Question 11 Mark
For any $2 \times 2$ matrix $P$, which of the following matrices can be $Q$ such that $P Q=Q P$ ?
Answer$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
View full question & answer→Question 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→Question 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$
\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}
$
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→Question 41 Mark
Given that $A=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]$ and $A^2=31$, then
AnswerWe have $, A=\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right] $
$\Rightarrow A^2=\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right]\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right]$
$=\left[\begin{array}{cc} \alpha^2+\beta \gamma & 0 \\
0 & \gamma \beta+\alpha^2 \end{array}\right]$
But $A^2=31$
$ \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] $
$\Rightarrow \alpha^2+\beta \gamma=3$
$\Rightarrow 3-\alpha^2-\beta \gamma=0$
View full question & answer→Question 51 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
AnswerWe have $, A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right] $
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 2 \cos \alpha=1 $
$\Rightarrow \cos \alpha=\frac{1}{2} $
$\Rightarrow \alpha=\frac{\pi}{3}$
View full question & answer→Question 61 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
AnswerWe 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] \ ($Given$)$
$ \Rightarrow-4 k=24,3 a=2 k, 2 b=3 k $
$ \Rightarrow k=-6, a=-4, b=-9$
View full question & answer→Question 71 Mark
If $A$ is square matrix such that $A^2=A$, then $(I+A)^3-7 A$ is equal to
AnswerWe have, $(I+A)^3-7 A$
$=I^3+A^3+3 I^2 A+3 I A^2-7 A$
$=1+A \cdot A+3 A+3 A-7 A$
$=1+A+3 A+3 A-7 A$
$=1$
View full question & answer→Question 81 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→Question 91 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
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→Question 101 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
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→Question 111 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
AnswerWe have,
$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] $
$ 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] $
$ =\left[\begin{array}{lll} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right]=61 $
$\Rightarrow B^{-1}=\frac{1}{6} A$
View full question & answer→Question 121 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
$2 a+b=4 .... (i)$
$5 c-d=11 ..... (iii)$
$a-2 b=-3..... (ii)$
$4 c+3 d=24 ...... (iv)$
Solving $(i)$ and $(ii),$ we get
$5 a=5 $
$\Rightarrow a=1, b=2$
Solving $(iii)$ and $(iv),$ we get
$19 c=57$
$ \Rightarrow c=3, d=4$
$\therefore a+b-c+2 d=1+2-3+8=8$
View full question & answer→Question 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→Question 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 $(i, j)$ :
Answer 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]$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
View full question & answer→Question 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, $[F(x)]^2=F(k x)$
$ \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] $
$=\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] $
$\Rightarrow k=2$
View full question & answer→Question 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 :
AnswerGiven $A=\left[\begin{array}{cc}2 & 1 \\ -4 & -2\end{array}\right]$, then
$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$
$\therefore I-A+A^2-A^3+\ldots=I-A$
$ =\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]$
View full question & answer→Question 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?
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→Question 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 :
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
$ 2 \tan x=2 \sqrt{3} $
$\Rightarrow \tan x=\sqrt{3}=\tan \frac{\pi}{3}$
$\Rightarrow x=\frac{\pi}{3} $
View full question & answer→Question 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→Question 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 :
AnswerWe have $\left[\begin{array}{ll}1 & x\end{array}\right]\left[\begin{array}{cc}4 & 0 \\ -2 & 0\end{array}\right]=0 $
$ \Rightarrow [1 \times 4+x \times(-2) 1 \times 0+x \times 0]=\left[0 0\right]$
$\Rightarrow 4-2 x=0 $
$\Rightarrow 4=2 x$
$ \Rightarrow x=2$
View full question & answer→Question 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.
