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.
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. $A=\left(\begin{smallmatrix}0&amp;-1\\1&amp;0\end{smallmatrix}\right)$.  To find them all use @Mitch Hughes' approach (which has been, unfortunately, deleted).  To summarize, calculate a system of equations from $$\left(\begin{smallmatrix}a&amp;b\\c&amp;d\end{smallmatrix}\right)^2=-I$$
There are essentially infinite solutions that satisfy A^2 = -I for 2x2 and 3x3 matrices.

They can be solved in similar manners of creating systems of equations and then solving those to produce general solutions. I will solve for 2x2 matrices. 
For standard square matrices we have a matrix A such that

A =\begin{bmatrix}a&amp;b\\c&amp;d\end{bmatrix}

by multiplying the matrix by itself, we get the following matrix

A^2 = \begin{bmatrix}a^2+bc&amp;ab+bd\\ac+cd&amp;bc+d^2\end{bmatrix}

so we have the following set of conditions that we must statisfy:


a^2+bc = -1
b(a+d) = 0
c(a+d) = 0
d^2 + bc = -1


By solving this system we will be able to see that conditions that a,b,c and d must satisfy

We will first begin by subtracting the first and last equations to get:

a^2-d^2 = 0

so therefore, we can see that a = +/- d

We will first explore the first case when a = d

We immediately see through the 2nd and 3 equations that b and c must equal 0 since a+d will never be 0, unless a = d = 0 (discussed later)

However, knowing that b=c=0, we see that that the 1st and 4th equations will never hold true since a^2 and d^2 will be positive, but it must equal -1.

Now we explore the case when a = -d

Say that a = an arbitrary constant t (then d = -t)

Through the 2nd and 3rd equation we see that c and d could essentially be anything too, since a+d = 0 and the equation would always be true.
Thus the only condition it needs to satisfy are:

bc + t^2 = -1 (since a^2 = d^2 = t^2)

so bc = -(1+t^2)

Now we pick another arbitrary constant s for c

so b = -(1+t^2)/s

As a result we have the following conditions for a,b,c, and d that satisfy A^2 = -I


a= t
b= -(1+t^2)/s
c= s
d= -t


which can also be written as:

A =\begin{bmatrix}t&amp;-(1+t^2)/s\\s&amp;-t\end{bmatrix}

for any values t and s (notice that case of a=d=0 is included here)

As a result, there are infinite a,b,c,d that satisfy A^2 = -I that all lie on a plane in space (since geometric multiplicity of 2)