MCQ 511 Mark
If $A$ is a square matrix, then $A-A^{\prime}$ is a
Answer(b) : As $\left(A-A^{\prime}\right)^{\prime}=A^{\prime}-\left(A^{\prime}\right)^{\prime}=A^{\prime}-A=-\left(A-A^{\prime}\right)$, therefore, $A-A^{\prime}$ is a skew-symmetric matrix.
View full question & answer→MCQ 521 Mark
Each diagonal element of a skew-symmetric matrix is
MCQ 531 Mark
If $A$ and $B$ are symmetric matrices of the same order, then
- A
$A B$ is a symmetric matrix
- B
$A-B$ is a skew-symmetric matrix
- C
$A B+B A$ is a symmetric matrix
- D
$A B-B A$ is a symmetric matrix
Answer$
\begin{array}{l}
\text {(c) : }(A B+B A)^T=(A B)^T+(B A)^T \\
=B^T A^T+A^T B^T=B A+A B=A B+B A \\
\qquad\left(\because A^T=A \text { and } B^T=B\right)
\end{array}
$
Hence, $A B+B A$ is a symmetric matrix.
View full question & answer→MCQ 541 Mark
For any square matrix $A, A A^T$ is a
Answer(b) : We have, $\left(A A^T\right)^T=\left(A^T\right)^T A^T=A A^T$
$\therefore \quad A A^T$ is a symmetric matrix.
View full question & answer→MCQ 551 Mark
The matrix $A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$ is a
Answer(c) : $A^{\prime}=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]=A$
Hence, $A$ is a symmetric matrix.
View full question & answer→MCQ 561 Mark
If $A=\left[\begin{array}{cc}3 & x-1 \\ 2 x+3 & x+2\end{array}\right]$ is a symmetric matrix, then $x=$
Answer(c) : $\because A$ is a symmetric matrix $\Rightarrow A^T=A$
$
\Rightarrow\left[\begin{array}{cc}
3 & 2 x+3 \\
x-1 & x+2
\end{array}\right]=\left[\begin{array}{cc}
3 & x-1 \\
2 x+3 & x+2
\end{array}\right]
$
$
\begin{array}{l}
\Rightarrow \quad x-1=2 x+3 \\
\Rightarrow \quad x=-4
\end{array}
$
View full question & answer→MCQ 571 Mark
The additive inverse of $A+B$, where $A$ and $B$ are given as $A=\left[\begin{array}{ll}2 & 5 \\ 9 & 3\end{array}\right], B=\left[\begin{array}{cc}-1 & 2 \\ 3 & -9\end{array}\right]$ is
- A
$\left[\begin{array}{cc}-1 & -7 \\ 12 & 6\end{array}\right]$
- B
$\left[\begin{array}{cc}-1 & 7 \\ -12 & -6\end{array}\right]$
- C
$\left[\begin{array}{cc}1 & -7 \\ -12 & 6\end{array}\right]$
- ✓
$\left[\begin{array}{cc}-1 & -7 \\ -12 & 6\end{array}\right]$
AnswerCorrect option: D. $\left[\begin{array}{cc}-1 & -7 \\ -12 & 6\end{array}\right]$
(d) : Let $C=A+B=\left[\begin{array}{cc}1 & 7 \\ 12 & -6\end{array}\right]$
Now, $(-C)=\left[\begin{array}{cc}-1 & -7 \\ -12 & 6\end{array}\right]$ is the additive inverse of $A+B$.
View full question & answer→MCQ 581 Mark
If $A=\left[\begin{array}{rrr}1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1\end{array}\right]$, then $(A-I)(A+I)=O$ for
- A
$a=b=0$ only
- B
$a=0$ only
- C
$b=0$ only
- ✓
any $a$ and $b$
AnswerCorrect option: D. any $a$ and $b$
(d) $:(A-I)(A+I)=O \Rightarrow A^2=I$
$\Rightarrow\left[\begin{array}{rrr}1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1\end{array}\right]\left[\begin{array}{rrr}1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1\end{array}\right]=\left[\begin{array}{rrr}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$, for any $a$ and $b$.
View full question & answer→MCQ 591 Mark
If $\left[\begin{array}{cc}a+b & 2 \\ 5 & a b\end{array}\right]=\left[\begin{array}{ll}6 & 2 \\ 5 & 8\end{array}\right]$, then find the values of $a$ and $b$ respectively.
Answer(c) : Since, $\left[\begin{array}{cc}a+b & 2 \\ 5 & a b\end{array}\right]=\left[\begin{array}{ll}6 & 2 \\ 5 & 8\end{array}\right]$
$\Rightarrow a+b=6$ and $a b=8$
$\Rightarrow \quad a+\frac{8}{a}=6$ $(\because a b=8 \Rightarrow b=8 / a)$
$\Rightarrow a^2-6 a+8=0 \Rightarrow(a-2)(a-4)=0 \Rightarrow a=2,4$
Hence, $a=2, b=4$ or $a=4, b=2$
View full question & answer→MCQ 601 Mark
Find the values of $a, b, c$ and $d$ respectively if $\left[\begin{array}{cc}2 a+b & a-2 b \\ 5 c-d & 4 c+3 d\end{array}\right]=\left[\begin{array}{cc}4 & -3 \\ 11 & 24\end{array}\right]$.
