= Solution
For any real <square matrix> $N$, the <symmetric and antisymmetric parts of a matrix> give $N=(N+N^T)/2+(N-N^T)/2$, whose summands are respectively a <symmetric matrix> and an <antisymmetric matrix>. Here it gives
$$
\boxed{\begin{pmatrix}a&b\\c&3\end{pmatrix}=\begin{pmatrix}a&(b+c)/2\\(b+c)/2&3\end{pmatrix}+\begin{pmatrix}0&(b-c)/2\\(c-b)/2&0\end{pmatrix}.}
$$
Transposing the first summand leaves it unchanged; transposing the second changes its sign. Entrywise addition recovers the original <matrix>. The decomposition is unique because a matrix that is both symmetric and antisymmetric must be zero.
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