MCQ 11 Mark
For any $2 \times 2$ matrix $P$, which of the following matrices can be $Q$ such that $P Q=Q P$ ?
AnswerCorrect option: B. $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
View full question & answer→MCQ 21 Mark
If $A=\left[\begin{array}{lll}1 & -1 & 1 \\ 1 & -1 & 1 \\ 1 & -1 & 1\end{array}\right]$, then $A^5-A^4-A^3+A^2$ is equal to
View full question & answer→MCQ 31 Mark
If order of matrix $A$ is $2 \times 3$, of matrix $B$ is $3 \times 2$, and of matrix $C$ is $3 \times 3$, then which one of the following is not defined?
Answer\[\begin{array}{l}
\text { (a) : Consider } C\left(A+B^{\prime}\right) \text { i.e., } C_{3 \times 3}\left(A_{2 \times 3}+B_{2 \times 3}^{\prime}\right) \\
=C_{3 \times 3}\left(A+B^{\prime}\right)_{2 \times 3}
\end{array}\]
Here, number of columns in the matrix $C$ is 3 and number of rows in the matrix $\left(A+B^{\prime}\right)$ is 2.so , it is not defined.
View full question & answer→MCQ 41 Mark
If $A=\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]$ and $B=\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]$, then
- A
$A^{-1}=B$
- B
$A^{-1}=6 B$
- C
$B^{-1}=B$
- D
$B^{-1}=\frac{1}{6} A$
AnswerWe have,
\[\begin{array}{l}
A B=\left[\begin{array}{ccc}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\left[\begin{array}{ccc}
2 & 2 & -4 \\
-4 & 2 & -4 \\
2 & -1 & 5
\end{array}\right] \\
=\left[\begin{array}{ccc}
2+4+0 & 2-2+0 & -4+4+0 \\
4-12+8 & 4+6-4 & -8-12+20 \\
0-4+4 & 0+2-2 & 0-4+10
\end{array}\right] \\
=\left[\begin{array}{lll}
6 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 6
\end{array}\right]=61 \Rightarrow B^{-1}=\frac{1}{6} A
\end{array}\]
View full question & answer→MCQ 51 Mark
If $\left[\begin{array}{cc}2 a+b & a-2 b \\ 5 c-d & 4 c+3 d\end{array}\right]=\left[\begin{array}{cc}4 & -3 \\ 11 & 24\end{array}\right]$, then value of $a+b-c+2 d$ is
AnswerFrom the definition of equality of two matrices, we have
$
\begin{array}{l}
2 a+b=4 ....(i)\\
5 c-d=11 .....(iii)
\end{array}
$ $
\begin{array}{l}
a-2 b=-3.....(ii) \\
4 c+3 d=24......(iv)
\end{array}
$
Solving (i) and (ii), we get
$
5 a=5 \Rightarrow a=1, b=2
$
Solving (iii) and (iv), we get
$
\begin{aligned}
& 19 c=57 \Rightarrow c=3, d=4 \\
\therefore \quad & a+b-c+2 d=1+2-3+8=8
\end{aligned}
$
View full question & answer→MCQ 61 Mark
Given that $A=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]$ and $A^2=31$, then
- A
$1+\alpha^2+\beta \gamma=0$
- B
$1-\alpha^2-\beta \gamma=0$
- C
$3-\alpha^2-\beta \gamma=0$
- D
$3+\alpha^2+\beta \gamma=0$
AnswerWe have,
\[\begin{array}{l}
\text { 8. (c) : We have, } A=\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right] \\
\Rightarrow A^2=\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]=\left[\begin{array}{cc}
\alpha^2+\beta \gamma & 0 \\
0 & \gamma \beta+\alpha^2
\end{array}\right]
\end{array}\]
But $A^2=31$
\[\begin{array}{l}
\Rightarrow\left[\begin{array}{cc}
\alpha^2+\beta \gamma & 0 \\
0 & \alpha^2+\beta \gamma
\end{array}\right]=\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right] \\
\Rightarrow \alpha^2+\beta \gamma=3 \\
\Rightarrow 3-\alpha^2-\beta \gamma=0
\end{array}\]
View full question & answer→MCQ 71 Mark
If $A=\left[\begin{array}{rr}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$ and $A+A^{\prime}=I$, then the value of $\alpha$ is
- A
$\frac{\pi}{6}$
- B
$\frac{\pi}{3}$
- C
$\pi$
- D
