If $$A=\begin{bmatrix} 0 & 0 \\ 0 & 5 \end{bmatrix}$$, then $${ A }^{ 12 }$$ is

$$\begin{bmatrix} 0 & 0 \\ 0 & 60 \end{bmatrix}$$

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

If $$A=\begin{bmatrix} 2 & 5 & -3 \\ -1 & 3 & 1 \\ 4 & 1 & 2 \end{bmatrix}$$ then $${A}^{2}$$ is

$$\begin{bmatrix} -13 & 22 & -7 \\ 15 & 25 & -7 \\ -1 & 5 & 8 \end{bmatrix}$$

$$\begin{bmatrix} -13 & 22 & -7 \\ 15 & 25 & 7 \\ 1 & 5 & 8 \end{bmatrix}$$

$$\begin{bmatrix} 13 & 22 & 7 \\ 15 & 25 & -7 \\ -1 & 5 & 8 \end{bmatrix}$$

MCQ Questions for Class 12 Commerce Maths Matrices Quiz 9

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

$A^{2} - 6A =$ _____.

0%
$27 I_{3}$
0%
$5 I_{3}$
0%
$20 I_{3}$
0%
$30 I_{3}$

Explanation

Given,

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

Therefore,

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

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

$=\begin {bmatrix} 1+16+16 & 4+4+16 & 4+16+4 \\ 4+4+16 & 16+1+16 & 16+4+4 \\ 4+16+4 & 16+4+4 & 16+16+1\end{bmatrix}$

$=\begin {bmatrix} 33 &24 &24\\24&33&24\\24&24&33 \end {bmatrix}$

and

$6A=\begin{bmatrix} 1& 4 & 4\\ 4 & 1 & 4\\ 4 & 4 & 1\end{bmatrix}$

$=\begin{bmatrix} 6& 24 & 24\\ 24 & 6 & 24\\ 24 & 24 & 6\end{bmatrix}$

Thus

$A^2-6A$

$=\begin {bmatrix} 33 &24 &24\\24&33&24\\24&24&33 \end {bmatrix}$

$-\begin{bmatrix} 6& 24 & 24\\ 24 & 6 & 24\\ 24 & 24 & 6\end{bmatrix}$

$=\begin {bmatrix} 33-6&24-24&24-24 \\24-24&33-6&24-24\\24-24 &24-24&33-6 \end {bmatrix}$

$=\begin {bmatrix} 27&0&0\\ 0&27&0\\ 0&0&27\end {bmatrix}$

$= 27I_{3}$

$A =\begin{pmatrix} -2& 2\\ 2 & -2\end{pmatrix}$ , then which one of the following is correct?

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

Explanation

Given

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

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

$=\begin{bmatrix} (-2)(-2)+2\cdot 2 & (-2)(2)+(2)(-2) \\ (2)(-2)+(-2)(2)& (2)(2)+(-2)(-2)\end{bmatrix}$

$=\begin{bmatrix} 8 & -8 \\ -8 & 8 \end{bmatrix}=-4\begin{bmatrix} -2 & 2 \\ 2 & -2 \end{bmatrix}=-4A$

$A=\begin{bmatrix} 0 & 0 \\ 0 & 5 \end{bmatrix}$ , then

${ A }^{ 12 }$ is

0%
$\begin{bmatrix} 0 & 0 \\ 0 & 60 \end{bmatrix}$
0%
$\begin{bmatrix} 0 & 0 \\ 0 & { 5 }^{ 12 } \end{bmatrix}$
0%
$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
0%
$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Explanation

We have

$A=\begin{bmatrix}0&0\\0&5\end{bmatrix}$

$A^2=A\times A=\begin{bmatrix}0&0\\0&5\end{bmatrix} \times \begin{bmatrix}0&0\\0&5\end{bmatrix}$

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

Similarly,

$A^3=A^2\times A=\begin{bmatrix}0&0\\0&5^2\end{bmatrix} \times \begin{bmatrix}0&0\\0&5\end{bmatrix}$

$=\begin{bmatrix}0&0\\0&5^3\end{bmatrix}$

Thus we can conclude

$A^{12}=\begin{bmatrix}0&0\\0&5^{12}\end{bmatrix}$

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

$A^3=$ ___________.

