Find the values for which $A^2 = I_2$, A is a matrix, with $A \neq I_2$ and $A \neq -I_2$ First I tried to find $A^2$ with
$$
    A=\begin{bmatrix}
    \alpha & \beta\\
    \delta & \gamma\\
    \end{bmatrix}
$$
I multiplied this by itself and got:
$$
    \begin{bmatrix}
    \alpha^2+\beta\delta& \beta(\alpha + \gamma)\\
    \delta (\alpha + \gamma) & \delta\beta+\gamma^2\\
    \end{bmatrix}
$$
I put this in a system:
$$
\left\{ 
\begin{array}{c}
\alpha^2+\beta\delta = 1 \\ 
\beta(\alpha + \gamma) = 0 \\ 
\delta (\alpha + \gamma) = 0  \\
\delta\beta+\gamma^2 = 1 \\
\end{array}
\right. 
$$
I tried to solve for $\beta$ first and right away got an issue:
$$\beta = \frac{1-\alpha^2}{\delta}$$
One solution given by my book is:
$$
    \begin{bmatrix}
    1& 0\\
    0 & -1\\
    \end{bmatrix}
$$
So $\delta$ can be zero but according to my system it can't. How is this possible?

You assumed $\delta \neq 0$ when you divided by it (when solving for $\beta$). The first system stills valid whether $\delta = 0$ or not.
In fact, when $\delta = 0, |\alpha| = |\gamma| = 1$. If $\gamma = -\alpha, \beta$ can be anything.
