Solution
= Solution
Multiply $A^2=-I/4$ by $A$ once more:
$$
\boxed{A^3=-\frac14A=\begin{pmatrix}0&-1/8\\1/8&0\end{pmatrix}}.
$$
No additional <diagonalization> is needed.
= Solution
Multiply $A^2=-I/4$ by $A$ once more:
$$
\boxed{A^3=-\frac14A=\begin{pmatrix}0&-1/8\\1/8&0\end{pmatrix}}.
$$
No additional <diagonalization> is needed.