0%
Vertices of a rectangle
0%
None of these

Explanation

Let

$A (-a, -b), B (0,0), C(a,b)$ and

$D({a}^{2},ab)$
Distance between two points

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

$\left( { x }_{ 2 },{ y }_{ 2 } \right)$ can be calculated
using the formula

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

Distance between the points A $(-a,-b)$ and B $(0,0) = \sqrt { \left( 0 + a \right) ^{ 2 }+\left( 0 + b \right) ^{ 2 } } = \sqrt { {a}^{2} + {b}^{2} }$

Distance between the points B $(0,0)$ and C $(a,b) = \sqrt { \left( a-0 \right) ^{ 2 }+\left( b-0 \right) ^{ 2 } } = \sqrt { {a}^{2} + {b}^{2} }$

Distance between the points C $(a,b)$ and D $({a}^{2}, ab) = \sqrt { \left( {a}^{2} - a \right) ^{ 2 }+\left(ab - b \right) ^{ 2 } } = (a-1)\sqrt { {a}^{2} + {b}^{2} }$

Distance between the points A $(-a,-b)$ and D $({a}^{2}, ab) = \sqrt { \left( {a}^{2} + a \right) ^{ 2 }+\left(ab + b \right) ^{ 2 } } = (a +1)\sqrt { {a}^{2} + {b}^{2} }$

Now. $AB+BC+CD = \sqrt { {a}^{2} + {b}^{2} } + \sqrt { {a}^{2} + {b}^{2} }+ (a-1)\sqrt { {a}^{2} + {b}^{2} } = (a +1)\sqrt { {a}^{2} + {b}^{2} } = AD$

Hence, the points are collinear.

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

$A(Adj. A)$ equals-

0%
$\begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix}$
0%
$\begin{bmatrix} 0 & 10 \\ 10 & 0 \end{bmatrix}$
0%
$\begin{bmatrix} 10 & 1 \\ 1 & 10 \end{bmatrix}$
0%
none of these

Explanation

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

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

$A\left( adjA \right) =\begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}\begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix}=\begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix}$

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

${(adj. A)}_{23}$ is equal to

0%
$13$
0%
$-13$
0%
$5$
0%
$-5$

Explanation

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

${(adj. A)}_{23}={C}_{32}$

So cofactor of

${a}_{32}$

${C}_{32}={(-1)}^{3+2}(-1-12)=13$

Ans: A

If the points

$(5, 1),\ (1, p)\ \& \ (4, 2)$ are collinear then the value of p will be

0%
$1$
0%
$5$
0%
$2$
0%
$-2$

Explanation

Area of a triangle with vertices $({ x }_{ 1 },{ y }_{ 1})$ ; $({ x }_{ 2 },{ y }_{ 2 })$ and $({ x }_{ 3 },{ y }_{ 3})$ is $\left| \dfrac { { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2}({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right|$

Since the given points are collinear, they do not form a triangle, which means area of the triangle is Zero.

Hence, substituting the points $({ x }_{ 1 },{ y }_{ 1 }) = (5,1)$ ; $({ x}_{ 2 },{ y }_{ 2 }) = (1,P)$ and $({ x }_{ 3 },{ y }_{ 3 }) = (4,2)$ in the area formula, we get $\left| \dfrac { 5(P-2)+1(2-1)+4(1-P) }{ 2 } \right| =0$

$=> 5P-10+1+4-4P = 0$

$=> P = 5$

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 5 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 5 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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