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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 $AB$: $AB = \begin{bmatrix} 1 & -6 & 3 \ -6 & 2 & 7 \ 3 & 7 & 4 \end{bmatrix} \begin{bmatrix} 0 & 5 & -6 \ -5 & 0 & -7 \ 6 & 7 & 0 \end{bmatrix} = \begin{bmatrix} 0+30+18 & 5+0+21 & -6+42+0 \ 0-10+42 & -30+0+49 & 36-14+0 \ 0-35+24 & 15+0+28 & -18-49+0 \end{bmatrix} = \begin{bmatrix} 48 & 26 & 36 \ 32 & 19 & 22 \ -11 & 43 & -67 \end{bmatrix}$.

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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