0%
$243$ A
0%
$81$ A
0%
$27$ A
0%
$729$ A

Explanation

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

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

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

$=3\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}\times3\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}$

$=9\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}\times\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}$

$=27\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}$

$A^3=A^2.A$

$=27\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix} \times 3\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}$

$=81\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix} \times \begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}$

$=81A$

The answer is option (B).

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

$B=\begin{bmatrix}a&h&g\\h&b&f\\g&f&c\end{bmatrix}, C=\begin{bmatrix}\alpha & \beta & \gamma \end{bmatrix}^{T}$ then

$ABC$ is

0%
Not defined
0%
a $1\times1$ matrix
0%
a $3\times3$ matrix
0%
a $3\times2$ matrix

Explanation

$A=\begin{bmatrix} x & y & z \end{bmatrix}\quad \quad B=\begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix},\quad \quad C=\begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix}\\ \quad \quad \quad 1\times 3\qquad \qquad \quad \quad \quad \ 3\times 3\qquad \qquad \qquad \ 3\times 1$

$ABC$ is a

$1 \times1$ matrix

Is it possible to define the matrix

$A + B$ when both

$A$ and

$B$ are square matrices of the same order?

0%
True
0%
False

Explanation

It is always possible to define the sum of two matrix

$X$ and

$Y$ if they are of same order.

So, If A and B are square matrices of same order, A + B is always definable.

Let A =

$\begin{bmatrix} 0 & 0 & -1 \\[0.3em] 0 & -1 & 0 \\[0.3em] -1 & 0 & 0 \end{bmatrix}.$ Then the only correct statement

$A$ is

0%
$A = 0$
0%
$A = (-1) I$
0%
$A^{-1}$ does not exist
0%
$A^2 = I$

Explanation

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

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

$\therefore A^2 = I$

Hence it is a identity matrix

If A =

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

$\begin{bmatrix} 1 & 0 \\ 5 & 1\end{bmatrix}$ whenever

$A^2 \, = \, B$
then values of

$\alpha$ is

0%
$1$
0%
$-1$
0%
$4$
0%
no real value of $\alpha$

Explanation

$A^2$ = B

$\Rightarrow \begin{bmatrix} \alpha & 0 \\ 1 & 1\end{bmatrix} \begin{bmatrix} \alpha & 0 \\ 1 & 1\end{bmatrix} \, = \, \begin{bmatrix} 1 & 0 \\ 5 & 1\end{bmatrix}$

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

$\Rightarrow \alpha^2$ = 1,

$\alpha$ + 1 = 5

$\therefore \alpha \, = \, \pm 1\, , \, \alpha \, = \, 4$
There is no real value of

$\alpha$ which satisfies both.

Using elementary row transformation find the inverse of the matrix A =

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

0%
$\left[\begin {array} {ll} \dfrac{-5}{8} & \dfrac{5}{4} & \dfrac{1}{8} \\ \dfrac{-3}{8} & \dfrac{3}{4} & \dfrac{-1}{8} \\ \dfrac{-5}{4} & \dfrac{3}{2} &\dfrac{1}{4} \end {array}\right]$
0%
$\dfrac {1}{8}\left [\begin {array} {ll} 5 & -5 & 1 \\ 3 & -3 & 1 \\ 0 & 3 & 1\end {array}\right]$
0%
$\dfrac{1}{8} \left[\begin {array} {ll} 5 & 5 & 1 \\ 3 & 6 & -1 \\ 10 & -12 & 2 \end {array}\right]$
0%
None of these

Obtain the inverse of the following matrix using elementary operation:

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

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

Explanation

GIven

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

Therefore,

$A^{-1}=\dfrac{1}{|A|}adj(A)$

$\implies A^{-1}=\dfrac{1}{3(3-0)-1(8-0)}\begin{bmatrix}3& -2 &8 \\ -4& 3&-12\\3 &-2&9\end{bmatrix}$

