In case of option (A),

$${({N^T}MN)^T} = {N^T}{M^T}{({N^T})^T} = {N^T}{M^T}N$$

Now, $${N^T}{M^T}N = \left\{ \matrix{ {N^T}MN,\,when\,{M^T} = M \hfill \cr - {N^T}MN,\,when\,{M^T} = - M \hfill \cr} \right.$$

$$\therefore$$ N^TMN is symmetric or skew symmetric according as M is symmetric or skew symmetric.

In case of option (B),

$${(MN - NM)^T} = {(MN)^T} - {(NM)^T}$$

$$ = {N^T}{M^T} - {M^T}{N^T}$$

$$ = NM - MN$$ [$$\because$$ $${M^T} = M,\,{N^T} = N$$]

$$ = - (MN - NM)$$

$$\therefore$$ $$MN - NM$$ is skew symmetric matrix.

In case of option (C),

$${(MN)^T} = {N^T}{M^T} = NM$$ [$$\because$$ $${M^T} = M,\,{N^T} = N$$]

Now, MN cannot always be equal to NM

$$\therefore$$ MN is not symmetric matrix.

In case of option (D),

$$(Adj\,M)(Adj\,N) = Adj(NM) \ne Adj(MN)$$

Therefore, (C) and (D) are the correct options.

Concept Involved If $$\left| {{A_{n \times n}}} \right| = \Delta $$, then $$\left| {adj\,A} \right| = {\Delta ^{n - 1}}$$

Here, $${P_{3 \times 3}} = \left[ {\matrix{ 1 & 4 & 4 \cr 2 & 1 & 7 \cr 1 & 1 & 3 \cr } } \right]$$

$$ \Rightarrow \left| {adj\,P} \right| = {\left| P \right|^2}$$

$$\therefore$$ $$\left| {adj\,P} \right| = \left| {\matrix{ 1 & 4 & 4 \cr 2 & 1 & 7 \cr 1 & 1 & 3 \cr } } \right|$$

$$ = 1(3 - 7) - 4(6 - 7) + 4(2 - 1)$$

$$ = - 4 + 4 + 4 = 4$$

$$ \Rightarrow \left| P \right| = \pm \,2$$

Given, $${M^T} = - M$$, $${N^T} = - N$$

and $$MN = NM$$ ..... (i)

$$\therefore$$ $${M^2}{N^2}{({M^T}N)^{ - 1}}{(M{N^{ - 1}})^T}$$

$$ = {M^2}{N^2}{N^{ - 1}}{({M^T})^{ - 1}}{({N^{ - 1}})^T}.{M^T}$$

$$ = {M^2}N(N{M^{ - 1}}){( - M)^{ - 1}}{({N^T})^{ - 1}}( - M)$$

$$ = {M^2}NI( - {M^{ - 1}}){( - N)^{ - 1}}( - M)$$

$$ = - {M^2}N{M^{ - 1}}{N^{ - 1}}M$$

$$ = - M.(MN){M^{ - 1}}{N^{ - 1}}M$$

$$ = - M(NM){M^{ - 1}}{N^{ - 1}}M$$

$$ = - MN(N{M^{ - 1}}){N^{ - 1}}M$$

$$ = - M(N{N^{ - 1}})M = - {M^2}$$

Note : This question is wrong, as given. An odd order skew symmetric matrix can't be invertible. Had the matrix be of even order, it could have been correct.