Question

Let M and N be two 3 $\times$ 3 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. $M^2$
• B. $-N^2$
• C. $-M^2$
• D. $MN$

Answer 1

The correct option is C $-M^2$

Given, $M^T=-M,N^T=-NandMN=NM$
$\therefore M^2{N}^2\left(M^{T}N{)}^{-1}\right(M{N}^{-1}{)}^T$
$\Rightarrow M^2{N}^2{N}^{-1}\left(M^T{)}^{-1}\right(N^{-1}{)}^T.M^T\Rightarrow M^{2}N\left(N{N}^{-1}\right)(-M{)}^{-1}(N^T{)}^{-1}(-M)\Rightarrow M^{2}NI(-M^{-1})(-N{)}^{-1}(-M)\Rightarrow -M^{2}N{M}^{-1}{N}^{-1}M\Rightarrow -M.(MN)M^{-1}{N}^{-1}M=-M(NM)M^{-1}{N}^{-1}M\Rightarrow -MN(MM{)}^{-1}{N}^{-1}M=-M\left(N{N}^{-1}\right)M\Rightarrow -M^2$