Question

If $A=\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$ is written as B + C, where B is a symmetric matrix and C is a skew-symmetric matrix, then B is equal to.

Answer 1

$$\begin{gathered} \mathrm{Given}:A=\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right] \\ \Rightarrow A^T=\left[\begin{array}{cc}1& 0\\ 2& 3\end{array}\right] \\ \mathrm{Let}B=\frac{1}{2}\left(A+A^T\right)=\frac{1}{2}\left(\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 2& 3\end{array}\right]\right) \\ =\frac{1}{2}\left[\begin{array}{cc}1+1& 2+0\\ 0+2& 3+3\end{array}\right] \\ =\frac{1}{2}\left[\begin{array}{cc}2& 2\\ 2& 6\end{array}\right] \\ =\left[\begin{array}{cc}1& 1\\ 1& 3\end{array}\right] \\ \mathrm{Now}, \\ {B}^T=\left[\begin{array}{cc}1& 1\\ 1& 3\end{array}\right]=B \\ \mathrm{Therefore},B\mathrm{is}\mathrm{symmetric}\mathrm{matrix}. \\ \mathrm{Let}C=\frac{1}{2}\left(A-A^T\right)=\frac{1}{2}\left(\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 2& 3\end{array}\right]\right) \\ =\frac{1}{2}\left[\begin{array}{cc}1-1& 2-0\\ 0-2& 3-3\end{array}\right] \\ =\frac{1}{2}\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right] \\ =\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right] \\ \therefore C^T={\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]}^T=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]=-\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]=C \\ \mathrm{So},C\mathrm{is}\mathrm{a}\mathrm{skew}-\mathrm{symmetric}\mathrm{matrix}. \\ \mathrm{Now}, \\ B+C=\left[\begin{array}{cc}1& 1\\ 1& 3\end{array}\right]+\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]=\left[\begin{array}{cc}1+0& 1+1\\ 1-1& 3+0\end{array}\right]=\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]=A \\ \therefore B=\left[\begin{array}{cc}1& 1\\ 1& 3\end{array}\right] \end{gathered}$$