# Linear Algebra - Finding the matrix for the transformation

1. Apr 2, 2013

### yesiammanu

1. The problem statement, all variables and given/known data
Find the matrix for the transformation which first reflects across the main diagnonal, then projects onto the line 2y+√3x=0, and then reflects about the line √3y=2x

2. Relevant equations
Reflection about the line y=x: T(x,y)=(y,x)
Orthogonal projection on the x-axis T(x,y)=(x,0)
Orthogonal projection on the y-axis T(x,y)=(0,y)

3. The attempt at a solution
Reflection about the line y=x: T(x,y)=(y,x), so the standard matrix for this would be the matrix {(0,1),(1,0)}

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=√3/2 x. Can you then separate these into a scalar operation (√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

2. Apr 3, 2013

### HallsofIvy

Staff Emeritus
Any vector that lies along the line $2y+ \sqrt{3}x= 0$ is mapped to itself while any vector perpendicular to that is mapped to 0. One vector along that line is $(-2, \sqrt{3})$ and, of course, $(\sqrt{3}, 2)$ is perpendicular to it. So
$$\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix} -2 \\ \sqrt{3} \end{bmatrix}= \begin{bmatrix} -2 \\ \sqrt{3} \end{bmatrix}$$
and
$$\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}\sqrt{3} \\ 2 \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
Those give you four equations to solve for a, b, c, and d.

Last edited: Apr 3, 2013