If $$D_{1}$$ and $$D_{2}$$ are two 3 x 3 diagonal matrices, then

$$D_{1}D_{2}$$ is a diagonal matrix

$$D_{1}+D_{2}$$ is a diagonal matrix

$$D_{1}^{2}+D_{2}^{2}$$ is a diagonal matrix

MCQ Questions for Class 12 Commerce Maths Matrices Quiz 3

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

$A^6 =$

0%
$6A$
0%
$12A$
0%
$16A$
0%
$32A$

Explanation

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

$\therefore A^6 = (2I)^6 = 2^6\times I^6 = 2^6I$

$A^6 = 2^6I = 2^5(2I) = 2^5A = 32A$

Consider the following statements in respect of the matrix

$A = \begin{bmatrix} 0 & 1 & 2\\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{bmatrix}$ :
The matrix A is skew-symmetric.
The matrix A is symmetric.
The matrix A is invertible.
Which of the above statements is/are correct ?

0%
1 only
0%
3 only
0%
1 and 3
0%
2 and 3

Explanation

For a matrix to be skew-symmetric

${ A }^{ T }=-A$

While for a matrix to be symmetric

${ A }^{ T }= A$

and for a matrix to be invertible

$\left| A \right| \neq 0$

$\Rightarrow$ so for

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

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

and

$\left| A \right| =0$

Hence it is skew symmetric but not invertible

If the sum of the matrices

$\begin{bmatrix} x \\ x \\ y \end{bmatrix},\begin{bmatrix} y \\ y \\ z \end{bmatrix}$ and

$\begin{bmatrix} z \\ 0 \\ 0 \end{bmatrix}$ is the matrix

$\begin{bmatrix} 10 \\ 5 \\ 5 \end{bmatrix}$ , then what is the value of

$y$ ?

0%
$-5$
0%
$0$
0%
$5$
0%
$10$

Explanation

Since, we have

$\left[ \begin{matrix} x \\ x \\ y \end{matrix} \right] +\left[ \begin{matrix} y \\ y \\ z \end{matrix} \right] +\left[ \begin{matrix} z \\ 0 \\ 0 \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \\ 5 \end{matrix} \right]$

$\therefore x+y+z=10,$

$x+y=5$ and

$y+z=5$

Replacing

$x+y=5$ in

$x+y+z=10$
We have,

$z=5$
Also,

$y+z=5$

$\therefore y=5-z=0$
Option B is correct.

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

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

$BA =$

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

Explanation

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

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

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

$A$ and

$B$ are square matrices such that

$AB = I$ and

$BA = I$ , then

$B$ is

0%
Unit matrix
0%
Null matrix
0%
Multiplicative inverse matrix of $A$
0%
$- A$

Explanation

$AB=I$

$\&$

$BA=I$ then

$B$ is the multiplicative inverse of

$A.$

Hence, the answer is multiplicative inverse matrix of

$A$ .

What is

$\begin{bmatrix} x & y & z \end{bmatrix} \begin{bmatrix} a& h & g\\ h & b & f\\ g & f & c\end{bmatrix}$ equal to?

0%
$\begin{bmatrix}ax + hy + gz & h + b + f & g + f + c\end{bmatrix}$
0%
$\begin{bmatrix}a & h & g\\ hx & by & fz\\ g & f & c\end{bmatrix}$
0%
$\begin{bmatrix}ax & hy & gz\\ hx & by & fz\\ gx & fy & cz\end{bmatrix}$
0%
$\begin{bmatrix} ax + hy + gz & hx + by + fz & gx + fy + cz\end{bmatrix}$

Explanation

Order of matrix

$\begin{bmatrix} x& y & z \end{bmatrix}$ is

$(1\times 2)$

Order of matrix

$\begin{bmatrix} a & h & g\\ h & b & f\\ g & f & c\end{bmatrix}$ is

$(3\times 3)$

So, the product

$\begin{bmatrix} x& y& z \end{bmatrix}$

$\begin{bmatrix} a & h & g\\ h & b & f\\ g & f & c\end{bmatrix} =\begin{bmatrix} ax+hy+gz & hx+by+fz & gx+fy+cz\end{bmatrix}$

$A$ is any matrix, then the product

$AA$ is defined only when A is a matrix of order

$m \times n$ where :

0%
$m > n$
0%
$m < n$
0%
$m=n$
0%
$m \leq n$

Explanation

Let two matrix be

$A$ and

$B$ of order

$m\times n$ and

$l\times k$ respectively.

