Let \(\displaystyle A = \[a \ b \\ c \ d \]\)

Then \(\displaystyle A^2 = \[a^2 + bc \ b(a + d) \\ c(a + d) \ bc + d^2 \] = \[4 \ 0 \\ 0 \ 4 \]\)

Which gives

a^2 + bc = 4 [1]

b(a + d) = 0 [2]

c(a + d) = 0 [3]

d^2 + bc = 4 [4]

[1]-[4]: a^2 - d^2 = 0 so \(\displaystyle a = d \vee a = -d\)

a = d gives with [2] and [3]: \(\displaystyle b = c = 0\)

\(\displaystyle a^2 = 4 \rightarrow a = -2 \vee 2\)

So \(\displaystyle A = \[-2 \ 0 \\ 0 \ -2 \] \vee A = \[2 \ 0 \\ 0 \ 2 \]\)

Which might also be found by using \(\displaystyle A^2 = 4I\)

Now we have a = -d

which gives 4-a^2 = bc (and a + d = 0)

So

\(\displaystyle \large A = \begin{bmatrix}

a & b\\

\frac{4 - a^2}{b} & -a

\end{bmatrix}\) (for non-zero b)

For a = -d, b = 0, we find

\(\displaystyle \large A = \begin{bmatrix}

2 & 0\\

c & -2

\end{bmatrix}\) ,\(\displaystyle \large A = \begin{bmatrix}

2 & c\\

0 & -2

\end{bmatrix}\), \(\displaystyle \large A = \begin{bmatrix}

-2 & 0\\

c & 2

\end{bmatrix}\) and \(\displaystyle \large A = \begin{bmatrix}

-2 & c\\

0 & 2

\end{bmatrix}\)