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?
 A: When you solved for $\beta$, you assumed that you could divide through by $\delta$, i.e., you assumed that $\delta \ne 0$. What you needed to do was this:
case 1: $\delta \ne 0$:
Then $\beta = \frac{1 - \alpha^2}{\delta}$
...
Case 2: $\delta = 0$: 
In this case, we have $\alpha = \pm 1,$ ...
(and you fill in the remaining details)
A: You made a mistake:
It’s like solving $x=y^2$ and then changing it to $1=\frac{y^2}{x}$ and the  saying that $(0,0)$ is not a solution since $x$ cannot be zero.
You assumed $x$ is not zero, the question did not assume anything. You have to solve for the case when $x=0$, same goes for your system.
A: 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.
A: Here is one way to find all such matrices $A$ without directly solving for the entries of $A$.  Let $\Bbb{K}$ be the base field.
First, suppose that $\Bbb{K}$ is of characteristic unequal to $2$.  For a $2\times 2$ matrix $A\neq \pm I$ to satisfy $A^2=I$, we have $(A-I)(A+I)=0$ but $A-I$ and $A+I$ are both non-zero.  That is, the dimensions of $\ker(A-I)$ and $\ker(A+I)$ must be both $1$.  Therefore, $\ker(A-I)$ and $\ker(A+I)$ is spanned by two non-zero vectors $u$ and $v$.    Therefore, if $M$ denotes the matrix
$$M=\begin{bmatrix}\vert &\vert\\ u &v\\\vert&\vert\end{bmatrix}.$$
Then, $A=MJM^{-1}$, where $$J=\begin{bmatrix}1&0\\0&-1\end{bmatrix}.$$
If $\Bbb K=\Bbb Q$, then we can take $u=\begin{bmatrix}\frac{2r}{1+r^2}\\\frac{1-r^2}{1+r^2}\end{bmatrix}$ and $v=\begin{bmatrix}\frac{2s}{1+s^2}\\\frac{1-s^2}{1+s^2}\end{bmatrix}$ with $r,s\in\left(-1,1\right]\cap\Bbb Q$ and $r\neq s$, so that
$$A=\frac{1}{(r-s)(1+rs)}\begin{bmatrix}(r+s)(1-rs)&-4rs\\(1-r^2)(1-s^2)&-(r+s)(1-rs)\end{bmatrix}.$$ 
 If $\Bbb K=\Bbb R$, then we can take $u=\begin{bmatrix}\cos x\\\sin x\end{bmatrix}$ and $v=\begin{bmatrix}\cos y\\\sin y\end{bmatrix}$ with $x,y\in\left[0,\pi\right)$ and $x\neq y$, so that
$$A=\frac{1}{\sin(x-y)}\begin{bmatrix}-\sin(x+y)&2\cos x\cos y\\-2\sin x\sin y&\sin(x+y)\end{bmatrix}.$$ 
If $\Bbb K=\Bbb C$, then we can take $u=\begin{bmatrix}e^{i\lambda}\cos x \\\sin x\end{bmatrix}$ and $v=\begin{bmatrix}e^{i\mu}\cos y \\\sin y\end{bmatrix}$ with $x,y\in\left[0,\pi\right)$ and $\lambda,\mu\in[0,2\pi)$ such that 


*

*if $x=0$ or $x=\pi/2$, then $\lambda=0$;

*if $y=0$ or $y=\pi/2$, then $\mu=0$;

*either $x\neq y$, or $x=y$ and $\lambda \neq \mu$.  


