If $$A = \begin{bmatrix} 4& -2\\ 6 & -3\end{bmatrix}$$, then $$A^2$$ is

$$\begin{bmatrix} 16 & 4\\ 36 & 9\end{bmatrix}$$

$$\begin{bmatrix} 8 & -4\\ 12 & -6\end{bmatrix}$$

MCQ Questions for Class 12 Commerce Maths Matrices Quiz 8

$A$ is a scalar matrix

$kI$ with scalar

$k\ne 0$ of order

$3$ , the

${A}^{-1}$ is:

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$\cfrac { 1 }{ { k }^{ 2 } } I$
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$\cfrac { 1 }{ { k }^{ 3 } } I$
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$\cfrac { 1 }{ { k}^{ } } I$
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$kI$

Explanation

$A$ is a scalar matrix with scalar

$k$ , then

$A=kI$

Thus

$A^{-1}=(kI)^{-1}=\dfrac{1}{k}I$

Note: Inverse of identity matrix is Identity matrix itself

And

$(kA)^{-1}=\dfrac{1}{k}I^{-1}=\dfrac1k I$

$A=\begin{bmatrix} 7 & 2 \\ 1 & 3 \end{bmatrix}$ and

$A+B=\begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix},$ then matrix

$B=$ ?

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$\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$
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$\begin{bmatrix} 6 & 2 \\ 3 & -1 \end{bmatrix}$
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$\begin{bmatrix} -8 & -2 \\ 1 & -7 \end{bmatrix}$
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$\begin{bmatrix} 8 & 2 \\ -1 & 7 \end{bmatrix}$

Explanation

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

$A= \begin{bmatrix} 7& 2 \\ 1& 3 \end{bmatrix}$

As we have learnt that we can subtract two matrices of the same order.

(Here, order of both the matrix is

$2 \times 2$ )

So, we will subtract matrix

$A$ from both sides of the equation.

We get,

$B= \begin{bmatrix} -1 & 0 \\ 2& -4 \end{bmatrix} - \begin{bmatrix} 7 & 2 \\ 1& 3 \end{bmatrix}$

$B = \begin{bmatrix} -1-7 & 0-2 \\ 2-1 & -4-3 \end{bmatrix}$

(To subtract 2 matrices, we have to subtract their corresponding elements.)

$B= \begin{bmatrix} -8 & -2 \\ 1 & -7 \end{bmatrix}$

Hence, Option C is the correct answer

$A = \begin{bmatrix} 4& -2\\ 6 & -3\end{bmatrix}$ , then

$A^2$ is

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$\begin{bmatrix} 16 & 4\\ 36 & 9\end{bmatrix}$
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$\begin{bmatrix} 8 & -4\\ 12 & -6\end{bmatrix}$
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$\begin{bmatrix} -4 & 2\\ -6 & 3\end{bmatrix}$
0%
$\begin{bmatrix} 4 & -2\\ 6 & -3\end{bmatrix}$

Explanation

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

${ A }^{ 2 }=\begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}\times\begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}$

using the multiplication properties of matrices, we get

${ A }^{ 2 }=\begin{bmatrix} 4\times4+(-2)\times6\quad \quad\quad\quad & 4\times(-2)+(-2)\times(-3) \\ 6\times4+(-3)\times6\quad \quad \quad \quad & 6\times(-2)+(-3)\times(-3) \end{bmatrix}$

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

$A+B=\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$ and

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

$B$ is

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$\begin{bmatrix} 1 & 1 \\ 4 & 2 \end{bmatrix}$
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$\begin{bmatrix} 1 & 4 \\ 1 & 2 \end{bmatrix}$
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$\begin{bmatrix} 2 & 4 \\ 1 & 1 \end{bmatrix}$
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$\begin{bmatrix} 4 & 2 \\ 1 & 1 \end{bmatrix}$

Explanation

As given

$A+B=\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$ and

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

Substitute

$A$ in

$A+B$ we get,

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

$\implies$

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

$\implies$

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

$\therefore$

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

Hence, option A is correct.

