MCQ 1511 Mark
If $\text{A}=\begin{bmatrix}a^2 &\text{amp; ab}&\text{amp; ac} \\\text{ab}&\text{amp; }\text{b}^2&\text{amp;}\text{ bc}\\\text{ac}&\text{amp;}\text{bc}&\text{amp;}\text{c}^2 \end{bmatrix}$and $\text{a}^2+\text{b}^2+\text{c}^3=1$ then $\text{A}^2=$
- A
$2\text{A}$
- ✓
$\text{A}$
- C
$3\text{A}$
- D
$\frac{1}{2}\text{A}$
AnswerCorrect option: B. $\text{A}$
$\text{A}^2=\begin{bmatrix}\text{a}^2 &\text{amp; ab}&\text{amp; ac} \\\text{ab}&\text{amp; }\text{b}^2&\text{amp;}\text{ bc}\\\text{ac}&\text{amp;}\text{bc}&\text{amp;}\text{c}^2 \end{bmatrix}\begin{bmatrix}\text{a}^2 &\text{amp; ab}&\text{amp; ac} \\\text{ab}&\text{amp; }\text{b}^2&\text{amp;}\text{ bc}\\\text{ac}&\text{amp;}\text{bc}&\text{amp;}\text{c}^2 \end{bmatrix}=\text{A}$
View full question & answer→MCQ 1521 Mark
The order the matrix is $\begin{bmatrix}2&\text{amp; }3&\text{amp; }4\\9&\text{amp; }8&\text{amp; }7\end{bmatrix}$ is :
- A
$4 \times 3$
- B
$3 \times 2$
- ✓
$2 \times 3$
- D
$3 \times 1$
AnswerCorrect option: C. $2 \times 3$
If $A$ is a matrix with mm rows and $n$ columns.
Then the order of a matrix is nothing but a size of a matrix, which is given by $m \times n.$
Since, in the given matrix, there are $2$ rows and $3$ columns.
So, order of given matrix will be $2 \times 3.$
View full question & answer→MCQ 1531 Mark
If $\begin{bmatrix}\text{r}+4&\text{amp; 6}\\3&\text{amp; 3}\end{bmatrix}=\begin{bmatrix}{5}&\text{amp;}\text{ r}+5\\\text{r+2}&\text{amp; 4}\end{bmatrix}$ then $\text{r}=$
AnswerWe know that two matrices are equal iff their corresponding elements are equal.
Thus comparing corresponding elements we get, for the first entry of.
the given matrices $r + 4 = 5$ and $r$ is satisfying other equations which are involving $r$
$ \Rightarrow r = 1$
View full question & answer→MCQ 1541 Mark
If a matrix is of order $2 \times 3,$ then the number of elements in the matrix is :
AnswerGiven a matrix $2\times3$
$\Rightarrow \begin{bmatrix} { \text{a} }_{11} &\text{amp; } {\text{a} }_{12} &\text{amp; } { \text{a} }_{13} \\ { \text{a} }_{21} &\text{amp; } {\text{a} }_{22} &\text{amp; } {\text{a} }_{23} \end{bmatrix}$ Clearly there are $6$ elements.
