Assume that $T$ is linear transformation. Find the matrix of $T$.
a) $T: R^2 $ → $ R^2 $ first rotates points through $ -\frac {3π}{4} $ radians (clockwise) and then reflects points through the horizontal $x_1$-axis.
b) $T: R^2 $ → $ R^2 $ first reflects points through the horizontal $x_1$-axis and then reflects points through the line $x_1=x_2$. Show that this transformation is merely a rotation about the origin. What is the angle of the rotation?
I am unsure where to start with this since i am new with this, so could anyone explain to me how to solve this? Thanks!
Best Answer
I'll do a and let you tackle the second. A rotation matrix in $\mathbb{R}^2$ representing rotation by angle $\theta$ is given by $$ \begin{bmatrix}\cos \theta&-\sin \theta\\ \sin\theta&\cos\theta\end{bmatrix} $$ reflection accross the $x$ axis is given by $$ \begin{bmatrix}1&0\\ 0&-1\end{bmatrix} $$ so in your case we have $$ \begin{bmatrix}1&0\\ 0&-1\end{bmatrix}\begin{bmatrix}\cos -\frac{3\pi}{4}&-\sin -\frac{3\pi}{4}\\ \sin-\frac{3\pi}{4}&\cos-\frac{3\pi}{4}\end{bmatrix}= \begin{bmatrix}1&0\\ 0&-1\end{bmatrix}\begin{bmatrix}-\frac{\sqrt{2}}{2}& -\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2}&-\frac{\sqrt{2}}{2}\end{bmatrix} $$ carrying out the matrix multiplication yields $$ \begin{bmatrix}-\frac{\sqrt{2}}{2}& -\frac{\sqrt{2}}{2}\\ -\frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}\end{bmatrix} $$