$\bigl(\begin{smallmatrix} 1& 2\\ 2 & 1\end{smallmatrix}\bigr) \bigl(\begin{smallmatrix} x \\ y \end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} 2 \\ 4 \end{smallmatrix}\bigr)$ , then the values of

$x$ and

$y$ respectively, are

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

Explanation

$\left[ \begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 4 \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} x & 2y \\ 2x & y \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 4 \end{matrix} \right]$

$\Rightarrow x+2y=2\longrightarrow (1)\;\; \& \;\;2x+y=4\longrightarrow (2)$

Multiply

$(1)$ by 2,$$

$\Rightarrow 2x+4y=4\longrightarrow(3)$

Solving

$(2)\;\&\;(3),[(2)-(3)],$ we get

$\Rightarrow -3y=0$

$\Rightarrow y=0$

Put

$y=0$ in

$(1),$ we get

$\Rightarrow x+0=2$

$\Rightarrow x=2$

$\Rightarrow y=0$

Hence, the answer is

$2,0.$

$\begin{bmatrix} 5 & x & 1 \end{bmatrix} \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} = (20)$ , then the value of

$x$ is

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$7$
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$-7$
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$\dfrac{1}{7}$
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$0$

Explanation

Given expression:

${ \begin{bmatrix} 5 &x&1\end{bmatrix} }_{ 1\times 3 }\times { \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} }_{ 3\times 1 }={ \begin{pmatrix} 20 \end{pmatrix} }_{ 1\times 1 }\\\Rightarrow { (10+(-x)+3) }_{ 1\times 1 }={ \begin{pmatrix} 20 \end{pmatrix} }_{ 1\times 1 }$

$\Rightarrow 13-x=20$ [by equality of matrix]

$\Rightarrow x=-7$

$A = \begin{bmatrix}7 &2 \\ 1 & 3\end{bmatrix}$ and

$A + B = \begin{bmatrix} -1& 0\\ 2 & -4\end{bmatrix}$ , then the matrix

$B =$

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$\left[\begin{matrix}1 &0 \\ 0 & 1\end{matrix}\right]$
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$\left[\begin{matrix} 6&2 \\ 3 & -1\end{matrix}\right]$
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$\left[\begin{matrix} -8& -2\\ 1 & -7\end{matrix}\right]$
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$\left[\begin{matrix} 8& 2\\ -1 & 7\end{matrix}\right]$

Explanation

Given,

$A=\begin{bmatrix} 7 & 2 \\ 1 & 3 \end{bmatrix}$

Also,

$A+B=\begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix}\\ \Rightarrow B=\begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix}-\begin{bmatrix} 7 & 2 \\ 1 & 3 \end{bmatrix}\\ \Rightarrow B=\begin{bmatrix} -1-7 & 0-2 \\ 2-1 & -4-3 \end{bmatrix}\\ \Rightarrow B=\begin{bmatrix} -8 & -2 \\ 1 & -7 \end{bmatrix}$

$\Rightarrow B=\begin{bmatrix} -8 & -2 \\ 1 & -7 \end{bmatrix}$

Choose the correct statement related to the matrices

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

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

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$A^3=A, B^3 \ne B$
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$A^3\ne A, B^3=B$
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$A^3=A, B^3 = B$
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$A^3\ne A, B^3 \ne B$

Explanation

Clearly,

$A$ is an identity matrix

Now,

${ A }^{ 3 }=$

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

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

$\implies A^3=A$

Also,

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

$\implies B^{3}=B$

Hence, option C is correct.

$A = \bigl(\begin{smallmatrix} 4& -2\\ 6 & -3\end{smallmatrix}\bigr)$ , then

$A^2$ is

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$\bigl(\begin{smallmatrix} 16 & 4\\ 36 & 9\end{smallmatrix}\bigr)$
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$\bigl(\begin{smallmatrix}8 & -4\\ 12 & -6\end{smallmatrix}\bigr)$
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$\bigl(\begin{smallmatrix} -4& 2\\ -6 & 3\end{smallmatrix}\bigr)$
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$\bigl(\begin{smallmatrix} 4& -2\\ 6 & -3\end{smallmatrix}\bigr)$

