(A) Let $A = \left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]$.
Then,the transpose $A^{\prime} = \left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]$.
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 $A + A^{\prime}$:
$A + A^{\prime} = \left[\begin{array}{ccc}6+6 & -2-2 & 2+2 \\ -2-2 & 3+3 & -1-1 \\ 2+2 & -1-1 & 3+3\end{array}\right] = \left[\begin{array}{ccc}12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6\end{array}\right]$.
Thus,$P = \frac{1}{2}(A + A^{\prime}) = \left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]$.
Since $P^{\prime} = P$,$P$ is a symmetric matrix.
Next,calculate $A - A^{\prime}$:
$A - A^{\prime} = \left[\begin{array}{ccc}6-6 & -2-(-2) & 2-2 \\ -2-(-2) & 3-3 & -1-(-1) \\ 2-2 & -1-(-1) & 3-3\end{array}\right] = \left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$.
Thus,$Q = \frac{1}{2}(A - A^{\prime}) = \left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$.
Since $Q^{\prime} = -Q$,$Q$ is a skew-symmetric matrix.
Finally,$A = P + Q = \left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right] + \left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$.