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?

One reason that it is hard to describe a systematic procedure that will let you find that solution to $A^2=-I$ is that there are so many other solutions to this equation. A solution cannot have real eigenvalues, so the image of the first standard basis vector $e_1$ (which give the firs column of your matrix) cannot be a multiple of $e_1$ itself, but otherwise it can be any vector$~v$ whatsoever. Then $[e_1,v]$ will be a basis of $\Bbb R^2$, and one can arrange that $A\cdot v=-e_1$ (since one can choose the images of a basis freely when defining a linear map), which will ensure that $A^2=-I$ (an easy check).
But the example you gave is so obtained for the simplest choice of$~v$, namely $v=e_2$.
