### Instruction: If the matrix $\mathbf{A}$ has an inverse and $(\mathbf{A} - 2 \mathbf{I})(\mathbf{A} - 4 \mathbf{I}) = \mathbf{0},$ then find
\[\mathbf{A} + 8 \mathbf{A}^{-1}.\].### Response: To solve this problem, I need to use some properties of matrix inverses and matrix multiplication.
First, I notice that the equation $(\mathbf{A} - 2 \mathbf{I})(\mathbf{A} - 4 \mathbf{I}) = \mathbf{0}$ can be rewritten as
\[\mathbf{A}^2 - 6 \mathbf{A} + 8 \mathbf{I} = \mathbf{0}.\]
Next, I multiply both sides of this equation by $\mathbf{A}^{-1}$, which is possible because $\mathbf{A}$ has an inverse.
This gives me
\[\mathbf{A} - 6 \mathbf{I} + 8 \mathbf{A}^{-1} = \mathbf{0}.\]
Now, I rearrange the terms and get
\[\mathbf{A} + 8 \mathbf{A}^{-1} = 6 \mathbf{I}.\]
This is exactly what I need to find, so I'm done.