They can be multiplied if

$n=l$

So, if a matrix

$A$ of order

$m\times n$ is to be multiplied by

$A$

Then,

$m=n$

Consider the following statements:
The product of two non-zero matrices can never be identity matrix.
The product of two non-zero matrices can never be zero matrix.
Which of the above statements is/are correct?

0%
1 only
0%
2 only
0%
Both 1 and 2
0%
Neither 1 nor 2

Explanation

Product of a matrix and it's inverse is always an identity matrix.

Example :-

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

And both are non-zero.

Thus, statement

$1$ is incorrect.

Product of two non-zero matrices can be a zero matrix. Example,

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

Thus, statement

$2$ is incorrect.

Hence, option D is correct.

$[2\ 3\ 4] \begin{bmatrix}1 & x &3 \\ 2 & 4 & 5\\ 3 & 2 &x \end{bmatrix} \begin{bmatrix} x\\ 2 \\ 0 \end{bmatrix} = 0$ , then

$x =$ ________.

0%
$\dfrac {7}{3}$
0%
$\dfrac {5}{3}$
0%
$-\dfrac {5}{3}$
0%
$-\dfrac {7}{3}$

Explanation

Given,

$[2\ 3\ 4] \begin{bmatrix}1 & x &3 \\ 2 & 4 & 5\\ 3 & 2 &x \end{bmatrix} \begin{bmatrix} x\\ 2 \\ 0 \end{bmatrix} = 0$

$\begin {bmatrix} 20 &20+2x&21+4x \end {bmatrix} \begin{bmatrix} x\\ 2 \\ 0 \end{bmatrix} = 0$

$\Rightarrow 20x+40+4x=0$

$\Rightarrow 24x + 40 = 0$

$\Rightarrow x = -\dfrac {5}{3}$

$A=\begin{bmatrix} 2 & x-3 & x-2 \\ 3 & -2 & -1 \\ 4 & -1 & -5 \end{bmatrix}$ is a symmetric matrices then

$x=$

0%
$0$
0%
$3$
0%
$6$
0%
$8$

Explanation

$A$ is a symmetric matrix then

$a_{ij}=a_{ji}$

i.e.

$a_{12}=a_{21}$ and

$a_{13}=a_{31}$

$\therefore x-3=3$ and

$x-2=4$

adding both we get

$2x-5=7$

$\Rightarrow 2x=12 \Rightarrow x=6$

Given

$A$ is a matrix of order

$3\times 2$ . If order of

$AB$ is

$3\times 3$ , then order of

$B$ will be

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

Explanation

Given

$A$ is a matrix of order

$3\times 2$ .

Now

$AB$ is of order

$3\times 3$ .

The product

$AB$ is possible if the number of columns of

$A(=2)$ is equal to the number of rows of

$B$ .

Since order of

$AB$ is

$3\times 3$ .

Also the number of columns of

$B$ is same as the number of columns of

$AB$ .

Now

$A$ is of order

$2\times 3$ .

The inverse of

$\begin{bmatrix} 1 & a & b \\ 0 & x & 0 \\ 0 & 0 & 1 \end{bmatrix}$ is

$\begin{bmatrix} 1 & -a & -b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ then

$x=$

0%
$a$
0%
$b$
0%
$0$
0%
$1$

Explanation

$\begin{array}{l} \left[ { \begin{array} { *{ 20 }{ c } }1 & a & b \\ 0 & x & 0 \\ 0 & 0 & 1 \end{array} } \right] \left[ { \begin{array} { *{ 20 }{ c } }1 & { -a } & { -b } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} } \right] =\left[ { \begin{array} { *{ 20 }{ c } }1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} } \right] \\ =\left[ { \begin{array} { *{ 20 }{ c } }1 & 0 & 0 \\ 0 & x & 0 \\ 0 & 0 & 1 \end{array} } \right] =\left[ { \begin{array} { *{ 20 }{ c } }1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} } \right] \\ \Rightarrow x=1 \end{array}$

Given

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

$B = \begin{bmatrix} 24 \\ 7\end{bmatrix},$ find the matrix

$X$ such that

$AX=B$ .

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

Explanation

Given,

$AX=B$

To perform matrix multiplication,

$AX$ the number of columns in A must equal the number of rows in

$X$

Hence,

$X$ should be matrix of order

$2\times1$

Say,

$X= \begin{bmatrix}x \\y\end{bmatrix}$

Consider,

$AX=B$

$\begin{bmatrix}3&4 \\4&-3\end{bmatrix}\begin{bmatrix}x \\y\end{bmatrix}=\begin{bmatrix}24 \\7\end{bmatrix}$

$\begin{bmatrix}3x+4y \\4x-3y\end{bmatrix}=\begin{bmatrix}24 \\7\end{bmatrix}$

$\therefore 3x+4y=24$ and

$4x-3y=7$

Solving the above equations, we get

$x=4 \ , \ y=3$

$\therefore X= \begin{bmatrix}4 \\3\end{bmatrix}$

Option B.

