### Instruction: Find the matrix that corresponds to reflecting over the vector $\begin{pmatrix} -1 \\ 3 \end{pmatrix}.$.### Response: To reflect over a vector, I need to find the unit vector in the direction of that vector and then use the formula for a reflection matrix.
The unit vector in the direction of $\begin{pmatrix} -1 \\ 3 \end{pmatrix}$ is $\frac{1}{\sqrt{10}}\begin{pmatrix} -1 \\ 3 \end{pmatrix},$ since the length of the original vector is $\sqrt{10}$.
The formula for a reflection matrix is $R = I - 2uu^T,$ where $u$ is the unit vector and $u^T$ is its transpose.
So, I need to plug in the unit vector I found and compute the matrix product.
I get $R = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - 2\frac{1}{\sqrt{10}}\begin{pmatrix} -1 \\ 3 \end{pmatrix}\frac{1}{\sqrt{10}}\begin{pmatrix} -1 & 3 \end{pmatrix}.$
Simplifying, I get $R = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \frac{1}{5}\begin{pmatrix} 1 & -3 \\ -3 & 9 \end{pmatrix}.$
Subtracting the matrices, I get $R = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 0.2 & -0.6 \\ -0.6 & 1.8 \end{pmatrix}.$
Finally, I get $R = \begin{pmatrix} 0.8 & 0.6 \\ 0.6 & -0.8 \end{pmatrix}.$