Explanation

$A=\left[ \begin{matrix} 4 & -2 \\ 6 & -3 \end{matrix} \right]$

$\Rightarrow { A }^{ 2 }=A.A=\left[ \begin{matrix} 4 & -2 \\ 6 & -3 \end{matrix} \right] \left[ \begin{matrix} 4 & -2 \\ 6 & -3 \end{matrix} \right]$

$=\left[ \begin{matrix} 16-12 & -8+6 \\ 24-18 & -12+9 \end{matrix} \right]$

$=\left[ \begin{matrix} 4 & -2 \\ 6 & -3 \end{matrix} \right]$

Hence, the answer is

$\left[ \begin{matrix} 4 & -2 \\ 6 & -3 \end{matrix} \right] .$

$\bigl(\begin{smallmatrix} -1& 0\\ 0 & 1\end{smallmatrix}\bigr) \bigl(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} 1& 0\\ 0 & -1\end{smallmatrix}\bigr)$ , then the values of

$a, b, c$ and

$d$ respectively are

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$-1, 0, 0, -1$
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$1, 0, 0, 1$
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$-1, 0, 1, 0$
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$1, 0, 0, 0$

Explanation

$\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \end{matrix} \right] \left[ \begin{matrix} a & b \\ c & d \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} -a & -b \\ c & d \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right]$

On comparing both sides, corresponding columns,

$\Rightarrow -a=1,$

$-b=0$ and

$c=0$

$a=-1,$

$b=0$ and

$d=-1$

Hence, the answer is

$-1,0,0,-1.$

$\bigl(\begin{smallmatrix}a & 3\\ 1 & 2\end{smallmatrix}\bigr) \bigl(\begin{smallmatrix} 2 \\ -1 \end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} 5\\ 0 \end{smallmatrix}\bigr)$ , then the value of

$a$ is

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$8$
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$4$
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$2$
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$11$

Explanation

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

$\Rightarrow \left[ \begin{matrix} 2a-3 \\ 0 \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 0 \end{matrix} \right]$

$\Rightarrow 2a-3=5$

$\Rightarrow 2a=8$

$\Rightarrow a=4$

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

$B=\left[ \begin{matrix} -1 \\ 2 \\ -3 \end{matrix} \right]$ , then

$A + B$ is

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$\begin{bmatrix} 0& 0 & 0 \end{bmatrix}$
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$\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$
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$\begin{bmatrix} -1& 4 \end{bmatrix}$
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not defined

Explanation

Given,

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

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

Here

$A+B$ is not possible because the order of the matrices is not same.

Hence, the answer is not defined.

If matrix

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

$Ax=I$ , then

$x=$ ................

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

Explanation

Given :

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

$Ax=I$ ...

$(i)$

Since,

$A$ and

$I$ are of order

$2\times 2$ . So,

$x$ will be a matrix of order

$2\times 2$

Let

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

From

$(i)$ , we get

$Ax=I$

$\implies \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\implies \begin{bmatrix} a+2c & b+2d \\ 4a+3c & 4b+3d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\implies a+2c=1$ ......

$(ii),$

$4a+3c=0$ ........

$(iii)$

$b+2d=0$ .......

$(iv)$ and

$4b+3d=1$ .......

$(v)$

Solving

$(ii)$ and

$(iii)$ simultaneously we get

$a=-\dfrac{3}{5}$ and

$c=\dfrac{4}{5}$

Then solving

$(iv)$ and

$(v)$ simultaneously we get

$b=\dfrac{2}{5}$ and

$d=-\dfrac{1}{5}$

Substituting all these values in

$x$ , we get

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

$\therefore x=\dfrac{1}{5} \begin{bmatrix} -3 & 2 \\ 4 & -1 \end{bmatrix}$

The matrix

$A = \begin{bmatrix} 0& 1 & -1\\ -1 & 0 & 1\\ 1 & -1 & 0\end{bmatrix}$ is a :

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Diagonal matrix
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Symmetric matrix
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Skew-symmetric matrix
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Identity matrix

