Matrices And Determinants Ques 86
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
(a) $M^{2}$
(b) $-N^{2}$
(c) $-M^{2}$
(d) $M N$
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Answer:
Correct Answer: 86.(c)
Solution:
- Given, $M^{T}=-M, N^{T}=-N$ and $M N=N M$
$\therefore M^{2} N^{2}\left(M^{T} N\right)^{-1}\left(M N^{-1}\right)^{T}$
$\Rightarrow \quad M^{2} N^{2} N^{-1}\left(M^{T}\right)^{-1}\left(N^{-1}\right)^{T} \cdot M^{T}$
$\Rightarrow \quad M^{2} N\left(N N^{-1}\right)(-M)^{-1}\left(N^{T}\right)^{-1}(-M)$
$\Rightarrow \quad M^{2} N I\left(-M^{-1}\right)(-N)^{-1}(-M)$
$\Rightarrow \quad -M^{2} N M^{-1} N^{-1} M$
$\Rightarrow \quad-M \cdot(M N) M^{-1} N^{-1} M=-M(N M) M^{-1} N^{-1} M$
$\Rightarrow \quad-M N\left(N M^{-1}\right) N^{-1} M=-M\left(N N^{-1}\right) M \quad \Rightarrow \quad-M^{2}$
NOTE: Here, non-singular word should not be used, since there is no non-singular $3 \times 3$ skew-symmetric matrix.
BETA
