I am asked to find the expression for reflection about the line $y=mx$. First, I wrote down the reflection matrix about the x-axis in the standard basis:
\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
Then I wrote down the rotation matrix to get from the standard to rotated frame:
\begin{pmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \end{pmatrix}
but now writing this in terms of $m$ we have
\begin{pmatrix} \frac{1}{\sqrt{1 + m^2}} & \frac{m}{\sqrt{1 + m^2}} \\ - \frac{m}{\sqrt{1 + m^2}} & \frac{1}{\sqrt{1 + m^2}} \end{pmatrix}
so total we have
\begin{gather*} \begin{pmatrix} \frac{1}{\sqrt{1 + m^2}} & \frac{m}{\sqrt{1 + m^2}} \\ - \frac{m}{\sqrt{1 + m^2}} & \frac{1}{\sqrt{1 + m^2}} \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{1 + m^2}} & \frac{m}{\sqrt{1 + m^2}} \\ - \frac{m}{\sqrt{1 + m^2}} & \frac{1}{\sqrt{1 + m^2}} \end{pmatrix}^{-1} \end{gather*}
But the answer first picks $\{(1,m), (-m,1)\}$ as a basis for the new frame and calculates the change of basis matrices as
\begin{pmatrix} 1 & -m \\ m & 1 \end{pmatrix}
and
\begin{gather*} \begin{pmatrix} 1 & -m \\ m & 1 \end{pmatrix}^{-1} = \begin{pmatrix} \frac{1}{1+m^2} & \frac{m}{1+m^2} \\ -\frac{m}{1+m^2} & \frac{1}{1+m^2} \end{pmatrix} \end{gather*}
I understand the method they used, but I don't understand why the first method doesn't give the same answer. There is a square root that shouldn't be there.