MCQ Questions for Class 12 Commerce Maths Matrices Quiz 5

$\mathrm{A}=\left\{\begin{array}{lll} 1 & 1 & 3\\ 5 & 2 & 6\\ -2 & -1 & -3 \end{array}\right\},$ then

$\mathrm{A}^{3}$ is a/an

0%
diagonal matrix
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square matrix
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null matrix
0%
unit Matrix

Explanation

$A^2$

$=$

$\left\{\begin{array}{lll} 1 & 1 & 3\\ 5 & 2 & 6\\ -2 & -1 & -3 \end{array}\right\}\times\left\{\begin{array}{lll} 1 & 1 & 3\\ 5 & 2 & 6\\ -2 & -1 & -3 \end{array}\right\}$

$=$

$\left\{\begin{array}{lll} 0 & 0 & 0\\ 3 & 3 & 9\\ -1 & -1 & -3 \end{array}\right\}$

$A^3$

$=$

${A^2}\times{A}$

$A^3$

$=$

$\left\{\begin{array}{lll} 0 & 0 & 0\\ 3 & 3 & 9\\ -1 & -1 & -3 \end{array}\right\}\times\left\{\begin{array}{lll} 1 & 1 & 3\\ 5 & 2 & 6\\ -2 & -1 & -3 \end{array}\right\}$

$=$

$\left\{\begin{array}{lll} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right\}$

$\begin{bmatrix} x & -3 \\ -9 & y \end{bmatrix}\begin{bmatrix} 4 & -3 \\ 9 & 7 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$ then

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

$, y$

$=$ ........

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$7,4$
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$4,7$
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$8,9$
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$9,8$

Explanation

$\begin{bmatrix} x & -3 \\ -9 & y \end{bmatrix}\begin{bmatrix} 4 & -3 \\ 9 & 7 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\\ \Rightarrow \begin{bmatrix} 4x-27 & -3x+(-21) \\ -36+9y & +27+7y \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Using conditions for equality of matrices we can write

$4x-27=1\\ \Rightarrow x=7\quad \quad -(i)\\ -(3x+21)=0\\ \Rightarrow x=-7\quad \quad -(ii)\\ -36+9y=0\\ \Rightarrow y=4\quad \quad -(iii)\\ 27+7y=1\\ \Rightarrow y=-4\quad \quad -(iv)$

Possible answer is

$x=7$ &

$y=4$

State true/false:

$A=\begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix},$ then solve is

${ A }^{ 3 }=I$ ?

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True
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False

Explanation

Given

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

Calculating

${ A }^{ 3 }$

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

${ A }^{ 2 }=\begin{bmatrix} \left( 1\times 1+2\times -1+1\times 1 \right) & \left( 1\times -1+\left( -1\times -1 \right) +\left( 1\times 0 \right) \right) & 1\times 1+-1\times 0+1\times 0 \\ 2\times 1+-1\times 2+0\times 1 & 2\times -1-1\times -1+0\times 0 & 1\times 1+1\times 0+0\times 0 \\ 1\times 1+2\times 0+0\times 1 & -1\times 1+-1\times 0+0\times 0 & 1\times 1+0\times 0+0\times 0 \end{bmatrix}$

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

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

${ A }^{ 3 }=\begin{bmatrix} 0\times 1+0\times 2+1\times 1 & 0\times -1+0\times -1+1\times 0 & 0\times 1+0\times 0+1\times 0 \\ 0\times 1+-1\times 2+2\times 1 & 0\times -1+-1\times -1+2\times 0 & 0\times 1+(-1\times 0)+2\times 0 \\ 1\times 1+2\times -1+1\times 1 & -1\times 1+(-1\times -1)+1\times 0 & 1\times 1+(-1\times 0)+(1\times 0) \end{bmatrix}$

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

$\Rightarrow { A }^{ 3 }=I$

Hence, given statement is true.