$\implies A^{-1}=\begin{bmatrix}3& -2 &8 \\ -4& 3&-12\\3 &-2&9\end{bmatrix}$

$I =\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & \quad \quad 0 & \quad 1 \end{pmatrix},$

$P=\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & \quad \quad 0 & \quad -2 \end{pmatrix},$ then the matrix

$P^3+2P^2$ is equal to

0%
$P$
0%
$1-P$
0%
$2I+P$
0%
$2I-P$

Explanation

Given the matrix

$P=\begin{pmatrix}1&0&0\\0&-1&0\\ 0&0&-2\end {pmatrix}$ .

Now,

$P^2=\begin{pmatrix}1^2&0&0\\0&(-1)^2&0\\ 0&0&(-2)^2\end {pmatrix}$ [ Since

$P$ is a diagonal matrix]

$\Rightarrow P^2=\begin{pmatrix}1&0&0\\0&1&0\\ 0&0&4\end {pmatrix}$ ......(1).

Again,

$P^3=\begin{pmatrix}1^3&0&0\\0&(-1)^3&0\\ 0&0&(-2)^3\end {pmatrix}$

$\Rightarrow P^3=\begin{pmatrix}1&0&0\\0&-1&0\\ 0&0&-8\end {pmatrix}$ .....(2).

So,

$P^3+2P^2$

$=\begin{pmatrix}1+2&0&0\\0&-1+2&0\\ 0&0&-8+8\end {pmatrix}$

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

$=2I+P$ .

Given

$A= \begin{bmatrix} 3&6 \\ -2&-8 \end{bmatrix}$ and

$B = \begin{bmatrix} 2 & 16\end{bmatrix}$ , find the matrix

$X$ such that

$XA=B$ .

0%
$\begin{bmatrix} -\dfrac{4}{3} & -3 \end{bmatrix}$
0%
$\begin{bmatrix} \dfrac{4}{3} & 3 \end{bmatrix}$
0%
$\begin{bmatrix} \dfrac{4}{3} & -3 \end{bmatrix}$
0%
$\begin{bmatrix} -\dfrac{4}{3} & 3 \end{bmatrix}$

Explanation

$XA=B$

${ X }_{ 1\times 2 }{ \begin{bmatrix} 3 & 6 \\ -2 & -8 \end{bmatrix} }_{ 2\times 2 }={ \begin{bmatrix} 2 & 16 \end{bmatrix} }_{ 1\times 2 }$

$X$ should be of order

$1 \times 2$

Let

$X$ be

$\begin{bmatrix} a & b \end{bmatrix}$

$\therefore \begin{bmatrix} a & b \end{bmatrix}\begin{bmatrix} 3 & 6 \\ -2 & -8 \end{bmatrix}=\begin{bmatrix} 2 & 16 \end{bmatrix}\\ \Rightarrow \begin{bmatrix} 3a-2b & 6a-8b \end{bmatrix}=\begin{bmatrix} 2 & 16 \end{bmatrix}\\ \therefore 3a-2b=2\longrightarrow (1)\times 2\\ \quad 6a-8b=16\longrightarrow (2)\times 1\\ \therefore 6a-4b=4\\ \quad 6a-8b=16\\ \quad \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \ 4b=-12\\ \quad \ \Rightarrow b=-3\\ \therefore 3a+6=2\\ \quad a=\dfrac{-4}{3}\\ \therefore X=\begin{bmatrix} \dfrac{-4}{3} & -3 \end{bmatrix}$

The matrix equation satisfied by

$A$ is ______

$,$ if

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

0%
$A^2-4A-5I=0$
0%
$A^2-4A-5=0$
0%
$A^2+4A-5I=0$
0%
$A^2+4A-5=0$

Explanation

Consider the given matrix,

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

Consider,

$A^2-4A-5I$

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

$=\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}-\begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix}-\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$

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

$=0$

State true/false:

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

$B=[-2-1-4],$ then

$(AB)^t\ne B^tA^t$ .

