Solve $x^2 = I_2$ where x is a 2 by 2 matrix I tried a basic approach and wrote x as a matrix of four unknown elements $\begin{pmatrix} a && b \\ c && d \end{pmatrix}$ and squared it when I obtained $\begin{pmatrix} a^2 + bc && ab + bd \\  ca + dc && cd + d^2\end{pmatrix}$ and by making it equal with $I_2$ I got the following system
$a^2 + bc = 1$
$ab +bd = 0$ 
$ca + dc = 0$
$cd + d^2 = 1$
I don't know how to proceed. (Also, if anyone knows of a better or simpler way of solving this matrix equation I'd be more than happy to know).

What you need to know about JCF is that for any $2\times 2$ matrix $x$, there is an invertible matrix $P$ such that  $P^{-1}xP=\left(\begin{smallmatrix}t&0\\0&t\end{smallmatrix}\right)$ or $P^{-1}xP=\left(\begin{smallmatrix}t&1\\0&t\end{smallmatrix}\right)$ or
$P^{-1}xP=\left(\begin{smallmatrix}t&0\\0&r\end{smallmatrix}\right)$.  Those three possibilities are the various Jordan forms you might get.
We have $x^2=I$.  Multiply on the left by $P^{-1}$ and on the right by $P$ and you get $P^{-1}x^2P=P^{-1}P=I$.  But then $P^{-1}x^2P=P^{-1}xIxP=P^{-1}x(PP^{-1})xP=(P^{-1}xP)(P^{-1}xP)=I$.
Hence if $x^2=I$, then the JCF also squares to $I$.  This means that the second case is out, and in the first or third case, $t^2=r^2=1$.  Hence the only possible JCF are the four matrices $\left(\begin{smallmatrix}\pm1 &0\\0&\pm1\end{smallmatrix}\right)$.
