If $$A=\left[ \begin{matrix} 6 & 9 \\ -4 & -6 \end{matrix} \right] $$, then $$A^{2}$$=

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

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

MCQ Questions for Class 12 Commerce Maths Matrices Quiz 10

$A = \left[ \begin{array} { c c } { a b } & { b ^ { 2 } } \\ { - a ^ { 2 } } & { - a b } \end{array} \right]$ and

$A ^ { n } = 0$ then the minimum value of

$n$ is

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

$A=\begin{bmatrix}ab&b^2\\-a^2&-ab\end{bmatrix}$

$A^2=\begin{bmatrix}ab&b^2\\-a^2&-ab\end{bmatrix}\begin{bmatrix}ab&b^2\\-a^z&-ab\end{bmatrix}=\begin{bmatrix}a^2b^2-a^2b^2&ab^3-ab^3\\-a^3b+a^3b&-a^2b^2+a^2b^2\end{bmatrix}$

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

$A^2=0$

$\therefore n=2$ .

$A=\left[ \begin{matrix} a & b \\ b & a \end{matrix} \right]$ and

$A^{2}=\left[ \begin{matrix} \alpha & \beta \\ \beta & \alpha \end{matrix} \right]$ then

0%
$\alpha=a^{2}+b^{2},\ \beta=2ab$
0%
$\alpha=a^{2}+b^{2},\ \beta=a^{2}b^{2}$
0%
$\alpha=2ab,\ \beta=a^{2}+b^{2}$
0%
$\alpha=a^{2}+b^{2},\ \beta=ab$

$A=\left[ \begin{matrix} 6 & 9 \\ -4 & -6 \end{matrix} \right]$ , then

$A^{2}$ =

0%
$\left[ \begin{matrix} 6 & 9 \\ -4 & 6 \end{matrix} \right]$
0%
$\left[ \begin{matrix} 6 & 9 \\ 4 & -6 \end{matrix} \right]$
0%
$\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right]$
0%
$\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right]$

If the order of

$A$ is

$4 \times 3$ , the order of

$B$ is

$4 \times 5$ and the order of

$C ,\ 7\times 3$ then the order of

$(A^{T}B^{T})^{T}C^{T}$ is

0%
$4 \times 5$
0%
$3 \times 7$
0%
$4 \times 3$
0%
$5 \times 7$

Explanation

Order of

$A=4\times 3$

$\therefore$ Order of

$A^{T}=3\times 4$

Order of

$B=4\times 5$

$\therefore$ Order of

$B^{T}=5\times 4$

Order of

$C=7\times 3$

$\therefore$ Order of

$C^{T}=3\times 7$

Order of

$(A^{T}B^{T})=4\times 5$

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

0%
$a =4$
0%
$a =1$
0%
$b =4$
0%
$b=1$

The inverse of a symmetric matrix is

0%
symmetric
0%
skew-symmetric
0%
diagonal matrix
0%
singular matrix

Explanation

$A^{T}=A$

$A^{-1}$ exists.

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

$=A(A^{-1})^{T}=I^{T}=I$

$A^{-1}A(A^{-1})^{T}=A^{-1}I$

$I(A^{-1})^{T}=(A^{-1})^{T}=A^{-1}$

$\therefore$ The inverse of a symmetric matrix is also symmetric.

$A = \left[ {\begin{array}{*{20}{c}}1&2\\3&4\end{array}} \right]$ , then

$8A^{-4}$ is equal to

0%
$145A^{-1}+27I$
0%
$145A^{-1}-27I$
0%
$27I - 145A^{-1}$
0%
$29A^{-1} +9I$

$A=\left[ \begin{matrix} 1 & 0 & -1 \\ 3 & 4 & 5 \\ 0 & 6 & 7 \end{matrix} \right]$ and