$\frac{3 \pi}{2}$
Answer$\begin{array}{l}\text {We have, } A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right] \\ \text { and } A+A^{\prime}=I \\ \Rightarrow\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]+\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\ \Rightarrow\left[\begin{array}{cc}2 \cos \alpha & 0 \\ 0 & 2 \cos \alpha\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\ \Rightarrow \quad 2 \cos \alpha=1 \Rightarrow \cos \alpha=\frac{1}{2} \Rightarrow \alpha=\frac{\pi}{3}\end{array}$
View full question & answer→MCQ 81 Mark
If $A=\left[\begin{array}{cc}0 & 2 \\ 3 & -4\end{array}\right]$ and $k A=\left[\begin{array}{cc}0 & 3 a \\ 2 b & 24\end{array}\right]$, then the values of $k$, $a$ and $b$ respectively are
- A
$-6,-12,-18$
- B
$-6,-4,-9$
- C
$-6,4,9$
- D
$-6,12,18$
Answer$\begin{array}{l}\text { We have, } A=\left[\begin{array}{cc}0 & 2 \\ 3 & -4\end{array}\right] \Rightarrow k A=\left[\begin{array}{cc}0 & 2 k \\ 3 k & -4 k\end{array}\right] \\ \Rightarrow\left[\begin{array}{cc}0 & 3 a \\ 2 b & 24\end{array}\right]=\left[\begin{array}{cc}0 & 2 k \\ 3 k & -4 k\end{array}\right] \text { (Given) } \\ \Rightarrow-4 k=24,3 a=2 k, 2 b=3 k \\ \Rightarrow k=-6, a=-4, b=-9\end{array}$
View full question & answer→MCQ 91 Mark
If $A$ is square matrix such that $A^2=A$, then $(I+A)^3-7 A$ is equal to
AnswerWe have, $(I+A)^3-7 A$
\[\begin{array}{l}
=I^3+A^3+3 I^2 A+3 I A^2-7 A=1+A \cdot A+3 A+3 A-7 A \\
=1+A+3 A+3 A-7 A=1
\end{array}\]
View full question & answer→MCQ 101 Mark
If $A=\left[\begin{array}{cc}4 & 2 \\ -1 & 1\end{array}\right]$, then $(A-2 l)(A-3 l)$ is equal to
View full question & answer→MCQ 111 Mark
Given that matrices $A$ and $B$ are of order $3 \times n$ and $m \times 5$ respectively, then the order of matrix $C=5 A+3 B$ is
- A
$3 \times 5$ and $m=n$
- B
$3 \times 5$
- C
$3 \times 3$
- D
$5 \times 5$
AnswerWe know that the sum of two matrices is defined only if both matrices have same order.
Here $5 A+3 B$ is defined if $A$ and $B$ have same order.
\[\Rightarrow 3 \times n=m \times 5 \Rightarrow n=5, m=3\]
So, order of matrix $C$ is $3 \times 5$ and $m \neq n$.
View full question & answer→MCQ 121 Mark
If $P$ is a $3 \times 3$ matrix such that $P^{\prime}=2 P+I$, where $P^{\prime}$ is the transpose of $P$, then
- A
$P=1$
- B
$P=-1$
- C
$P=21$
- D
$P=-21$
AnswerWe have, $P^{\prime}=2 P+I$
Now, $\left(P^{\prime}\right)^{\prime}=(2 P+l)^{\prime}=2 P^{\prime}+I$
$\Rightarrow P=2(2 P+1)+1$
[Using (i)]
$\Rightarrow P=4 P+3 I \Rightarrow P=-I$
View full question & answer→MCQ 131 Mark
If $A=\left[\begin{array}{ll}x & 0 \\ 1 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}4 & 0 \\ -1 & 1\end{array}\right]$, then value of $x$ for which $A^2=B$ is
AnswerGiven, $A=\left[\begin{array}{ll}x & 0 \\ 1 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}4 & 0 \\ -1 & 1\end{array}\right]$
Now, $A^2=\left[\begin{array}{ll}x & 0 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}x & 0 \\ 1 & 1\end{array}\right]=\left[\begin{array}{cc}x^2 & 0 \\ x+1 & 1\end{array}\right]$
\[\begin{array}{c}
A^2=B \\
\Rightarrow\left[\begin{array}{cc}
x^2 & 0 \\
x+1 & 1
\end{array}\right]=\left[\begin{array}{cc}
4 & 0 \\
-1 & 1
\end{array}\right]
\end{array}\]
On comparing, we get $x^2=4$ and $x+1=-1$
$\Rightarrow x= \pm 2$ and $x=-2$
$\therefore \quad$ Required value of $x$ is -2 .
