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Question
Solution
\[Given: A = \begin{bmatrix}3 & 2 & 7 \\ 1 & 4 & 3 \\ - 2 & 5 & 8\end{bmatrix} \]
\[ \Rightarrow A^T = \begin{bmatrix}3 & 1 & - 2 \\ 2 & 4 & 5 \\ 7 & 3 & 8\end{bmatrix}\]
\[\text{Let X} = \frac{1}{2}\left( A + A^T \right) = \frac{1}{2}\left( \begin{bmatrix}3 & 2 & 7 \\ 1 & 4 & 3 \\ - 2 & 5 & 8\end{bmatrix} + \begin{bmatrix}3 & 1 & - 2 \\ 2 & 4 & 5 \\ 7 & 3 & 8\end{bmatrix} \right) = \begin{bmatrix}3 & \frac{3}{2} & \frac{5}{2} \\ \frac{3}{2} & 4 & 4 \\ \frac{5}{2} & 4 & 8\end{bmatrix}\]
\[\text{Let Y} = \frac{1}{2}\left( A - A^T \right) = \frac{1}{2}\left( \begin{bmatrix}3 & 2 & 7 \\ 1 & 4 & 3 \\ - 2 & 5 & 8\end{bmatrix} - \begin{bmatrix}3 & 1 & - 2 \\ 2 & 4 & 5 \\ 7 & 3 & 8\end{bmatrix} \right) = \begin{bmatrix}0 & \frac{1}{2} & \frac{9}{2} \\ \frac{- 1}{2} & 0 & - 1 \\ \frac{- 9}{2} & 1 & 0\end{bmatrix}\]
\[ X^T = \begin{bmatrix}3 & \frac{3}{2} & \frac{5}{2} \\ \frac{3}{2} & 4 & 4 \\ \frac{5}{2} & 4 & 8\end{bmatrix}^T = \begin{bmatrix}3 & \frac{3}{2} & \frac{5}{2} \\ \frac{3}{2} & 4 & 4 \\ \frac{5}{2} & 4 & 8\end{bmatrix}^T = X\]
\[ Y^T = \begin{bmatrix}0 & \frac{1}{2} & \frac{9}{2} \\ \frac{- 1}{2} & 0 & - 1 \\ \frac{- 9}{2} & 1 & 0\end{bmatrix}^T = \begin{bmatrix}0 & \frac{- 1}{2} & \frac{- 9}{2} \\ \frac{1}{2} & 0 & 1 \\ \frac{9}{2} & - 1 & 0\end{bmatrix} = - \begin{bmatrix}0 & \frac{1}{2} & \frac{9}{2} \\ \frac{- 1}{2} & 0 & - 1 \\ \frac{- 9}{2} & 1 & 0\end{bmatrix} = Y\]
Thus, X is a symmetric matrix and Y is skew - symmetric matrix .
\[Now, \]
\[X + Y = \begin{bmatrix}3 & \frac{3}{2} & \frac{5}{2} \\ \frac{3}{2} & 4 & 4 \\ \frac{5}{2} & 4 & 8\end{bmatrix} + \begin{bmatrix}0 & \frac{1}{2} & \frac{9}{2} \\ \frac{- 1}{2} & 0 & - 1 \\ \frac{- 9}{2} & 1 & 0\end{bmatrix} = \begin{bmatrix}3 & 2 & 7 \\ 1 & 4 & 3 \\ - 2 & 5 & 8\end{bmatrix} = A\]
\[ \therefore X = \begin{bmatrix}3 & \frac{3}{2} & \frac{5}{2} \\ \frac{3}{2} & 4 & 4 \\ \frac{5}{2} & 4 & 8\end{bmatrix} \text{and Y} = \begin{bmatrix}0 & \frac{1}{2} & \frac{9}{2} \\ \frac{- 1}{2} & 0 & - 1 \\ \frac{- 9}{2} & 1 & 0\end{bmatrix}\]
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