Matrices And Determinants Ques 11
If $A$ is a symmetric matrix and $B$ is a skew-symmetric matrix such that $A+B=\begin{bmatrix}2 & 3 \\ 5 & -1\end{bmatrix}$, then $A B$ is equal to
(a) $\begin{bmatrix}-4 & -2 \\ -1 & 4\end{bmatrix}$
(b) $\begin{bmatrix}4 & -2 \\ -1 & -4\end{bmatrix}$
(c) $\begin{bmatrix}4 & -2 \\ 1 & -4\end{bmatrix}$
(d) $\begin{bmatrix}-4 & 2 \\ 1 & 4\end{bmatrix}$
(2019 Main, 12 April I)
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Answer:
Correct Answer: 11.(b)
Solution:
Formula:
- Given matrix $A$ is a symmetric and matrix $B$ is a skew-symmetric.
$\therefore \quad A^{T}=A$ and $B^{T}=-B$
Since, $A+B=\begin{bmatrix}2 & 3 \\ 5 & -1\end{bmatrix}$ (given)… (i)
On taking transpose both sides, we get
$ \begin{aligned} (A+B)^{T}& =\begin{bmatrix} 2 & 3\\ 5 & -1 \end{bmatrix}^T \\ \Rightarrow \quad A^{T}+B^{T}& =\begin{bmatrix} 2 & 5 \\ 3 & -1 \end{bmatrix} \quad …….(i) \end{aligned} $
Given, $A^{T}=A$ and $B^{T}=-B$
$ \Rightarrow \quad A-B=\begin{bmatrix} 2 & 5 \\ 3 & -1 \end{bmatrix} $
On solving Eqs. (i) and (ii), we get
$ \begin{aligned} A & =\begin{bmatrix} 2 & 4 \\ 4 & -1 \end{bmatrix} \text { and } B=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \end{aligned} $
So, $A B =$ $\begin{bmatrix} 2 & 4 \\ 4 & -1 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$
$\quad \quad \quad=$ $\begin{bmatrix} 4 & -2 \\ -1 & -4 \end{bmatrix}$
BETA
