Let $A=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}$ be a $2\times 2$ matrix satisfying the conditions. Then as $A$ is symmetric, we have $A^{\trans}=A$. This yields that $b=c$.
So, we find all matrices $A=\begin{bmatrix}
a & b\\
b& d
\end{bmatrix}$ satisfying $A\begin{bmatrix}
1 \\
-1
\end{bmatrix}
=
\begin{bmatrix}
2 \\
3
\end{bmatrix}$.
We have
\begin{align*}
\begin{bmatrix}
2 \\
3
\end{bmatrix}=\begin{bmatrix}
a & b\\
b& d
\end{bmatrix}\begin{bmatrix}
1 \\
-1
\end{bmatrix}
=\begin{bmatrix}
a-b \\
b-d
\end{bmatrix}.
\end{align*}
Hence, we need $a-b=2$ and $b-d=3$.
Equivalently, $a=b+2, d=b-3$. So, we have
\begin{align*}
A=\begin{bmatrix}
a & b\\
b& d
\end{bmatrix}=\begin{bmatrix}
b+2 & b\\
b& b-3
\end{bmatrix}=b\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix}+\begin{bmatrix}
2 & 0\\
0& -3
\end{bmatrix},
\end{align*}
where $b$ is a free variable.

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

One common mistake is not using the assumption that $A$ is symmetric or using wrongly.
A matrix $A$ is symmetric if $A^{\trans}=A$. For a 2 by 2 matrix, this yields that the off-diagonal entries must be the same.
However, note that the diagonal entries can be distinct. Some students assumed the same diagonal entries and concluded that there are no matrices satisfying the conditions.

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