View full question & answer→MCQ 141 Mark
Find the matrix $A^2$, where $A=\left[a_{i j}\right]$ is a $2 \times 2$ matrix whose elements are given by $a_{i j}=$ maximum $(i, j)-$ minimum $(i, j)$ :
- A
$\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
- B
$\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
- C
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
- D
$\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$
Answer$\begin{array}{l}\text {(c) : Let } A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right] \\ a_{11}=\max (1,1)-\min (1,1)=1-1=0 \\ a_{12}=\max (1,2)-\min (1,2)=2-1=1 \\ a_{21}=\max (2,1)-\min (2,1)=2-1=1 \\ a_{22}=\max (2,2)-\min (2,2)=2-2=0 \\ \therefore A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] A^2=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\end{array}$
View full question & answer→MCQ 151 Mark
If $F(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$ and $[F(x)]^2=F(k x)$, then the value of $k$ is :
AnswerWe have, $[F(x)]^2=F(k x)$
\[\begin{array}{l}
\Rightarrow\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right] \\
\quad=\left[\begin{array}{ccc}
\cos k x & -\sin k x & 0 \\
\sin k x & \cos k x & 0 \\
0 & 0 & 1
\end{array}\right] \\
\Rightarrow\left[\begin{array}{ccc}
\cos 2 x & -\sin 2 x & 0 \\
\sin 2 x & \cos 2 x & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
\cos k x & -\sin k x & 0 \\
\sin k x & \cos k x & 0 \\
0 & 0 & 1
\end{array}\right] \\
\Rightarrow k=2
\end{array}\]
View full question & answer→MCQ 161 Mark
If $A=\left[\begin{array}{cc}2 & 1 \\ -4 & -2\end{array}\right]$, then the value of $I-A+A^2-A^3+\ldots$ is :
- A
$\left[\begin{array}{cc}-1 & -1 \\ 4 & 3\end{array}\right]$
- B
$\left[\begin{array}{cc}3 & 1 \\ -4 & -1\end{array}\right]$
- C
$\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
- D
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
AnswerGiven $A=\left[\begin{array}{cc}2 & 1 \\ -4 & -2\end{array}\right]$, then
\[\begin{array}{l}
A^2=\left[\begin{array}{cc}
2 & 1 \\
-4 & -2
\end{array}\right]\left[\begin{array}{cc}
2 & 1 \\
-4 & -2
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=0 \\
\Rightarrow A^n=0 \forall n \geq 2 \\
\therefore \quad I-A+A^2-A^3+\ldots=I-A \\
=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{cc}
2 & 1 \\
-4 & -2
\end{array}\right]=\left[\begin{array}{cc}
-1 & -1 \\
4 & 3
\end{array}\right]
\end{array}\]
View full question & answer→MCQ 171 Mark
If $A=\left[a_{i j}\right]$ be a $3 \times 3$ matrix, where $a_{i j}=i-3 j$, then which of the following is false?
- A
$a_{11}<0$
- B
$a_{12}+a_{21}=-6$
- C
$a_{13}>a_{31}$
- D
$a_{31}=0$
AnswerWe have, $a_{i j}=i-3 j$
(a) $a_{11}=1-3 \times 1=-2<0$
(b) $a_{12}+a_{21}=(1-3 \times 2)+(2-3 \times 1)=(-5)+(-1)=-6$
(c) $a_{13}=1-3 \times 3=-8$ and $a_{31}=3-3 \times 1=0>-8$
$\Rightarrow a_{31}>a_{13}$
(d) $a_{31}=0$
View full question & answer→MCQ 181 Mark
If for the matrix $A=\left[\begin{array}{cc}\tan x & 1 \\ -1 & \tan x\end{array}\right], A+A^{\prime}=2 \sqrt{3}$, , then the value of $x \in\left[0, \frac{\pi}{2}\right]$ is :
- A
- B
$\frac{\pi}{4}$
- C
$\frac{\pi}{3}$
- D
$\frac{\pi}{6}$
AnswerWe have, $A=\left[\begin{array}{cc}\tan x & 1 \\ -1 & \tan x\end{array}\right]$ $A+A^{\prime}=2 \sqrt{3}$ I
$\Rightarrow\left[\begin{array}{cc}\tan x & 1 \\ -1 & \tan x\end{array}\right]+\left[\begin{array}{cc}\tan x & -1 \\ 1 & \tan x\end{array}\right]=2 \sqrt{3}\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}2 \tan x & 0 \\ 0 & 2 \tan x\end{array}\right]=\left[\begin{array}{cc}2 \sqrt{3} & 0 \\ 0 & 2 \sqrt{3}\end{array}\right]$
On comparing, we get
\[\begin{array}{l}
2 \tan x=2 \sqrt{3} \Rightarrow \tan x=\sqrt{3}=\tan \frac{\pi}{3} \\
\Rightarrow \quad x=\frac{\pi}{3}
\end{array}\]
View full question & answer→MCQ 191 Mark
If the sum of all elements of a $3 \times 3$ scalar matrix is 9 , then the product of all its elements is :
AnswerAs we know that in a scalar matrix, every nondiagonal element is zero.