0%
True
0%
False

Explanation

Given,

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

Now,

$AB={ \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} }\times \begin{bmatrix} -2 & -1 & -4 \end{bmatrix} =\begin{bmatrix} 2 & 1 & 4 \\ -4 & -2 & -8 \\ -6 & -3 & -12 \end{bmatrix}\\ \therefore { \left( AB \right) }^{ t }=\begin{bmatrix} 2 & -4 & -6 \\ 1 & -2 & -3 \\ 4 & -8 & -12 \end{bmatrix}\\ Consider, \quad { B }^{ t }\times { A }^{ t }={ \begin{bmatrix} -2 \\ -1 \\ -4 \end{bmatrix} }\times \begin{bmatrix} -1 & 2 & 3 \end{bmatrix} =\begin{bmatrix} 2 & -4 & -6 \\ 1 & -2 & -3 \\ 4 & -8 & -12 \end{bmatrix}\\ \therefore { \left( AB \right) }^{ t }={ B }^{ t }\times { A }^{ t }$

Hence, given statement is false.

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

$B = \begin{bmatrix}3&1&2\\2&0&5\\1&2&0\end{bmatrix}$ , find

$AB$ .

0%
$\begin{bmatrix}5&6&-9\\4&3&-9\\-1&3&-2\end{bmatrix}$
0%
$\begin{bmatrix}2&4&7\\1&5&3\\-1&0&-5\end{bmatrix}$
0%
$\begin{bmatrix}7&11&-9\\4&10&-11\\-1&2&-5\end{bmatrix}$
0%
$\begin{bmatrix}7&8&-19\\4&1&-5\\-1&12&-5\end{bmatrix}$

Explanation

$A=\begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}\quad B=\begin{bmatrix} 3 & 1 & 2 \\ 2 & 0 & 5 \\ 1 & 2 & 0 \end{bmatrix}\\ AB=\begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}\times \begin{bmatrix} 3 & 1 & 2 \\ 2 & 0 & 5 \\ 1 & 2 & 0 \end{bmatrix}\\ \quad =\begin{bmatrix} 9-6+4 & 3+0+8 & 6-15+0 \\ 6-6+4 & 2+0+8 & 4-15+0 \\ 0-2+1 & 0+0+2 & 0-5+0 \end{bmatrix}\\ \quad =\begin{bmatrix} 7 & 11 & -9 \\ 4 & 10 & -11 \\ -1 & 2 & -5 \end{bmatrix}$

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

What will be the sum of the diagonal elements of the resultant matrix?

0%
10
0%
6
0%
9
0%
7

Explanation

$\begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix}+ \begin{bmatrix} 5 & 1 & -2 \\ 1 & 1 & 0 \\ -2 & -2 & 1 \end{bmatrix}= \begin{bmatrix} 6 & 1 & 0 \\ 0 & 2 & -2 \\ -2 & 0 & 2 \end{bmatrix}$

Sum of diagonal

$=6+2+2 = 10$

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

$A^{100}$ is equal to

0%
$\begin{bmatrix} 1 & 0 \\ \left( \frac { 1 }{ 2 } \right) \times { 100 } & 1 \end{bmatrix}$
0%
$\begin{bmatrix} 1 & 0 \\ 25 & 1 \end{bmatrix}$
0%
$\begin{bmatrix} 1 & 0 \\ 50 & 1 \end{bmatrix}$
0%
$\begin{bmatrix} 1 & 0 \\ 100 & 1 \end{bmatrix}$

Explanation

Given

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

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

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

$A^4=\begin{bmatrix} 1& 0\\ \dfrac{3}{2}& 1\end{bmatrix}\times \begin{bmatrix} 1& 0\\ \dfrac{1}{2}& 1\end{bmatrix}=\begin{bmatrix} 1& 0\\ \dfrac{4}{2}& 1\end{bmatrix}$