- A
$1,3,9,8$
- ✓
- C
$1,4,8,10$
- D
$1,5,6,7$
Answer(b) : Since, $\left[\begin{array}{cc}2 a+b & a-2 b \\ 5 c-d & 4 c+3 d\end{array}\right]=\left[\begin{array}{cc}4 & -3 \\ 11 & 24\end{array}\right]$
$\therefore \quad 2 a+b=4 \ldots$..(i), $a-2 b=-3 \ldots$..ii), $5 c-d=11 \ldots$ (iii) and $4 c+3 d=24 \ldots$ (iv)
On solving (i) and (ii), we get $a=1, b=2$
On solving (iii) and (iv), we get $c=3, d=4$
View full question & answer→MCQ 611 Mark
Two matrices of same order are said to be equal if the ______ of the two matrices are equal.
- ✓
- B
- C
only non-diagonal elements
- D
View full question & answer→MCQ 621 Mark
Find the values of $x, y, z$ and $w$ respectively such that $\left[\begin{array}{cc}x-y & 2 z+w \\ 2 x-y & 2 x+w\end{array}\right]=\left[\begin{array}{cc}5 & 3 \\ 12 & 15\end{array}\right]$.
View full question & answer→MCQ 631 Mark
If $\left[\begin{array}{ll}x+y & 2 x+z \\ x-y & 2 z+w\end{array}\right]=\left[\begin{array}{cc}4 & 7 \\ 0 & 10\end{array}\right]$, then the values of $x, y, z$ and $w$ respectively are
Answer(a) : Since, $\left[\begin{array}{ll}x+y & 2 x+z \\ x-y & 2 z+w\end{array}\right]=\left[\begin{array}{cc}4 & 7 \\ 0 & 10\end{array}\right]$
$\Rightarrow x+y=4$ ...(i)
$
\begin{array}{l}
x-y=0 && ...(ii) \\
2 x+z=7 && ...(iii) \\
\end{array}
$
and $2 z+w=10$ ...(iv)
On solving these equations, we get $x=2, y=2, z=3$ and $w=4$
View full question & answer→MCQ 641 Mark
If matrix $A=\left[a_{i j}\right]_{2 \times 2}$, where $a_{i j}=\left\{\begin{array}{l}1, \text { if } i \neq j \\ 0, \text { if } i=j\end{array}\right.$ then $A^2$ is equal to
Answer(a) : $a_{11}=0, a_{12}=1, a_{21}=1, a_{22}=0$
$
\begin{array}{l}
\therefore 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]=I
\end{array}
$
View full question & answer→MCQ 651 Mark
If $A$ and $B$ are two matrices of the order $3 \times m$ and $3 \times n$, respectively, and $m=n$, then the order of matrix $(5 A-2 B)$ is
- A
$m \times 3$
- B
$3 \times 3$
- C
$m \times n$
- ✓
$3 \times n$
AnswerCorrect option: D. $3 \times n$
(d) : $A$ is of order $3 \times m$ and $B$ is of order $3 \times n$ and $m=n$
So, $5 A-2 B$ is of order $3 \times m$ or $3 \times n$
View full question & answer→MCQ 661 Mark
If $A=\left[a_{i j}\right]_{2 \times 2}$, where $a_{i j}=\frac{(i+2 j)^2}{2}$, then $A$ is equal to
- A
$\left[\begin{array}{ll}9 & 25 \\ 8 & 18\end{array}\right]$
- ✓
$\left[\begin{array}{cc}9 / 2 & 25 / 2 \\ 8 & 18\end{array}\right]$
- C
$\left[\begin{array}{cc}9 & 25 \\ 4 & 9\end{array}\right]$
- D
$\left[\begin{array}{cc}9 / 2 & 15 / 2 \\ 4 & 9\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{cc}9 / 2 & 25 / 2 \\ 8 & 18\end{array}\right]$
(b): Here,
$
\begin{array}{l}
a_{11}=\frac{(1+2 \times 1)^2}{2}=\frac{9}{2}, a_{12}=\frac{(1+2 \times 2)^2}{2}=\frac{25}{2}, \\
a_{21}=\frac{(2+2 \times 1)^2}{2}=8 \text { and } a_{22}=\frac{(2+2 \times 2)^2}{2}=18
\end{array}
$
So, the required matrix $A=\left[\begin{array}{cc}9 / 2 & 25 / 2 \\ 8 & 18\end{array}\right]$
View full question & answer→MCQ 671 Mark
If $A=\left[\begin{array}{r}1 \\ -4 \\ 3\end{array}\right]$ and $B=\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right]$, then $(A B)^{\prime}$ is equal to
- ✓
$\left[\begin{array}{rrr}-1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3\end{array}\right]$
- B
$\left[\begin{array}{rrr}-1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3\end{array}\right]$
- C
$\left[\begin{array}{rrr}1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & 4 & 3\end{array}\right]$
- D
$\left[\begin{array}{rrr}-1 & 4 & -3 \\ 2 & 8 & 6 \\ 1 & -4 & 3\end{array}\right]$
AnswerCorrect option: A. $\left[\begin{array}{rrr}-1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3\end{array}\right]$
