# Show by using matrix method that a reflection about the line y=x followed by rotation about origin through 90° +ve is equivalent to reflection about y-axis.?

May 22, 2018

See below

#### Explanation:

Reflection about the line $y = x$

The effect of this reflection is to switch the x and y values of the reflected point. The matrix is:

• $A = \left(\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right)$

CCW rotation of a point

For CCW rotations about origin by angle $\alpha$:

• $R \left(\alpha\right) = \left(\begin{matrix}\cos \alpha & - \sin \alpha \\ \sin \alpha & \cos \alpha\end{matrix}\right)$

If we combine these in the order suggested:

$\boldsymbol{x} ' = A \setminus R \left({90}^{o}\right) \setminus \boldsymbol{x}$

$\boldsymbol{x} ' = \left(\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right) \left(\begin{matrix}0 & - 1 \\ 1 & 0\end{matrix}\right) \boldsymbol{x}$

$= \left(\begin{matrix}1 & 0 \\ 0 & - 1\end{matrix}\right) \boldsymbol{x}$

$\implies \left(\begin{matrix}x ' \\ y '\end{matrix}\right) = \left(\begin{matrix}1 & 0 \\ 0 & - 1\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}x \\ - y\end{matrix}\right)$

That is equivalent to a reflection in x-axis .

Making it a CW rotation:

$\left(\begin{matrix}x ' \\ y '\end{matrix}\right) = \left(\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right) \left(\begin{matrix}0 & 1 \\ - 1 & 0\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right)$

$= \left(\begin{matrix}- 1 & 0 \\ 0 & 1\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}- x \\ y\end{matrix}\right)$

That is a reflection in the y-axis.