Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 2 / 16 / a

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 $A^2=\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right)$ and multiplication once more gives the identity matrix.

Using $A^3=I$, the matrix power reduces modulo three:

Since $A^{2}A=A^3=I=A{A}^2$, the inverse matrix is

Here $A^{3n+1}=A$, so its inverse matrix is