### Instruction: Let
\[\mathbf{A} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}.\]Compute $\mathbf{A}^{95}.$.### Response: This problem involves matrix exponentiation, which is usually done by diagonalizing the matrix or finding its eigenvalues and eigenvectors.
However, this matrix has a special structure that makes it easier to find its powers without doing those calculations.
Notice that the matrix $\mathbf{A}$ is skew-symmetric, meaning that $\mathbf{A}^T = -\mathbf{A}.$
This implies that $\mathbf{A}$ is also orthogonal, meaning that $\mathbf{A}^T\mathbf{A} = \mathbf{A}\mathbf{A}^T = \mathbf{I},$ where $\mathbf{I}$ is the identity matrix.
Therefore, any power of $\mathbf{A}$ can be expressed as a product of $\mathbf{A}$ and its transpose, which are inverses of each other.
For example, $\mathbf{A}^2 = \mathbf{A}\mathbf{A} = -\mathbf{A}^T\mathbf{A} = -\mathbf{I}.$
Similarly, $\mathbf{A}^3 = \mathbf{A}\mathbf{A}^2 = -\mathbf{A}.$
In general, we can see that $\mathbf{A}^{2n} = (-1)^n\mathbf{I}$ and $\mathbf{A}^{2n+1} = (-1)^n\mathbf{A}$ for any integer $n.$
Therefore, we can find $\mathbf{A}^{95}$ by writing $95 = 2\times 47 + 1$ and using the formula above.