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 $M^2{N}^2{\left(M^{T}N\right)}^{-1}{\left(M{N}^{-1}\right)}^T$ is equal to?

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

Answer 1

The correct option is C

$-M^2$

Explanation of correct option:

We know that, $M$ and $N$ are two$3\times 3$ non-singular skew-symmetric matrices. This means that

$$\begin{gathered} M^T=-M \\ {N}^T=-N \end{gathered}$$

Now,

$$\begin{gathered} M^2{N}^2{\left(M^{T}N\right)}^{-1}{\left(M{N}^{-1}\right)}^T \\ =M^2{N}^2{N}^{-1}{\left(M^T\right)}^{-1}{\left(N^T\right)}^{-1}{\left(M\right)}^T\mathbf{[}\mathbf{\because }{\mathbf{\left(}\mathbf{A}\mathbf{B}\mathbf{\right)}}^{\mathbf{-}\mathbf{1}}\mathbf{=}{\mathbf{B}}^{\mathbf{-}\mathbf{1}}{\mathbf{A}}^{\mathbf{-}\mathbf{1}}\mathbf{,}{\mathbf{\left(}\mathbf{A}\mathbf{B}\mathbf{\right)}}^{\mathbf{T}}\mathbf{=}{\mathbf{B}}^{\mathbf{T}}{\mathbf{A}}^{\mathbf{T}}\mathbf{,}{\mathbf{\left(}{\mathbf{A}}^{\mathbf{-}\mathbf{1}}\mathbf{\right)}}^{\mathbf{T}}\mathbf{=}{\mathbf{\left(}{\mathbf{A}}^{\mathbf{T}}\mathbf{\right)}}^{\mathbf{-}\mathbf{1}}\mathbf{]} \\ =M^{2}N\left(N{N}^{-1}\right){\left(M^T\right)}^{-1}{\left(N^T\right)}^{-1}{\left(M\right)}^T \\ =M^{2}N{(-M)}^{-1}{(-N)}^{-1}(-M)\mathbf{[}\mathbf{\because }\mathbf{N}{\mathbf{N}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\mathit{I}\mathbf{]} \\ =-M\left(MN\right)M^{-1}{N}^{-1}M \\ =-MN{N}^{-1}M\mathbf{[}\mathbf{\because }{\mathbf{MM}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\mathit{I}\mathbf{,}\mathit{M}\mathit{N}\mathbf{=}\mathit{N}\mathit{M}\mathbf{]} \\ =-MM \\ =-M^2 \end{gathered}$$

Hence, option(C) is correct.