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

$B= \begin{bmatrix} 2 & 1 \\ 3 & 5 \end{bmatrix}$ and

$C= \begin{bmatrix} 0 & 1 \\ 1 & 2 \end{bmatrix},$ then

$\left ( AB \right )\times C=$

0%
$\begin{bmatrix} 8 & 11 \\ 18 & 23 \end{bmatrix}$
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$\begin{bmatrix} 11 & 30 \\ 23 & 60 \end{bmatrix}$
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$\begin{bmatrix} 17 & 47 \\ 13 & 38 \end{bmatrix}$
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$\begin{bmatrix} 0 & 1 \\ 1 &2 \end{bmatrix}$

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

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

$AB=$ is

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$\begin{bmatrix} 6 & 2 & 5 \\ 7 & 2 & 3 \end{bmatrix}$
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$\begin{bmatrix} 2 & 1 & 5 \\ 0 & 1 & 7 \end{bmatrix}$
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$\begin{bmatrix} -5 & 2 & 5 \\ 1 & 2 & 11 \end{bmatrix}$
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$\begin{bmatrix} 7 & 4 & 4 \\ 6 & 2 & 12 \end{bmatrix}$

Explanation

Given

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

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

$AB=\begin{bmatrix} 2\times 3 +1\times1 & 2\times 2+1\times0 & 2\times0+1\times4 \\ 1\times3+3\times1 & 1\times2+3\times0 & 1\times0+3\times4 \end{bmatrix}$

$=\begin{bmatrix} 6+1 & 4+0 & 0+4 \\ 3+3 & 2+0 & 0+12 \end{bmatrix}$

$=\begin{bmatrix} 7 & 4 & 4 \\ 6 & 2 & 12 \end{bmatrix}$

Therefore the correct option is

$(D)$

If A =

$\begin{bmatrix} 2\ \ \ 4 \\ 3\ \ 5 \end{bmatrix},$

$B= \begin{bmatrix} x\ \ \ y \\ 6\ \ \ 5 \end{bmatrix}$ and

$AB= \begin{bmatrix}8\ \ \ 2 \\ 6\ \ \ -2 \end{bmatrix}$
then

$x=$ ______, and

$y =$ _____.

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$- 9, -8$
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$8, 9$
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$- 8, -9$
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$9, 8$

Explanation

$A = \begin{bmatrix} 3 & 4 \\ 3 & 5 \end{bmatrix}$ ;

$B = \begin{bmatrix} x & y \\ 6 & 5 \end{bmatrix}$

and

$AB = \begin{bmatrix} 8 & 2 \\ 6 & -2 \end{bmatrix}$

Now,

$AB = \begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix} \begin{bmatrix} x & y \\ 6 & 5 \end{bmatrix} \\ = \begin{bmatrix} 2x+24 & 2y + 20 \\ 3x + 30 & 3y + 25 \end{bmatrix} \, \, \textrm{ [By the property of multiplication of matrix] }$

Now, comparing the given value we get,

$\begin{bmatrix} 8 & 2 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 2x+24 & 2y + 20 \\ 3x+30 & 3y + 25 \end{bmatrix}$

Equating each element we get,

$2x + 24 = 8 \Rightarrow x = -8 \\ 2y + 20 = 2 \\ \Rightarrow y = -9$

$\therefore$ option C is correct.

$P=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and if

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

$Q=$

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

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

$AB=$

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$\begin{bmatrix} 5 & 3 & 11 \\ 1 & 2 & 2 \\ 1 & 3 & 5 \end{bmatrix}$
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$\begin{bmatrix} 5 & 9 & 13 \\ -1 & 2 & 4 \\ -2 & 2 & 4 \end{bmatrix}$
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$\begin{bmatrix} 5 & 8 & 11 \\ 1 & 2 & 3 \\ 2 & 2 & -3 \end{bmatrix}$
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$\begin{bmatrix} 5 & 3 & 13 \\ -1 & 2 & 4 \\ -2 & 3 & -5 \end{bmatrix}$