$\therefore \quad$ Product of all elements of the given scalar matrix $=0$.
View full question & answer→MCQ 201 Mark
Given that $\left[\begin{array}{ll}1 & x\end{array}\right]\left[\begin{array}{cc}4 & 0 \\ -2 & 0\end{array}\right]=0$, the value of $x$ is :
Answer$\begin{array}{l}\text {We have, }\left[\begin{array}{ll}1 & x\end{array}\right]\left[\begin{array}{cc}4 & 0 \\ -2 & 0\end{array}\right]=0 \\ \Rightarrow \quad[1 \times 4+x \times(-2) \quad 1 \times 0+x \times 0]=\left[\begin{array}{ll}0 & 0\end{array}\right] \\ \Rightarrow \quad 4-2 x=0 \Rightarrow 4=2 x \Rightarrow x=2\end{array}$
View full question & answer→MCQ 211 Mark
If $\left[\begin{array}{lll}a & c & 0 \\ b & d & 0 \\ 0 & 0 & 5\end{array}\right]$ is a scalar matrix, then the value of $a+2 b+3 c+4 d$ is :
AnswerWe have, $\left[\begin{array}{lll}a & c & 0 \\ b & d & 0 \\ 0 & 0 & 5\end{array}\right]$ is a scalar matrix.
\[\begin{array}{l}
\therefore \quad a=d=5, b=c=0 \\
a+2 b+3 c+4 d=5+2 \times 0+3 \times 0+4 \times 5=5+20=25
\end{array}\]
View full question & answer→MCQ 221 Mark
If $A=\left[a_{i j}\right]$ is a square matrix of order 2 such that $a_{i j}=\left\{\begin{array}{l}1, \text { when } i \neq j \\ 0, \text { when } i=j\end{array}\right.$, then $A^2$ is
- A
$\left[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right]$
- B
$\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]$
- C
$\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]$
- D
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Answer$\begin{array}{l}\text {We have, } A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \\ \therefore \quad A^2=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{ll}0+1 & 0+0 \\ 0+0 & 1+0\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\end{array}$
View full question & answer→MCQ 231 Mark
If $A=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]$ and $(3 I+4 A)(3 I-4 A)=x^2 I$, then the value(s) $x$ is/are:
- A
$\pm \sqrt{7}$
- B
- C
$\pm 5$
- D
AnswerGiven, $A=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]$
Now, $3 I+4 A=3\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]+4\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]$
$
\begin{array}{l}
=\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]+\left[\begin{array}{cc}
0 & 4 \\
-4 & 0
\end{array}\right]=\left[\begin{array}{cc}
3 & 4 \\
-4 & 3
\end{array}\right] \\
\text { and } 3 I-4 A=3\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-4\left[\begin{array}{cc}
0 & 1 \\
-1 & 0
\end{array}\right] \\
=\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]-\left[\begin{array}{cc}
0 & 4 \\
-4 & 0
\end{array}\right]=\left[\begin{array}{cc}
3 & -4 \\
4 & 3
\end{array}\right]
\end{array}$
Consider, $(3 I+4 A) \cdot(3 I-4 A)=\left[\begin{array}{cc}3 & 4 \\ -4 & 3\end{array}\right]\left[\begin{array}{cc}3 & -4 \\ 4 & 3\end{array}\right]$
$
=\left[\begin{array}{cc}
25 & 0 \\
0 & 25
\end{array}\right]=25\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=25 I
$
As, $(3 I+4 A)(3 I-4 A)=x^2 I$
$
\Rightarrow \quad 25 I=x^2 I \Rightarrow x^2=25 \Rightarrow x= \pm 5
$
View full question & answer→MCQ 241 Mark
If $A=\left[\begin{array}{ll}3 & 4 \\ 5 & 2\end{array}\right]$ and $2 A+B$ is a null matrix, then $B$ is equal to:
- A
$\left[\begin{array}{cc}6 & 8 \\ 10 & 4\end{array}\right]$
- B
$\left[\begin{array}{cc}-6 & -8 \\ -10 & -4\end{array}\right]$
- C
$\left[\begin{array}{cc}5 & 8 \\ 10 & 3\end{array}\right]$
- D
$\left[\begin{array}{cc}-5 & -8 \\ -10 & -3\end{array}\right]$
AnswerGiven, $A=\left[\begin{array}{ll}3 & 4 \\ 5 & 2\end{array}\right]$ and $2 A+B=O=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