$A^{-1}=[\alpha_{ij}]_{3\times 3}$ then

$\alpha_{23}=$

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

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

$I$ is the unit matrix of order

$2$ , then

$A^{2}$ equals

0%
$4A-3I$
0%
$3A-4I$
0%
$A-I$
0%
$A+I$

Explanation

$\begin{array}{l} A=\left( { \begin{array} { *{ 20 }{ c } }2 & { -1 } \\ { -1 } & 2 \end{array} } \right) \\ I=\left( { \begin{array} { *{ 20 }{ c } }1 & 0 \\ 0 & 1 \end{array} } \right) \\ { A^{ 2 } }=A\cdot A \\ =\left( { \begin{array} { *{ 20 }{ c } }2 & { -1 } \\ { -1 } & 2 \end{array} } \right) \left( { \begin{array} { *{ 20 }{ c } }2 & { -1 } \\ { -1 } & 2 \end{array} } \right) \\ =\left( { \begin{array} { *{ 20 }{ c } }{ 2\times 2+1 } & { -2-2 } \\ { -2-2 } & { 1+4 } \end{array} } \right) \\ =\left( { \begin{array} { *{ 20 }{ c } }5 & { -4 } \\ { -4 } & 5 \end{array} } \right) \\ Now, \quad 4A-3I \\= 4\left( { \begin{array} { *{ 20 }{ c } }8 & { -4 } \\ { -4 } & 8 \end{array} } \right) -\left( { \begin{array} { *{ 20 }{ c } }3 & 0 \\ 0 & 3 \end{array} } \right) \\ = \left( { \begin{array} { *{ 20 }{ c } }{ 8-3 } & { -4-0 } \\ { -4-0 } & { 8-3 } \end{array} } \right) \\ = \left( { \begin{array} { *{ 20 }{ c } }5 & { -4 } \\ { -4 } & 5 \end{array} } \right) ={ A^{ 2 } } \end{array}$

$\begin{bmatrix} 3 & 2 & -1 \\ 4 & 9 & 2 \\ 5 & 0 & -2 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0 \\ 7 \\ 2 \end{bmatrix}$ , then

$(x,\ y,\ z)=$

0%
$(1,\ -1,\ 1)$
0%
$(2,\ -1,\ -4)$
0%
$(3,\ 0,\ 6)$
0%
$(2,\ -1,\ 4)$

Explanation

$\begin{bmatrix} 3 & 2 & -1\\ 4 & 9 & 2\\ 5 & 0 & -2\end{bmatrix}\begin{bmatrix} x \\ y\\ z\end{bmatrix}=\begin{bmatrix} 0\\ 7\\ 2\end{bmatrix}$

Thus, multiplying

$3x+2y-z=0$ …………..

$(1)$

$4x+9y+2z=7$ ………..

$(2)$

$5x-2z=2$ …………

$(3)$

Thus

$z=\dfrac{5x-2}{2}$

$\therefore 3x+2y=\dfrac{5x-2}{2}$

$\therefore 6x+4y=5x-2$

$\therefore x+4y=-2$ …………

$(4)$

Also,

$4x+9y+2\left(\dfrac{5x-2}{2}\right)=7$

$9x+9y=9$

$\therefore x+y=1$ ………..

$(5)$

Solving

$(4)$ and

$(5)$

$3y=-3$

$y=-1$

$\therefore x=1-y$

$=1-(-1)=2$

$\therefore x=2$

$\therefore z=\dfrac{5x-2}{2}=\dfrac{5(2)-2}{2}$

$\therefore z=4$

$=4$

$(x, y, z)=(2, -1, 4)$ .

If A is a 2 X 2 matrix such that

$A^{2009} + A^{2008}$ = I, then :

$(A^{2008})^{-1}$ =

0%
$A^{2008} + I$
0%
$A^{2009} + 1$
0%
A + I
0%
A

$\left[ \begin{matrix} 2 & -3 \\ 1 & \lambda \end{matrix} \right] \times \left[ \begin{matrix} 1 & 5 & \mu \\ 0 & 2 & -3 \end{matrix} \right] =\left[ \begin{matrix} 2 & 4 & 1 \\ 1 & -1 & 13 \end{matrix} \right],$ then

0%
$\lambda=3, \mu= -4$
0%
$\lambda=4, \mu=-3$
0%
no real values of $\lambda, \mu$ are possible
0%
none of these

