Let $A=\begin{pmatrix}a&b\\b&c\end{pmatrix}$. Then $A^2=\begin{pmatrix}a^2+b^2&ab+bc\\ab+bc&b^2+c^2\end{pmatrix}$.

Sum of diagonal of $A^2 = a^2+2b^2+c^2 = 1$.

Since $a,b,c$ are integers and $a^2+2b^2+c^2=1\geq0$: $b=0$ (since $2b^2\leq1$ with integer $b$ forces $b=0$), then $a^2+c^2=1$.

Solutions: $(a,c)\in\{(1,0),(−1,0),(0,1),(0,−1)\}$ and $b=0$: 4 matrices. But we should check $b=\pm1$: $2(1)=2>1$, not possible. Hmm — that gives only 4. But checking $b=0$, $a^2+c^2=1$: integer solutions are $(\pm1,0)$ and $(0,\pm1)$ — exactly 4 pairs, each with $b=0$. So $\mathbf{4}$ matrices... but wait: $b$ can also vary if we allow it. Total = $\mathbf{4}$ matrices.