Explanation

The matrix

$A=\quad \begin{bmatrix} 0 & \quad \quad 1 & \quad \quad -1 \\ -1 & \quad \quad 0 & \quad \quad 1 \\ 1 & \quad \quad -1 & \quad \quad 0 \end{bmatrix}$ is a

${ \quad A }^{ T\quad }=\quad \begin{bmatrix} 0 & \quad -1 & \quad \quad 1 \\ 1 & \quad \quad 0 & \quad \quad -1 \\ -1 & \quad \quad 1 & \quad \quad 0 \end{bmatrix}$

$Hence \ \ \ { -A }^{ T }\quad =\quad \begin{bmatrix} 0 & \quad \quad 1 & \quad \quad -1 \\ -1 & \quad \quad 0 & \quad \quad 1 \\ 1 & \quad \quad -1 & \quad \quad 0 \end{bmatrix}\quad \\ \\ \quad A\quad =\quad { -A }^{ T }$
The matrix A is skew symmetric matrix

The inverse of a diagonal matrix is a :

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Symmetric matrix
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Skew-symmetric matrix
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Diagonal matrix
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None of the above

$\begin{bmatrix} 3 & -1 \\ 0 & 6 \end{bmatrix}\begin{bmatrix} 3x \\ 1 \end{bmatrix}+\begin{bmatrix} -2x \\ 3 \end{bmatrix}=\begin{bmatrix} 8 \\ 9 \end{bmatrix}$ , then the value of

$x$ is

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$-\dfrac { 3 }{ 8 }$
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$7$
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$-\dfrac { 2 }{ 9 }$
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None of these

Explanation

Given,

$\begin{bmatrix} 3 & -1 \\ 0 & 6 \end{bmatrix}\begin{bmatrix} 3x \\ 1 \end{bmatrix}+\begin{bmatrix} -2x \\ 3 \end{bmatrix}=\begin{bmatrix} 8 \\ 9 \end{bmatrix}$
LHS

$=\begin{bmatrix} 3 & -1 \\ 0 & 6 \end{bmatrix}\begin{bmatrix} 3x \\ 1 \end{bmatrix}+\begin{bmatrix} -2x \\ 3 \end{bmatrix}$

$=\begin{bmatrix} 9x+\left( -1 \right) \\ 0+6 \end{bmatrix}+\begin{bmatrix} -2x \\ 3 \end{bmatrix}=\begin{bmatrix} 9x-1 \\ 6 \end{bmatrix}+\begin{bmatrix} -2x \\ 3 \end{bmatrix}$

$=\begin{bmatrix} 9x-1-2x \\ 6+3 \end{bmatrix}=\begin{bmatrix} 7x-1 \\ 9 \end{bmatrix}$
Now,

$\begin{bmatrix} 7x-1 \\ 9 \end{bmatrix}=\begin{bmatrix} 8 \\ 9 \end{bmatrix}$

$\therefore 7x-1=8$

$\Rightarrow 7x=9$

$\Rightarrow x=\dfrac { 9 }{ 7 }$ .

Let

$A = \begin{bmatrix}x + y & y\\ 2x & x - y\end{bmatrix}, B = \begin{bmatrix} 2& -1\end{bmatrix}$ and

$C = \begin{bmatrix} 3& 2\end{bmatrix}.$ If

$AB = C$ , then

$A^{2}$ is equal to

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$\begin{bmatrix} 6 & -10\\ 4 & 26 \end{bmatrix}$
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$\begin{bmatrix} -10 & 5\\ 4 & 24 \end{bmatrix}$
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$\begin{bmatrix} -5 & -6\\ -4 & -20 \end{bmatrix}$
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None of these.

Explanation

$A=\begin{bmatrix} x+y & y \\ 2x & x-y \end{bmatrix}, B=\begin{bmatrix} 2 & -1 \end{bmatrix}$ and

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

Since,

$AB=C\\ \Longrightarrow \begin{bmatrix} x+y & y \\ 2x & x-y \end{bmatrix}\begin{bmatrix} 2 & -1 \end{bmatrix}=\begin{bmatrix} 3 & 2 \end{bmatrix}$

The multiplication of

$A$ and

$B$ can't be done because it doesn't satisfy the conditions of matrix multipication

If the matrix A is such that

$\begin{bmatrix} 1 & 3\\ 0 & 1\end{bmatrix} A=\begin{bmatrix} 1 & 1 \\ 0 & -1\end{bmatrix}$ , then what is equal to A?

