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Let $A$ and $B$ be two symmetric matrices of same order. Then, the matrix $A B-B A$ is
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Verified Answer
The correct answer is:
a skew-symmetric matrix
Given, $A=A^{\prime}, B=B^{\prime}$
Now, $(A B-B A)^{\prime}=(A B)^{\prime}-(B A)^{\prime}$
$\begin{aligned}
& =B^{\prime} A^{\prime}-A^{\prime} B^{\prime} \\
& =B A-A B \\
& =-(A B-B A)
\end{aligned}$
$\therefore A B-B A$ is a skew-symmetric matrix.
Now, $(A B-B A)^{\prime}=(A B)^{\prime}-(B A)^{\prime}$
$\begin{aligned}
& =B^{\prime} A^{\prime}-A^{\prime} B^{\prime} \\
& =B A-A B \\
& =-(A B-B A)
\end{aligned}$
$\therefore A B-B A$ is a skew-symmetric matrix.
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