### Instruction: If $\mathbf{A}^{-1} = \begin{pmatrix} 4 & 6 \\ -2 & 10 \end{pmatrix},$ and $\mathbf{B} = \frac{1}{2} \mathbf{A},$ then find $\mathbf{B}^{-1}.$.### Response: I want to find the inverse of $\mathbf{B},$ which is a matrix that satisfies $\mathbf{B} \mathbf{B}^{-1} = \mathbf{I},$ where $\mathbf{I}$ is the identity matrix.
I notice that $\mathbf{B}$ is related to $\mathbf{A}$ by a scalar multiplication, so I wonder if there is a way to use the inverse of $\mathbf{A}$ to find the inverse of $\mathbf{B}.$
I recall that for any nonzero scalar $k,$ the inverse of $k \mathbf{A}$ is $\frac{1}{k} \mathbf{A}^{-1},$ since $(k \mathbf{A}) (\frac{1}{k} \mathbf{A}^{-1}) = \mathbf{I}.$
I apply this rule to $\mathbf{B} = \frac{1}{2} \mathbf{A},$ and conclude that $\mathbf{B}^{-1} = 2 \mathbf{A}^{-1}.$
I multiply the matrix $\mathbf{A}^{-1}$ by the scalar 2 to get $\mathbf{B}^{-1}.$