$A=\begin{bmatrix} \cos { x } & \sin { x } \\ -\sin { x } & \cos { x } \end{bmatrix}$ and

$A(AdjA)=k\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ then the value of

$k$ is

0%
$\sin{x}\cos{x}$
0%
$1$
0%
$-1$
0%
$2$

Explanation

$A(AdjA)=\begin{bmatrix}c^{2}+s^{2}&cs-cs\\-cs+cs&c^{2}+s^{2}\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\implies k=1$

$A=\begin{bmatrix} 0 & a+1 & b-2 \\ 2a-1 & 0 & c-2 \\ 2b+1 & 2+c & 0 \end{bmatrix}$ is skew symmetric then

$a+b+c$ =

0%
$3$
0%
$-3$
0%
$\cfrac{1}{3}$
0%
$-\cfrac{1}{3}$

Explanation

$A^{T}+A=0\implies a+1+2a-1=0;b-2+2b+1=0;c+2+c-2=0\implies a=c=0,$

$b=\displaystyle \frac{1}{3}\implies a+b+c=\frac{1}{3}$

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

$\left[ {3{A^2} + 12A} \right]$ is equal to

0%
$\left[ {\begin{array}{*{20}{c}} {72}&{ - 84} \\ { - 63}&{51} \end{array}} \right]$
0%
$\left[ {\begin{array}{*{20}{c}} {51}&{63} \\ {84}&{72} \end{array}} \right]$
0%
$\left[ {\begin{array}{*{20}{c}} {51}&{84} \\ {63}&{72} \end{array}} \right]$
0%
$\left[ {\begin{array}{*{20}{c}} {72}&{ - 63} \\ { - 84}&{51} \end{array}} \right]$

Explanation

We have,

$A=\left[ \begin{matrix} 2 & -3 \\ -4 & 1 \end{matrix} \right]$

$A^2 = A \times A =\left[ \begin{matrix} 2 & -3 \\ -4 & 1 \end{matrix} \right] \times \left[ \begin{matrix} 2 & -3 \\ -4 & 1 \end{matrix} \right]$

$A^2=\left[ \begin{matrix} 16 & -9 \\ -12 & 13 \end{matrix} \right]$

Thus,

$3A^2=\left[ \begin{matrix} 48 & -27 \\ -36 & 39 \end{matrix} \right]$

Now,

$12A=\left[ \begin{matrix} 24 & -36 \\ -48 & 12 \end{matrix} \right]$

Therefore,

$[3A^2 + 12A]=\left[ \begin{matrix} 48 & -27 \\ -36 & 39 \end{matrix} \right] +\left[ \begin{matrix} 24 & -36 \\ -48 & 12 \end{matrix} \right] \\ [3A^2 +12A]= \left[ \begin{matrix} 72 & -63 \\ -84 & 51 \end{matrix} \right]$

The Inverse of a square matrix, if it exist is unique.

0%
True
0%
False

Explanation

$Inverse\>of\>any\>square\>matrix\>is\>always\>unique\>as\>for\>it\\AA^{-1}=I\\where\>I\>=\>identity\>matrix$

Hence True

$A=\begin{bmatrix} a & b \\ 0 & c \end{bmatrix}$ then

${A}^{-1}+(A-aI)(A-cI)=$

0%
$\cfrac { 1 }{ ac } \begin{bmatrix} a & b \\ 0 & -c \end{bmatrix}$
0%
$\cfrac { 1 }{ ac } \begin{bmatrix} -a & b \\ 0 & c \end{bmatrix}$
0%
$\cfrac { 1 }{ ac } \begin{bmatrix} c & -b \\ 0 & a \end{bmatrix}$
0%
$\cfrac { 1 }{ ac } \begin{bmatrix} c & b \\ 0 & a \end{bmatrix}$

Explanation

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

$A^{-1}=\dfrac{1}{ac-0(b)}\begin{bmatrix}c&-b\\0&a\end{bmatrix}=\dfrac{1}{ac}\begin{bmatrix}c&-b\\0&a\end{bmatrix}$

