Express the matrix $A = \left[\begin{array}{cc}1 & 5 \\ -1 & 2\end{array}\right]$ as the sum of a symmetric and a skew-symmetric matrix.

(N/A) Let $A = \left[\begin{array}{cc}1 & 5 \\ -1 & 2\end{array}\right]$. Then the transpose of $A$ is $A^{\prime} = \left[\begin{array}{cc}1 & -1 \\ 5 & 2\end{array}\right]$.
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 $A + A^{\prime} = \left[\begin{array}{cc}1 & 5 \\ -1 & 2\end{array}\right] + \left[\begin{array}{cc}1 & -1 \\ 5 & 2\end{array}\right] = \left[\begin{array}{cc}2 & 4 \\ 4 & 4\end{array}\right]$.
Thus,$P = \frac{1}{2}(A + A^{\prime}) = \left[\begin{array}{cc}1 & 2 \\ 2 & 2\end{array}\right]$. Since $P^{\prime} = P$,$P$ is symmetric.
Next,calculate $A - A^{\prime} = \left[\begin{array}{cc}1 & 5 \\ -1 & 2\end{array}\right] - \left[\begin{array}{cc}1 & -1 \\ 5 & 2\end{array}\right] = \left[\begin{array}{cc}0 & 6 \\ -6 & 0\end{array}\right]$.
Thus,$Q = \frac{1}{2}(A - A^{\prime}) = \left[\begin{array}{cc}0 & 3 \\ -3 & 0\end{array}\right]$. Since $Q^{\prime} = -Q$,$Q$ is skew-symmetric.
Therefore,$A = P + Q = \left[\begin{array}{cc}1 & 2 \\ 2 & 2\end{array}\right] + \left[\begin{array}{cc}0 & 3 \\ -3 & 0\end{array}\right]$.