Let M and N be two 3 times 3 non-singular ske | vedclass

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(C) Given that $M$ and $N$ are skew-symmetric matrices,we have $M^T = -M$ and $N^T = -N$. Since $M$ and $N$ are $3 	imes 3$ matrices,they are non-sin

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$-M^2$

Vedclass · Answer

(C) Given that $M$ and $N$ are skew-symmetric matrices,we have $M^T = -M$ and $N^T = -N$. Since $M$ and $N$ are $3 	imes 3$ matrices,they are non-singular,and $MN = NM$. We need to simplify the expression $E = M^2 N^2 (M^T N)^{-1} (M N^{-1})^T$. First,substitute $M^T = -M$ into the expression: $E = M^2 N^2 (-M N)^{-1} (M N^{-1})^T$. Using the property $(AB)^{-1} = B^{-1} A^{-1}$,we get $(-MN)^{-1} = N^{-1} (-M)^{-1} = -N^{-1} M^{-1}$. Using the property $(AB)^T = B^T A^T$,we get $(M N^{-1})^T = (N^{-1})^T M^T = (N^T)^{-1} M^T = (-N)^{-1} (-M) = (-N^{-1}) (-M) = N^{-1} M$. Now substitute these back into the expression: $E = M^2 N^2 (-N^{-1} M^{-1}) (N^{-1} M)$. Since $MN = NM$,it follows that $M^{-1} N^{-1} = N^{-1} M^{-1}$ and $M N^{-1} = N^{-1} M$. $E = -M^2 N^2 N^{-1} M^{-1} N^{-1} M$. $E = -M^2 N (N N^{-1}) M^{-1} N^{-1} M$. $E = -M^2 N I M^{-1} N^{-1} M$. $E = -M^2 (N M^{-1}) N^{-1} M$. Since $N M^{-1} = M^{-1} N$,we have: $E = -M^2 M^{-1} N N^{-1} M$. $E = -M (N N^{-1}) M$. $E = -M I M = -M^2$. Thus,the correct option is $(C)$.

Column $I$	Column $II$
$(A)$ The set $\{\operatorname{Re}(\frac{2 i z}{1-z^2}): \|z\|=1, z \neq \pm 1\}$ is	$(p)$ $(-\infty,-1) \cup(1, \infty)$
$(B)$ The domain of $f(x)=\sin ^{-1}(\frac{8(3)^{x-2}}{1-3^{2(x-1)}})$ is	$(q)$ $(-\infty, 0) \cup(0, \infty)$
$(C)$ If $f(\theta)=\left\|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right\|$,then the set $\{f(\theta): 0 \leq \theta < \frac{\pi}{2}\}$ is	$(r)$ $[2, \infty)$
$(D)$ If $f(x)=x^{3 / 2}(3 x-10), x \geq 0$,then $f(x)$ is increasing in	$(s)$ $(-\infty,-1] \cup[1, \infty)$
	$(t)$ $(-\infty, 0] \cup[2, \infty)$

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

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