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?

I found a much simpler way to solve this (probably the one intended by the authors).
Since 
$$AA = I_2 \Leftrightarrow A = I_2.A^{-1} \Leftrightarrow A = A^{-1}$$
So,
$$A=\begin{bmatrix}
    a & b\\
    c & d\\
    \end{bmatrix} = A^{-1} = \det(A)^{-1}*\begin{bmatrix}
    d & -b\\
    -c & a\\
    \end{bmatrix} $$
$\det(A)^{-1} = \frac{1}{ad-bc} = \delta$
This means that $A=A^{-1}$ equals
$$\begin{bmatrix}
    d\delta & -b\delta\\
    -c\delta & a\delta\\
    \end{bmatrix}$$
Putting this all in a system:
$$
\left\{ 
\begin{array}{c}
a = d\delta \\ 
b = -b\delta \\ 
c = -c\delta  \\
d = a\delta \\
\end{array}
\right.
$$
Now there's 2 possibilities: $b=c=0$ or $b,c \neq 0$
Starting with $b=c=0$:
$$
\left\{ 
\begin{array}{c}
b = c = 0 \\
a = d\delta \\
d = d\delta^2 \Leftrightarrow \pm 1 = \delta \\
\end{array}
\right.
$$
Now for the values of $\delta$ we get:
$\frac{1}{ad} = 1 \lor \frac{1}{ad} = -1 \Leftrightarrow a = 1/d \lor a = -1/d$
$$
\left\{ 
\begin{array}{c}
b = c = 0 \\
a = 1/d \\
d \in \mathbb{R}\\
\end{array}
\right.
$$
$$
\left\{ 
\begin{array}{c}
b = c = 0 \\
a = -1/d \\
d \in \mathbb{R}\\
\end{array}
\right.
$$
Now for the case of $b,c \neq 0$ we have:
$$\left\{ 
\begin{array}{c}
a = -d \\ 
\delta=-1  \\
\end{array}
\right.$$
Solving for the value of d in this case:
$$\delta = -1 \Leftrightarrow \frac{1}{ad-bc} \Leftrightarrow d^2+bc = 1 \Leftrightarrow \\ \pm d = \sqrt{1-bc}$$
Finally the system becomes:
$$\left\{ 
\begin{array}{c}
a = -d \\ 
d = \mp \sqrt{1-bc}  \\
b,c \in \mathbb{R} \setminus \{0\}
\end{array}
\right.$$
I think that covers them all... how'd I do?
