"Simpler" geometrical description So i was asked to find:
Find the matrix that represents the linear transformation of the plane obtained by:


*

*reflecting in the line y = x, $\begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}$

*then rotating anticlockwise through an angle of 45 degrees, $\begin{bmatrix} \frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \end{bmatrix}$

*and finally reflecting in the y axis.  $\begin{bmatrix} -1&0 \\ 0&1 \end{bmatrix}$

*Give a simpler geometrical description of what this transformation does.  


Which i guess is just multiplying all the steps
$$\begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}  \begin{bmatrix} \frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \end{bmatrix}  \begin{bmatrix} -1&0 \\ 0&1 \end{bmatrix} = \begin{bmatrix} -\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \\ \frac{1}{-\sqrt{2}}&-\frac{1}{\sqrt{2}} \end{bmatrix}  $$
How do i give a simpler geometrical description of this other than what's already stated in the question: reflect in the line y=x, rotate through angle of 45 degree and reflect by the y axis. 
Is this some kind of inverse project it feels a lot like the geometric aspect of an inverse to its main function. 
 A: Think about what the composition does to things in $\Bbb R^2$, when all is said and done.
In particular, 


*

*Where does the $x$-axis get sent?

*Where does the line $y = x$ get sent? And finally,

*where does the $y$-axis get sent?


I'm noticing your matrix may be a bit off. Let's look at what those operations do geometrically to the point $(1, 0)$ for example:
\begin{align*}(1, 0) 
&\mapsto (0, 1)  &\text{ reflection about }y = x
\\ \\ 
(0, 1) &\mapsto \left(\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) & 45^\circ \text{rotation counter-clockwise} 
\\ \\
\left(\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)&\mapsto \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) & \text{reflection about the $y$-axis}
\end{align*}
Think about this for a few minutes and draw some pictures. Once you've decided what's happening, here are some extra tidbits:
In general, it turns out that all rotations and reflections get along very nicely. Here we have a product that looks like
$$\operatorname{Reflection}_1 \circ \operatorname{Rotation} \circ \operatorname{Reflection}_2,$$
and the result will always be a rotation.
You should have come to a conclusion right now, but to make things even more concrete...
Rotations have a very specific form. An anti-clockwise rotation through an angle $\theta$ have the form.
$$\begin{pmatrix} \cos\theta  &  -\sin\theta \\ \sin\theta  & \cos\theta  \end{pmatrix}$$
Does your result look like that, for any $\theta$ in particular?
A: there is a result from geometry that says any rotation by angle $2\theta$ is equivalent to two reflections on mirrors separated by an angle $\theta.$  the two mirrors can be placed on ant line through the point of rotation. 
we will also use the fact that reflections are involuntary; that they are their own inverses.
let $R_{x = 0}, R_{y= \tan(3\pi/8) x},$ and $ R_{y = x}$ represent the reflections on the lines $x = 0, y= \tan(3\pi/8) x$ and $y = x.$ by the geometry argument we have $$Rot_{\pi/4} = R_{y= \tan(3\pi/8) x} R_{y=x}$$ where $Rot_{\pi/4}$ represents the rotation by $45^\circ$ counterclockwise about the origin.
we can compute $$\begin{align}R_{x = 0} Rot_{\pi/4} R_{y=x} &= R_{x = 0} R_{y= \tan(3\pi/8) x} R_{y=x} R_{y=x}\\
&=  R_{x = 0} R_{y= \tan(3\pi/8) x}\\
&=Rot_{\pi/4}\end{align}$$
the there transformations result in a rotation by $45^\circ$ counter clockwise. this is represented by the matrix $\pmatrix{1/\sqrt 2& -1/\sqrt 2\\1/\sqrt 2&1\sqrt 2}.$