Explanation

Given,

$\pmatrix{2&-3\\1&\lambda}\times \pmatrix{1&5& \mu\\0&2&-3}$ =

$\pmatrix{2&4&1\\1&-1&13}$

$\pmatrix{2&4&2\mu-9\\1&5+2\lambda&\mu-3\lambda}=\pmatrix{2&4&1\\1&-1&13}$

Comparing both sides we get

$2\mu+9=1\implies 2\mu=-8\implies \mu=-4$

Also

$5+2\lambda=-1\implies 2\lambda=6\implies \lambda=3$

Hence, we get

$\mu=-4,\lambda=3.$

Let p be a non-singular matrix,

$1+p+p^{2}+....+p^{n}=0$ (0 denotes the null matrix) then

$p^{-1}=$

0%
$p^{n}$
0%
- $p^{n}$
0%
-(1+p+...+ $p^{n}$ )
0%
none

$A=\begin{bmatrix} 4 & -1 \\ -1 & k \end{bmatrix}$ such that

$A^{2}-6A+7I=0$ , then

$k=$

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

Explanation

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

${A}^{2} = \begin{bmatrix} 4 & -1 \\ -1 & k \end{bmatrix} \times \begin{bmatrix} 4 & -1 \\ -1 & k \end{bmatrix}$

$\Rightarrow {A}^{2} = \begin{bmatrix} 16+ 1 & -4 - k \\ -4 - k & 1 + {k}^{2} \end{bmatrix} = \begin{bmatrix} 17 & - \left( 4 + k \right) \\ - \left( 4 + k \right) & 1 + {k}^{2} \end{bmatrix}$

$6A = 6 \times \begin{bmatrix} 4 & -1 \\ -1 & k \end{bmatrix} = \begin{bmatrix} 24 & -6 \\ -6 & 6k \end{bmatrix}$

$7I = 7 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

Therefore,

${A}^{2} - 6A + 7I = 0$

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

$\begin{bmatrix} 17 & - \left( 4 + k \right) \\ - \left( 4 + k \right) & 1 + {k}^{2} \end{bmatrix} = \begin{bmatrix} 24 & -6 \\ -6 & 6k \end{bmatrix} - \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

$\begin{bmatrix} 17 & - \left( 4 + k \right) \\ - \left( 4 + k \right) & 1 + {k}^{2} \end{bmatrix} = \begin{bmatrix} 17 & -6 \\ -6 & 6k - 7 \end{bmatrix}$

Comparing both sides, we get

$- \left( 4 + k \right) = -6$

$\Rightarrow k = 6 - 4 = 2$

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

$A^{-1}$ is given by?

0%
$\begin{bmatrix} 0 & -1\\ 2 & -4\end{bmatrix}$
0%
$\begin{bmatrix} 0 & -1\\ -2 & -4\end{bmatrix}$
0%
$\begin{bmatrix} 0 & 1\\ 2 & -4\end{bmatrix}$
0%
None of these

A is an involuntary matrix given by

$A=\begin{bmatrix} 0 & 1 & -1\\ 4 & -3 & 4\\ 3 & -3 & 4\end{bmatrix}$ then the inverse of

$\dfrac{A}{2}$ will be?

0%
$2A$
0%
$\dfrac{A^{-1}}{2}$
0%
$\dfrac{A}{2}$
0%
$A^{-2}$

A is an involutory matrix given by

$A=\begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix}$ then the inverse of

$\dfrac{A}{2}$ will be

0%
$2A$
0%
$\dfrac{A^{-1}}{2}$
0%
$\dfrac{A}{2}$
0%
$A^{2}$

Explanation

Given

$A$ to be involutory matrix, then according to the definition of involutory matrix we have,

$A^2 =I$ . [

$I$ being identity matrix of order

$3$ ].

So,

$A^2=I$

or,

$A=A^{-1}$ [ Since involutory matrix is always invertible]

or,

$\dfrac{A}{2}=\dfrac{A^{-1}}{2}$ .

Let A=

$\left[ \begin{matrix} 1 \\ 2 \end{matrix}\begin{matrix} 2 \\ 1 \end{matrix} \right] and\quad B=\left[ \begin{matrix} 4 \\ 5 \\ 0 \end{matrix}\begin{matrix} -3 \\ 6 \\ 1 \end{matrix} \right]$ then

0%
$AB$ exists
0%
$AB$ and $BA$ both exists
0%
Neither $AB$ nor $BA$ exists
0%
$BA$ exists, but $$AB$ does not exists

Explanation

Given,

Order of

$A=2\times 2$ matrix

Order of

$B=3\times 2$ matrix

$AB$ does not exist, as the number of columns of matrix

$A$ is not equal to number of rows of

$B$

But,

$BA$ exists, as the number of columns of matrix

$B$ is equal to number of rows of

$A$

$\left[ \begin{matrix} 1 & x & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{matrix} \right] \left[ \begin{matrix} 1 \\ 1 \\ x \end{matrix} \right] =0,$ then

