### Instruction: Let
\[\mathbf{A} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}.\]Compute $\mathbf{A}^{100}.$.### Response: I notice that $\mathbf{A}$ is a permutation matrix, which means it rearranges the entries of any vector it multiplies by.
Specifically, $\mathbf{A}$ swaps the first and second components of a vector, and then swaps the second and third components of the resulting vector.
For example, $\mathbf{A} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} z \\ x \\ y \end{pmatrix}.$
I wonder what happens when I apply $\mathbf{A}$ multiple times to a vector.
It seems like it will cycle through the three components in a cyclic order.
For example, $\mathbf{A}^2 \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{A} \begin{pmatrix} z \\ x \\ y \end{pmatrix} = \begin{pmatrix} y \\ z \\ x \end{pmatrix}.$
I realize that this means that $\mathbf{A}$ has order 3, which means that $\mathbf{A}^3 = \mathbf{I}$, the identity matrix.
This can also be verified by direct multiplication.
Therefore, I can use modular arithmetic to simplify the exponent of $\mathbf{A}$.
Since $100 \equiv 1 \pmod{3}$, I have $\mathbf{A}^{100} = \mathbf{A}^{3 \times 33 + 1} = (\mathbf{A}^3)^{33} \mathbf{A} = \mathbf{I}^{33} \mathbf{A} = \mathbf{A}.$
So the answer is just $\mathbf{A}$ itself.