In this case,
$$A=\frac{1}{\cos\frac{\lambda-\mu}{2}\sin(x-y)-i\sin\frac{\lambda-\mu}{2}\sin(x+y)}\begin{bmatrix}-\cos\frac{\lambda-\mu}{2}\sin(x+y)+i\sin\frac{\lambda-\mu}{2}\sin(x-y)&2e^{i\left(\frac{\lambda+\mu}{2}\right)}\cos x\cos y \\2e^{-i\left(\frac{\lambda+\mu}{2}\right)}\sin x\sin y &\cos\frac{\lambda-\mu}{2}\sin(x+y)-i\sin\frac{\lambda-\mu}{2}\sin(x-y)\end{bmatrix}$$
Note that the set of such matrices $A$ is in a $1$-to-$1$ correspondence with the right-coset space of $\operatorname{GL}_2(\Bbb{K})$ modulo the subgroup of diagonal matrices (isomorphic to $\Bbb{K}^\times \times \Bbb{K}^\times$).  Particularly, if $\mathbb{K}$ is a finite field with $q$ elements, then there are exactly $$\frac{(q^2-1)(q^2-q)}{(q-1)(q-1)}=q(q+1)=q^2+q$$ such matrices $A$.  (For instance, when $q=3$, there are $12$ possible choices of $A$: $\pm\begin{bmatrix}1&0\\0&-1\end{bmatrix}$, $\pm\begin{bmatrix}0&1\\1&0\end{bmatrix}$, $\pm\begin{bmatrix}1&1\\0&-1\end{bmatrix}$, $\pm\begin{bmatrix}1&-1\\0&-1\end{bmatrix}$, $\pm\begin{bmatrix}1&0\\1&-1\end{bmatrix}$, $\pm\begin{bmatrix}1&0\\-1&-1\end{bmatrix}$.)
In characteristic $2$, we note that $A^2-I=(A-I)^2$.  Since $A\neq I$, the Jordan canonical form of $A$ is $$J=\begin{bmatrix}1&1\\0&1\end{bmatrix}.$$  We start with an arbitrary ordered basis $(u,v)$ of $\Bbb{K}^2$, and then declare that $(A-I)v=u$.  That is, with $$M=\begin{bmatrix}\vert &\vert\\ u &v\\\vert&\vert\end{bmatrix},$$ we have $A=MJM^{-1}$.
Note that the set of such matrices $A$ is in a $1$-to-$1$ correspondence with the right-coset space of $\operatorname{GL}_2(\Bbb{K})$ modulo the subgroup of upper-diagonal matrices with identical diagonal entries (isomorphic to $\Bbb{K}^\times \times\Bbb K$).  Particularly, if $\mathbb{K}$ is a finite field with $q$ elements, then there are exactly $$\frac{(q^2-1)(q^2-q)}{(q-1)q}=(q+1)(q-1)=q^2-1$$ such matrices $A$.  (For instance, when $q=2$, there are only three possibilities for $A$: $\begin{bmatrix}0&1\\1&0\end{bmatrix}$, $\begin{bmatrix}1&1\\0&1\end{bmatrix}$, and $\begin{bmatrix}1&0\\1&1\end{bmatrix}$.)
In general, all possible $A$ (regardless of the characteristic of $\Bbb K$) are given by the two parametrizations below.


*

*$A=\begin{bmatrix}\alpha&\beta\\\frac{1-\alpha^2}{\beta}&-\alpha
\end{bmatrix}$ with $\alpha\in \Bbb{K}$ and $\beta\in \Bbb K\setminus\{0\}$.

*$A=\begin{bmatrix}\alpha&0\\\delta&-\alpha
\end{bmatrix}$ with $\alpha=\pm 1$ and $\delta\in\Bbb K$.  (If $\Bbb K$ has characteristic $2$, then $\alpha=1$ and $\delta$ must be taken from $\Bbb K\setminus\{0\}$.)

A: I know, because $\alpha ^2+\beta\delta =1$, so you think $\beta=\frac{1-\alpha ^2}{\delta}$
But for $\alpha^2+\beta\delta =1$, you can only know $\beta\delta=1-\alpha ^2$.
when $\delta =0$, now $0=1-\alpha ^2$; 
when $\delta\neq 0$, now $\beta=\frac{\beta\delta}{\delta}=\frac{1-\alpha ^2}{\delta}$
A: 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?