$\therefore a=d=5, b=c=0$
$a+2 b+3 c+4 d$
$=5+2 \times 0+3 \times 0+4 \times 5$
$=5+20$
$=25$
View full question & answer→Question 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
AnswerWe have $, A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] $
$\therefore A^2=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
$=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
$=\left[\begin{array}{ll}0+1 & 0+0 \\ 0+0 & 1+0\end{array}\right]$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
View full question & answer→Question 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:
AnswerGiven, $A=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]$
Now, $3 I+4 A=3\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]+4\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]$
$ =\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] $
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]$
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 25 I=x^2 I$
$ \Rightarrow x^2=25 $
$\Rightarrow x= \pm 5 $
View full question & answer→Question 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:
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
$a+6=0 $
$\Rightarrow a=-6 ; b+8=0 $
$\Rightarrow b=-8$
$c+10=0$
$ \Rightarrow c=-10 $ and $ d+4=0 $
$\Rightarrow d=-4 . $
$ \therefore$ Required matrix, $ B=\left[\begin{array}{cc} -6 & -8 \\ -10 & -4 \end{array}\right]$
View full question & answer→Question 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 :
AnswerLet $X=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
We have, $x\left[\begin{array}{cc}3 & 2 \\ 1 & -1\end{array}\right]=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]$
$ \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]$
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→Question 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→Question 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
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→Question 281 Mark
If $A$ is a square matrix and $A^2=A$, then $(I+A)^2-3 A$ is equal to
AnswerGiven that $A^2=A$
$\text { Consider }(I+A)^2-3 A$
$=I^2+A^2+2 A I-3 A$
$=I+A+2 A-3 A \quad\left[\because I^2=I, A^2=A \text { (given) }\right]$
$=1$
View full question & answer→Question 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$\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] $
$\therefore x+y+z=6 .....(i)$
$y+z=3........(ii)$
$ z=2.......(iii)$
$\Rightarrow y+2=3$
$[$Using $(ii)$ and $(iii)]$
$\Rightarrow y=1$
$\Rightarrow x+1+2=6$
$\Rightarrow x=3$
$[$Using $(i), (iii)$ and $(iv)]$
So, $2 x+y-z$
$=(2 \times 3)+1-2$
$=6+1-2$
$=5$
View full question & answer→Question 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 :
AnswerSince $, A=\left[\begin{array}{cc} 1 & -1 \\ -1 & 1\end{array}\right] $
$\Rightarrow A^2=A \cdot A=\left[\begin{array}{cc} 1 & -1 \\-1 & 1 \end{array}\right]$
$\left[\begin{array}{cc} 1 & -1 \\-1 & 1 \end{array}\right]=\left[\begin{array}{cc} 2 & -2 \\ -2 & 2 \end{array}\right]$
$=2\left[\begin{array}{cc}1 & -1 \\-1 & 1 \end{array}\right]$
$=2 A$ On comparing, $k=2$
View full question & answer→Question 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
$\left(A^{-1} A\right) \cdot A-A^{-1} \cdot A+A^{-1} \cdot I=A^{-1} \cdot O$
$\Rightarrow I \cdot A-I+A^{-1}=O$
$\Rightarrow A^{-1}=-(A-I)=I-A$
View full question & answer→Question 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
$A^{-1} A^2-3 A^{-1} A+A^{-1} I=A^{-1} O$
$\Rightarrow A-3 I+A^{-1}=O$
$\Rightarrow A^{-1}=-A+3 I$
On comparing with equation $(ii),$ we get
$x=-1 \text { and } y=3$
$\therefore x+y=-1+3=2$
View full question & answer→Question 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$A=\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]$
$A^{\prime}=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$
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→Question 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
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→Question 351 Mark
If $A, B$ are non-singular square matrices of the same order, then $\left(A B^{-1}\right)^{-1}=$
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→Question 361 Mark
If $A=\left[a_{i j}\right]$ is a skew-symmetric matrix of order $n$, then
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→Question 371 Mark
If $A$ is a square matrix such that $A^2=A$, then $(I-A)^3+A$ is equal to
AnswerWe have, $A^2=A$
$\text { Now, }(I-A)^3+A=(I-A)(I-A)(I-A)+A$
$=(I \cdot I-I \cdot A-A \cdot I+A \cdot A)(I-A)+A$
$=(I-A-A+A)(I-A)+A \quad\left[\because I \cdot A=A \cdot I=A \text { and } A^2=A\right]$
$=(I-A)(I-A)+A=(I \cdot I-I \cdot A-A \cdot I+A \cdot A)+A$
$=(I-A-A+A)+A$
$=(I-A)+A$
$=I$
View full question & answer→Question 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→Question 391 Mark
If $A$ and $B$ are square matrices of same order and $A^{\prime}$ denotes the transpose of $A$, then
Answer(a): $(A B)^{\prime}=B^{\prime} A^{\prime}$ is true. This result is a standard result called "reversal law of transposes."
View full question & answer→Question 401 Mark
If $A$ and $B$ are $2 \times 2$ matrices, then which of the following is true?
Answer(c) : Given that, $A$ and $B$ are $2 \times 2$ matrices.
$
\begin{aligned}
\therefore \quad & (A-B) \times(A+B)=A \times A+A \times B-B \times A-B \times B \\
& =A^2+A B-B A-B^2
\end{aligned}
$
View full question & answer→Question 411 Mark
If $A=\left[\begin{array}{ll}-2 & 4 \\ -1 & 2\end{array}\right]$, then $A^2$ is
Answer(a) : Given that, $A=\left[\begin{array}{ll}-2 & 4 \\ -1 & 2\end{array}\right]$
$\therefore \quad A^2=\left[\begin{array}{ll}-2 & 4 \\ -1 & 2\end{array}\right]\left[\begin{array}{ll}-2 & 4 \\ -1 & 2\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$\Rightarrow A^2$ is a null matrix.