Let $B=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \Rightarrow 2\left[\begin{array}{ll}3 & 4 \\ 5 & 2\end{array}\right]+\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
On comparing, we get
\[\begin{array}{l}
a+6=0 \Rightarrow a=-6 ; b+8=0 \Rightarrow b=-8 \\
c+10=0 \Rightarrow c=-10 \text { and } d+4=0 \Rightarrow d=-4 . \\
\therefore \quad \text { Required matrix, } B=\left[\begin{array}{cc}
-6 & -8 \\
-10 & -4
\end{array}\right]
\end{array}\]
View full question & answer→MCQ 251 Mark
It is given that $X\left[\begin{array}{cc}3 & 2 \\ 1 & -1\end{array}\right]=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]$. Then matrix $X$ is :
- A
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
- B
$\left[\begin{array}{cc}0 & -1 \\ 1 & 1\end{array}\right]$
- C
$\left[\begin{array}{cc}1 & 1 \\ 1 & -1\end{array}\right]$
- D
$\left[\begin{array}{ll}1 & -1 \\ 1 & -1\end{array}\right] \quad$
AnswerLet $X=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
We have, $x\left[\begin{array}{cc}3 & 2 \\ 1 & -1\end{array}\right]=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]$
$
\begin{array}{l}
\Rightarrow\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{cc}
3 & 2 \\
1 & -1
\end{array}\right]=\left[\begin{array}{ll}
4 & 1 \\
2 & 3
\end{array}\right] \\
\Rightarrow\left[\begin{array}{ll}
3 a+b & 2 a-b \\
3 c+d & 2 c-d
\end{array}\right]=\left[\begin{array}{ll}
4 & 1 \\
2 & 3
\end{array}\right]
\end{array}
$
On comparing the element of matrices, we get
$
\begin{array}{l}
3 a+b=4 \\
2 a-b=1 \\
3 c+d=2 \\
2 c-d=3
\end{array}
$
Adding (i) and (ii), we get $5 a=5 \Rightarrow a=1$
Putting $a=1$ in (i), we get $3(1)+b=4 \Rightarrow b=1$
Adding (iii) and (iv), we get $5 c=5 \Rightarrow c=1$
Putting $c=1$ in (iii), we get $3+d=2 \Rightarrow d=-1$
$
\therefore X=\left[\begin{array}{cc}
1 & 1 \\
1 & -1
\end{array}\right]
$
View full question & answer→MCQ 261 Mark
If $A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right], B=\left[\begin{array}{ll}x & 0 \\ 1 & 1\end{array}\right]$ and $A=B^2$, then $x$ equals
AnswerWehave, $A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}x & 0 \\ 1 & 1\end{array}\right]$
\[\therefore B^2=\left[\begin{array}{ll}
x & 0 \\
1 & 1
\end{array}\right]\left[\begin{array}{ll}
x & 0 \\
1 & 1
\end{array}\right]=\left[\begin{array}{cc}
x^2 & 0 \\
x+1 & 1
\end{array}\right]\]
Now, it is given that $A=B^2$
\[\Rightarrow\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right]=\left[\begin{array}{cc}
x^2 & 0 \\
x+1 & 1
\end{array}\right]\]
On comparing, we get
\[\begin{array}{ll}
& x^2=1 \text { and } x+1=2 \Rightarrow x= \pm 1 \text { and } x=1 \\
\therefore & x=1
\end{array}\]
View full question & answer→MCQ 271 Mark
If $x\left[\begin{array}{l}1 \\ 2\end{array}\right]+y\left[\begin{array}{l}2 \\ 5\end{array}\right]=\left[\begin{array}{l}4 \\ 9\end{array}\right]$, then
- A
$x=1, y=2$
- B
$x=2, y=1$
- C
$x=1, y=-1$
- D
$x=3, y=2$
AnswerWe have, $x\left[\begin{array}{l}1 \\ 2\end{array}\right]+y\left[\begin{array}{l}2 \\ 5\end{array}\right]=\left[\begin{array}{l}4 \\ 9\end{array}\right]$
\[\Rightarrow\left[\begin{array}{c}
x+2 y \\
2 x+5 y
\end{array}\right]=\left[\begin{array}{l}
4 \\
9
\end{array}\right]
\]$\Rightarrow x+2 y=4$ ...(i) and $2 x+5 y=9$ ....(ii)
Solving (i) and (ii), we get $x=2, y=1$
View full question & answer→MCQ 281 Mark
If $A$ is a square matrix and $A^2=A$, then $(I+A)^2-3 A$ is equal to
AnswerGiven that $A^2=A$
\[\begin{array}{l}
\text { Consider }(I+A)^2-3 A=I^2+A^2+2 A I-3 A \\
=I+A+2 A-3 A \quad\left[\because I^2=I, A^2=A \text { (given) }\right] \\
=1 \\
\end{array}\]
