Assertion (A) & Reason (B) MCQ

Question 11 Mark

Assertion (A): For any symmetric matrix A, B'AB is a skew-symmetric matrix.
Reason (R): A square matrix P is skew-symmetric if P'$P^{\prime}=-P$.

Answer

$\because A$ is symmetric matrix
$\Rightarrow A^{\prime}=A$ ......(i)
Now, $\left(B^{\prime} A B\right)^{\prime}=B^{\prime} A^{\prime}\left(B^{\prime}\right)^{\prime}=B^{\prime} A B$ (using (i)) $\Rightarrow B^{\prime} A B$ is a symmetric matrix
So, assertion is false but reason is true.

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Question 21 Mark

Assertion $(A) : A=\left[\begin{array}{ll}2 & 3 \\ 1 & 4\end{array}\right], B=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, then $(A+B)^2=A^2+B^2+2 A B$.
Reason $(R)$: For the matrices $A$ and $B$ given in assertion, $A B=B A$.

Answer

$ A=\left[\begin{array}{ll}2 & 3 \\ 1 & 4\end{array}\right] $ and $B=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=I $
$ A B=A I=A $ and $ B A=I A=A $
$ \Rightarrow A B=B A$
Consequently, $(A+B)^2=(A+B)(A+B) $
$=A(A+B)+B(A+B)=A^2+A B+B A+B^2 $
$=A^2+A B+A B+B^2=A^2+2 A B+B^2$

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Question 31 Mark

Assertion $(A)$ : If $\left[\begin{array}{ll}x & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -2 & 3\end{array}\right]\left[\begin{array}{c}x \\ -5\end{array}\right]=0$, then value of $x$ is either $-3$ or $5$ .
Reason $(R)$ : Two matrices $\left(\begin{array}{ll}x & y \\ u & v\end{array}\right)$ and $\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)$ are equal if and only if their corresponding entries are equal.

Answer

Given $\left[\begin{array}{ll}x & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -2 & 3\end{array}\right]\left[\begin{array}{c}x \\ -5\end{array}\right]=0$
$ \begin{array}{l} \Rightarrow [(x-2) 3]\left[\begin{array}{c} x \\ -5 \end{array}\right]\end{array}=0 $
$ \Rightarrow x(x-2)-15=0$
$ \Rightarrow x^2-2 x-15=0 $
$\Rightarrow x=-3,5$
$\therefore$ Assertion and Reason both are true and Reason is the correct explanation of Assertion.

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Question 41 Mark

Assertion (A): The matrix $A=\left(\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right)$ is a skew-symmetric matrix.
Reason (R) : A square matrix $A=\left(a_{i j}\right)$ of order $m$ is said to be skew-symmetric if $A^T=-A$.

Answer

(a) : $A=\left(\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right) \therefore A^T=\left(\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right)$
$
\Rightarrow \quad A^T=-A
$
$\therefore \quad$ Assertion and Reason both are true and Reason is correct explanation of Assertion.

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Question 51 Mark

Assertion $(A)$ : If $A=\frac{1}{3}\left[\begin{array}{ccc}1 & -2 & 2 \\ -2 & 1 & 2 \\ -2 & -2 & -1\end{array}\right]$, then $A\left(A^T\right)=I$
Reason $(R)$ : For any square matrix $A,\left(A^T\right)^T=A$

Answer

$\because A A^T=\frac{1}{3}\left[\begin{array}{ccc}1 & -2 & 2 \\ -2 & 1 & 2 \\ -2 & -2 & -1\end{array}\right] \cdot \frac{1}{3}\left[\begin{array}{ccc}1 & -2 & -2 \\ -2 & 1 & -2 \\ 2 & 2 & -1\end{array}\right] $
$ =\frac{1}{9}\left[\begin{array}{lll}9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I$

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Question 61 Mark

Assertion (A) : $B=\left[\begin{array}{llll}-\frac{1}{2} & \sqrt{5} & 2 & 3\end{array}\right]_{1 \times 4}$ is a row matrix.
Reason (R): If $B=\left[b_{i j}\right]_{1 \times n}$ is a row matrix, then its order is $n \times 1$.

