MCQ 11 Mark
If matrix $A$ is given by $A=\left[a_{i j}\right]_{2 \times 2}$, where $a_{i j}=i+j$, then $A$ is equal to:
- A
$\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
- ✓
$\left[\begin{array}{ll}2 & 3 \\ 3 & 4\end{array}\right]$
- C
$\left[\begin{array}{ll}1 & 1 \\ 2 & 2\end{array}\right]$
- D
$\left[\begin{array}{ll}1 & 2 \\ 1 & 2\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{ll}2 & 3 \\ 3 & 4\end{array}\right]$
(B) $\left[\begin{array}{ll}2 & 3 \\ 3 & 4\end{array}\right]$
Explanation: Here,
$\begin{array}{l}a_{11}=1+1=2 \\a_{12}=1+2=3 \\a_{21}=2+1=3 \\\text{and}\quad a_{22}=2+2=4\end{array}$
$\therefore A=\left[\begin{array}{ll}2 & 3 \\3 & 4\end{array}\right]$
View full question & answer→MCQ 21 Mark
If matrix $A=\left(\begin{array}{ccc}a & b & -5 \\ c & d & 0 \\ 5 & 0 & 0\end{array}\right)$ is skew symmetric, then value of $2 a+b+c-3 d$ is:
Answer( C ) $0$
Explanation:
$
\begin{array}{rlrl}
2 a+b+c-3 d & =b+c & (\because a=d=0) \\
& =b+(-b) & (\because c=-b) \\
& =0
\end{array}
$
View full question & answer→MCQ 31 Mark
If $A$ is matrix of order $m \times n$ and $B$ is a matrix such that $A B^{\prime}$ and $B^{\prime} A$ are both defined, then order of matrix $B$ is:
- A
$m \times m$
- B
$n \times n$
- C
$n \times m$
- ✓
$m \times n$
AnswerCorrect option: D. $m \times n$
(D) $m \times n$
Explanation: Let, $A=\left[a_{i j}\right]_{m \times n}$ and $B=\left[b_{i j}\right]_{p \times q}$
$B^{\prime}=\left[b_{j i}\right]_{q \times p}$
Now, $AB ^{\prime}$ is defined, so $n=q$
and $B^{\prime} A$ is also defined, so $p=m$
$\therefore$ Order of $B^{\prime}=\left[b_{j i}\right]_{n \times m}$
And order of $B=\left[b_{i j}\right]_{m \times n}$
View full question & answer→MCQ 41 Mark
The matrix $\left[\begin{array}{ccc}0 & -5 & 8 \\ 5 & 0 & 12 \\ -8 & -12 & 0\end{array}\right]$ is a:
Answer(C) Skew symmetric matrix
Explanation: We know that, in a square matrix, if $b_{i j}$, when $i \neq j$ then it is said to be a diagonal matrix. Here, $b_{12}, b_{13} \ldots \neq 0$ so the given matrix is not a diagonal matrix.
$\begin{aligned}\text{Now, } B & =\left[\begin{array}{ccc}0 & -5 & 8 \\ 5 & 0 & 12 \\ -8 & -12 & 0\end{array}\right] \\ B^{\prime} & =\left[\begin{array}{ccc}0 & 5 & -8 \\ -5 & 0 & -12 \\ -8 & 12 & 0\end{array}\right] \\ & =-\left[\begin{array}{ccc}0 & -5 & 8 \\ 5 & 0 & 12 \\ -8 & -12 & 0\end{array}\right] \\ & =-B\end{aligned}$
So, the given matrix is a skew - symmetric matrix, since we know that in a square matrix $B$, if $B^{\prime}=-B$, then it is called skew-symmetric matrix.
View full question & answer→MCQ 51 Mark
The matrix $\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4\end{array}\right]$ is a:
Answer(B) symmetric matrix
Explanation :
$\begin{aligned}A & =\left[\begin{array}{lll}1 & 0 & 0 \\0 & 2 & 0 \\0 & 0 & 4\end{array}\right] \\\therefore A^{\prime} & =\left[\begin{array}{lll}1 & 0 & 0 \\0 & 2 & 0 \\0 & 0 & 4\end{array}\right]=A\end{aligned}$
So, the given matrix is a symmetric matrix.
[Since, in a square matrix $A$, if $A^{\prime}=A$, then $A$ is called symmetric matrix.]