View full question & answer→MCQ 291 Mark
If $\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}6 \\ 3 \\ 2\end{array}\right]$, then the value of $(2 x+y-z)$ is
Answer$
\begin{array}{l}\text {}\left[\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
6 \\
3 \\
2
\end{array}\right] \\
\therefore \quad x+y+z=6 .....(i)\\
\quad y+z=3........(ii) \\
\quad z=2.......(iii) \\
\Rightarrow \quad y+2=3
\end{array}
$
[Using (ii) and (iii)]
$
\begin{array}{l}
\Rightarrow \quad y=1 \\
\Rightarrow \quad x+1+2=6 \\
\Rightarrow \quad x=3
\end{array}
$
[Using (i), (iii) and (iv)]
So, $2 x+y-z=(2 \times 3)+1-2=6+1-2=5$
View full question & answer→MCQ 301 Mark
If matrix $A=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]$ and $A^2=k A$, then the value of k is :
Answer\[\begin{array}{l}\text {Since, } A=\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right] \\
\Rightarrow A^2=A \cdot A=\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]=\left[\begin{array}{cc}
2 & -2 \\
-2 & 2
\end{array}\right]=2\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]=2 A
\end{array}\]
On comparing, $k=2$
View full question & answer→MCQ 311 Mark
If for a square matrix $A, A^2-A+I=0$, then $A^{-1}$ equals
AnswerWe have, $A^2-A+I=O$
Pre-multiplying with $A^{-1}$ on both sides, we get
$
\begin{array}{l}
\left(A^{-1} A\right) \cdot A-A^{-1} \cdot A+A^{-1} \cdot I=A^{-1} \cdot O \\
\Rightarrow \quad I \cdot A-I+A^{-1}=O \\
\Rightarrow \quad A^{-1}=-(A-I)=I-A
\end{array}
$
View full question & answer→MCQ 321 Mark
If for a square matrix $A, A^2-3 A+I=O$ and $A^{-1}=x A+y l$, then the value of $x+y$ is:
AnswerGiven, $A^2-3 A+I=0$
and $A^{-1}=x A+y l$
where $A$ is a square matrix
Now, pre-multiply (i) by $A^{-1}$ on both sides, we get
$
\begin{array}{l}
A^{-1} A^2-3 A^{-1} A+A^{-1} I=A^{-1} O \\
\Rightarrow A-3 I+A^{-1}=O \\
\Rightarrow A^{-1}=-A+3 I
\end{array}
$
On comparing with equation (ii), we get
$
\begin{array}{l}
x=-1 \text { and } y=3 \\
\therefore \quad x+y=-1+3=2
\end{array}
$
View full question & answer→MCQ 331 Mark
If a matrix $A=\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]$, then the matrix $A A^{\prime}$ (where $A^{\prime}$ is the transpose of $A$ ) is
Answer$(d): A=\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]$
$A^{\prime}=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$
So, $A A^{\prime}=\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]=[1+4+9]=[14]$
View full question & answer→MCQ 341 Mark
If $A=\left[\begin{array}{ll}5 & x \\ y & 0\end{array}\right]$ and $A=A^{\top}$, where $A^T$ is the transpose of the matrix $A$, then
- A
$x=0, y=5$
- B
$x=y$
- C
$x+y=5$
- D
$x=5, y=0$
AnswerWe have, $A=A^T$
$
\Rightarrow\left[\begin{array}{ll}
5 & x \\
y & 0
\end{array}\right]=\left[\begin{array}{ll}
5 & y \\
x & 0
\end{array}\right]
$
View full question & answer→MCQ 351 Mark
If $A, B$ are non-singular square matrices of the same order, then $\left(A B^{-1}\right)^{-1}=$
- A
$A^{-1} B$
- B
$A^{-1} B^{-1}$
- C
$B A^{-1}$
- D
$A B$
AnswerWe know that, if $A$ and $B$ are non-singular matrices of same order, then
$
(A B)^{-1}=B^{-1} A^{-1} ;\left(A B^{-1}\right)^{-1}=\left(B^{-1}\right)^{-1} A^{-1}=B A^{-1}
$
View full question & answer→MCQ 361 Mark
If $A=\left[a_{i j}\right]$ is a skew-symmetric matrix of order $n$, then
- A
$a_{i j}=\frac{1}{a_{j i}} \forall i, j$
- B
$a_{i j} \neq 0 \forall i, j$
- C
$a_{i j}=0$, where $i=j$
- D
$\quad a_{i j} \neq 0$ where $i=j$
AnswerIn a skew-symmetric matrix, the $(i, j)^{\text {th }}$ element is negative of the $(i, i)^{\text {th }}$ element. Hence the $(i, i)^{\text {th }}$ element $=0$.
