Solution
= Solution
The <permutation matrix> acts by $A(x,y,z)^T=(y,z,x)^T$. Applying this cyclic permutation three times returns every <vector> to itself, proving $A^3=I$. Explicitly $A^2=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}$ and multiplication once more gives the <identity matrix>.