View full question & answer→MCQ 61 Mark
If matrix $A=\left[a_{i j}\right]_2 \times{ }_2$, where $a_{i j}=1$ if $i \neq j$, and 0 if $i=j$ then $A ^2$ is equal to
Answer(A) $I$
Explanation: We have, $A=\left[a_{i j}\right]_{2 \times 2'}$
where $a_{i j}=1$ if $i \neq j$ and 0 if $i=j$.
$\therefore A=\left[\begin{array}{ll}0 & 1 \\1 & 0\end{array}\right]$
And
$\begin{aligned}A^2 & =\left[\begin{array}{ll}0 & 1 \\1 & 0\end{array}\right] \cdot\left[\begin{array}{ll}0 & 1 \\1 & 0\end{array}\right] \\& =\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right] \\& =I\end{aligned}$
View full question & answer→MCQ 71 Mark
If $A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$ then $A^2$ is equal to
- A
$\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
- B
$\left[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right]$
- C
$\left[\begin{array}{ll}0 & 1 \\ 0 & 1\end{array}\right]$
- ✓
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
AnswerCorrect option: D. $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
(D) $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Explanation:
$\begin{aligned}A^2 & =A \cdot A \\& =\left[\begin{array}{ll}0 & 1 \\1 & 0\end{array}\right] \cdot\left[\begin{array}{ll}0 & 1 \\1 & 0\end{array}\right] \\& =\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]\end{aligned}$
View full question & answer→MCQ 81 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) $3 \times n$
Explanation: $A _{3 \times m}$ and $B _{3 \times n}$ are two matrices. If $m=n$, then A and B have same orders $3 \times n$ as each, so the order of $(5 A-2 B)$ should be same as $3 \times n$.
View full question & answer→MCQ 91 Mark
If $\left[\begin{array}{ll}2 x+y & 4 x \\ 5 x-7 & 4 x\end{array}\right]=\left[\begin{array}{cc}7 & 7 y-13 \\ y & x+6\end{array}\right]$, then the value of $x+y$ is:
Answer(B) 5
Explanation: We have,
$4x=x+6$
$\Rightarrow x=2$
$\text{And}\quad 4x=7y-13$
$\Rightarrow 8=7y-13$
$\Rightarrow 7y=21$
$\Rightarrow y=3$
$\therefore x+y=2+3=5$
View full question & answer→MCQ 101 Mark
Total number of possible matrices of order $3 \times 3$ with each entry 2 or 0 is:
Answer(D) 512
Explanation: Total number of possible matrices of order $3 \times 3$ with each entry 2 or 0 is $2^9$ i.e., 512.
View full question & answer→MCQ 111 Mark
The matrix $P=\left[\begin{array}{lll}0 & 0 & 4 \\ 0 & 4 & 0 \\ 4 & 0 & 0\end{array}\right]$ is a:
Answer(A) square matrix
Explanation : We know that, in a square matrix number of rows is equal to the number of columns.
So, the matrix $P=\left[\begin{array}{lll}0 & 0 & 4 \\ 0 & 4 & 0 \\ 4 & 0 & 0\end{array}\right]$ is a square matrix.
View full question & answer→MCQ 121 Mark
If $A$ is a square matrix such that $A^2=A$, then $(I+A)^3-7 A$ is equal to
Answer(C) $I$
Explanation:
$\begin{aligned}(I+A)^3-7 A & =I^3+A^3+3 I^2 A+3 A^2 I-7 A \\& =I+A^3+3 A+3 A^2-7 A \\& =I+A^2 \cdot A+3 A+3 A-7 A \quad\left[\because A^2=A\right] \\& =I+A \cdot A-A \\& =I+A^2-A \\& =I+A-A \\& =I\end{aligned}$
$\therefore(I+A)^3-7 A=I$
View full question & answer→MCQ 131 Mark
If the matrix $A$ is both symmetric and skew symmetric, then
Answer(B) A is a zero matrix
Explanation: If $A$ is both symmetric and skew-symmetric matrices, then we should have,
$A'=A$ and $A'=-A$
$\Rightarrow A=-A$
$\Rightarrow A+A=0$
$\Rightarrow 2A=0$
$\Rightarrow A=0$
Therefore, $A$ is a zero matrix.