Answer

(a) : $B=\left[\begin{array}{llll}-\frac{1}{2} & \sqrt{5} & 2 & 3\end{array}\right]_{1 \times 4}$ is a row matrix. In general, $B=\left[b_{i j}\right]_{1 \times n}$ is a row matrix of order $1 \times n$.

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Question 71 Mark

Assertion (A) : If $\left[\begin{array}{cc}x y & 4 \\ z+5 & x+y\end{array}\right]=\left[\begin{array}{cc}4 & w \\ 0 & 4\end{array}\right]$, then $x=2, y=2, z=-5$ and $w=4$.
Reason (R) : Two matrices are equal, if their orders are same and their corresponding elements are equal.

Answer

(a) : $\left[\begin{array}{cc}x y & 4 \\ z+5 & x+y\end{array}\right]=\left[\begin{array}{cc}4 & w \\ 0 & 4\end{array}\right]$
On equating the corresponding elements, we get
$
x y=4, w=4, z+5=0 \text { and } x+y=4
$
On solving these equations, we get $x=2, y=2, z=-5$ and $w=4$. Also, the two matrices are equal, if their orders are same and their corresponding elements are equal.

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Question 81 Mark

For any square matrix $A$ with real number entries, consider the following statements.
Assertion (A) : $A+A^{\prime}$ is a symmetric matrix.
Reason (R): $A-A^{\prime}$ is a skew-symmetric matrix.

Answer

(b) : Let $B=A+A^{\prime}$, then
$
B^{\prime}=\left(A+A^{\prime}\right)^{\prime}=A^{\prime}+\left(A^{\prime}\right)^{\prime}=A^{\prime}+A=A+A^{\prime}=B
$
Therefore, $B=A+A^{\prime}$ is a symmetric matrix.
Now, let $C=A-A^{\prime}$
$
C^{\prime}=\left(A-A^{\prime}\right)^{\prime}=A^{\prime}-\left(A^{\prime}\right)^{\prime}=A^{\prime}-A=-\left(A-A^{\prime}\right)=-C
$
Therefore, $C=A-A^{\prime}$ is a skew-symmetric matrix.

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Question 91 Mark

Assertion (A): $\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 7\end{array}\right]$ is a diagonal matrix.
Reason (R): $A=\left[a_{i j}\right]_{n \times n}$ is a square matrix such that $a_{i j}=0, \forall i \neq j$, then $A$ is called diagonal matrix.

Answer

(a): If $A=\left[a_{i j}\right]_{n \times n}$ is a square matrix such that $a_{i j}=0$ for $i \neq j$, then $A$ is called diagonal matrix. Thus, the given statement is true and $A=\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 7\end{array}\right]$ is a diagonal matrix.

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Question 101 Mark

Assertion (A) : The matrix $\left(\begin{array}{llll}1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0\end{array}\right)$ is a diagonal matrix.
Reason (R) : $A=\left(a_{i j}\right)_{m \times m}$ is a square matrix such that entry $a_{i j}=0 \forall i, j$, then $A$ is called diagonal matrix.

Answer

(d) : The given matrix having order $3 \times 4$.
$\therefore \quad$ Given matrix is not a square matrix. Diagonal exist only in the square matrix.
$\therefore \quad$ Assertion is false.
On the other side, Reason satisfies the condition of diagonal matrix.
$\therefore \quad$ Assertion is false but Reason is true.

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Question 111 Mark

Assertion (A) : Scalar matrix $A=\left[a_{i j}\right]$ $=\left\{\begin{array}{ll}k ; & i=j \\ 0 ; & i \neq j\end{array}\right.$ where $k$ is a scalar, is an identity matrix when $k=1$.
Reason (R) : Every identity matrix is not a scalar matrix.

Answer

(c) : A scalar matrix $A=\left[a_{i j}\right]=\left\{\begin{array}{ll}k, & i=j \\ 0, & i \neq j\end{array}\right.$ is an identity matrix when $k=1$. But every identity matrix is clearly a scalar matrix.

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