A, B, C, D are square matrices such that A+B | vedclass

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(B) Given that $A+B$ is symmetric, $(A+B)^T = A+B \Rightarrow A^T+B^T = A+B$. Given that $A-B$ is skew-symmetric, $(A-B)^T = -(A-B) \Rightarrow A^T-B^

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$\left[\begin{array}{ccc}-1 & 6 & 3 \ 3 & 2 & -2 \ 1 & -2 & 6\end{array}ight]$

Vedclass · Answer

(B) Given that $A+B$ is symmetric, $(A+B)^T = A+B \Rightarrow A^T+B^T = A+B$. Given that $A-B$ is skew-symmetric, $(A-B)^T = -(A-B) \Rightarrow A^T-B^T = -A+B$. Adding these two equations: $2A^T = 2B \Rightarrow B = A^T$. Subtracting these two equations: $2B^T = 2A \Rightarrow B^T = A$. Since $A = \left[\begin{array}{ccc}-1 & 2 & 3 \ 4 & 3 & -2 \ 3 & -4 & 5\end{array}ight]$, then $B = A^T = \left[\begin{array}{ccc}-1 & 4 & 3 \ 2 & 3 & -4 \ 3 & -2 & 5\end{array}ight]$. Given $D = C^T$, and $C = \left[\begin{array}{ccc}0 & 1 & -2 \ 2 & -1 & 0 \ 0 & 2 & 1\end{array}ight]$, then $D = \left[\begin{array}{ccc}0 & 2 & 0 \ 1 & -1 & 2 \ -2 & 0 & 1\end{array}ight]$. Now, $B+D = \left[\begin{array}{ccc}-1 & 4 & 3 \ 2 & 3 & -4 \ 3 & -2 & 5\end{array}ight] + \left[\begin{array}{ccc}0 & 2 & 0 \ 1 & -1 & 2 \ -2 & 0 & 1\end{array}ight] = \left[\begin{array}{ccc}-1 & 6 & 3 \ 3 & 2 & -2 \ 1 & -2 & 6\end{array}ight]$.

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