Logic behind finding a $ (2 \times 2) $-matrix $ A $ such that $ A^{2} = - \mathsf{I} $. I know the following matrix "$A$" results in the negative identity matrix when you take $A*A$ (same for $B*B$, where $B=-A$):
$$A=\begin{pmatrix}0 & -1\cr 1 & 0\end{pmatrix}$$
However, I am not certain how one would go about finding this matrix without guessing and checking. Is there some systematic way of doing so? I have tried by assuming the following:
$$\begin{pmatrix}a & b\cr c & d\end{pmatrix}^2 = \begin{pmatrix}-1 & 0\cr 0 & -1\end{pmatrix}$$
...and then get the following equations:
$$a*a+b*c=-1$$
$$a*b+b*d=0$$
$$a*c+c*d=0$$
$$b*c+d*d=-1$$
This only tells me that $a=-d$, and then as best I can tell leaves both $c$ and $d$ without a solution, so perhaps this isn't the best method, but its the only approach that's coming to mind.
Any suggestions?

If you assume that $A$ is invertible, then you can write:
$$
AA = -\mathbb{I}_2 \rightarrow A^{-1}AA = -A^{-1}\mathbb{I}_2 \rightarrow A = -A^{-1}\mathbb{I}_2
$$
But we know the inverse for a $2\times2$ matrix--and it's going to create a set of four linear equations for four variables:
$$
\begin{pmatrix}
a& b \\
c & d
\end{pmatrix} = \begin{pmatrix}
-d & b \\
c & -a
\end{pmatrix}
$$
This creates four equations:
\begin{align}
a = -d \rightarrow a + d = 0 \\
b = b \\
c = c \\
d = -a \rightarrow a + d = 0
\end{align}
This suggests that this is trues for all values of $a$ and $d$ such that $a + d = 0$ and any values of $b$ and $c$.  Let's check:
$$
\begin{pmatrix}
a & x \\
y & -a 
\end{pmatrix}
\times \begin{pmatrix}
a & x \\
y & -a 
\end{pmatrix}
=
\begin{pmatrix}
a^2 + xy & ax - ax \\
ay - ay & xy + a^2 
\end{pmatrix}
$$
Then it's a matter of solving the equation $a^2 + xy = -1$.  For instance, we could choose $a = 2$ and $xy = -5$:
$$
\begin{pmatrix}
2 & -5 \\
1 & -2
\end{pmatrix}
\text{ and its inverse }
\begin{pmatrix}
-2 & -5 \\
1 & 2
\end{pmatrix}
\text{ both work}
$$
...obviously there are many more you could choose.  A general way would be:
$$
\begin{pmatrix}
a & b \\
-\frac{a^2 + 1}{b} & -a
\end{pmatrix}
\text{ or }
\begin{pmatrix}
a & -\frac{a^2 + 1}{b} \\
b & -a
\end{pmatrix}
$$
It may be worth noting that if $b = 0$ then $a^2 + 1 = 0$ must also be true so that the other value, $\frac{a^2 + 1}{b}$, can take on any value.  In this case we get: $a^2 = -1 \rightarrow a = \pm i$ and gives the equation $a^2 + 2xy = -1 \rightarrow -1 + 2*x * 0 = -1 \rightarrow -1 = -1$ (which is true for any value of $x$--the non-zero off diagonal).  So this gives the other answer as well:
$$
\begin{pmatrix}
\pm i & 0 \\
x & \mp i
\end{pmatrix}
\text{ or }
\begin{pmatrix}
\pm i & x \\
0 & \mp i
\end{pmatrix}
$$
You can have arbitrarily complex components.  For instance choose $a = 5i$.  This gives: $a^2 + xy = -1 \rightarrow -25 + xy = -1 \rightarrow xy = 24$.  So we could choose 6 and 4, 3 and 8, 2 and 12...or any two real values (or any two complex values--although they would have to be conjugates).  Another complex example:
$$
\begin{pmatrix}
5i & 3 \\
8 & -5i
\end{pmatrix}
$$
or perhaps:
$$
\begin{pmatrix}
3i+2 & -4 \\
3i - 1 & -3i - 2
\end{pmatrix}
$$
