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Let $M$ and $N$ be two $3 \times 3$ non-singular skew-symmetric matrices such that $M N=N M$. If $P^T$ denotes the transpose of $P$, then $M^2 N^2\left(M^T N\right)^{-1}\left(M N^{-1}\right)^T$ is equal to
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Verified Answer
The correct answers are:
$-M^2$
$-M^2$
Given, $M^T=-M, N^T=-N$
and $\quad M N=N M$
$$
\begin{aligned}
\therefore & M^2 N^2\left(M^T N\right)^{-1}\left(M N^{-1}\right)^T \\
& =M^2 N^2 N^{-1}\left(M^T\right)^{-1}\left(N^{-1}\right)^T \cdot M^T \\
& =M^2 N\left(N N^{-1}\right)(-M)^{-1}\left(N^T\right)^{-1}(-M) \\
& =M^2 N\left(-M^{-1}\right)(-N)^{-1}(-M) \\
& =-M^2 N M^{-1} N^{-1} M \\
& =-M \cdot(M N) M^{-1} N^{-1} M \\
& =-M(N M) M^{-1} N^{-1} M \\
& =-M N\left(N M^{-1}\right) N^{-1} M \\
& =-M\left(N N^{-1}\right) M=-M^2
\end{aligned}
$$
Note Here, non-singular word should not be used, since there is no non-singular $3 \times 3$ skew-symmetric matrix.
and $\quad M N=N M$
$$
\begin{aligned}
\therefore & M^2 N^2\left(M^T N\right)^{-1}\left(M N^{-1}\right)^T \\
& =M^2 N^2 N^{-1}\left(M^T\right)^{-1}\left(N^{-1}\right)^T \cdot M^T \\
& =M^2 N\left(N N^{-1}\right)(-M)^{-1}\left(N^T\right)^{-1}(-M) \\
& =M^2 N\left(-M^{-1}\right)(-N)^{-1}(-M) \\
& =-M^2 N M^{-1} N^{-1} M \\
& =-M \cdot(M N) M^{-1} N^{-1} M \\
& =-M(N M) M^{-1} N^{-1} M \\
& =-M N\left(N M^{-1}\right) N^{-1} M \\
& =-M\left(N N^{-1}\right) M=-M^2
\end{aligned}
$$
Note Here, non-singular word should not be used, since there is no non-singular $3 \times 3$ skew-symmetric matrix.
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