$(A-aI)(A-cI)=A^{2}-(a+c)A+acI=\begin{bmatrix}a&b\\0&c\end{bmatrix}\begin{bmatrix}a&b\\0&c\end{bmatrix}-(a+c)\begin{bmatrix}a&b\\0&c\end{bmatrix}+acI=\begin{bmatrix}0&0\\0&0\end{bmatrix}$

$A^{-1}+(A-{a}I)(A-{c}I)=\dfrac{1}{ac}\begin{bmatrix}c&-b\\0&a\end{bmatrix}$

$A$ is square matrix such that

${A^2} = 1$ then

${A^{ - 1}} = ?$

0%
$2$ A
0%
A
0%
$0$
0%
A+ $1$

Explanation

${ A }^{ 2 }={ 1 }$ can be witten as

${ A }^{ 2 }={ I }$

multiply

${ { A }^{ -1 } }$ on both side

$\\ { { A }^{ -1 } }\times ({ A }\times { A })={ A }^{ -1 }\times { I }\\ ({ A }^{ -1 }\times { A })\times { A }={ A }^{ -1 }\\ { I }\times { A }={ { A }^{ -1 } }\\ { A }^{ -1 }={ A }\\$

Corrrect option is B

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

$A^ {2}$ is equal to

0%
$A$
0%
$-A$
0%
$2A$
0%
$-2A$

Explanation

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

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

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

$\therefore A^2=2A$ .

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

$A^{100}$ =..............

0%
100A
0%
$2^{99}A$
0%
$2^{100}A$
0%
99A

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

$A^3-2A^2$ is

0%
a null matrix
0%
an identity matrix
0%
equal to A
0%
equal to -A

Explanation

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

$\Rightarrow$

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

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

$=\begin{bmatrix}1+1&-1-1\\-1-1&1+1\end{bmatrix}$

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

$\Rightarrow$

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

$\Rightarrow$

$A^3=$

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

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

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

$=\begin{bmatrix}4&-4\\-4&4\end{bmatrix}$

$\Rightarrow$

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

$\Rightarrow$

$A^3-2A^2=$

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

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

$=\begin{bmatrix}4&-4\\-4&4\end{bmatrix}-\begin{bmatrix}4&-4\\-4&4\end{bmatrix}$

$=\begin{bmatrix}4-4&-4+4\\-4+4&4-4\end{bmatrix}$

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

$\therefore$ The expression

$A63-2A^2$ is null matrix.

State true/false:

Matrices of any order can be added.

0%
True
0%
False

$\left[\begin{array}{ll} x & 1\\ 1 & y \end{array}\right]\left[\begin{array}{ll} 1 & 4\\ 2 & 6 \end{array}\right] =\left[\begin{array}{ll} 4 & 14\\ 7 & 22 \end{array}\right]$ , then

$(x,y)$

$=$

0%
$(1,-2)$
0%
$( 2,1)$
0%
$(3,2)$
0%
$(2,3)$

Explanation

We have,

$\left[\begin{array}{ll} x & 1\\ 1 & y \end{array}\right]\left[\begin{array}{ll} 1 & 4\\ 2 & 6 \end{array}\right] =\left[\begin{array}{ll} 4 & 14\\ 7 & 22 \end{array}\right]$

$\Rightarrow \left[\begin{array}{ll} x\times1+1\times 2 & x\times4+1\times 6 \\ 1\times 1+y\times 2 & 1\times 4+y\times 6 \end{array}\right]=\left[\begin{array}{ll} 4 & 14\\ 7 & 22 \end{array}\right]$

$\Rightarrow \left[\begin{array}{ll} x+2 & 4x+6\\ 2y+1 & 6y+4 \end{array}\right]=\left[\begin{array}{ll} 4 & 14\\ 7 & 22 \end{array}\right]$

Now equating corresponding elements,
we get,

$(x,y)=(2,3)$

$D_{1}$ and

$D_{2}$ are two 3 x 3 diagonal matrices, then

0%
$D_{1}D_{2}$ is a diagonal matrix
0%
$D_{1}+D_{2}$ is a diagonal matrix
0%
$D_{1}^{2}+D_{2}^{2}$ is a diagonal matrix
0%
1, 2, 3 are correct

Explanation

$D_{1}$ and

$D_{2}$ are diagonal matrix then

$D_{1}D_{2},D_{1}+D_{2}$ is also a diagonal matrix.