$x=$

0%
$\cfrac{-9\pm\sqrt{35}}{2}$
0%
$\cfrac{-7\pm\sqrt{53}}{2}$
0%
$\cfrac{-9\pm\sqrt{53}}{2}$
0%
$\cfrac{-7\pm\sqrt{35}}{2}$

Explanation

Given,

$\begin{pmatrix}1&x&1\end{pmatrix}\begin{pmatrix}1&3&2\\ 0&5&1\\ 0&3&2\end{pmatrix}\begin{pmatrix}1\\ 1\\ x\end{pmatrix}=0$

Upon multiplying the first 2 matrices, we get,

$\begin{pmatrix}1\cdot \:1+x\cdot \:0+1\cdot \:0&1\cdot \:3+x\cdot \:5+1\cdot \:3&1\cdot \:2+x\cdot \:1+1\cdot \:2\end{pmatrix}\begin{pmatrix}1\\ 1\\ x\end{pmatrix}=0$

Upon further simplification, we get,

$\begin{pmatrix}1&6+5x&4+x\end{pmatrix}\begin{pmatrix}1\\ 1\\ x\end{pmatrix}=0$

Again multiplying the above matrices, we get,

$\begin{pmatrix}1\cdot \:1+\left(6+5x\right)\cdot \:1+\left(4+x\right)x\end{pmatrix}=0$

Upon further simplification, we get,

$\begin{pmatrix}x^2+9x+7\end{pmatrix}=0$

Solving the above quadratic equation, we get,

$x=\dfrac{-9\pm \sqrt{53}}{2}$

$A$ and

$B$ are two matrices such that

$AB$ and

$A+B$ are both defined then

$A$ and

$B$ are

0%
Square matrices of the same order
0%
Square matrices of different order
0%
Rectangular matrices of same order
0%
Rectangular matrices of different order

Explanation

For the addition of two matrices,it is required that

the matrices should be of the same order.

For multiplication of two matrices A of order

$m\times$ n and

B of order

$p\times q,$ it requires that

$n=p----(1)$

$m = p----(2)$

$n = q----(3)$

form (1) and (3)

$p = q-------(4)$

from (1)and (2)

$n = m-----(5)$

From (4)and (5),we can conclude that

A and B are square matrices and their orders

are one and same .

$\therefore$ The addition and multiplication of two matrices

should be of same order and they should be

square matrices.

$\therefore$ so, the answer is A. square matrices of the same

order.

if A=

$\left[ \begin{matrix} 2 & 3 \\ 5 & -7 \end{matrix} \right] then\quad \left( { A }^{ '} \right) ^{ 2 }=$

0%
$\left[ \begin{matrix} 5 & -7 & 12 \\ 1 & 4 & 22 \end{matrix} \right]$
0%
$\left[ \begin{matrix} 1 & 17 \\ 1 & -4 \\ 0 & 2 \end{matrix} \right]$
0%
$\left[ \begin{matrix} -19 & -25 \\ -15 & 64 \end{matrix} \right]$
0%
$\left[ \begin{matrix} 19 & -25 \\ -15 & 64 \end{matrix} \right]$

Explanation

Given,

$A=\begin{pmatrix}2&3\\ \:5&-7\end{pmatrix}$

$\Rightarrow A'=\begin{pmatrix}2&5\\ 3&-7\end{pmatrix}$

Now,

$(A')^2$

$=\begin{pmatrix}2&5\\ 3&-7\end{pmatrix}^2$

$=\begin{pmatrix}2&5\\ 3&-7\end{pmatrix}\begin{pmatrix}2&5\\ 3&-7\end{pmatrix}$

$=\begin{pmatrix}2\cdot \:2+5\cdot \:3&2\cdot \:5+5\left(-7\right)\\ 3\cdot \:2+\left(-7\right)\cdot \:3&3\cdot \:5+\left(-7\right)\left(-7\right)\end{pmatrix}$

$=\begin{pmatrix}19&-25\\ -15&64\end{pmatrix}$

$A=\left[ \begin{matrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{matrix} \right]$ , then value of

$A^{-1}$ is equal to

0%
$A$
0%
$A^{2}$
0%
$A^{3}$
0%
$A^{4}$

$U = [ 2, -3 , 4 ]$ ,

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

$X = [0 , 2 , 3]$ and

$Y = \begin{bmatrix} 2 \\ 2 \\ 4 \end{bmatrix}$ , then

$UV + XY$ =

0%
$20$
0%
$[ -20 ]$
0%
$-20$
0%
$[ 20 ]$

$A=\begin{bmatrix} 2 & -1\\ -7 & 4\end{bmatrix}$ and

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

$B^TA^T$ is equal to?

