Solve this

Given : $A=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]$

$A^{T}=\left[\begin{array}{cc}3 & 1 \\ -4 & -1\end{array}\right]$

Let $X=\frac{1}{2}\left(A+A^{T}\right)=\frac{1}{2}\left(\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]+\left[\begin{array}{cc}3 & 1 \\ -4 & -1\end{array}\right]\right)=\left[\begin{array}{cc}3 & \frac{-3}{2} \\ \frac{-3}{2} & -1\end{array}\right]$

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

Let $Y=\frac{1}{2}\left(A-A^{T}\right)=\frac{1}{2}\left(\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]-\left[\begin{array}{cc}3 & 1 \\ -4 & -1\end{array}\right]\right)=\left[\begin{array}{ll}0 & \frac{-5}{2} \\ \frac{5}{2} & 0\end{array}\right]$

$Y^{T}=\left[\begin{array}{cc}0 & \frac{-5}{2} \\ \frac{5}{2} & 0\end{array}\right]^{T}=\left[\begin{array}{cc}0 & \frac{5}{2} \\ \frac{-5}{2} & 0\end{array}\right]=-\left[\begin{array}{cc}0 & \frac{-5}{2} \\ \frac{5}{2} & 0\end{array}\right]=Y$

$\therefore X$ is a symmetric matrix and $Y$ is a skew $-$ symmetric matrix.

$X+Y=\left[\begin{array}{cc}3 & \frac{-3}{2} \\ \frac{-3}{2} & -1\end{array}\right]+\left[\begin{array}{cc}0 & \frac{-5}{2} \\ \frac{5}{2} & 0\end{array}\right]=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]=A$