If $$A=[x \quad y],B=[_{ h } ^{ a }\quad _{ b } ^{ h }],C=[_{ y } ^{ x }]$$, then $$ABC = $$ ___.

$$({ ax }^{ 2 }+2hxy+{ by }^{ 2 })$$

$$({ ax }^{ 2 }-2hxy+{ by }^{ 2 })$$

$$({ bx }^{ 2 }-2hxy+{ ay }^{ 2 })$$

MCQ Questions for Class 12 Commerce Maths Matrices Quiz 12

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

$A^{2}-4\ A$ is equal to

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$2\ I_3$
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$3\ I_3$
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$4\ I_3$
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$5\ I_3$

$A$ and

$B$ are square matrices such that

$B=-A^{-1}BA$ , then

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$AB+BA=0$
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$(A+B)^{o}=A^{2}+B^{2}$
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$(A+B)^{2}=A^{2}+2AB+B^{2}$
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$(A+B)^{2}=A+B$

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

$10B=\left[ \begin{matrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{matrix} \right]$ where

$B=A^{-1}$ then

$\alpha$ is equal to-

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$2$
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$-1$
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$-2$
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$5$

The inverse of the matrix

$\left[ \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { 3 } & { 3 } & { 0 } \\ { 5 } & { 2 } & { - 1 } \end{array} \right]$ is

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$- \dfrac { 1 } { 3 } \left[ \begin{array} { c c c } { - 3 } & { 0 } & { 0 } \\ { 3 } & { 1 } & { 0 } \\ { 9 } & { 2 } & { - 3 } \end{array} \right]$
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$- \dfrac { 1 } { 3 } \left[ \begin{array} { c c c } { - 3 } & { 0 } & { 0 } \\ { 3 } & { - 1 } & { 0 } \\ { - 9 } & { - 2 } & { 3 } \end{array} \right]$
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$- \dfrac { 1 } { 3 } \left[ \begin{array} { c c c } { 3 } & { 0 } & { 0 } \\ { 3 } & { - 1 } & { 0 } \\ { - 9 } & { - 2 } & { 3 } \end{array} \right]$
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$- \dfrac { 1 } { 3 } \left[ \begin{array} { c c c } { - 3 } & { 0 } & { 0 } \\ { - 3 } & { - 1 } & { 0 } \\ { - 9 } & { - 2 } & { 3 } \end{array} \right]$

$A$ is a

$2\times 2$ matrix such that

$A^{2}-4A+3I=0$ , then the inverse of

$A+3I$ is equal to

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$\dfrac{1}{24}S-\dfrac{7}{24}I$
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$\dfrac{1}{21} A-\dfrac{7}{21}I$
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$\dfrac{7}{24}I+\dfrac{1}{24}A$
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$A-3I$ `

Let A be a

$3\times 3$ matrix such that

$A\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} \right]$ Then

${ A }^{ -1 }$ .

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$\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 2 & 3 \end{matrix} \right]$
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$\left[ \begin{matrix} 3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1 \end{matrix} \right]$
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$\left[ \begin{matrix} 3 & 2 & 1 \\ 3 & 2 & 0 \\ 1 & 1 & 0 \end{matrix} \right]$
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$\left[ \begin{matrix} 0 & 1 & 3 \\ 0 & 2 & 3 \\ 1 & 1 & 1 \end{matrix} \right]$

Inverse of

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

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$\begin{bmatrix} 2/13 & -5/13 \\ 3/13 & -1/13 \end{bmatrix}$
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$\begin{bmatrix} -2/13 & 5/13 \\ -3/13 & 1/13 \end{bmatrix}$
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$\begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix}$
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$Cannot\ be\ determined$

$A=\left[ \begin{matrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{matrix} \right] { A }^{ 1 }$ is given by:

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$-A$
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${ A }^{ 1 }$
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${ -A }^{ 1 }$
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$A$

$\begin{bmatrix} 1 & -\tan { \theta } \\ \tan { \theta } & 1 \end{bmatrix}{ \begin{bmatrix} 1 & -\tan { \theta } \\ \tan { \theta } & 1 \end{bmatrix} }^{ -1 }={ \begin{bmatrix} \cos { \alpha } & -\sin { \alpha } \\ \sin { \alpha } & \cos { \alpha } \end{bmatrix} }^{ -1 }$ , then

$\alpha$ =

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$0$
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$\frac { \pi }{ 2 }$
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$\frac { \pi }{ 4 }$
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$\frac { \pi }{ 6 }$

A is a

$2\times 2$ matrix such that

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

${ A }^{ 2 }\begin{bmatrix} 1 & \\ & -1 \end{bmatrix}=\begin{bmatrix} 1 & \\ 0 & \end{bmatrix}$ . The sum of the elements f A, is

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-1
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0
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2
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5

Let A be a

$3 \times 3$ matrix such that is:

$A\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{matrix} \right]=\left[ \begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} \right]$ Then

$A^{-1}$ is

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$\left[ \begin{matrix} 0 & 1 & 3 \\ 0 & 2 & 3 \\ 1 & 1 & 1 \end{matrix} \right]$
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$\left[ \begin{matrix} 3 & 2 & 1 \\ 3 & 2 & 0 \\ 1 & 1 & 0 \end{matrix} \right]$
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$\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 2 & 3 \end{matrix} \right]$
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$\left[ \begin{matrix} 3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1 \end{matrix} \right]$

$A=[x \quad y],B=[_{ h } ^{ a }\quad _{ b } ^{ h }],C=[_{ y } ^{ x }]$ , then

$ABC =$ ___.

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$({ ax }+hy+{ bxy })$
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$({ ax }^{ 2 }+2hxy+{ by }^{ 2 })$
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$({ ax }^{ 2 }-2hxy+{ by }^{ 2 })$
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$({ bx }^{ 2 }-2hxy+{ ay }^{ 2 })$

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

$B=\begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$ then the value of

$\alpha$ for which

$A^2=b$ is

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1
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-1
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4
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no real value

If a , b and c are all different from zero such that

$\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 0$ , then the matrix A =

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

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symmetric
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non-singular
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can be written as sum of a symmetric and a skew symmetric matrix
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none of these

$\displaystyle \begin{bmatrix} a & b \\ -a & 2b \end{bmatrix}\left[ \begin{matrix} 2 \\ -1 \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 4 \end{matrix} \right]$ then

$(a,b)$ is

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$(1,-3)$
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$(-3,1)$
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$(1,3)$
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$(-1,3)$

Matrix A when multiplied with Matrix C gives the Identity matrix I, what is C?

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Identity matrix
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Inverse of A
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Square of A
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Transpose of A

For each real

$x, -1 < x < 1 .$ let A (x) be the matrix

$(1-x)^{-1} \begin{bmatrix} 1 & -x \\ -x & 1 \end{bmatrix}$ and z =

$\dfrac { x +y }{ 1 +xy}$ . then

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$A (z) = A(x) A(y)$
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$A (z) = A(x) - A(y)$
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$A(z) = A(x) + A(y )$
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$A(z) = A(x) [ A (y)]^{-1}$

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

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

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