Is there a mistake to the answer given to this problem: Find the matrix of the reflection through the line y=−2x/3. This is the answer given:

I think this answer is not correct because the line given by y=-2x/3 makes an angle that is below the x axis, so the order is incorrect right? We need to start  with a counterclockwise rotation first and end with the clockwise rotation, but here, we started with the clockwise rotation first.
 A: You are right. The given answer is not correct.
But what is more:
The way how to find the reflection matrix is far too cumbersome.
The standard way is as follows:
Let $d=  \begin{pmatrix} 3\\ -2 \end{pmatrix}$ be the direction vector of the line. Then, the projection $P_d$ of any point with positional vector $v$ onto the line is given by:
$$P_d(v) = \frac{d\cdot v}{|d|^2}v$$
So, you get the reflection $T$ of $v$  in your line by flipping the sign of the part of $v$ which is orthogonal to $d$:
$$v = P_dv \color{blue}{+} (v-P_dv) {\longrightarrow} Tv = P_dv \color{blue}{-} (v-P_dv) = 2P_dv - v = (2P_d-I)v$$
So, you only need the matrix representation of $P_d$ which is quickly done by plugging in the unit vectors $e_1 = \begin{pmatrix} 1\\ 0 \end{pmatrix}$ and $ e_2 =\begin{pmatrix} 0\\ 1 \end{pmatrix}$:
$$P_de_1 = \frac 1{13}\begin{pmatrix} 9\\ -6 \end{pmatrix},P_de_2 = \frac 1{13}\begin{pmatrix} -6\\ 4\end{pmatrix} \Rightarrow P_d =\frac 1{13}\begin{pmatrix} 9 & -6\\ -6 & 4\end{pmatrix} $$
Plug this into
$$T = 2P_d - I \Rightarrow T = \frac 1{13}\begin{pmatrix} 5 & -12\\ -12 & -5\end{pmatrix}$$
I recommend to check - for example - with Geogebra that this matrix nicely reflects points in the given line.
A: Here the slope of the line = $m = tan \gamma = -\frac{2}{3}$.
$\therefore sin \gamma = \frac{m}{\sqrt{1+m^2}},\; cos \gamma = \frac{1}{\sqrt{1+m^2}}$.
The transformation matrix for reflection around the line is given by the product of 3 matrices (rotate by angle -$\gamma$ to make it horizontal, reflect w.r.t the horizontal line and rotate back by angle $\gamma$)
\begin{align}T&=T_{-\gamma}.R_0.T_{\gamma}\\
&=\begin{pmatrix}cos \gamma & -sin\gamma\\ sin\gamma  & cos \gamma\end{pmatrix} 
\begin{pmatrix}1&0\\ 0  & -1\end{pmatrix} 
\begin{pmatrix}cos \gamma & sin\gamma\\ -sin\gamma  & cos \gamma\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} 
\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} \\
&=\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}\frac{1}{\sqrt{1+m^2}} & \frac{m}{\sqrt{1+m^2}}\\ \frac{m}{\sqrt{1+m^2}}  & -\frac{m}{\sqrt{1+m^2}}\end{pmatrix} \\
&=\frac{1}{1 + m^2}\begin{pmatrix}1-m^2&2m\\2m&m^2-1\end{pmatrix} \\
&=\begin{pmatrix}\frac{5}{13}&-\frac{12}{13}\\ -\frac{12}{13}&-\frac{5}{13}\end{pmatrix}
\end{align}
The following example in R shows how the black point (1,1) is reflected w.r.t. the life to new location (red point)
x <- seq(-10,10,0.01)
y <- -2*x/3
m <- -2/3
T <- matrix(c((1-m^2)/(1+m^2),2*m/(1+m^2),2*m/(1+m^2),(m^2-1)/(1+m^2)), nrow=2, byrow=T)
p <- c(1,2)
p1 <- T %*% p
plot(x,y, type='l')
points(p[1],p[2],pch=19)
points(p1[1],p1[2],pch=19, col='red')


