Matrices and Determinants 1 Question 1
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1. If $A$ is a symmetric matrix and $B$ is a skew-symmetric matrix such that $A+B=\begin{array}{cc}2 & 3 \ 5 & -1\end{array}$, then $A B$ is equal to
======= ####1. 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
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed
(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: 1. (b)
Solution:
- 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} \\\ \Rightarrow \quad A^{T}+B^{T}& =\begin{bmatrix} 2 & 5 \\\ 3 & -1 \end{bmatrix} \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} \\\ \text { So, } A B & =\begin{bmatrix} 2 & 4 & 0 & -1 \\\ 4 & -1 & 1 & 0 \end{bmatrix}=\begin{bmatrix} 4 & -2 \\\ -1 & -4 \end{bmatrix} \end{aligned} $