Explanation

Given,

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

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

Then,

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

$= \begin{bmatrix} 1\times2+3\times1+0\times(-1) & 1\times3+3\times2+0\times1& 1\times4+3\times3+0\times2 \\ (-1)\times2+2\times1+1\times(-1) & (-1)\times3+2\times2+1\times1 & (-1)\times4+2\times3+1\times2 \\ 0\times2+0\times1+2\times(-1) & 0\times3+0\times2+2\times1 & 0\times4+0\times3+2\times2 \end{bmatrix}$

$= \begin{bmatrix} 5 & 9 & 13 \\ -1 & 2 & 4 \\ -2 & 2 & 4\end{bmatrix}$

Hence, the correct option is

$(B)$

$\displaystyle A=\left [ a_{ij} \right ]$ is a square matrix of even order such that

$\displaystyle \left [ a_{ij} \right ]=i^{2}-j^{2}$ , then

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$A$ is a skew-symmetric matrix and $\displaystyle \left | A \right |=0$
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$A$ is symmetric matrix and $\displaystyle \left | A \right |$ is a square
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$A$ is symmetric matrix and $\displaystyle \left | A \right |=0$
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none of these

Explanation

Given

$\displaystyle \left [ a_{ij} \right ]=i^{2}-j^{2}$

$\Rightarrow a_{ii}=0$ for all i.

Also, for

$i \ne j, a_{ij}=i^{2}-j^{2}=-(j^{2}-i^{2})=-a_{ji}$

$\Rightarrow a_{ij}=-a_{ji}$

Hence, A is skew-symmetric matrix.
For order 2,

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

$|A|=9\ne 0$

Hence, A is a skew-symmetric matrix and |A| is a square

$A=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}, \text{then} \,A^{3}-4A^{2}-6A=$

0%
$0$
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$A$
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$-A$
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$I$

Explanation

$A^{2}=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}=\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}$

$A^{3}=A^{2}\times A=\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}=\begin{bmatrix} 41 & 42 & 42 \\ 42 & 41 & 42 \\ 42 & 42 & 41 \end{bmatrix}$

$A^{3}-4A^{2}-6A=\begin{bmatrix} 41 & 42 & 42 \\ 42 & 41 & 42 \\ 42 & 42 & 41 \end{bmatrix}-\begin{bmatrix} 36 & 32 & 32 \\ 32 & 36 & 32 \\ 32 & 32 & 36 \end{bmatrix}\begin{bmatrix} 6 & 12 & 12 \\ 12 & 6 & 12 \\ 12 & 12 & 6 \end{bmatrix}$

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

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

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

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$\begin{bmatrix} -cos2x & sin2x \\ -sin2x & cos2x \end{bmatrix}$
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$\begin{bmatrix} cos2x & -sin2x \\ sin2x & cos2x \end{bmatrix}$
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$\begin{bmatrix} cos2x & cos2x \\ cos2x & sin2x \end{bmatrix}$
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none of these

Explanation

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

$\Rightarrow |A|=\sec^{2}x$

$adj A=C^{T}={\begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}}^{T}$

$\Rightarrow adj A=\begin{bmatrix} \cos ^{ 2 }{ x } & -\sin { x } \cos { x } \\ \sin { x } \cos { x } & \cos ^{ 2 }{ x } \end{bmatrix}$

$A^{-1}=\displaystyle \frac {adj A}{|A|}$

$=\begin{bmatrix} \cos ^{ 2 }{ x } & -\sin { x } \cos { x } \\ \sin { x } \cos { x } & \cos ^{ 2 }{ x } \end{bmatrix}$

Now,

$A^{T}A^{-1}=\begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \begin{bmatrix} \cos ^{ 2 }{ x } & -\sin { x } \cos { x } \\ \sin { x } \cos { x } & \cos ^{ 2 }{ x } \end{bmatrix}$

