Question:

The transpose of a matrix $\displaystyle M=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is the matrix $\displaystyle M^{T}=\begin{pmatrix} a & c \\ b & d \end{pmatrix} $ and M is said to be orthogonal when $\displaystyle M^{T}M=I$, where I is the unit matrix. Given that the matrix

$\displaystyle N=\begin{pmatrix} \frac{2}{\sqrt{5}} &\frac{1}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}} & k \end{pmatrix} $

is orthogonal, find the value of k. Describe geometrically the transformation of x-y plane which is represented by N. Under a transformation S of the real plane into a itself, a point $\displaystyle P = (x,y) $ is mapped onto the point $\displaystyle S(P) = (ax+by, cx+dy) $. Show that when M is orthogonal, the distance between any two points P and Q is the same as the distance between their images S(P) and S(Q).

My attempt:

$\displaystyle N=\begin{pmatrix} \frac{2}{\sqrt{5}} &\frac{1}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}} & k \end{pmatrix} $

$\displaystyle N^{T}=\begin{pmatrix} \frac{2}{\sqrt{5}} &-\frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & k \end{pmatrix} $

$\displaystyle N^{T}N=I$

$\displaystyle \begin{pmatrix} \frac{2}{\sqrt{5}} &\frac{1}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}} & k \end{pmatrix}\begin{pmatrix} \frac{2}{\sqrt{5}} &-\frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & k \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} $

$\displaystyle \begin{pmatrix} 5 &\frac{2}{5}- \frac{k}{\sqrt{5}} \\ \frac{2}{5}- \frac{k}{\sqrt{5}} & k^{2} - \frac{1}{\sqrt{5}} \end{pmatrix}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $

$\displaystyle \frac{2}{5}- \frac{k}{\sqrt{5}} =0 $

$\displaystyle k = \frac{2\sqrt{5}}{5} $

The transformation represented by N is a rotation by angle 26.6 (degrees) clockwise

I am having difficulty answering the last part of the question

Please I would like some help with the last part of the question

$\begin{pmatrix}a &b \\ c & d\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} =\begin{pmatrix}ax+by\\ cx+dy \end{pmatrix}$