0%
$\begin{bmatrix} 1 & 0\\ -12 & 1\end{bmatrix}$
0%
$\begin{bmatrix} 1 & 1\\ 1 & 1\end{bmatrix}$
0%
$\begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix}$
0%
$\begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix}$

Adjacency matrix of all graphs are symmetric.

0%
True
0%
False

$A^2-A+1=0$ , then the inverse of A is?

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

$\begin{bmatrix} 1 & 1\\ -1 & 1\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}=\begin{bmatrix} 2 \\ 4\end{bmatrix}$ , then the values of

$x$ and

$y$ respectively are?

0%
$-3, -1$
0%
$1, 3$
0%
$3, 1$
0%
$-1, 3$

Explanation

On equating given two matrices, we get

$x+y=2$ and

$-x+y=4$

Solving for

$x, y : x=-1, y=3$ .

Let

$\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 3\\ 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & n-1\\ 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & 78\\ 0 & 1\end{bmatrix}$
If

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

$A^{-1}=?$

0%
$\begin{bmatrix} 1 & 12\\ 0 & 1\end{bmatrix}$
0%
$\begin{bmatrix} 1 & -13\\ 0 & 1\end{bmatrix}$
0%
$\begin{bmatrix} 1 & -12\\ 0 & 1\end{bmatrix}$
0%
$\begin{bmatrix} 1 & 0\\ -13 & 1\end{bmatrix}$

Explanation

$\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 3\\ 0 & 1\end{bmatrix}..\begin{bmatrix} 1 & n-1\\ 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & 78\\ 0 & 1\end{bmatrix}$

$\Rightarrow \dfrac{n(n-1)}{2}=78\Rightarrow n=13$

$A=\begin{bmatrix} 1 & 13\\ 0 & 1\end{bmatrix}$
so

$A^{-1}=\begin{bmatrix} 1 & -13\\ 0 & 1\end{bmatrix}$ .

$\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 3\\ 0 & 1\end{bmatrix} ..\begin{bmatrix} 1 & n-1\\ 0 & 1\end{bmatrix} =\begin{bmatrix} 1 & 78\\ 0 & 1\end{bmatrix}$ , then the inverse of

$\begin{bmatrix} 1 & n\\ 0 & 1\end{bmatrix}$ is?

0%
$\begin{bmatrix} 1 & -13\\ 0 & 1\end{bmatrix}$
0%
$\begin{bmatrix} 1 & 0\\ 12 & 1\end{bmatrix}$
0%
$\begin{bmatrix} 1 & -12\\ 0 & 1\end{bmatrix}$
0%
$\begin{bmatrix} 1 & 0\\ 13 & 1\end{bmatrix}$

Explanation

$\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 3\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & n-1\\ 0 & 1\end{bmatrix} =\begin{bmatrix} 1 & 78\\ 0 & 1\end{bmatrix}$

$\Rightarrow \begin{bmatrix} 1 & 1+2+3+...+n-1\\ 0 & 1\end{bmatrix}=$

$\begin{bmatrix} 1 & 78\\ 0 & 1\end{bmatrix}$

$\Rightarrow \dfrac{n(n-1)}{2}=78\Rightarrow n=13$ ,

$-12$ (reject)

$\therefore$ We have to find inverse of

$\begin{bmatrix} 1 & 13\\ 0 & 1\end{bmatrix}$

$\therefore \begin{bmatrix} 1 & -13\\ 0 & 1\end{bmatrix}$

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

$B = \begin{bmatrix}a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3}\end{bmatrix}$ , then

$AB$ is equal to

0%
$B$
0%
$nB$
0%
$B^{n}$
0%
$A + B$

Explanation

Let,

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

$B = \begin{bmatrix}a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3}\end{bmatrix}$ .

Now,

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

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

or,

$A B= \begin{bmatrix}na_{1} & na_{2} & na_{3}\\ nb_{1} & nb_{2} & nb_{3}\\ nc_{1} & nc_{2} & nc_{3}\end{bmatrix}$

or,

$AB=nB$ .

CBSE Questions for Class 12 Commerce Maths Matrices Quiz 10 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Matrices Quiz 10 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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