Let
\[\mathbf{A} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}.\]Compute $\mathbf{A}^{100}.$
OK, let's compute the first few powers of A. The first power is just the matrix A. The second power is A * A. The third power is A * A * A. \[\mathbf{A}^3 = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\] Oh, so A cubed is just the identity matrix. I think so. That means A to any multiple of 3 is the identity matrix. But we know that A cubed is the identity matrix, and 99 is a multiple of 3, so \[\mathbf{A}^{99} = \mathbf{A}^{98}\mathbf{A} = \dots = \mathbf{A}^3 = \mathbf{I}\] Then \[\mathbf{A}^{100} = \mathbf{A}^{99}\mathbf{A} = \mathbf{I}\mathbf{A} = \mathbf{A}\]

