Question

Let M and N be two $3\times 3$ non-singular skew-symmetric matrices such that MN = NM. If $P^T$ denotes the transpose of P, then $M^2{N}^2\left(M^{T}N{)}^1\right(M{N}^{-1}{)}^T$ is equal to

• A. $MN$
• B. $-N^2$
• C. $M^2$
• D. $-M^2$

Answer 1

Answer: (A), (B)

$-N^2$

Answer (Statement of the question is incorrect)
Every skew-symmetric matrix of odd order is always singular and inverse of a singular matrix does not exist. The statement of the given question seems incorrect.
If the matrix is given to be symmetric and non-singular, then
$M^T=M$
$(MN{)}^T=MN(\text{as MN = NM given})$
$M^2{N}^2\left(M^{T}N{)}^{-1}\right(M{N}^{-1}{)}^T$
$=M^2{N}^2\left(MN{)}^{-1}\right(\left(N^T{)}^{-1}{M}^T\right)$
$=M^2{N}^2\left(N^{-1}{M}^{-1}\right)\left(N^{-1}M\right)$
$=M^{2}N{M}^{-1}{N}^{-1}M$
$=M^2$
If there may exist a non-singular skew-symmetric matrix (which does not exist), then
$M^2{N}^2\left(M^{T}N{)}^{-1}\right(M{N}^{-1}{)}^T=M^2{N}^2(-MN{)}^{-1}(\left(N^{-1}{)}^T{M}^T\right)$
$=-M^{2}N{M}^{-1}(-N^{-1}(-M\left)\right)=-M^2$