View full question & answer→MCQ 371 Mark
If $A$ is a square matrix such that $A^2=A$, then $(I-A)^3+A$ is equal to
AnswerWe have, $A^2=A$
$
\begin{array}{l}
\text { Now, }(I-A)^3+A=(I-A)(I-A)(I-A)+A \\
=(I \cdot I-I \cdot A-A \cdot I+A \cdot A)(I-A)+A \\
=(I-A-A+A)(I-A)+A \quad\left[\because I \cdot A=A \cdot I=A \text { and } A^2=A\right] \\
=(I-A)(I-A)+A=(I \cdot I-I \cdot A-A \cdot I+A \cdot A)+A \\
=(I-A-A+A)+A=(I-A)+A=I
\end{array}
$
View full question & answer→MCQ 381 Mark
If $A=\left[\begin{array}{lll}2 & -3 & 4\end{array}\right], B=\left[\begin{array}{l}3 \\ 2 \\ 2\end{array}\right], X=\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]$ and $Y=\left[\begin{array}{l}2 \\ 3 \\ 4\end{array}\right]$, then $A B+X Y$ equals
AnswerConsider, $A B=\left[\begin{array}{lll}2 & -3 & 4\end{array}\right]\left[\begin{array}{l}3 \\ 2 \\ 2\end{array}\right]=[6-6+8]=[8]$
and $X Y=\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]\left[\begin{array}{l}2 \\ 3 \\ 4\end{array}\right]=[2+6+12]=[20]$
$A B+X Y=[8]+[20]=[28]
$
View full question & answer→MCQ 391 Mark
If $A=\left[\begin{array}{rr}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$, then for what value of $\alpha, A$ is an identity matrix?
- ✓
$0^{\circ}$
- B
$90^{\circ}$
- C
$45^{\circ}$
- D
$30^{\circ}$
AnswerCorrect option: A. $0^{\circ}$
(a) : $A=\left[\begin{array}{rr}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$ is an identity matrix if, $\left[\begin{array}{rr}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\therefore \cos \alpha=1$ and $\sin \alpha=0 \Rightarrow \alpha=0^{\circ}$.
View full question & answer→MCQ 401 Mark
If the matrix $A=\left[\begin{array}{rrr}5 & 2 & x \\ y & 2 & -3 \\ 4 & t & -7\end{array}\right]$ is a symmetric matrix, then find the values of $x, y$ and $t$ respectively.
- A
$4,2,3$
- ✓
$4,2,-3$
- C
$4,2,-7$
- D
$2,4,-7$
AnswerCorrect option: B. $4,2,-3$
(b): $A$ is a symmetric matrix $\therefore A=A^T$
$
\Rightarrow\left[\begin{array}{rrr}
5 & 2 & x \\
y & 2 & -3 \\
4 & t & -7
\end{array}\right]=\left[\begin{array}{rrr}
5 & y & 4 \\
2 & 2 & t \\
x & -3 & -7
\end{array}\right]
$
On comparing, we get $y=2, x=4, t=-3$
View full question & answer→MCQ 411 Mark
$\left[\begin{array}{lll}7 & 1 & 2 \\ 9 & 2 & 1\end{array}\right]\left[\begin{array}{l}3 \\ 4 \\ 5\end{array}\right]+2\left[\begin{array}{l}4 \\ 2\end{array}\right]$ is equal to
- ✓
$\left[\begin{array}{l}43 \\ 44\end{array}\right]$
- B
$\left[\begin{array}{l}43 \\ 45\end{array}\right]$
- C
$\left[\begin{array}{l}45 \\ 44\end{array}\right]$
- D
$\left[\begin{array}{l}44 \\ 45\end{array}\right]$
AnswerCorrect option: A. $\left[\begin{array}{l}43 \\ 44\end{array}\right]$
(a) :
$
\begin{aligned}
{\left[\begin{array}{lll}
7 & 1 & 2 \\
9 & 2 & 1
\end{array}\right]\left[\begin{array}{l}
3 \\
4 \\
5
\end{array}\right]+2\left[\begin{array}{l}
4 \\
2
\end{array}\right] } & =\left[\begin{array}{c}
21+4+10 \\
27+8+5
\end{array}\right]+\left[\begin{array}{l}
8 \\
4
\end{array}\right] \\
& =\left[\begin{array}{l}
35 \\
40
\end{array}\right]+\left[\begin{array}{l}
8 \\
4
\end{array}\right]=\left[\begin{array}{l}
43 \\
44
\end{array}\right]
\end{aligned}
$
View full question & answer→MCQ 421 Mark
If $A$ is a $m \times n$ matrix such that $A B$ and $B A$ are both defined, then $B$ is an
- A
$m \times n$ matrix
- ✓
$n \times m$ matrix
- C
$n \times n$ matrix
- D
$m \times m$ matrix
AnswerCorrect option: B. $n \times m$ matrix
(b) : Since $A B$ exists, therefore, number of columns in $A=$ number of rows in $B$. So, $B$ has $n$ rows. Since $B A$ exists, number of columns in $B=$ number of rows in $A$. So, $B$ has $m$ columns.
