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(B) Given that $A$ is a symmetric matrix,$A^T = A$. Given that $B$ is a skew-symmetric matrix,$B^T = -B$. We are given $A + B = \begin{bmatrix} 2 & 3

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$\begin{bmatrix} 4 & -2 \ -1 & -4 \end{bmatrix}$

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(B) Given that $A$ is a symmetric matrix,$A^T = A$. Given that $B$ is a skew-symmetric matrix,$B^T = -B$. We are given $A + B = \begin{bmatrix} 2 & 3 \ 5 & -1 \end{bmatrix} \quad (1)$. Taking the transpose of both sides: $(A + B)^T = \begin{bmatrix} 2 & 3 \ 5 & -1 \end{bmatrix}^T \implies A^T + B^T = \begin{bmatrix} 2 & 5 \ 3 & -1 \end{bmatrix}$. Substituting $A^T = A$ and $B^T = -B$,we get $A - B = \begin{bmatrix} 2 & 5 \ 3 & -1 \end{bmatrix} \quad (2)$. Adding equations $(1)$ and $(2)$: $2A = \begin{bmatrix} 2+2 & 3+5 \ 5+3 & -1-1 \end{bmatrix} = \begin{bmatrix} 4 & 8 \ 8 & -2 \end{bmatrix} \implies A = \begin{bmatrix} 2 & 4 \ 4 & -1 \end{bmatrix}$. Subtracting equation $(2)$ from $(1)$: $2B = \begin{bmatrix} 2-2 & 3-5 \ 5-3 & -1-(-1) \end{bmatrix} = \begin{bmatrix} 0 & -2 \ 2 & 0 \end{bmatrix} \implies B = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$. Now,calculating $AB$: $AB = \begin{bmatrix} 2 & 4 \ 4 & -1 \end{bmatrix} \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix} = \begin{bmatrix} (2)(0) + (4)(1) & (2)(-1) + (4)(0) \ (4)(0) + (-1)(1) & (4)(-1) + (-1)(0) \end{bmatrix} = \begin{bmatrix} 4 & -2 \ -1 & -4 \end{bmatrix}$.

If $A$ is a symmetric matrix and $B$ is a skew-symmetric matrix such that $A + B = \begin{bmatrix} 2 & 3 \\ 5 & -1 \end{bmatrix}$,then $AB$ is equal to

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