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

$A^{100}=\begin{bmatrix} 1& 0 \\ \dfrac{100}{2}&1 \end{bmatrix}=\begin{bmatrix} 1& 0\\ 50& 1\end{bmatrix}$

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

$B=\begin{bmatrix}4&7\\5&6\end{bmatrix},$ then

$AB=$

0%
$\begin{bmatrix}19&25\\28&38\end {bmatrix}$
0%
$\begin{bmatrix}25&8\\26&14\end {bmatrix}$
0%
$\begin{bmatrix}34&8\\58&23\end {bmatrix}$
0%
$\begin{bmatrix}5&10\\7&10\end {bmatrix}$

Explanation

Given,

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

$B=\begin{bmatrix}4&7\\5&6\end{bmatrix}$

$AB=\begin{bmatrix}1&3\\2&4\end{bmatrix}\begin{bmatrix}4&7\\5&6\end{bmatrix}$

$=\begin{bmatrix}1(4)+3(5)&1(7)+3(6)\\2(4)+4(5)& 2(7)+4(6)\end{bmatrix}$

$=\begin{bmatrix}19&25\\28&38\end{bmatrix}$

$B=A+A^{2}+A^{3}+A^{4}$

If order of

$A$ is

$3$ then order of

$B$ is

0%
$3$
0%
$6$
0%
$2$
0%
$9$

Explanation

The order of matrix does not change when the operation are done on it.

So, the order of

$B$ remains same as the order of

$A$ .

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

${A}^{2}$ is

0%
$\begin{bmatrix} -13 & 22 & -7 \\ 15 & 25 & -7 \\ -1 & 5 & 8 \end{bmatrix}$
0%
$\begin{bmatrix} -13 & 22 & -7 \\ -1 & 5 & 8 \\ 15 & 25 & -7 \end{bmatrix}$
0%
$\begin{bmatrix} -13 & 22 & -7 \\ 15 & 25 & 7 \\ 1 & 5 & 8 \end{bmatrix}$
0%
$\begin{bmatrix} 13 & 22 & 7 \\ 15 & 25 & -7 \\ -1 & 5 & 8 \end{bmatrix}$

Explanation

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

$=\begin{bmatrix} -13\quad \quad & 22\quad \quad & 2\times -3+5\times 1+(-3)\times 2 \\ -1 & 5 & 8 \\ 15 & 25 & -7 \end{bmatrix}$

$\begin{bmatrix} -13 & 22 & -7 \\ -1 & 5 & 8 \\ -15 & 25 & -7 \end{bmatrix}$

Hence, B will be correct answer.

If the graph of

$y = f(x)$ is symmetrical about the lines

$x = 1$ and

$x = 2$ , then which of the following is true?

0%
$f(x + 1) = f(x)$
0%
$f(x + 3) = f(x)$
0%
$f(x + 2) = f(x)$
0%
None of these

Explanation

$f(x)$ is symmetrical about the line

$x=1$

So,

$f(1-x)=f(1+x)$ .................... (1)

$f(x)$ is symmetrical about the line

$x=2$

So,

$f(2-x)=f(2+x)$ .................... (2)

Replace

$x$ with

$1-x$ in equation (1) ,

$f(1-1+x)=f(1+1-x)$

$f(x)=f(2-x)$ ..........................(3)

From (2) and (3) ,

$f(x)=f(2+x)$

Find the inverse of the following matrix.

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

0%
$\displaystyle \left[ {\begin{array}{ccccccccccccccc}{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{ 4}&3&{ 1}\\{ - \frac{5}{2}}&{ - \frac{3}{2}}&{\frac{1}{2}}\end{array}} \right]$
0%
$\displaystyle \left[ {\begin{array}{ccccccccccccccc}{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{ - 4}&-3&{ - 1}\\{ - \frac{5}{2}}&{ - \frac{3}{2}}&{\frac{1}{2}}\end{array}} \right]$
0%
$\displaystyle \left[ {\begin{array}{ccccccccccccccc}{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{ - 4}&3&{ - 1}\\{ - \frac{5}{2}}&{ - \frac{3}{2}}&{\frac{1}{2}}\end{array}} \right]$
0%
$\left[ {\begin{array}{ccccccccccccccc}0&1&2\\1&-2&3\\3&1&-1\end{array}} \right]$

State true or false:

The product matrix of

$\left[ \begin{array}{l}1\,\,\,\,\,\,0\\0\,\,\,\,\,\,1\end{array} \right]$ and

$\left[ \begin{array}{l}0\,\,\,\,\,\,\,\,1\\1\,\,\,\,\,\,\,\,0\end{array} \right]$ is an identity matrix.

