When is a matrix equal to its own inverse? When is a matrix equal to its own inverse?
If you have a $2\times2$ matrix and one if the entries is equal to $x$, for what values of $x$ is this matrix equal to its own matrix? And why?
 A: Suppose $\;A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\;$ is invertible, then
$$A^{-1}=\frac1{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$$
Now calculate stuff.
A: The easiest way is just to write that
$$
A^2=\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)=\left(\begin{matrix}a^2+bc & (a+d)b\\(a+d)c & d^2+bc\end{matrix}\right)=I.
$$
Since $(a+d)b=(a+d)c=0$, either $a+d=0$ or $b=c=0$.  In the latter case, we must have $a^2=d^2=1$.  In the former case, we must have $a^2+bc=1$... either $a^2=1$ and $bc=0$, or else $a^2\neq 1$ and $bc=1-a^2$.  So
$$
A=\left(\begin{matrix}\pm 1 & 0 \\ 0 & \pm 1\end{matrix}\right),
$$
or else
$$
A=\left(\begin{matrix}\pm 1 & b \\ 0 & \mp 1\end{matrix}\right) \text{ or } \left(\begin{matrix}\pm 1 & 0 \\ c & \mp 1\end{matrix}\right),
$$
or
$$
A=\left(\begin{matrix} a & b \\ (1-a^2)/b & -a\end{matrix}\right) \qquad (a^2\neq 1, b\neq 0).
$$
A: By Cayley-Hamilton we have $A^2-tr(A)A+\det(A)I=0$. Now $A=A^{-1}$ implies $A^2=I$. So we obtain $tr(A)A=(\det(A)+1)I$.  Now the rest is easy.
