The question asked us to find the matrix of the reflection on the plane with equation $x+2y-2z=0$. From what I have learnt, the (unit) normal to the plane is $[\frac{1}{3},\frac{2}{3},-\frac{2}{3}]$. We can treat this as a quaternion, $q=(\frac{1}{3},\frac{2}{3},-\frac{2}{3},0)$, where $0$ is the real part of the quaternion.
We also learnt that to obtain the matrix of a reflection, the formula is given to be $R(x)=-qxq^{-1}=qxq$.
$$R(x)=qxq=\begin{pmatrix} 0 & \frac{2}{3} & \frac{2}{3} & \frac{1}{3}\\ -\frac{2}{3} & 0 & -\frac{1}{3} & \frac{2}{3}\\ -\frac{2}{3} & \frac{1}{3} & 0 & -\frac{2}{3} \\ -\frac{1}{3} & -\frac{2}{3} & \frac{2}{3} & 0 \end{pmatrix}xq=\begin{pmatrix} 0 & \frac{2}{3} & \frac{2}{3} & \frac{1}{3}\\ -\frac{2}{3} & 0 & -\frac{1}{3} & \frac{2}{3}\\ -\frac{2}{3} & \frac{1}{3} & 0 & -\frac{2}{3} \\ -\frac{1}{3} & -\frac{2}{3} & \frac{2}{3} & 0 \end{pmatrix}\begin{pmatrix} 0 & -\frac{2}{3} & -\frac{2}{3} & \frac{1}{3}\\ \frac{2}{3} & 0 & \frac{1}{3} & \frac{2}{3}\\ \frac{2}{3} & -\frac{1}{3} & 0 & -\frac{2}{3} \\ -\frac{1}{3} & -\frac{2}{3} & \frac{2}{3} & 0 \end{pmatrix}x$$ The problem comes in when I evaluate the product of matrix. The product is \begin{pmatrix} \frac{7}{9} & -\frac{4}{9} & \frac{4}{9} & 0\\ -\frac{4}{9} & \frac{1}{9} & \frac{8}{9} & 0\\ \frac{4}{9} & \frac{8}{9} & \frac{1}{9} & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}
Why is the element in the bottom row and column $-1$? My teacher taught us that it should be a positive $1$. Is there something wrong with the formula that I used?