View full question & answer→MCQ 141 Mark
If $A=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]$ is such that $A^2=I$, then
- A
$1+\alpha^2+\beta \gamma=0$
- B
$1-\alpha^2+\beta \gamma=0$
- ✓
$1-\alpha^2-\beta \gamma=0$
- D
$1+\alpha^2-\beta \gamma=0$
AnswerCorrect option: C. $1-\alpha^2-\beta \gamma=0$
(C) $1-\alpha^2-\beta \gamma=0$
Explanation:
$\begin{aligned}A & =\left[\begin{array}{cc}\alpha & \beta \\\gamma & -\alpha\end{array}\right] \\\therefore A^2 & =A \cdot A \\& =\left[\begin{array}{cc}\alpha & \beta \\\gamma & -\alpha\end{array}\right] \cdot\left[\begin{array}{cc}\alpha & \beta \\\gamma & -\alpha\end{array}\right] \\& =\left[\begin{array}{cc}\alpha^2+\beta \gamma & \alpha \beta-\alpha \beta \\\alpha \gamma-\alpha \gamma & \beta \gamma+\alpha^2\end{array}\right] \\& =\left[\begin{array}{cc}\alpha^2+\beta \gamma & 0 \\0 & \beta \gamma+\alpha^2\end{array}\right]\end{aligned}$
Now,
$\begin{array}{c}A^2=I \\{\left[\begin{array}{cc}\alpha^2+\beta \gamma & 0 \\0 & \beta \gamma+\alpha^2\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]}\end{array}$
On comparing the corresponding elements, we have:
$\begin{aligned}\alpha^2+\beta \gamma & =1 \\\Rightarrow \alpha^2+\beta \gamma-1 & =0 \\\Rightarrow 1-\alpha^2-\beta\gamma & =0\end{aligned}$
View full question & answer→MCQ 151 Mark
If A and B are symmetric matrices of same order, $A B-B A$ is a:
Answer(A) Skew-symmetric matrix
Explanation: Given that, $A$ and $B$ are symmetric matrices.
$\Rightarrow \quad A=A^{\prime} \text { and } B=B^{\prime}$
$\begin{array}{ll}\text { Now, } & (A B-B A)^{\prime}=(A B)^{\prime}-(B A)^{\prime}\quad\quad\ldots\text{(i)} \\\Rightarrow & (A B-B A)^{\prime}=B^{\prime} A^{\prime}-A^{\prime} B^{\prime}\quad\text{[By reversal law]}\end{array}$
$\begin{array}{ll}\Rightarrow & (A B-B A)^{\prime}=B A-A B \quad \text { [From Eq. (i)] } \\\Rightarrow & (AB-BA)^{\prime}=-(A B-B A)\end{array}$
View full question & answer→MCQ 161 Mark
Assume $X , Y , Z~ W$ and $P$ are matrices of order $2 \times n$, $3 \times k, 2 \times p, n \times 3$ and $p \times k$, respectively. If $n=p$, then the order of the matrix $7 X -5 Z$ is:
- A
$p \times 2$
- ✓
$2 \times n$
- C
$n \times 3$
- D
$p \times n$
AnswerCorrect option: B. $2 \times n$
(B) $2 \times n$
Explanation: Matrix X is of the order $2 \times n$.
Therefore, matrix 7X is also of the same order.
Matrix Z is of the order $2 \times p$, i.e., $2 \times n\quad$ [Since $n=p$ ]
Therefore, matrix 5Z is also of the same order.
Now, both the matrices 7X and 5Z are of the order $2 \times n$.
Thus, matrix $7 X -5 Z$ is well-defined and is of the order $2 \times n.$
View full question & answer→MCQ 171 Mark
Assume $X , Y , Z , W$ and $P$ are matrices of order $2 \times n, 3 \times k, 2 \times p, n \times 3$ and $p \times k$, respectively. The restriction on $n, k$ and $p$ so that $PY + WY$ will be defined are:
- ✓
$k=3, p=n$
- B
$k$ is arbitrary, $p=2$
- C
$p$ is arbitrary, $k=3$
- D
$k=2, p=3$
AnswerCorrect option: A. $k=3, p=n$
(A) $k=3, p=n$
Explanation: Matrices P and Y are of the orders $p \times k$ and $3 \times k$, respectively. Therefore, matrix PY will be defined if $k=3$. Consequently, PY will be of the order $p \times k$. Matrices W and Y are of the orders $n$ $\times 3$ and $3 \times k$ respectively.
Since the number of columns in $W$ is equal to the number of rows in Y, matrix WY is well-defined and is of the order $n \times k$. Matrices PY and WY can be added only when their orders are the same.
However, PY is of the order $p \times k$ and WY is of the order $n \times k$. Therefore, we must have $p=n$. Thus, $k$ $=3$ and $p=n$ are the restrictions on $n, k$, and $p$ so that PY + WY will be defined.
