How can you find a reflection matrix about a given line, using matrix multiplication and the idea of composition of transformations? I'm reading Linear Algebra Done Wrong, and now there have been not 1 but 2 exercises that are extremely suggestive and indicative that there should be a way to do this. This being, to find a matrix that encodes a linear transformation of a vector reflected across the line of $y = -2x/3$, all in $\Bbb R^2$.
I have basically lost myself at this point in the weeds. I was able, for example, to in an earlier exercise multiply 2 rotation matrices and derive the trig identities for $\sin(a + b)$ and $\cos(a + b)$, which I then thought the similar idea of matrix multiplication would be applied right here. I still believe it has to be so! But I cannot for the life of me figure out actually which transformations to do to result in the final desired one.
Your help is greatly appreciated, thank you
 A: So you have a vector $(x_1,y_1)^T$, and you want to find a matrix $M$ such that $(x_2,y_2)^T=M(x_1,y_1)^T$ is the reflection across $y=-2x/3$. So what does the reflection means? It means that the middle of the two points is on the line of reflection, and the line between those points is perpendicular to the reflection line.
The first condition can be written as $$\frac{y_1+y_2}2=-\frac 23\frac{x_1+x_2}2$$
The second condition means that the slope of the line between the two points is $-1/m$, where $m$ is the slope of the reflection line:$$\frac{y_2-y_1}{x_2-x_1}=-\frac 1{-2/3}=\frac32$$
You now write $x_2$ and $y_2$ in terms of $x_1$ and $y_1$:$$x_2=m_{11}x_1+m_{12}y_1\\y_2=m_{21}x_1+m_{22}y_1$$
Then the matrix you are looking for has these coefficients.
$$M=\begin{pmatrix}m_{11}&m_{12}\\m_{21}&m_{22}\end{pmatrix}$$
A: One strategy is to rotate the plane so that the line becomes $y=0$, apply the reflection in the $x$-axis, and then rotate back.
Let $\tan\theta=2/3$. Then the required matrix is $$\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}=\begin{pmatrix}\cos2\theta&-\sin2\theta\\-\sin2\theta&-\cos2\theta\end{pmatrix}$$
