Question:
Find the matrix for the transformation which first reflects across the main diagonal, then projects onto the line $2y+\sqrt{3}x=0$, and then reflects about the line $\sqrt{3}y=2x$.
Attempt to Question:
Reflection about the line $y=x: T(x,y)=(y,x)$, so the standard matrix for this would be the matrix
$$\begin{pmatrix}0 &1 \\ 1 &0 \end{pmatrix}.$$
However I'm not sure how to deal with equations rather than axis. I assume in the second projection, you can simplify it to $y=\frac{-\sqrt{3}}{2} x$. Can you then separate these into a scalar operation $\frac{-\sqrt{3}}{2}$ and orthogonal operation $y=x$? Even so, I wouldn't know how to go further than this since I only know how to do it among the axis.
Edit: $T_1=\left[\begin{array}{cc}0&1\\1&0\end{array}\right]$ by reflecting across main diagonal
$2y+\sqrt{3}x=0\rightarrow y=\frac{-\sqrt{3}}{2}x$
$T_2=\frac{-\sqrt{3}}{2}\left[\begin{array}{cc}\cos-45&-\sin-45\\\sin-45&\cos-45\end{array}\right]\left[\begin{array}{cc}0&0\\0&1\end{array}\right]\left[\begin{array}{cc}\cos45&-\sin45\\\sin45&\cos45\end{array}\right]$ by rotating $45^\circ$, projecting along the y axis, then rotating $-45^\circ$ (transformations are applied right to left)
$\sqrt{3}y=2x\rightarrow y=\frac{2}{\sqrt{3}}x$
$T_3=\frac{2}{\sqrt{3}}\left[\begin{array}{cc}0&1\\1&0\end{array}\right]$ by separating fraction and then reflecting along $y=x$