$\therefore \quad B$ is an $n \times m$ matrix.
View full question & answer→MCQ 431 Mark
The matrix $\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0\end{array}\right]$ is a/an
Answer(b) : Let $A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0\end{array}\right] \Rightarrow A^T=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0\end{array}\right]=A$
$\therefore \quad A$ is a symmetric matrix.
View full question & answer→MCQ 441 Mark
If $A$ and $B$ are matrices of same order, then $\left(A B^{\prime}-B A^{\prime}\right)$ is a
Answer$
\begin{array}{l}
\text { (a): }\left(A B^{\prime}-B A^{\prime}\right)^{\prime}=\left(A B^{\prime}\right)^{\prime}-\left(B A^{\prime}\right)^{\prime} \\
=\left(B^{\prime}\right)^{\prime} A^{\prime}-\left(A^{\prime}\right)^{\prime} B^{\prime}=B A^{\prime}-A B^{\prime}=-\left(A B^{\prime}-B A^{\prime}\right)
\end{array}
$
Hence, $\left(A B^{\prime}-B A^{\prime}\right)$ is a skew symmetric matrix.
View full question & answer→MCQ 451 Mark
For any two matrices $A$ and $B$, we have
- A
$A B=B A$
- B
$A B \neq B A$
- C
$A B=O$
- ✓
View full question & answer→MCQ 461 Mark
If $\left[\begin{array}{cc}2 x+y & 4 x \\ 5 x-7 & 4 x\end{array}\right]=\left[\begin{array}{cc}7 & 7 y-13 \\ y & x+6\end{array}\right]$, then the values of $x, y$ respectively are
Answer(b) : $\left[\begin{array}{cc}2 x+y & 4 x \\ 5 x-7 & 4 x\end{array}\right]=\left[\begin{array}{cc}7 & 7 y-13 \\ y & x+6\end{array}\right]$
On comparing, we get
$
4 x=x+6 \Rightarrow x=2 \text { and } 2 x+y=7 \Rightarrow y=7-4=3
$
View full question & answer→MCQ 471 Mark
If $A^3=O$, then $A^2+A+I=$
- A
$I-A$
- ✓
$(I-A)^{-1}$
- C
$(I+A)^{-1}$
- D
$I+A$
AnswerCorrect option: B. $(I-A)^{-1}$
(b) $: I-A^3=I \Rightarrow(I-A)\left(I+A+A^2\right)=I$
$\therefore \quad I+A+A^2=(I-A)^{-1}$.
View full question & answer→MCQ 481 Mark
If $A$ is a symmetric matrix and $n \in N$, then $A^n$ is a
Answer(a) : Since, $A$ is a symmetric matrix $\Rightarrow A^{\prime}=A$
Now, $\left(A^n\right)^{\prime}=\left(A^{\prime}\right)^n=A^n$
Therefore, $A^n$ is a symmetric matrix.
View full question & answer→MCQ 491 Mark
If a matrix $A$ is both symmetric and skewsymmetric, then
- A
$A$ is a diagonal matrix
- ✓
$A$ is a zero matrix
- C
$A$ is a scalar matrix
- D
$A$ is a square matrix
AnswerCorrect option: B. $A$ is a zero matrix
(b) : $A$ is a symmetric matrix
$
\therefore \quad A^T=A
$
$A$ is also a skew-symmetric matrix
$
\therefore \quad A^T=-A
$
From (i) and (ii), we get
$
A=-A \Rightarrow A=O
$
Hence, $A$ is zero matrix.
View full question & answer→MCQ 501 Mark
Total number of possible matrices of order $3 \times 3$ with each entry 2 or 0 is
Answer(d) : Required number of matrices $=2^9=512$
View full question & answer→