0%
True
0%
False

Explanation

Consider,

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

$=\begin{bmatrix}1(0)+0(1) & 1(1)+0(0)\\0(0)+1(1) & 0(1)+1(0)\end{bmatrix}$

$=\begin {bmatrix} 0&1\\1&0\end{bmatrix},$ which is not an identify matrix of order

$2 \times 2$ .

Hence given statement is false.

$A$ and

$B$ are two non-singular matrices and both are symmetric and commute each other, then

0%
Both $A^{-1}B$ and $A^{-1}B^{-1}$ are symmetric
0%
$A^{-1}B$ is symmetric but $A^{-1}B^{-1}$ is not symmetric
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$A^{-1}B^{-1}$ is symmetric but $A^{-1}B$ is not symmetric
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Neither $A^{-1}B$ nor $A^{-1}B^{-1}$ are symmetric

Explanation

$\begin{array}{l} \left| A \right| \ne 0\, \, \& \, \, \left| B \right| \ne 0 \\ { \left( { { A^{ -1 } }{ B^{ -1 } } } \right) ^{ T } }={ \left( { { B^{ -1 } } } \right) ^{ T } }{ \left( { { A^{ -1 } } } \right) ^{ T } } \\ { \left( { { A^{ -1 } }{ B^{ -1 } } } \right) ^{ T } }={ \left( { { B^{ T } } } \right) ^{ -1 } }{ \left( { { A^{ T } } } \right) ^{ -1 } }\, \, \, \, \left[ \begin{array}{l} { A^{ T } }=A \\ { B^{ T } }=B \end{array} \right] \\ { \left( { { A^{ -1 } }{ B^{ -1 } } } \right) ^{ T } }={ B^{ -1 } }{ A^{ -1 } } \\ { \left( { { A^{ -1 } }{ B^{ -1 } } } \right) ^{ T } }={ \left( { AB } \right) ^{ -1 } }\, \, \, \, \, \, \left[ { AB=BA } \right] \\ { \left( { { A^{ -1 } }{ B^{ -1 } } } \right) ^{ T } }={ \left( { BA } \right) ^{ -1 } } \\ { \left( { { A^{ -1 } }{ B^{ -1 } } } \right) ^{ T } }={ A^{ -1 } }{ B^{ -1 } } \\ { A^{ -1 } }{ B^{ -1 } }\, is\, \, a\, symmetric\, \, matrix \end{array}$

similarly

$A^{-1}B$ is also symmetric matrix.

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

$kA = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$ , then the value of

$k, a, b$ , are respectively.

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$-6, -12, -18$
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$-6, 4, 9$
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$-6, -4, -9$
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$-6, 12, 18$

Explanation

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

then,

$kA=\begin{bmatrix} 0 & 2k \\ 3k & -4k \end{bmatrix}$

But it is given that

$kA=\begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$

So,

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

$=\begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$

$\Rightarrow -4k=24$

$k=-6$

$3a=2k$

$3a=-12$

$a=-4$

$2b=3k$

$2b=-18$

$b=-9$

$A = \left[ {\begin{array}{*{20}{c}}2 & 0 & 0\\0 & 2 & 0\\0 & 0 & 2\end{array}} \right]$ then

$A^6$ = _____________.