$\Rightarrow A^{T}A^{-1}=\begin{bmatrix} \cos { 2x } & -\sin { 2x } \\ \sin { 2x } & \cos { 2x } \end{bmatrix}$

$\begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix}A=\begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}$ , then sum of all the elements of matrix

$A$ is

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

Explanation

$\begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix}A=\begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}$
let

$A=\begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix}\begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix}=\begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 2a-d & 2b-e & 2c-f \\ a & b & c \\ -3a+4d & -3b+4e & -3c+4f \end{bmatrix}=\begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}$

$\Rightarrow 2a-d=-1,2b-e=-8,2c-f=-10$ -----(1)
and

$a=1,b=-2,c=-5$ ----(2)
from (1) and (2)

$a+b+c=-6$ and

$d+e+f=7$

$\therefore$ Sum of all the elements in the matrix

$a+b+c+d+e+f=1$
Hence, option B.

Let

$A=\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\ and\ B=\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}, a,b\in N.$ Then:

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there exists exactly one B such that $AB=BA$
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there exist exactly infinitely many B's such that $AB=BA$
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there cannot exist any B such that $AB=BA$
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there exist more than one but finite number of B's such that $AB=BA$

Explanation

Given

$A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}B=\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}$

Finding AB

$AB=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}=\begin{bmatrix} a+2\times 0 & 1\times 0+2\times b \\ 3\times a+4\times 0 & 3\times 0+4\times b \end{bmatrix}$

$AB=\begin{bmatrix} a & 2b \\ 3a & 4b \end{bmatrix}$

Finding BA

$BA=\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}=\begin{bmatrix} a+0\times 3 & 2a+0\times 4 \\ 0\times 1+3b & 0\times 2+4b \end{bmatrix}$

$BA=\begin{bmatrix} a & 2a \\ 3b & 4b \end{bmatrix}$

$AB=BA$

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

Comparing each element we get

$a=a,2b=2a$

$\Rightarrow a=b$

$\therefore$ There are infinitely many b's

for which

$AB=BA$

Answer=B

$\begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}$ and

$A^{2}=I$ , then

$x=$

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

Explanation

Given

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

Also,

$A^2=I$

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

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

For the above two matrices to be equal each of the first matrix must be equal to the corresponding element of the second matrix.

Therefore,

$x^2+1=1 ...(i)$ and

$x=0...(ii)$ .

From equation

$(i)$ and equation

$(ii)$ we get,

$x=0$

Hence, the correct option is

$(A)$

$[1\ 2\ 3] B = [3\ 4],$ then order of the matrix

$B$ is

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$3 \times 1$
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$1 \times 3$
0%
$2 \times 3$
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$3 \times 2$

Explanation

Given,

$[1 \ 2\ 3]$

$B = [3 \ 4]$
or

${[1 \ 2\ 3]}$

$_{1\times3}$

$B={[3\ 4]}$

$_{1\times2}$
So, order of

$B$ should be

$3\times2.$
Hence, option 'D' is correct.

The inverse of the matrix

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

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

Explanation

Given,

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

$|A|=5$

Now,

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

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

$A^{-1}=\displaystyle \frac{1}{5}\begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}$

$\displaystyle \:A= \left [ \begin{matrix}1 &2 &x \\0 &1 &0 \\0 &0 &1 \end{matrix} \right ]and \ B\left [ \begin{matrix}1 &-2 &y \\0 &1 &0 \\0 &0 &1 \end{matrix} \right ]$ and

$\displaystyle \:AB= I,$ then

$x+y$ equals

0%
$0$
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$-1$
0%
$2$
0%
none of these

Explanation

$\displaystyle \:A= \left [ \begin{matrix}1 &2 &x \\0 &1 &0 \\0 &0 &1 \end{matrix} \right ]$ and

$B=\left [ \begin{matrix}1 &-2 &y \\0 &1 &0 \\0 &0 &1 \end{matrix} \right ]$
Also given,

$\displaystyle \:AB= I$

$\left[ \begin{matrix} 1 & 2 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & -2 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] =\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

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

$\Rightarrow x+y=0$
Hence, option 'A' is correct.