(a) : $(A B)=\left[\begin{array}{c}1 \\ -4 \\ 3\end{array}\right]\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right]=\left[\begin{array}{ccc}-1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3\end{array}\right]$
$
\therefore \quad(A B)^{\prime}=\left[\begin{array}{ccc}
-1 & 4 & -3 \\
2 & -8 & 6 \\
1 & -4 & 3
\end{array}\right]
$
View full question & answer→MCQ 681 Mark
If $A=\left[\begin{array}{rrr}0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0\end{array}\right]$, then $A+2 A^T$ equals
- A
$A$
- B
$-A^T$
- ✓
$A^T$
- D
$2 A^2$
Answer(c) : $A^T=\left[\begin{array}{ccc}0 & 1 & -2 \\ -1 & 0 & -3 \\ 2 & 3 & 0\end{array}\right]=-\left[\begin{array}{ccc}0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0\end{array}\right]=-A$
So, $A^T=-A \Rightarrow A=-A^T$
Hence, $A+2 A^T=-A^T+2 A^T=A^T$
View full question & answer→MCQ 691 Mark
If $A$ and $B$ are square matrices of same order and $A^{\prime}$ denotes the transpose of $A$, then
- ✓
$(A B)^{\prime}=B^{\prime} A^{\prime}$
- B
$(A B)^{\prime}=A^{\prime} B^{\prime}$
- C
$A B=O \Rightarrow|A|=0$ and $|B|=0$
- D
$A B=O \Rightarrow A=O$ or $B=O$
AnswerCorrect option: A. $(A B)^{\prime}=B^{\prime} A^{\prime}$
(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→MCQ 701 Mark
If $A$ and $B$ are $2 \times 2$ matrices, then which of the following is true?
AnswerCorrect option: C. $(A-B)(A+B)=A^2+A B-B A-B^2$
(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→MCQ 711 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→MCQ 721 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
AnswerCorrect option: A. $\begin{array}{l}A+B=B+A \text { and } A+(B+C)=(A+B)+C\end{array}$
(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→MCQ 731 Mark
If $A=\left[\begin{array}{ll}0 & 2 \\ 2 & 0\end{array}\right]$, then $A^2$ is equal to
- A
$\left[\begin{array}{ll}0 & 2 \\ 2 & 0\end{array}\right]$
- B
$\left[\begin{array}{ll}4 & 0 \\ 4 & 0\end{array}\right]$
- C
$\left[\begin{array}{ll}0 & 2 \\ 0 & 4\end{array}\right]$
- ✓
$\left[\begin{array}{ll}4 & 0 \\ 0 & 4\end{array}\right]$
AnswerCorrect option: D. $\left[\begin{array}{ll}4 & 0 \\ 0 & 4\end{array}\right]$
(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→MCQ 741 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→MCQ 751 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 }
$
- A
$2 \times 3$
- B
$2 \times 2$
- C
$3 \times 2$
- ✓
$3 \times 3$
AnswerCorrect option: D. $3 \times 3$
(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→MCQ 761 Mark
If $A B=A$ and $B A=B$, then
- A
$B=I$
- B
$A=I$
- C
$A^2=A$
- D
$B^2=I$
Answer$
\begin{array}{l}
\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
\end{array}
$
View full question & answer→MCQ 771 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 $A^T-B^T$.
- ✓
$\left[\begin{array}{rr}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]$
- B
$\left[\begin{array}{ll}4 & 3 \\ 3 & 0 \\ 1 & 2\end{array}\right]$
- C
$\left[\begin{array}{rr}4 & 0 \\ -1 & -3 \\ 3 & -2\end{array}\right]$
- D
$\left[\begin{array}{rr}1 & -3 \\ 2 & 4 \\ 3 & 5\end{array}\right]$
AnswerCorrect option: A. $\left[\begin{array}{rr}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]$
(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→MCQ 781 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
- A
$\left[\begin{array}{rr}-3 & -2 \\ 10 & 7\end{array}\right]$
- ✓
$\left[\begin{array}{rr}-3 & 10 \\ -2 & 7\end{array}\right]$
- C
$\left[\begin{array}{ll}-3 & 7 \\ 10 & 2\end{array}\right]$
- D
AnswerCorrect option: B. $\left[\begin{array}{rr}-3 & 10 \\ -2 & 7\end{array}\right]$
(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→