Solution
= Solution
Write the general real <symmetric matrix> as $E_2=\begin{pmatrix}r&t\\t&u\end{pmatrix}$. With the specified $E_1$, anticommutation reads
$$
E_1E_2+E_2E_1=\begin{pmatrix}2t&r+u\\r+u&2t\end{pmatrix}=0.
$$
Hence $t=0$ and $u=-r$. The remaining condition $E_2^2=I$ gives $r^2=1$, leaving exactly
$$
\boxed{E_2=\pm\begin{pmatrix}1&0\\0&-1\end{pmatrix}.}
$$
Both matrices directly satisfy all the conditions.