Let A be a square matrix. Which of the following is/are not symmetric matrix/matrices?

0%
$A+A^{T}$
0%
$AA^{T}$
0%
$A-A^{T}$
0%
$A^{T}A$

Explanation

We know a square matrix

$A$ is called symmetric if

$A^T = A$
. Let

$P = A-A^T \Rightarrow P^T = (A-A^T)^T=A^T-A=-P\Rightarrow$ This is not a symmetric matrix, it is skew symmetric matrix.
All other matrices in the other option are symmetric.

$A = \begin{bmatrix}1 & 2 & 2\\ 2 & 1 & -2\\ a & 2 & b\end{bmatrix}$ is a matrix satisfying

$AA^T = 9 I_3$ , then the values of

$a$ and

$b$ are

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$a = -2, b = - 1$
0%
$a = -2, b = 1$
0%
$a = 2, b = - 1$
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No values of a,b satisfy given conditions

Explanation

We have,

$A = \begin{bmatrix}1 & 2 & 2\\ 2 & 1 & -2\\ a & 2 & b\end{bmatrix} \Rightarrow A^T = \begin{bmatrix}1 & 2 & a\\ 2 & 1 & 2\\ 2 & -2 & b\end{bmatrix}$

$\therefore AA^T = 9I_3$

$\Rightarrow \begin{bmatrix}1 & 2 & 2\\ 2 & 1 & -2\\ a & 2 & b\end{bmatrix} \begin{bmatrix}1 & 2 & a\\ 2 & 1 & 2\\ 2 & -2 & b\end{bmatrix} = 9 \begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$
or

$\begin{bmatrix}9 & 0 & a+2b + 4\\ 0 & 9 & 2a + 2 - 2b\\ a+2b +4 & 2a + 2 - 2b & a^2 + 4 + b^2\end{bmatrix} = \begin{bmatrix}9 & 0 & 0\\ 0 & 9 & 0\\ 0 & 0 & 9\end{bmatrix}$

On comparing above matrices, we get

$a + 2b + 4 = 0, 2a + 2 - 2b = 0$ and

$a^2 + 4 + b^2 = 9$
or

$a + 2b + 4 = 0, a - b + 1= 0$ and

$a^2 + b^2 = 5$

Solving

$a + 2b + 4 =0$ and

$a - b + 1 = 0$ , we get

$a = -2, b = -1$ . Clearly, these values satisfy

$a^2 + b^2 = 5$ .

$\therefore$

$a = - 2$ and

$b = - 1$ .

Using elementary transformation, find the inverse of the matrix

$A =\begin{bmatrix}a & b\\ c & \left ( \frac{1 + bc}{a} \right )\end{bmatrix}$ .

0%
$\Rightarrow A^{-1} = \begin{bmatrix} \frac{1 + bc}{a}& b\\ -c & a\end{bmatrix}$
0%
$\Rightarrow A^{-1} = \begin{bmatrix} \frac{1 + bc}{a}& - b\\ c & a\end{bmatrix}$
0%
$\Rightarrow A^{-1} = \begin{bmatrix} \frac{1 + bc}{a}& b\\ c & a\end{bmatrix}$
0%
None of these.

