### Instruction: Let $P$ be the plane passing through the origin with normal vector $\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}.$  Find the matrix $\mathbf{R}$ such that for any vector $\mathbf{v},$ $\mathbf{R} \mathbf{v}$ is the reflection of $\mathbf{v}$ through plane $P.$.### Response: To reflect a vector $\mathbf{v}$ through a plane $P,$ we can subtract its projection onto the normal vector of $P$ twice.
That is, $\mathbf{R} \mathbf{v} = \mathbf{v} - 2 \left(\frac{\mathbf{v} \cdot \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}}{\left\| \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \right\|^2}\right) \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}.$
To find the matrix $\mathbf{R},$ we can apply this formula to the standard basis vectors $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},$ and $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ and see what their images are.
For $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},$ we have $\mathbf{R} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} - 2 \left(\frac{1}{3}\right) \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} \frac{1}{3} \\ -\frac{2}{3} \\ \frac{2}{3} \end{pmatrix}.$
Similarly, for $\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},$ we have $\mathbf{R} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} - 2 \left(\frac{1}{3}\right) \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} -\frac{2}{3} \\ \frac{1}{3} \\ \frac{2}{3} \end{pmatrix}.$
And for $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix},$ we have $\mathbf{R} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} - 2 \left(\frac{-1}{3}\right) \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} \frac{2}{3} \\ \frac{2}{3} \\ \frac{1}{3} \end{pmatrix}.$
Therefore, the matrix $\mathbf{R}$ has these three vectors as its columns, and we can write $\mathbf{R} = \frac{1}{3} \begin{pmatrix} 1 & -2 & 2 \\ -2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix}.$