Determine all $2 \times 2$ real matrices $A$ such that $(1) \ \ A^2=I$, $(2) \ \ A^2=0$ 
Determine all $2 \times 2$ real matrices $A$ such that $(1) \ \ A^2=I$, $(2) \ \ A^2=0$

I came across this problem recently where I have to determine all the $2\times2$ matrices satisfying the aforementioned criteria. The problem is that the equations which I get after multiplication are not so easy to work with and I'm unable to proceed. 
If someone could help me with these equations and solve for the various cases, that'd be really helpful.  
Answer to part one: 
$$A=I$$ $$A=-I$$ $$a_{11}=a_{22}=0  \ , \ a_{12}a_{21}=1 $$$$a_{11}=-a_{22}\neq0, \ a_{11}^2+a_{12}a_{21}=1$$ 
Answer to part two:
$$a_{11}=a_{12}=a_{22}=0$$$$a_{11}=a_{21}=a_{22}=0$$$$a_{11}=-a_{22}\neq0, \ a_{11}^2+a_{12}a_{21}=0$$
 A: Case $A^{2} = I$.
Two possibilities are of course $A = \pm I$. If $A \ne \pm I$, then the characteristic polynomial of $A$ is $x^{2} - 1$, which means $A$ has trace $0$ and determinant $-1$. Therefore
\begin{equation*}
\begin{bmatrix}
a & b\\
c & -a
\end{bmatrix},
\end{equation*}
where $a^{2} + b c = 1$.
Case $A^{2} = 0$.
One possibility is of course $A = 0$. If $A \ne 0$, then the characteristic polynomial of $A$ is $x^{2}$, which means $A$ has trace and determinant $0$. Therefore
\begin{equation*}
\begin{bmatrix}
a & b\\
c & -a
\end{bmatrix},
\end{equation*}
where $a^{2} + b c = 0$, and $a, b, c$ are not all zero.
In part one, distinguish now 


*

*the case when $a = \pm 1$, and then either $b = 0$ and $b$ is arbitrary, or vice versa;

*the case when $a \ne \pm 1$, and then $b \ne 0$, say, can be chosen arbitrarily, and $c = (1 - a^{2})/b \ne 0$.


In part two, distinguish now 


*

*the case when $a = 0$, and then either $b = 0$ and $c$ is arbitrary, or vice versa;

*the case when $a \ne 0$, and then $b \ne 0$, say, can be chosen arbitrarily, and $c = -a^{2}/b\ne 0$.

A: $$A = \begin{bmatrix}
a & b\\
c & d
\end{bmatrix}
\Rightarrow A^2 = \begin{bmatrix}
a^2+bc & (a+d)b\\
(a+d)c & bc +d^2
\end{bmatrix}$$
In both cases, you have to impose $(a+d)b = (a+d)c = 0$.


*

*Case $a+d \neq 0$: then $b = c = 0$ and $A^2 = \begin{bmatrix}
a^2 & 0\\
0 & d^2
\end{bmatrix}$ and it's easy to continue.

*Case $a+d = 0$: $A^2 = \begin{bmatrix}
a^2+bc & 0\\
0 & bc + d^2
\end{bmatrix}$ and it's easy to continue.