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

Explanation

Given

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

Let

$A=\begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}$

$\begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix}$

$\begin{bmatrix} \alpha +3\gamma & \beta +4\delta \\ \gamma & \delta \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix}$

Two matrices are equal if corresponding elements are equal

$\Rightarrow \alpha +3\gamma =1, \gamma =0,$

$\beta +3\delta =1$ and

$\delta =-1$

$\Rightarrow \alpha =1,$

$\beta =4$

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

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

$B=\begin{bmatrix}-1&-2&-1\\6&12&6\\5&10&5\end{bmatrix}$ , then which of the following is/are correct?
A and B commute.
AB is null matrix.
Select the correct answer using the code given below :

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1 only
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2 only
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Both 1 and 2
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Neither 1 nor 2

Explanation

We have,

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

$B=\begin{bmatrix}-1&-2&-1\\6&12&6\\5&10&5\end{bmatrix}$

$AB=\begin{bmatrix}-1+6-5&-2+12-10&-1+6-5\\-2-18+20&-4-36+40&-2-18+20\\-3-12+15&-6-24+30&-3-12+15\end{bmatrix}$

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

Hence,

$AB$ is a null matrix.

and

$AB\neq BA$

Hence, B is the correct option.

$\begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix}\times \begin{pmatrix} 5 & -2 \\ -3 & 1 \end{pmatrix}=\begin{pmatrix} 1 & -1 \\ 17 & \lambda \end{pmatrix}$ then what is

$\lambda$ equal to?

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$7$
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$-7$
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$9$
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$-9$

Explanation

GIven,

$\begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix}\times \begin{pmatrix} 5 & -2 \\ -3 & 1 \end{pmatrix}=\begin{pmatrix} 1 & -1 \\ 17 & \lambda \end{pmatrix}$

Now,

$\begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix}\times \begin{pmatrix} 5 & -2 \\ -3 & 1 \end{pmatrix}=\begin{pmatrix} 10-9 & -4+3 \\ 20-3 & -8+1 \end{pmatrix}=\begin{pmatrix} 1 & -1 \\ 17 & -7 \end{pmatrix}$

So comparing both, we get

$\lambda =-7$

$\quad A=\begin{pmatrix} 1 & 3 \\ 4 & 5 \end{pmatrix}$ then

${ A }^{ -1 }$ equals

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$\cfrac { 1 }{ 7 } \left( A+6I \right)$
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$\cfrac { 1 }{ 7 } \left( A-6I \right)$
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$\cfrac { 1 }{ 7 } \left( 6I-A \right)$
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None of these

Explanation

$A=\begin{pmatrix} 1 & 3 \\ 4 & 5 \end{pmatrix}$
for every square matrix

$A$ , the characterisitc of

$A$ is given

$\left| A-\lambda I \right| =0$

$\Rightarrow \begin{vmatrix} 1-\lambda & 3 \\ 4 & 5-\lambda \end{vmatrix}=0$

$\Rightarrow { \lambda }^{ 2 }-6\lambda -7=0\quad \Rightarrow { A }^{ 2 }-6A-7I=0\Rightarrow 7I={ A }^{ 2 }-6A\quad \quad$

$\Rightarrow { A }^{ -1 }=1/7(A-6I)\quad$

$[1\,x\,1] \begin{bmatrix} 1&3&2 \\ 0&5&1\\0&2&0 \end{bmatrix}$

$\begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix}=0$ , then the values of

$x$ are:

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$1,5$
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$-1,-5$
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$1,6$
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$-1,-6$
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$3,3$

Explanation

Since,

$[1\,x\,1] \begin{bmatrix} 1&3&2 \\ 0&5&1\\0&2&0 \end{bmatrix}$

$\begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix}=0$

$\Rightarrow \begin{bmatrix} 1 & 3+5x+2 & 2+x+0 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix}=0$

