Solve this

Solution:

Given : $A=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]$

$\Rightarrow A^{T}=\left[\begin{array}{ll}1 & 0 \\ 2 & 3\end{array}\right]$

Let $B=\frac{1}{2}\left(A+A^{T}\right)=\frac{1}{2}\left(\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 2 & 3\end{array}\right]\right)$

$=\frac{1}{2}\left[\begin{array}{ll}1+1 & 2+0 \\ 0+2 & 3+3\end{array}\right]$

$=\frac{1}{2}\left[\begin{array}{ll}2 & 2 \\ 2 & 6\end{array}\right]$

$=\left[\begin{array}{ll}1 & 1 \\ 1 & 3\end{array}\right]$

Now,

$B^{T}=\left[\begin{array}{ll}1 & 1 \\ 1 & 3\end{array}\right]=B$

Therefore, $B$ is symmetric matrix.

Let $C=\frac{1}{2}\left(A-A^{T}\right)=\frac{1}{2}\left(\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 2 & 3\end{array}\right]\right)$

$=\frac{1}{2}\left[\begin{array}{cc}1-1 & 2-0 \\ 0-2 & 3-3\end{array}\right]$

$=\frac{1}{2}\left[\begin{array}{cc}0 & 2 \\ -2 & 0\end{array}\right]$

$=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]$

$\therefore C^{T}=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]^{T}=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]=-\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]=C$

So, $C$ is a skew - symmetric matrix.

Now,

$B+C=\left[\begin{array}{ll}1 & 1 \\ 1 & 3\end{array}\right]+\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}1+0 & 1+1 \\ 1-1 & 3+0\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]=A$\

$\therefore B=\left[\begin{array}{ll}1 & 1 \\ 1 & 3\end{array}\right]$