View full question & answer→MCQ 1551 Mark
If $A$ is a matrix of order m×n and B is a matrix such that $AB ^{\top}$ and $B ^{\top} A$are both defined, then the order of matrix $B$ is:
- ✓
$m \times n$
- B
$n \times n$
- C
$n \times m$
- D
$m \times n$
AnswerCorrect option: A. $m \times n$
a. $m \times n$
Solution:
$A$ is $m \times n$ matrix and $A B^{\top}$ is defined then number of columns in $A=$ number of rows in $B^{\top}=n$
$B^{\top} A$ is also defined then number of columns in $B^{\top}=$ number of rows in $A=m$ Order of $B$ is $m \times n$
View full question & answer→MCQ 1561 Mark
Choose the correct answer from the given four options. If $\begin{bmatrix}2\text{x}+\text{y}&4\text{x}\\5\text{x}-7&4\text{x}\end{bmatrix}=\begin{bmatrix}7&7\text{y}-13\\\text{y}&\text{x}+6\end{bmatrix},$ then the value of $x + y$ is :
- A
$x = 3, y = 1$
- ✓
$x = 2, y = 3$
- C
$x = 2, y = 4$
- D
$x = 3, y = 3$
AnswerCorrect option: B. $x = 2, y = 3$
We have, $\begin{bmatrix}2\text{x}+\text{y}&4\text{x}\\5\text{x}-7&4\text{x}\end{bmatrix}=\begin{bmatrix}7&7\text{y}-13\\\text{y}&\text{x}+6\end{bmatrix}$
$\Rightarrow 4x = x + 6$
$ \Rightarrow x = 2$
and $4x = 7y - 13$
$\Rightarrow 8 = 7y - 13$
$\Rightarrow y = 3$
$\therefore x + y = 2 + 3 = 5$
View full question & answer→MCQ 1571 Mark
If $\text{A}=\begin{bmatrix}1&2&\text{x}\\0&1&0\\0&0&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&-2&\text{y}\\0&1&0\\0&0&1\end{bmatrix}$ and $A B=I_3$, then $x+y$ equals :
AnswerGiven : $A B=I_3$
$\Rightarrow\begin{bmatrix}1&2&\text{x}\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}1&-2&\text{y}\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1&0&\text{y+x}\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\therefore\ \text{y}+\text{x}=0$
View full question & answer→MCQ 1581 Mark
Choose the correct answer from the given four options. If $\text{A}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\text{x}\pi)&\tan^{-1}\Big(\frac{\text{x}}{\pi}\Big)\\\sin^{-1}\Big(\frac{\text{x}}{\pi}\Big)&\cot^{-1}(\pi\text{x})\end{bmatrix}$ and $\text{B}=\frac{1}{\pi}\begin{bmatrix}-\cos^{-1}(\text{x}\pi)&\tan^{-1}\Big(\frac{\text{x}}{\pi}\Big)\\\sin^{-1}\Big(\frac{\text{x}}{\pi}\Big)&\tan^{-1}(\pi\text{x})\end{bmatrix}$ then $A - B$ is :
- A
$\text{I}$
- B
$0$
- C
$2\text{I}$
- ✓
$\frac{1}{2}\text{I}$
AnswerCorrect option: D. $\frac{1}{2}\text{I}$
We have, $\text{B}=\begin{bmatrix}-\frac{1}{\pi}\cos^{-1}\text{x}\pi&\frac{1}{\pi}\tan^{-1}\frac{\text{x}}{\pi}\\\frac{1}{\pi}\sin^{-1}\frac{\text{x}}{\pi}&-\frac{1}{\pi}\tan^{-1}\pi\text{x}\end{bmatrix}$
and $\text{A}=\begin{bmatrix}\frac{1}{\pi}\sin^{-1}\text{x}\pi&\frac{1}{\pi}\tan^{-1}\frac{\text{x}}{\pi}\\\frac{1}{\pi}\sin^{-1}\frac{\text{x}}{\pi}&\frac{1}{\pi}\cot^{-1}\pi\text{x}\end{bmatrix}$
$\therefore\ \text{A}-\text{B}=\begin{bmatrix}\frac{1}{\pi}(\sin^{-1}\text{x}\pi+\cos^{-1}\text{x}\pi)&0\\0&\frac{1}{\pi}\big(\cot^{-1}\text{x}\pi+\tan^{-1}\pi\text{x}\big)\end{bmatrix}$
$=\begin{bmatrix}\frac{1}{\pi}\Big(\frac{\pi}{2}\Big)&0\\0&\frac{1}{\pi}\Big(\frac{\pi}{2}\Big)\end{bmatrix}$
$=\frac{1}{2}\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$=\frac{1}{2}\text{I}$
View full question & answer→MCQ 1591 Mark
The total number of matrices formed with the help of $6$ different numbers are :
- A
$6$
- B
$6!$
- C
$2(6!)$
- ✓
$4(6!)$
AnswerCorrect option: D. $4(6!)$
No. of numbers in Matrix is $6$
The possible orientations of Matrix is.