$\Rightarrow \begin{bmatrix} 1 &5+5x & 2+x \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix}=0$

$\Rightarrow \left[ 1+5+5x+2x+{ x }^{ 2 } \right] =0$

$\Rightarrow { x }^{ 2 }+7x+6=0$

$\Rightarrow x^2+6x+x+6=0$

$\Rightarrow (x+1)(x+6)=0$

$\therefore x=-1,-6$

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

$A^{3} + 3A^{2} - 4A + 1$ is equal to

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$\begin{bmatrix} 1& 1\\ 1 & 0\end{bmatrix}$
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$\begin{bmatrix} -14& 51\\ -17 & -14\end{bmatrix}$
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$\begin{bmatrix} -14& -51\\ -17 & -14\end{bmatrix}$
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$\begin{bmatrix} -1& -1\\ -1 & 0\end{bmatrix}$

Explanation

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

$\therefore A^{2} = \begin{pmatrix} 1& 12\\ -4 & 1\end{pmatrix}, A^{3} = \begin{pmatrix} -10& 27\\ -9 & -10\end{pmatrix}$

$\therefore A^{3} + 3A^{2} - 4A + t = \begin{pmatrix} -10& 27\\ -9 & -10\end{pmatrix} + \begin{pmatrix} 3& 36\\ -12 & 3\end{pmatrix} + \begin{pmatrix} -8& -12\\ 4 & -8\end{pmatrix} + \begin{pmatrix} 1& 0\\ 0 & 1\end{pmatrix}$

$= \begin{pmatrix} -14& 51\\ -17 & -14\end{pmatrix}$
Hence (b) is correct choice.

$A=\begin{pmatrix} 3 & 1 \\ -9 & -3 \end{pmatrix}$ then

${ \left( 1+2A+3{ A }^{ 2 }+....\infty \right) }^{ -1 }$ equals

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$\begin{pmatrix} -5 & -2 \\ 18 & 7 \end{pmatrix}$
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$\begin{pmatrix} -5 & 18 \\ -2 & 7 \end{pmatrix}$
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$\begin{pmatrix} 7 & -2 \\ 18 & -5 \end{pmatrix}$
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None of these

Explanation

$A=\begin{pmatrix} 3 & 1 \\ -9 & -3 \end{pmatrix}$

$\quad \therefore { A }^{ 2 }=\begin{pmatrix} 3 & 1 \\ -9 & -3 \end{pmatrix}\begin{pmatrix} 3 & 1 \\ -9 & -3 \end{pmatrix}=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$

$\therefore 1+2A+3{ A }^{ 2 }+....\infty =1+2A$

$=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}+\begin{pmatrix} 6 & 2 \\ -18 & -6 \end{pmatrix}=\begin{pmatrix} 7 & 2 \\ -18 & -5 \end{pmatrix}$

$\therefore (1+2A+3{ A }^{ 2 }+....\infty)^{-1}=(1+2A)^{-1}$

$\quad ={ \begin{pmatrix} 7 & 2 \\ -18 & -5 \end{pmatrix} }^{ -1 }=\begin{pmatrix} -5 & -2 \\ 18 & 7 \end{pmatrix}$

$\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$ to the square is two rowed unit matrix, then

$\alpha ,\beta ,\gamma$ should satisfy the relation

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$1+{ \alpha }^{ 2 }+\beta \gamma =0$
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$1-{ \alpha }^{ 2 }-\beta \gamma =0$
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$1-{ \alpha }^{ 2 }+\beta \gamma =0$
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${ \alpha }^{ 2 }+\beta \gamma -1=0$

Explanation

since

$\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$ is a square root of

${I}_{2}$ i.e., two rowed unit matrix

$\therefore { \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} }^{ 2 }=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} { \alpha }^{ 2 }+\beta \gamma & 0 \\ 0 & { \alpha }^{ 2 }+\beta \gamma \end{bmatrix}$

