Question 511 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
Answer$\left[\begin{array}{lll}7 & 1 & 2 \\ 9 & 2 & 1\end{array}\right]\left[\begin{array}{l}3 \\ 4 \\ 5\end{array}\right]+2\left[\begin{array}{l}4 \\ 2\end{array}\right]$
$=\left[\begin{array}{c}21+4+10 \\ 27+8+5\end{array}\right]+\left[\begin{array}{l}8 \\ 4\end{array}\right]$
$=\left[\begin{array}{l}35 \\ 40\end{array}\right]+\left[\begin{array}{l}8 \\ 4\end{array}\right]$
$=\left[\begin{array}{l}43 \\ 44\end{array}\right]$
View full question & answer→Question 521 Mark
If $A$ is a $m \times n$ matrix such that $A B$ and $B A$ are both defined, then $B$ is an
Answer(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→Question 531 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→Question 541 Mark
If $A$ and $B$ are matrices of same order, then $\left(A B\ ^{\prime}-B A\ ^{\prime}\right)$ is a
Answer$\text { (a): }\left(A B\ ^{\prime}-B A\ ^{\prime}\right)^{\prime}=\left(A B\ ^{\prime}\right)\ ^{\prime}-\left(B A\ ^{\prime}\right)\ ^{\prime}$
$=\left(B\ ^{\prime}\right)\ ^{\prime} A\ ^{\prime}-\left(A\ ^{\prime}\right)\ ^{\prime} B\ ^{\prime}=B A\ ^{\prime}-A B\ ^{\prime}=-\left(A B\ ^{\prime}-B A\ ^{\prime}\right)$
Hence, $\left(A B\ ^{\prime}-B A\ ^{\prime}\right)$ is a skew symmetric matrix.
View full question & answer→Question 551 Mark
For any two matrices $A$ and $B$, we have
View full question & answer→Question 561 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→Question 571 Mark
If $A^3=O$, then $A^2+A+I=$
Answer(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→Question 581 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→Question 591 Mark
If a matrix $A$ is both symmetric and skewsymmetric, then
Answer(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→Question 601 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→Question 611 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→Question 621 Mark
Each diagonal element of a skew-symmetric matrix is
Question 631 Mark
If $A$ and $B$ are symmetric matrices of the same order, then
Answer$\text {(c) : }(A B+B A)^T$$=(A B)^T+(B A)^T$
$=B^T A^T+A^T B^T=B A+A B=A B+B A$
$\qquad\left(\because A^T=A \text { and } B^T=B\right)$
Hence, $A B+B A$ is a symmetric matrix.
View full question & answer→Question 641 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→Question 651 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→Question 661 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] $
$\Rightarrow x-1=2 x+3$
$\Rightarrow x=-4 $
View full question & answer→Question 671 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
Answer(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→Question 681 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
Answer(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→Question 691 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→Question 701 Mark
Find the values of $a, b, c$ and $d$ respectively if $\left[\begin{array}{cc}2 a+b & a-2 b \\ 5 c-d & 4 c+3 d\end{array}\right]=\left[\begin{array}{cc}4 & -3 \\ 11 & 24\end{array}\right]$.
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→Question 711 Mark
Two matrices of same order are said to be equal if the ______ of the two matrices are equal.
Question 721 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→Question 731 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, $\left[\begin{array}{ll}x+y & 2 x+z \\ x-y & 2 z+w\end{array}\right]=\left[\begin{array}{cc}4 & 7 \\ 0 & 10\end{array}\right]$
$\Rightarrow x+y=4 ...(i)$
$x-y=0 ...(ii)$
$2 x+z=7 ...(iii)$
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→Question 741 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$a_{11}=0, a_{12}=1, a_{21}=1, a_{22}=0$
$\therefore A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
$\therefore A^2=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
$=\left[\begin{array}{ll}0+1 & 0+0 \\ 0+0 & 1+0\end{array}\right]$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=I$
View full question & answer→Question 751 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
Answer(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→Question 761 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
Answer$a_{11}=\frac{(1+2 \times 1)^2}{2}=\frac{9}{2}, a_{12}=\frac{(1+2 \times 2)^2}{2}=\frac{25}{2},$
$a_{21}=\frac{(2+2 \times 1)^2}{2}=8$ and $a_{22}=\frac{(2+2 \times 2)^2}{2}=18$
So, the required matrix $A=\left[\begin{array}{cc}9 / 2 & 25 / 2 \\ 8 & 18\end{array}\right]$
View full question & answer→Question 771 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
Answer(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→Question 781 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
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 791 Mark
Final adjoint of matrix $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 4\end{array}\right]$
- ✓
$\left[\begin{array}{cc}4 & -3 \\ -1 & 2\end{array}\right]$
- B
$\left[\begin{array}{ll}-2 & -3 \\ -1 & -4\end{array}\right]$
- C
$\left[\begin{array}{rr}-4 & 3 \\ 1 & -2\end{array}\right]$
- D
$\left[\begin{array}{ll}4 & 1 \\ 3 & 2\end{array}\right]$
AnswerCorrect option: A. $\left[\begin{array}{cc}4 & -3 \\ -1 & 2\end{array}\right]$
View full question & answer→MCQ 801 Mark
If $A=\left[\begin{array}{rr}0 & -1 \\ 0 & 2\end{array}\right]$ and $B=\left[\begin{array}{ll}3 & 5 \\ 0 & 0\end{array}\right]$ then value of $A B$ is-
- ✓
$\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
- B
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
- C
$\left[\begin{array}{cc}0 & 0 \\ 1 & 1\end{array}\right]$
- D
$\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
AnswerCorrect option: A. $\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
View full question & answer→