$1 \times 6, 2 \times 3, 3 \times 2, 6 \times 1$
The numbers in each orientation can be arranged in $6!$ ways.
$\implies$ The total possibilities are $4(6!).$
View full question & answer→MCQ 1601 Mark
If order of a matrix is $3 \times 3,$ then it is a?
AnswerSince, order of given matrix is $3 \times 3.$
$\therefore$ No of rows $=$ No. of columns
So, given matrix is a square matrix.
View full question & answer→MCQ 1611 Mark
A matrix having $mm$ rows and $nn$ columns with $m = n$ is said to be a?
AnswerA matrix having $mm$ rows and $nn$ columns with $m = n,$ means number of rows are equal to number of columns.
$\therefore$ given matrix is square matrix.
View full question & answer→MCQ 1621 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 1631 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 1641 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?
AnswerCorrect option: A. $C\left(A+B^{\prime}\right)$
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→MCQ 1651 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
$B^{-1}=B$
- ✓
$B^{-1}=\frac{1}{6} A$
AnswerCorrect option: D. $B^{-1}=\frac{1}{6} A$
We 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] \end{array} $
$ =\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 $
$\Rightarrow B^{-1}=\frac{1}{6} A$
View full question & answer→MCQ 1661 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
$1-\alpha^2-\beta \gamma=0$
- ✓
$3-\alpha^2-\beta \gamma=0$
- D
$3+\alpha^2+\beta \gamma=0$
AnswerCorrect option: C. $3-\alpha^2-\beta \gamma=0$
We have,
We have, $ A=\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right] $
$\Rightarrow A^2=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right]\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right]=\left[\begin{array}{cc} \alpha^2+\beta \gamma & 0 \\
0 & \gamma \beta+\alpha^2 \end{array}\right]$
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→MCQ 1671 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→MCQ 1681 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
$-6,-12,-18$
- ✓
$-6,-4,-9$
- C
$-6,4,9$
- D
$-6,12,18$
AnswerCorrect option: B. $-6,-4,-9$
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] $
Given $\Rightarrow-4 k=24,3 a=2 k, 2 b=3 k $
$ \Rightarrow k=-6, a=-4, b=-9$
View full question & answer→MCQ 1691 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 1701 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→MCQ 1711 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
$\frac{\pi}{6}$
- ✓
$\frac{\pi}{3}$
- C
$\pi$
- D
$\frac{3 \pi}{2}$
AnswerCorrect option: B. $\frac{\pi}{3}$
We 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 \quad 2 \cos \alpha=1 $
$\Rightarrow \cos \alpha=\frac{1}{2} $
$\Rightarrow \alpha=\frac{\pi}{3}$
View full question & answer→MCQ 1721 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 1731 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 1741 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
$0$
- B
$\frac{\pi}{4}$
- ✓
$\frac{\pi}{3}$
- D
$\frac{\pi}{6}$
AnswerCorrect option: C. $\frac{\pi}{3}$
We 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→MCQ 1751 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→MCQ 1761 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
$\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
- ✓
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
- D
$\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$
AnswerCorrect option: C. $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
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→MCQ 1771 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 1781 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→MCQ 1791 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 :
- ✓
$\left[\begin{array}{cc}-1 & -1 \\ 4 & 3\end{array}\right]$
- 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
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
AnswerCorrect option: A. $\left[\begin{array}{cc}-1 & -1 \\ 4 & 3\end{array}\right]$