$\therefore { \alpha }^{ 2 }+\beta \gamma =1$

$\Rightarrow 1-{ \alpha }^{ 2 }-\beta \gamma =0$

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

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

$\alpha$ for which

${A}^{2}=B$ , is

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$1$
0%
$-1$
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$4$
0%
No real value

Explanation

Given :

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

$B=\begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$

${ A }^{ 2 }=\begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} { \alpha }^{ 2 } & 0 \\ \alpha +1 & 1 \end{bmatrix}$

Now

${ A }^{ 2 }=B\quad$

$\begin{bmatrix} { \alpha }^{ 2 } & 0 \\ \alpha +1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$

$\Rightarrow { \alpha }^{ 2 }=1$ and

$\alpha +1=5$

Clearly, no real value of

$\alpha$ exist.

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

$A^{2} - kA - 5I_{2} = 0$ , then the value of

$k$ is

0%
$3$
0%
$5$
0%
$7$
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$-7$

Explanation

Given,

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

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

$= \begin{bmatrix}1 + 9 & 3 + 12 \\ 3 + 12 & 9 + 16\end{bmatrix} - \begin{bmatrix}5 &0 \\ 0 & 5\end{bmatrix}$

$= \begin{bmatrix}5 &15 \\ 15 & 20\end{bmatrix} = 5\begin{bmatrix}1 &3 \\ 3 & 4\end{bmatrix}$

$\Rightarrow A^{2} - 5I_{2} = 5A$
But it is given that

$A^{2} - 5I_{2} = kA$

$k = 5$ .

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

${ A }^{ 2 }-4A+10I=A$ , then

$k$ is equal to

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$0$
0%
$-4$
0%
$4$ and not $1$
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$1$ or $4$

Explanation

Given,

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

${ A }^{ 2 }-4A+10I=A$

$\Rightarrow \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix} \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix} -4 \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix} + 10 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix}$

$\Rightarrow \begin{bmatrix} -5 & -3-3k \\ 2+2k & -6+{ k }^{ 2 } \end{bmatrix} - \begin{bmatrix} 4 & -12 \\ 8 & 4k \end{bmatrix} + \begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix} = \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 1 & 9-3k \\ -6+2k & 4+{ k }^{ 2 }-4k \end{bmatrix} = \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix}$

$\Rightarrow 9-3k=-3, -6+2k=2$ ....(i)
and

$4+{ k }^{ 2 }-4k=k$

${ k }^{ 2 }-5k+4=0$

$\Rightarrow \left( k-4 \right) \left( k-1 \right) =0 \Rightarrow k=4,1$
But

$k=1$ is not satisfied the equation (i).

$A = \begin{vmatrix} 5 & x-2 \\ 2x+3 & x+1 \end{vmatrix}$ is symmetric, then

$x =$ _____

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$4$
0%
$5$
0%
$-5$
0%
$-4$

Explanation

$\begin {vmatrix} 5&x-2\\2x-3&x+1\end{vmatrix}$

$\implies A^T=$

$\begin {vmatrix} 5&2x-3\\x-2&x+1\end{vmatrix}$

Since

$A$ is symmetric,

$\implies A=A^T$

$\implies x-2=2x+3$

$\implies x=-5$

$A$ is a non zero square matrix of order

$n$ with

$det\left( I+A \right) \neq 0$ , and

${A}^{3}=0$ , where

$I,O$ are unit and null matrices of order

$n\times n$ respectively, then

${ \left( I+A \right) }^{ -1 }=$

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$I-A+{ A }^{ 2 }$
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$I+A+{ A }^{ 2 }$
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$I+{ A }^{ 2 }$
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$I+A$

Explanation

$det(I+A)\neq 0$

$A^3=0$ where

$0$ is null matrix,

$I$ is the identity matrix

$A^3+I=I$ [adding

$I$ on both sides]

$(A+I)(A^2-IA+I^2)=I$ [by the formula of

$a^3+b^3$ ]

$(A+I)(A^2-A+I)=I$

$(I+A)(I+A)^{-1}=I$ [by the rule of inverse matrix]

hence

$(I+A)^{-1}=(A^2-A+I)$

Ans:

$I-A+A^2$

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

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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