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Question
If \[A = \begin{bmatrix}1 & 2 \\ 0 & 3\end{bmatrix}\] is written as B + C, where B is a symmetric matrix and C is a skew-symmetric matrix, then B is equal to.
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Solution
\[Given: A = \begin{bmatrix}1 & 2 \\ 0 & 3\end{bmatrix}\]
\[ \Rightarrow A^T = \begin{bmatrix}1 & 0 \\ 2 & 3\end{bmatrix}\]
\[\text{Let B} = \frac{1}{2}\left( A + A^T \right) = \frac{1}{2}\left( \begin{bmatrix}1 & 2 \\ 0 & 3\end{bmatrix} + \begin{bmatrix}1 & 0 \\ 2 & 3\end{bmatrix} \right)\]
\[ = \frac{1}{2}\begin{bmatrix}1 + 1 & 2 + 0 \\ 0 + 2 & 3 + 3\end{bmatrix}\]
\[ = \frac{1}{2}\begin{bmatrix}2 & 2 \\ 2 & 6\end{bmatrix}\]
\[ = \begin{bmatrix}1 & 1 \\ 1 & 3\end{bmatrix}\]
\[Now, \]
\[ B^T = \begin{bmatrix}1 & 1 \\ 1 & 3\end{bmatrix} = B\]
\[ \text{Therefore, B is symmetric matrix }. \]
\[Let C = \frac{1}{2}\left( A - A^T \right) = \frac{1}{2}\left( \begin{bmatrix}1 & 2 \\ 0 & 3\end{bmatrix} - \begin{bmatrix}1 & 0 \\ 2 & 3\end{bmatrix} \right)\]
\[ = \frac{1}{2}\begin{bmatrix}1 - 1 & 2 - 0 \\ 0 - 2 & 3 - 3\end{bmatrix}\]
\[ = \frac{1}{2}\begin{bmatrix}0 & 2 \\ - 2 & 0\end{bmatrix}\]
\[ = \begin{bmatrix}0 & 1 \\ - 1 & 0\end{bmatrix}\]
\[ \therefore C^T = \begin{bmatrix}0 & 1 \\ - 1 & 0\end{bmatrix}^T = \begin{bmatrix}0 & - 1 \\ 1 & 0\end{bmatrix} = - \begin{bmatrix}0 & 1 \\ - 1 & 0\end{bmatrix} = C\]
\[So, \text{C is a skew - symmetric matrix }. \]
\[Now, \]
\[B + C = \begin{bmatrix}1 & 1 \\ 1 & 3\end{bmatrix} + \begin{bmatrix}0 & 1 \\ - 1 & 0\end{bmatrix} = \begin{bmatrix}1 + 0 & 1 + 1 \\ 1 - 1 & 3 + 0\end{bmatrix} = \begin{bmatrix}1 & 2 \\ 0 & 3\end{bmatrix} = A\]
\[ \therefore B = \begin{bmatrix}1 & 1 \\ 1 & 3\end{bmatrix}\]
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