Question

If $f\left(\alpha \right)=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0\\ \sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]$ find
$f(-\alpha )+f\left(\alpha \right)$

Answer 1

$f\left(\alpha \right)=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0\\ \sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]$

$f(-\alpha )=\left[\begin{array}{ccc}\cos (-\alpha )& -\sin (-\alpha )& 0\\ \sin (-\alpha )& \cos (-\alpha )& 0\\ 0& 0& 1\end{array}\right]$

$f(-\alpha )=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0\\ -\sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]$

Therefore,

$f\left(\alpha \right)+f(-\alpha )=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0\\ \sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]+\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0\\ -\sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]$

$=\left[\begin{array}{ccc}2\cos \alpha & 0& 0\\ 0& 2\cos \alpha & 0\\ 0& 0& 2\end{array}\right]$