Q:

3$\times$3 and 2$\times$2 matrices that satisfy $A^2 = -I$, $I$ being the identity matrix.

I am looking for all square matrices $A$ of order 2 and 3 that satisfy $A^2 = - I$, $I$ being the identity matrix of the corresponding order.

A:

There is no $3\times 3$ matrix $A$ with real entries that satisfies the equation $A^2=-I$.  Suppose there were.  We take determinants, to get $$\left\lvert A\right\rvert^2=\left\lvert A^2\right\rvert=\left\lvert -I\right\rvert=-1$$
But $\left\lvert A\right\rvert$ is real, and no real number squares to $-1$.
In the $2\times 2$ case, by a similar argument we must have $\left\lvert A\right\rvert=\pm1$.  There are solutions, e.g. $A=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$.  To find them all use @Mitch Hughes' approach (which has been, unfortunately, deleted).  To summarize, calculate a system of equations from $$\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)^2=-I$$