Explanation

$A = \begin{bmatrix}a & b\\ c & \left ( \frac{1 + bc}{a} \right ) \end{bmatrix}$
We

write,

$A = I.A$

$\Rightarrow \begin{bmatrix}a & b\\ c & \left ( \frac{1 + bc}{a} \right ) \end{bmatrix} = \begin{bmatrix}1 & 0\\ 0 & 1 \end{bmatrix}A$

$\Rightarrow \begin{bmatrix}1 & \frac{b}{a}\\ c & \left ( \frac{1 + bc}{a} \right ) \end{bmatrix} = \begin{bmatrix}\frac{1}{a} & 0\\ 0 & 1 \end{bmatrix} A$

$\left ( R_1 \rightarrow \frac{R_1}{a} \right )$
or

$\begin{bmatrix}1 & \frac{b}{a}\\ 0 & \frac{1}{a} \end{bmatrix} = \begin{bmatrix} \frac{1}{a} & 0\\ \frac{-c}{a} & 1\end{bmatrix} A$

$(R_2 \rightarrow R_2 - c R_1)$
or

$\begin{bmatrix}1 & \frac{b}{a}\\ 0 & 1\end{bmatrix} = \begin{bmatrix}\frac{1}{a} & 0\\ -c & a\end{bmatrix}A$

$(R_2 \rightarrow aR_2)$
or

$\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix} = \begin{bmatrix} \frac{1 + bc}{a}& -b\\ -c & a\end{bmatrix} A$

$\left ( R_1 \rightarrow R_1 - \frac{b}{a} R_2 \right )$

$\Rightarrow A^{-1} = \begin{bmatrix} \frac{1 + bc}{a}& - b\\ -c & a\end{bmatrix}$

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

$B = \begin{bmatrix}2 & 3\\ 4 & 5\\ 2 & 1\end{bmatrix}$ , then the product of AB and BA is

0%
$\begin{bmatrix}-10 & 2 & 21\\ -16 & 2 & 37\\ -2 & -2 & 11\end{bmatrix}$
0%
$\begin{bmatrix}0 & -4\\ 10 & 3\end{bmatrix}$
0%
$\begin{bmatrix}-64 & -8\\ -148 & 26\end{bmatrix}$
0%
$Cannot\space be\space computed$

Explanation

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

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

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

$\begin{bmatrix}2 - 12 & -4 + 6 & 6 + 15\\ 4 - 20 & - 8 + 10 & 12 + 25\\ 2 - 4 & -4 + 2 & 6 + 5\end{bmatrix}= \begin{bmatrix}-10 & 2 & 21\\ -16 & 2 & 37\\ -2 & -2 & 11\end{bmatrix}$

Thus, order of AB is

$2\times2$ and order of

$BA$ is $$3\times3.$$

So, the product AB and BA can not be computed.

The value of x is such that matrix product

$\begin{bmatrix}2 & 0 & 7\\ 0 & 1 & 0\\ 1 & -2 & 1\end{bmatrix} \begin{bmatrix}-x & 14x & 7x\\ 0 & 1 & 0\\ x & -4x & -2x\end{bmatrix}$ equals an identity matrix. Then the value of 20x is

0%
4
0%
5
0%
1
0%
7

Explanation

$\begin{bmatrix}2 & 0 & 7\\ 0 & 1 & 0\\ 1 & -2 & 1\end{bmatrix} \begin{bmatrix}-x & 14x & 7x\\ 0 & 1 & 0\\ x & -4x & -2x\end{bmatrix}$

$=\begin{bmatrix}5x & 0 & 0\\ 0 & 1 & 0\\ 0 & 10x-2 & 5x\end{bmatrix}= \begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$ (given)
Comparing elements we get,

$\Rightarrow \displaystyle 5x = 1, 10 x - 2 = 0, \therefore x = \frac{1}{5}$

If A is a skew-symmetric matrix and n is odd positive integer, then

$A^n$ is

0%
a skew-symmetric matrix
0%
a symmetric matrix
0%
a diagonal matrix
0%
none of these

Explanation

Given,

$A$ is a skew symmetric matrix. Therefore,

$A^T=-A$ .

$n$ is an odd positive integer.

Now, to check whether

$A^n$ is a skew symmetric matrix or not.

