Let $A = \left[\begin{array}{cc}3 & 5 \\ 1 & -1\end{array}\right]$.
Then,the transpose $A^{\prime} = \left[\begin{array}{cc}3 & 1 \\ 5 & -1\end{array}\right]$.
Any square matrix $A$ can be written as $A = P + Q$,where $P = \frac{1}{2}(A + A^{\prime})$ is a symmetric matrix and $Q = \frac{1}{2}(A - A^{\prime})$ is a skew-symmetric matrix.
First,calculate $P = \frac{1}{2}(A + A^{\prime})$:
$A + A^{\prime} = \left[\begin{array}{cc}3+3 & 5+1 \\ 1+5 & -1-1\end{array}\right] = \left[\begin{array}{cc}6 & 6 \\ 6 & -2\end{array}\right]$
$P = \frac{1}{2} \left[\begin{array}{cc}6 & 6 \\ 6 & -2\end{array}\right] = \left[\begin{array}{cc}3 & 3 \\ 3 & -1\end{array}\right]$
Since $P^{\prime} = P$,$P$ is symmetric.
Next,calculate $Q = \frac{1}{2}(A - A^{\prime})$:
$A - A^{\prime} = \left[\begin{array}{cc}3-3 & 5-1 \\ 1-5 & -1-(-1)\end{array}\right] = \left[\begin{array}{cc}0 & 4 \\ -4 & 0\end{array}\right]$
$Q = \frac{1}{2} \left[\begin{array}{cc}0 & 4 \\ -4 & 0\end{array}\right] = \left[\begin{array}{cc}0 & 2 \\ -2 & 0\end{array}\right]$
Since $Q^{\prime} = -Q$,$Q$ is skew-symmetric.
Thus,$A = P + Q = \left[\begin{array}{cc}3 & 3 \\ 3 & -1\end{array}\right] + \left[\begin{array}{cc}0 & 2 \\ -2 & 0\end{array}\right]$.