$A=\begin{bmatrix} a_{11} & & & \\ & a_{22} & & \\ & & . & \\ & & &a_{nn} \end{bmatrix}$

$B=\begin{bmatrix} l_{11} & & & \\ & l_{22} & & \\ & & . & \\ & & &l_{nn} \end{bmatrix}$

then,

$AB=\begin{bmatrix} a_{11}l_{11} & & & \\ & a_{22}l_{22} & & \\ & & . & \\ & & &a_{nn}l_{nn} \end{bmatrix}$

$A+B=\begin{bmatrix} a_{11}+l_{11} & & & \\ & a_{22}+l_{22} & & \\ & & . & \\ & & &a_{nn}+l_{nn} \end{bmatrix}$

$A^2=\begin{bmatrix} a_{11}^{2} & & & \\ & a_{22}^{2} & & \\ & & . & \\ & & &a_{nn}^{2} \end{bmatrix}$

$\therefore D_{1}^{2}+D_{2}^{2}$ is also a diagonal matrix.

$\left[\begin{array}{lll} x & 0 & 0\\ y & \mathrm{z} & 0\\ l & m & n \end{array}\right]\left[\begin{array}{lll} a & 0 & 0\\ 0 & b & 0\\ 0 & 0 & c \end{array}\right] =$

0%
$\left[\begin{array}{lll} ax & 0 & 0\\ ay & b_{Z} & 0\\ al & mb & nc \end{array}\right]$
0%
$\left[\begin{array}{lll} 0 & 0 & nc\\ 0 & b_{Z} & mb\\ a\mathrm{x} & ab & al \end{array}\right]$
0%
$\left[\begin{array}{lll} a\mathrm{x} & ab & al\\ 0 & b_{Z} & mb\\ 0 & 0 & nc \end{array}\right]$
0%
None of these

Explanation

We know that for the multiplication of matrices

$AB=C$
So,

$C_{ij}=a_{ir} b_{rj}$

$\therefore$

$\begin{bmatrix} x & 0 & 0\\ y & z & 0\\ l & m & n \end{bmatrix}\begin{bmatrix} a & 0& 0\\ 0 & b & 0\\ 0 & 0 & c \end{bmatrix}$

$=\begin{bmatrix} ax & 0 & 0\\ ay & bz& 0\\ al & mb & nc \end{bmatrix}$

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

$A^{2}=$

0%
$\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$
0%
$\begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix}$
0%
$\begin{bmatrix} { 3 } & { -4 } \\ { 2 } & -1 \end{bmatrix}$
0%
None of these

Explanation

Given,

$A=\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}\\ A^{ 2 }=\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}\times \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}\\ =\begin{bmatrix} 9-4 & -12+4 \\ 3-1 & -4+1 \end{bmatrix}=\begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix}$

$A=\begin{bmatrix} a & b \\ b & a \end{bmatrix}$ and

${ A }^{ 2 }=\begin{bmatrix} \alpha & \beta \\ \beta & \alpha \end{bmatrix}$ , 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%
$\alpha ={ a }^{ 2 }+{ b }^{ 2 },\beta =ab$

Explanation

We are given

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

Now we find

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

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

Compare it with the given

${ A }^{ 2 }$ matrix. we get

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

$A=[a_{ij}]_{3\times 3}$ is a square matrix so that

$a_{ij}=i^{2}-j^{2}$ , then

$A$ is a

0%
unit matrix
0%
symmetric marix
0%
skew symmetric matrix
0%
orthogonal matrix

Explanation

Given:

$A = [a_{ij}]_{(3\times3)}$

where,

$a_{ij} = i^2-j^2$

$\therefore a_{ij}=0$ if

$i=j$

Now,

$a_{12}=1^{2}-2^{2}=-3$

$a_{13}=1^{2}-3^{2}=-8$

$a_{21}=2^2-1^2 = 3$

$a_{23}=2^{2}-3^{2}=-5$

$a_{31}=3^2 - 1^2 = 8$

$a_{32}=3^2-2^2 = 5$

$\therefore A=\begin{bmatrix} 0 &-3 &-8 \\ 3 & 0&-5 \\ 8& 5 &0 \end{bmatrix}$

Here,

$A^T=-A$

$\therefore\ A$ is a skew-symmetric matrix.

Hence, option C.

$\mathrm{A}= \left[\begin{array}{lll} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{array}\right]$ , then

$\mathrm{A}^{4}$ is equal to

0%
$16\mathrm{A}$
0%
$32 \mathrm A$
0%
$4\mathrm{A}$
0%
$8\mathrm{A}$

Explanation

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

$A^{4} = 16 \begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 &0 &1 \end{bmatrix}$

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

$\therefore A^{4} = 8 \times A$

CBSE Questions for Class 12 Commerce Maths Matrices Quiz 3 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Matrices Quiz 3 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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