View full question & answer→MCQ 181 Mark
The number of all possible matrices of order $3 \times 3$ with each entry 0 or 1 is:
Answer(D) 512
Explanation: The given matrix of the order $3 \times 3$ has $9$ elements and each of these elements can be either 0 or 1.
Now, each of the 9 elements can be filled in two possible ways.
Therefore, by the multiplication principle, the required number of possible matrices is $2^9=512$.
View full question & answer→MCQ 191 Mark
Which of the given values of $x$ and $y$ make the following pair of matrices equal?
$\left[\begin{array}{cc}3 x+7 & 5 \\y+1 & 2-3 x\end{array}\right],\left[\begin{array}{cc}0 & y-2 \\8 & 4\end{array}\right]$
Answer(B) Not possible to find
Explanation: It is given that
$\left[\begin{array}{cc}3 x+7 & 5 \\y+1 & 2-3 x\end{array}\right]=\left[\begin{array}{cc}0 & y-2 \\8 & 4\end{array}\right]$
Equating the corresponding elements, we get:
$\begin{aligned}3 x+7 & =0 \\\Rightarrow \quad x & =-\frac{7}{3} \\5 & =y-2 \\\Rightarrow \quad y & =7 \\y+1 & =8 \\\Rightarrow \quad y & =7 \\2-3 x & =4 \\\Rightarrow \quad x & =-\frac{2}{3}\end{aligned}$
We find that on comparing the corresponding elements of the two matrices, we get two different values of $x$, which is not possible.
Hence, it is not possible to find the values of $x$ and $y$ for which the given matrices are equal.
View full question & answer→MCQ 201 Mark
$A=\left[a_{\text {if }}\right]_{\operatorname{m \times n}}$ is a square matrix, if
Answer(C) $m = n$
Explanation : It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns.
Therefore,
$A=\left[a_{i f}\right]_{m \times x}$ is a square matrix, if $m=n$.
View full question & answer→MCQ 211 Mark
If $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & 2\end{array}\right]$, then $a I+b A+2 A^2$ equals:
Answer(D) None of these
Explanation: Given, $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & 2\end{array}\right]$
$\therefore \quad A^2=\left[\begin{array}{lll}1 & 0 & 1 \\0 & 0 & 1 \\a & b & 2\end{array}\right]\left[\begin{array}{lll}1 & 0 & 1 \\0 &0 & 1 \\a & b & 2\end{array}\right]$
$=\left[\begin{array}{ccc}1+a & b & 3 \\ a & b & 2 \\ 3 a & 2 b & a+b+4\end{array}\right]$
Now, $a I+b A+2 A^2$
$=a\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]+b\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & 2\end{array}\right]$
$+2\left[\begin{array}{ccc}1+a & b & 3 \\ a & b & 2 \\ 3 a & 2 b & a+b+4\end{array}\right]$
$=\left[\begin{array}{ccc}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right]+\left[\begin{array}{ccc}b & 0 & b \\ 0 & 0 & b \\ a b & b^2 & 2 b\end{array}\right]$
$+\left[\begin{array}{ccc}2+2 a & 2 b & 6 \\ 2 a & 2 b & 4 \\ 6 a & 4 b & 2 a+2 b+8\end{array}\right]$
$=\left[\begin{array}{ccc}3 a+b+2 & 2 b & b+6 \\ 2 a & a+2 b & b+4 \\ 6 a+a b & 4 b+b^2 & 3 a+4 b+8\end{array}\right]$
View full question & answer→MCQ 221 Mark
If $A=\left[\begin{array}{lll}1 & 2 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & -2 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$ and $A B=I_{3'}$ then $(x+y)$ equals:
Answer(A) 0
$\begin{aligned}\text{Explanation:}\quad A B & =\left[\begin{array}{lll}1 & 2 & x \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}1 & -2 & y \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right] \\& =\left[\begin{array}{ccc}1 & 0 & y+x \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]\end{aligned}$
Since given $\quad A B=I_3$
$\therefore \quad\left[\begin{array}{ccc}1 & 0 & y+x \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]$
On comparing, we get $y+x=0 \text { or } x+y=0$
View full question & answer→MCQ 231 Mark
If $A$ is a square matrix such that $A^2=A$, then $(I+A)^2-3 A$ is equal to:
Answer(A) $I$
Explanation: We have, $(I+A)^2-3 A$
$\begin{array}{l}=I^2+2 I A+A^2-3 A \\=I+2 A+A-3 A\quad\quad\quad[\text{Since given}~ A^2=A ]\end{array}$
$=I$
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