Given $A=\left[\begin{array}{cc}2 & 1 \\ -4 & -2\end{array}\right]$, 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→MCQ 1801 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 1811 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→MCQ 1821 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 1831 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
$\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]$
- ✓
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
AnswerCorrect option: D. $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
We 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→MCQ 1841 Mark
If for a square matrix $A, A^2-A+I=0$, then $A^{-1}$ equals
AnswerCorrect option: C. $1- A$
We 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→MCQ 1851 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 $ and $y=3$
$\therefore x+y$
$=-1+3$
$=2$
View full question & answer→MCQ 1861 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
AnswerCorrect option: D. $[14]$
$A=\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]$
$A^{\prime}=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$
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 1871 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 1881 Mark
If $A$ is a square matrix and $A^2=A,$ then $(I+A)^2-3 A$ is equal to
AnswerGiven that $A^2=A$
Consider $(I+A)^2-3 A=I^2+A^2+2 A I-3 A$
$=I+A+2 A-3 A \ \left[\because I^2=I, A^2=A \text { (given) }\right]$
$=I$
View full question & answer→MCQ 1891 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
$x=1, y=2$
- ✓
$x=2, y=1$
- C
$x=1, y=-1$
- D
$x=3, y=2$
AnswerCorrect option: B. $x=2, y=1$
We 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 1901 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→MCQ 1911 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 1921 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
$0$
- ✓
$\pm 5$
- D
$25$
AnswerCorrect option: C. $\pm 5$
Given, $A=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]$
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→MCQ 1931 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
$\left[\begin{array}{cc}6 & 8 \\ 10 & 4\end{array}\right]$
- ✓
$\left[\begin{array}{cc}-6 & -8 \\ -10 & -4\end{array}\right]$
- C
$\left[\begin{array}{cc}5 & 8 \\ 10 & 3\end{array}\right]$
- D
$\left[\begin{array}{cc}-5 & -8 \\ -10 & -3\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{cc}-6 & -8 \\ -10 & -4\end{array}\right]$
Given, $A=\left[\begin{array}{ll}3 & 4 \\ 5 & 2\end{array}\right]$ 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 $
$\Rightarrow d=-4 . \\
\therefore \text { Required matrix, } B=\left[\begin{array}{cc} -6 & -8 \\ -10 & -4 \end{array}\right]$
View full question & answer→MCQ 1941 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
$\left[\begin{array}{cc}0 & -1 \\ 1 & 1\end{array}\right]$
- ✓
$\left[\begin{array}{cc}1 & 1 \\ 1 & -1\end{array}\right]$
- D
$\left[\begin{array}{ll}1 & -1 \\ 1 & -1\end{array}\right] \quad$
AnswerCorrect option: C. $\left[\begin{array}{cc}1 & 1 \\ 1 & -1\end{array}\right]$
Let $X=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
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
$ 3 a+b=4$
$2 a-b=1$
$3 c+d=2$
$2 c-d=3 $
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 1951 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→MCQ 1961 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 1971 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 1981 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
AnswerCorrect option: A. $[28]$
Consider, $A B=\left[\begin{array}{lll}2 & -3 & 4\end{array}\right]\left[\begin{array}{l}3 \\ 2 \\ 2\end{array}\right]$
$=[6-6+8]=[8]$
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 1991 Mark
If $A$ is a square matrix such that $A^2=A$, then $(I-A)^3+A$ is equal to
AnswerWe have, $A^2=A$
$\text { Now, }(I-A)^3+A=(I-A)(I-A)(I-A)+A$
$=(I \cdot I-I \cdot A-A \cdot I+A \cdot A)(I-A)+A$
$=(I-A-A+A)(I-A)+A \left[\because I \cdot A=A \cdot I=A \text { and } A^2=A\right]$
$=(I-A)(I-A)+A$
$=(I \cdot I-I \cdot A-A \cdot I+A \cdot A)+A$
$=(I-A-A+A)+A$
$=(I-A)+A$
$=I$
View full question & answer→MCQ 2001 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→