View full question & answer→Question 421 Mark
If $A=\left[\begin{array}{rr}2 & 1 \\ -1 & 2\end{array}\right], B=\left[\begin{array}{rr}1 & -2 \\ 2 & 1\end{array}\right], C=\left[\begin{array}{rr}1 & -3 \\ 2 & 1\end{array}\right]$,then
Answer(a) : In option (a), there are two laws, commutative law and associative law, which are satisfied by all matrices. Thus, option (a) is correct.
View full question & answer→Question 431 Mark
If $A=\left[\begin{array}{ll}0 & 2 \\ 2 & 0\end{array}\right]$, then $A^2$ is equal to
Answer(d) : We have, $A=\left[\begin{array}{ll}0 & 2 \\ 2 & 0\end{array}\right]$
$
\Rightarrow \quad A^2=\left[\begin{array}{ll}
0 & 2 \\
2 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 2 \\
2 & 0
\end{array}\right]=\left[\begin{array}{ll}
0+4 & 0+0 \\
0+0 & 4+0
\end{array}\right]=\left[\begin{array}{ll}
4 & 0 \\
0 & 4
\end{array}\right]
$
View full question & answer→Question 441 Mark
If $A=\left[a_{i j}\right]=\left[\begin{array}{cc}2 & -1 \\ -3 & 4 \\ 1 & 2\end{array}\right]$ and $B=\left[b_{i j}\right]=\left[\begin{array}{ccc}2 & 3 & -5 \\ 1 & 4 & 9 \\ 0 & 7 & -2\end{array}\right]$, then value of $a_{11} b_{11}+a_{22} b_{22}$ is
Answer(b) : We have, $A=\left[\begin{array}{cc}2 & -1 \\ -3 & 4 \\ 1 & 2\end{array}\right]$ and $B=\left[\begin{array}{ccc}2 & 3 & -5 \\ 1 & 4 & 9 \\ 0 & 7 & -2\end{array}\right]$
Here, $a_{11}=2, a_{22}=4, b_{11}=2, b_{22}=4$
$
\therefore \quad a_{11} b_{11}+a_{22} b_{22}=2(2)+4(4)=4+16=20
$
View full question & answer→Question 451 Mark
The order of the single matrix obtained from
$
\left[\begin{array}{rr}
1 & -1 \\
0 & 2 \\
2 & 3
\end{array}\right]\left\{\left[\begin{array}{rrr}
-1 & 0 & 2 \\
2 & 0 & 1
\end{array}\right]-\left[\begin{array}{lll}
0 & 1 & 23 \\
1 & 0 & 21
\end{array}\right]\right\} \text { is }
$
Answer(d) : When a $3 \times 2$ matrix is post multiplied by a $2 \times 3$ matrix, then the product is a $3 \times 3$ matrix.
View full question & answer→Question 461 Mark
If $A B=A$ and $B A=B$, then
Answer$\text { (c) }: A=A B=A(B A)=(A B) A=A \cdot A=A^2$
$B=B A=B(A B)=(B A) B=B \cdot B=B^2$
View full question & answer→Question 471 Mark
If $A^T=\left[\begin{array}{rr}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{rrr}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right]$, then find $A^T-B^T$.
Answer(a) : $B^T=\left[\begin{array}{rr}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]$
So, $\quad A^T-B^T=\left[\begin{array}{cc}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]-\left[\begin{array}{cc}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]=\left[\begin{array}{cc}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]$
View full question & answer→Question 481 Mark
If $A=\left[\begin{array}{rrr}1 & -2 & 1 \\ 2 & 1 & 3\end{array}\right]$ and $B=\left[\begin{array}{ll}2 & 1 \\ 3 & 2 \\ 1 & 1\end{array}\right]$, then $(A B)^T$ is equal to
Answer(b) : $A B=\left[\begin{array}{rrr}1 & -2 & 1 \\ 2 & 1 & 3\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 3 & 2 \\ 1 & 1\end{array}\right]=\left[\begin{array}{rr}-3 & -2 \\ 10 & 7\end{array}\right]$
$
\Rightarrow \quad(A B)^T=\left[\begin{array}{rr}
-3 & 10 \\
-2 & 7
\end{array}\right]
$
View full question & answer→Question 491 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?
Answer(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→Question 501 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.
Answer(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→