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$6A$
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$12A$
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$16A$
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$32A$

Explanation

$A=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}\\ A^{ 2 }=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}=\begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}\\ A^{ 4 }=A^{ 2 }A^{ 2 }=\begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}\begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}=\begin{bmatrix} 16 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 16 \end{bmatrix}\\ A^{ 6 }=A^{ 4 }A^{ 2 }=\begin{bmatrix} 16 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 16 \end{bmatrix}\begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}=\begin{bmatrix} 64 & 0 & 0 \\ 0 & 64 & 0 \\ 0 & 0 & 64 \end{bmatrix}\\ taking\quad 32\quad out\quad we\quad get\\ A^{ 6 }=32\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}=32A$

$\left[ {\begin{array}{*{20}{c}}x & 4 & { - 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2 & 1 & 0\\1 & 0 & 2\\0 & 2 & 4\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\4\\{ - 1}\end{array}} \right] = 0$ then

$x$ is equal to

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$- 1 + \sqrt 6$
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$8 \pm \sqrt 5$
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$- 2 \pm \sqrt {10}$
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$3 \pm \sqrt 6$

Explanation

$\left[ \begin{matrix} x & 4 & 1 \end{matrix} \right]_{1\times 3}\left[ \begin{matrix}2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 4 \end{matrix} \right]_{3\times 3}\left[ \begin{matrix} x \\ 4 \\ -1 \end{matrix} \right]_{3\times 1}=0$

$\left[ \begin{matrix} x & 4 & 1 \end{matrix} \right]_{1\times 3}\left[ \begin{matrix} 2x+4+0 \\ x+0-2 \\ 0+8-4 \end{matrix} \right]_{3\times 1}=0$

$\Rightarrow\,2{x}^{2}+4x+4x-8-4=0$

$\Rightarrow\,2{x}^{2}+8x-12=0$

$\Rightarrow\,{x}^{2}+4x-6=0$

$\Rightarrow\,{x}^{2}+2\times 2x+4-4-6=0$

$\Rightarrow\,{\left(x+2\right)}^{2}=10$

$\Rightarrow\,x+2=\pm\sqrt{10}$

$\therefore\,x=-2\pm\sqrt{10}$

$A=\begin{bmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{bmatrix},B=\begin{bmatrix} { a }^{ 2 } & ab & ac \\ ba & { b }^{ 2 } & bc \\ ca & cb & { c }^{ 2 } \end{bmatrix}$ then

$AB=$

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$O$
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$I$
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$2I$
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None of these

Explanation

$AB=\begin{bmatrix} 0 & c & -b \\ -c & b & a \\ b & -a & 0 \end{bmatrix}\begin{bmatrix} a^{ 2 } & ab & ac \\ ba & b^{ 2 } & bc \\ ca & cb & c^{ 2 } \end{bmatrix}\\ =\begin{bmatrix} abc-abc & b^{ 2 }c^{ 1 }-b^{ 2 }c & bc^{ 2 }-bc^{ 2 } \\ -a^{ 2 }c+a^{ 2 }c & -abc+abc & -ac^{ 2 }+ac^{ 2 } \\ a^{ 2 }b-a^{ 2 }b & ab^{ 2 }-ab^{ 2 } & abc-abc \end{bmatrix}\\ =\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}=0\Rightarrow (A)$

The value of

$x$ , so that

$\left\lceil 1\quad x\quad 1 \right\rceil \begin{bmatrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix}=0$ is

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$\cfrac{-7\pm \sqrt{35}}{2}$
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$\cfrac{-9\pm \sqrt{53}}{2}$
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$\pm 2$
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$\dfrac{-7}{10}$

Explanation

Consider,

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

$\begin{bmatrix} 1 & 6+5x & 5\end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ x\end{bmatrix}=7+10x$

By given, we have

$7+10x =0$

$\Rightarrow x=-\dfrac{7}{10}$ .

If A be square matrix of order n and k is a scalar, then adj (KA) is:

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$K^{n}(adjA)$
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K (adj A)
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$K^{n-1}(adjA)$
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$K^{n+1}(adjA)$

CBSE Questions for Class 12 Commerce Maths Matrices Quiz 9 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Matrices Quiz 9 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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