Solution
= Solution
Repeatedly using the quadratic identity gives
$$
\boxed{A^{2n}=\left(-\frac14\right)^nI,\qquad
A^{2n+1}=\left(-\frac14\right)^nA}.
$$
In entries these are $(-1)^n4^{-n}\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and $(-1)^n2^{-(2n+1)}\begin{pmatrix}0&1\\-1&0\end{pmatrix}$, respectively. This is the basic reduction for <functions of a two-dimensional antisymmetric matrix>.