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