matrices $A$ and $B$ such that $AB = -BA$? I was trying to find matrices non-singular $A$ and $B$ such that $AB = -BA$.I tried taking $A$ and $B$ to be general matrices and started with an order of the matrix as $2$ but I go into a bit of lengthy calculation.
This made me think while it was intuitive for me to calculate the inverse of a $2 \times 2$ , $3 \times 3$ matrix for simple matrices so is it intuitive to find matrices say $A$ such that $A^2 = 0$ or $AB = BA$ or similar type of questions?.
I think such type of interesting generalizations and results can be done and found out?
EDIT - 
From the answer's below and comments we see that taking the determinants simplifies the problem a  bit that it can work only for even order square matrices but still a way/ hint to guessing it would help?

For $2 \times 2$ matrices you can use $$A=
\begin{pmatrix}
a & 0 \\
0 & -a  \\
\end{pmatrix},$$ 
with $a\not =0$ and
$$B=
\begin{pmatrix}
0 & x \\
y & 0  \\
\end{pmatrix}.$$
Then $AB=-BA$ for arbitrary $x,y$.