$(A^n)^T=(A.A^{n-1})^T$

$= (A^{n-1})^T.A^T$

$= (A.A^{n-2})^T.(-A)$

$= (A^{n-2})^T.A^T.(-A)$

$= (A.A^{n-3})^T(-A)(-A)$

$= (A^{n-3})^T.A^T.(-A)^2$ and so on....

$= (-A)^n$

$= (-1)^nA^n$

$= -A^n$

Therefore,

$A^n$ is also a skew symmetric matrix if

$A$ is a skew symmetric matrix

$\displaystyle A=\begin{bmatrix} 1 & 0 \\ \cfrac { 1 }{ 2 } & 1 \end{bmatrix},$ then

${A}^{50}$ is

0%
$\begin{bmatrix} 1 & 0 \\ 0 & 50 \end{bmatrix}$
0%
$\begin{bmatrix} 1 & 0 \\ 50 & 1 \end{bmatrix}$
0%
$\begin{bmatrix} 1 & 25 \\ 0 & 1 \end{bmatrix}$
0%
None of these

Explanation

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

$\displaystyle{ A }^{ 2 }=\begin{bmatrix} 1 & 0 \\ \cfrac { 1 }{ 2 } & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ \cfrac { 1 }{ 2 } & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2\left( \cfrac { 1 }{ 2 } \right) & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$

${ A }^{ 4 }=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\\ { A }^{ 8 }=\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}$

Similarly

$\displaystyle{ A }^{ n }=\begin{bmatrix} 1 & 0 \\ \cfrac { n }{ 2 } & 1 \end{bmatrix}$ for even

$n$

$\Rightarrow { A }^{ 50 }=\begin{bmatrix} 1 & 0 \\ 25 & 1 \end{bmatrix}$

State true or false:

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

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

$(A+B)^2 = A^2 + AB + BA + B^2 = A^2 + 2AB + B^2$

0%
True
0%
False

Explanation

Given

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

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

$(A+B)^2 = A^2 + AB + BA + B^2 = A^2 + 2AB + B^2$ is true only when

$AB=BA$

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

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

$AB\neq BA$

$\therefore$ Given statement is false.

$\displaystyle \begin{bmatrix} 2 & -3 \\ 1 & \lambda \end{bmatrix} \times \begin{bmatrix} 1 & 5 & \mu \\ 0 & 2 & -3 \end{bmatrix} = \begin{bmatrix} 2 & 4 & 1 \\ 1 & -1 & 13 \end{bmatrix}$ then

0%
$\displaystyle \lambda = 3, \mu = 4$
0%
$\displaystyle \lambda = 4, \mu = 3$
0%
$\lambda = 2 , \mu = 5$
0%
none of these

Explanation

Since,

$\begin{bmatrix} 2 & -3 \\ 1 & \lambda \end{bmatrix}\times \begin{bmatrix} 1 & 5 & \mu \\ 0 & 2 & -3 \end{bmatrix}=\begin{bmatrix} 2 & 4 & 1 \\ 1 & -1 & 13 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 2-0 & 10-6 & 2\mu +9 \\ 1+0 & 5+2\lambda & \mu -3\lambda \end{bmatrix}=\begin{bmatrix} 2 & 4 & 1 \\ 1 & -1 & 13 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 2 & 4 & 2\mu +9 \\ 1 & 5+2\lambda & \mu -3\lambda \end{bmatrix}=\begin{bmatrix} 2 & 4 & 1 \\ 1 & -1 & 13 \end{bmatrix}$

$\Rightarrow 2\mu +9=1$ and

$5+2\lambda =-1$

$\Rightarrow \mu =-4,\lambda =-3$
But these value does not satisfy

$\mu -3\lambda =13$

If the matrix

$\displaystyle A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ then

$\displaystyle A^2$ is

0%
$\displaystyle \begin{bmatrix} a^2 & b^2 \\ c^2 & d^2 \end{bmatrix}$
0%
$\displaystyle \begin{bmatrix} a^2 + bc & ab + bd \\ ac + dc & bc + d^2 \end{bmatrix}$
0%
nonexistent
0%
none of these

Explanation

Given

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

$A^{2}=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

$=\displaystyle \begin{bmatrix} a^2 + bc & ab + bd \\ ac + dc & bc + d^2 \end{bmatrix}$

State true or false.

