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 \[{M^2}{N^2}{({M^T}N)^{ - 1}}{(M{N^{ - 1}})^T}\] is equal to
A. \[{M^2}\]
B. \[ - {N^2}\]
C. \[ - {M^2}\]
D. \[MN\]

Answer 1

Hint: We use the formula for skew-symmetric matrix and write the value of transpose of M and N. Now define the formula for inverse of product of two matrices and transpose of product of two matrices and apply on the given equation. In the end combine powers of the same matrix and solve.
* A matrix is said to be skew-symmetric if \[{A^T} = - A\]
* If A and B are two matrices then\[{(AB)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}\] and \[{(AB)^T} = {B^T}{A^T}\]

Complete step by step answer:
We are given M and N are skew-symmetric matrices.
We write the value of transpose of both matrices using the definition of skew-symmetric matrices.
We know M is skew-symmetric matrix
\[ \Rightarrow {M^T} = - M\] … (1)
Similarly, if N is skew-symmetric matrix
\[ \Rightarrow {N^T} = - N\] … (2)
Now we also write the value of inverse of product of matrices, if M and N are two matrices then
\[{(MN)^{ - 1}} = {N^{ - 1}}{M^{ - 1}}\] … (3)
Similarly when we are taking transpose of product of matrices M and N, we can write
\[{(MN)^T} = {N^T}{M^T}\] … (4)
We have to find the value of \[{M^2}{N^2}{({M^T}N)^{ - 1}}{(M{N^{ - 1}})^T}\]. Substitute the obtained values step-by-step to obtain the answer.
Use equation (1) to change the value of transpose of M in RHS of the equation
\[ \Rightarrow {M^2}{N^2}{({M^T}N)^{ - 1}}{(M{N^{ - 1}})^T} = {M^2}{N^2}{( - MN)^{ - 1}}{(M{N^{ - 1}})^T}\]
Take negative sign outside of the bracket in RHS of the equation
\[ \Rightarrow {M^2}{N^2}{({M^T}N)^{ - 1}}{(M{N^{ - 1}})^T} = - \left\{ {{M^2}{N^2}{{(MN)}^{ - 1}}{{(M{N^{ - 1}})}^T}} \right\}\]
Now use equation (3) to write inverse of product of matrices M and N in RHS of the equation
\[ \Rightarrow {M^2}{N^2}{({M^T}N)^{ - 1}}{(M{N^{ - 1}})^T} = - \left\{ {{M^2}{N^2}{N^{ - 1}}{M^{ - 1}}{{(M{N^{ - 1}})}^T}} \right\}\]
Now use equation (4) to write transpose of the product of matrices M and inverse of N in RHS of the equation
\[ \Rightarrow {M^2}{N^2}{({M^T}N)^{ - 1}}{(M{N^{ - 1}})^T} = - \left\{ {{M^2}{N^2}{N^{ - 1}}{M^{ - 1}}{{({N^{ - 1}})}^T}{M^T}} \right\}\]
Use equation (1) to change the value of transpose of M in RHS of the equation
\[ \Rightarrow {M^2}{N^2}{({M^T}N)^{ - 1}}{(M{N^{ - 1}})^T} = - \left\{ {{M^2}{N^2}{N^{ - 1}}{M^{ - 1}}{{({N^{ - 1}})}^T}( - M)} \right\}\]
Bring negative sign out of the bracket in RHS of the equation
\[ \Rightarrow {M^2}{N^2}{({M^T}N)^{ - 1}}{(M{N^{ - 1}})^T} = - 1 \times - 1\left\{ {{M^2}{N^2}{N^{ - 1}}{M^{ - 1}}{{({N^{ - 1}})}^T}M} \right\}\]
Interchange the powers of inverse and transpose on N and multiply both negative signs in RHs of the equation.
\[ \Rightarrow {M^2}{N^2}{({M^T}N)^{ - 1}}{(M{N^{ - 1}})^T} = \left\{ {{M^2}{N^2}{N^{ - 1}}{M^{ - 1}}{{({N^T})}^{ - 1}}M} \right\}\]
Use equation (2) to change the value of transpose of N in RHS of the equation
\[ \Rightarrow {M^2}{N^2}{({M^T}N)^{ - 1}}{(M{N^{ - 1}})^T} = \left\{ {{M^2}{N^2}{N^{ - 1}}{M^{ - 1}}{{( - N)}^{ - 1}}M} \right\}\]
Bring negative sign out of the bracket in RHS of the equation
\[ \Rightarrow {M^2}{N^2}{({M^T}N)^{ - 1}}{(M{N^{ - 1}})^T} = - \left\{ {{M^2}{N^2}{N^{ - 1}}{M^{ - 1}}{N^{ - 1}}M} \right\}\]
Collect powers of each matrix
\[ \Rightarrow {M^2}{N^2}{({M^T}N)^{ - 1}}{(M{N^{ - 1}})^T} = - \left\{ {{M^{2 - 1 + 1}}{N^{2 - 1 - 1}}} \right\}\]
\[ \Rightarrow {M^2}{N^2}{({M^T}N)^{ - 1}}{(M{N^{ - 1}})^T} = - {M^2}\]
\[\therefore \]The value of \[{M^2}{N^2}{({M^T}N)^{ - 1}}{(M{N^{ - 1}})^T}\] is \[ - {M^2}\]

\[\therefore \]Correct option is C.

Note: Many students make mistake of writing the inverse of transpose of a matrix and inverse of matrix, keep in mind we can change the values in power using the rule of exponent i.e. \[{({x^m})^n} = {({x^n})^m}\].