Determining the Automorphism Group of a Complex-Valued Group Let $U$ be the set of matrices $$\left(\begin{array}{cc} a&b \\ c&d
\end{array}\right)$$ where $a,b,c,d$ are integers where $ad-bc = 1.$
I would like to find the set of all matrices that are their own inverse.
We see that we want to find $a,b,c,d$ such that
$$\left(\begin{array}{cc} a^2 + bc&ab +bd \\ ac + dc& cb + d^2 \end{array}
\right) = \left(\begin{array}{cc} 1&0\\ 0&1 \end{array}\right).$$
This implies that
$a^2 + bc = 1$ and $ab = -bd$ and $ac = -dc$ and $cb + d^2 = 1.$ This
seems like many cases to handle in order to handle this system of
non-linear equations. Is there a simpler method to solving this system of
equations? If so, how would I implement that method on this system?
 A: It seems that separating into cases may be of help.

Case $(1)$: $b = 0, c=0$

Then $a^2=d^2=1$, so the solutions are $a=\pm 1$ and $d=\pm1$.

Case $(2)$: $b = 0, c\neq0$

Then $a^2=d^2=1$ and $a=-d$, so the solutions are:


*

*$a= 1, d=-1, c\in \mathbb{C}$

*$a=-1, d=1, c\in \mathbb{C}$



Case $(3)$: $b \neq 0, c=0$

Like $(2)$.

Case $(4)$: $b \neq 0, c\neq0$

Then $a=-d$, so it suffices to find solutions of $a^2+bc=1$ with $bc\neq 0$.
Whenever $bc\neq 1$ we have that $1-bc \neq 0$, so it has two complex square roots $\rho$ and $-\rho$: one is $a$, the other is $d$.
When $bc=1$, then $a=d=0$.
Hence, the solutions are


*

*$b,c\in \mathbb{C}\setminus\{0\}$, $a$ is a square root of $1-bc$ and $d=-a$.

A: $ad−bc=1$ or $|ad-bc| = 1$?
For any $2\times2$ matrix.  $ A = \begin{bmatrix} a&b\\c&d\end{bmatrix},$
$A^{-1} = \frac 1{ad-bc} \begin{bmatrix} d&-b\\-c&a\end{bmatrix}$ 
If we limit our selves such that $ad -bc = 1$ then $a = d$ and $b = -b$ and $c = -c$
and that leaves the identity matrix.
but if $ad-bc = -1$ then there is a set of matrices  $ U = \begin{bmatrix} a&b\\\frac {1-a^2}{b}&-a\end{bmatrix}$
and a subset of those would be
$ U = \begin{bmatrix} a&1-a\\1+a&-a\end{bmatrix}$