$A, B, C$ are three matrices such that

$A = \begin{bmatrix}x & y & z\end{bmatrix},$

$B = \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c\end{bmatrix}$ and

$C = \begin{bmatrix}x \\ y \\ z\end{bmatrix},$ then

$ABC = \begin{bmatrix}ax^2 + by^2 + cz^2 + 2hxy + 2gzx + 2fyz\end{bmatrix}$

0%
True
0%
False

Explanation

Given,

$A, B, C$ are three matrices such that

$A = \begin{bmatrix}x & y & z\end{bmatrix},$

$B = \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c\end{bmatrix}$ and

$C = \begin{bmatrix}x \\ y \\ z\end{bmatrix}$

Consider,

$AB=\left[ xyz \right] \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}$

$\Rightarrow$

$= \left[ ax+yh+zg\quad xh+yb+zc\quad xg+yf+zc \right]$

Now,

$ABC=\left[ ax+yh+zg\quad xh+yb+zc\quad xg+yf+zc \right] \begin{bmatrix} x \\ y \\ z \end{bmatrix}$

$\Rightarrow$

$=\left[ { ax }^{ 2 }+{ xy }h+zgx+xhy+{ y }^{ 2 }h+zcy+xgz+yfz+{ z }^{ 2 }c \right]$

Hence it's true

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

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

$C = \begin{bmatrix} -1 & -1 & 1 \\ 2 & 2 & -2 \\ -3 & -3 & 3 \end{bmatrix},$
which if the following are null matrices ?

0%
$CA$
0%
$AB$
0%
$BA$
0%
$AC$

Explanation

Given

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

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

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

$CA=\begin{bmatrix} -1 & -1 & 1 \\ 2 & 2 & -2 \\ -3 & -3 & 3 \end{bmatrix}\begin{bmatrix}1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{bmatrix}=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}=O$

$AB=\begin{bmatrix}1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{bmatrix}\begin{bmatrix} -1 & -2 & -1 \\ 6 & 12 & 6 \\ 5 & 10 & 5 \end{bmatrix}=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}=O$

$BA=\begin{bmatrix} -1 & -2 & -1 \\ 6 & 12 & 6 \\ 5 & 10 & 5 \end{bmatrix}\begin{bmatrix}1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{bmatrix}=\begin{bmatrix} -8 & 7 & -10 \\ 48 & -42 & 60 \\ 40 & -35 & 50 \end{bmatrix}\neq O$

$AC=\begin{bmatrix}1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{bmatrix}\begin{bmatrix} -1 & -1 & 1 \\ 2 & 2 & -2 \\ -3 & -3 & 3 \end{bmatrix}=\begin{bmatrix} 4 & 4 & -4 \\ -20 & -20 & 20 \\ -16 & -16 & 16 \end{bmatrix}\neq O$

Hence, options A and B.

$A = \begin{bmatrix} 0 & 2 & 3 \\ 3 & 5 & 7 \end{bmatrix},$

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

$A+B = \begin{bmatrix} 1 & 5 & 10 \\ 5 & k & 8 \end{bmatrix},\\$ then find the value of

$k .$

0%
9
0%
4
0%
5
0%
1

Explanation

Given

$A = \begin{bmatrix} 0 & 2 & 3 \\ 3 & 5 & 7 \end{bmatrix}$ and

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

$\therefore A+B = \mbox{sum of respective elements of A and B}=\begin{bmatrix} 1 & 5 & 10 \\ 5 & 9 & 8 \end{bmatrix}$

Hence the value of

$k$ is

$9$

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

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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