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(B) For matrix $A$ to be symmetric, $A = A^T$, which implies $a=b$, $c=d$, and $3=3$. Thus, $A = \begin{bmatrix} 1 & a & 3 \ a & 2 & c \ 3 & c & 4 \

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$\begin{bmatrix} 48 & 26 & 36 \ 32 & 19 & 22 \ -11 & 43 & -67 \end{bmatrix}$

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(B) For matrix $A$ to be symmetric, $A = A^T$, which implies $a=b$, $c=d$, and $3=3$. Thus, $A = \begin{bmatrix} 1 & a & 3 \ a & 2 & c \ 3 & c & 4 \end{bmatrix}$. For matrix $B$ to be skew-symmetric, $B = -B^T$, which implies diagonal elements are $0$. From $B_{13} = -B_{31}$, we get $b = -6$. From $B_{23} = -B_{32}$, we get $-7 = -c$, so $c = 7$. Substituting these values, we get $a = b = -6$ and $d = c = 7$. So, $A = \begin{bmatrix} 1 & -6 & 3 \ -6 & 2 & 7 \ 3 & 7 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 5 & -6 \ -5 & 0 & -7 \ 6 & 7 & 0 \end{bmatrix}$. Now, calculating the product $AB$: $AB = \begin{bmatrix} 1 & -6 & 3 \ -6 & 2 & 7 \ 3 & 7 & 4 \end{bmatrix} \begin{bmatrix} 0 & 5 & -6 \ -5 & 0 & -7 \ 6 & 7 & 0 \end{bmatrix} = \begin{bmatrix} 0+30+18 & 5+0+21 & -6+42+0 \ 0-10+42 & -30+0+49 & 36-14+0 \ 0-35+24 & 15+0+28 & -18-49+0 \end{bmatrix} = \begin{bmatrix} 48 & 26 & 36 \ 32 & 19 & 22 \ -11 & 43 & -67 \end{bmatrix}$.

List-$I$	List-$II$
$A$. $\|k^{-1} A^{-1}\|$	$I$. $BA^k + A^kB$
$B$. $\|\text{Adj}(A^{-1})\|$	$II$. $\frac{B\text{Adj}(B)}{\|B\|}$
$C$. $BAB^{-1} = I \Rightarrow BA^kB^{-1} =$	$III$. $\frac{1}{\|B\|^3\|A\|}$
$D$. $\text{Adj}(\text{Adj}(A^{-1})) =$	$IV$. $\frac{1}{\|A\|}(A^{-1})$
	$V$. $\frac{1}{\|A\|^2}$

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	$(s)$ $(-\infty,-1] \cup[1, \infty)$
	$(t)$ $(-\infty, 0] \cup[2, \infty)$

If $A = \begin{bmatrix} 1 & a & 3 \\ b & 2 & c \\ 3 & d & 4 \end{bmatrix}$ is a symmetric matrix and $B = \begin{bmatrix} 0 & 5 & b \\ -5 & 0 & -7 \\ 6 & c & 0 \end{bmatrix}$ is a skew-symmetric matrix, then $AB = $

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