Question

Simplify: $\cos \theta \left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]+\sin \theta \left[\begin{array}{cc}\sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{array}\right]$

Answer 1

$\cos \theta \left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]+\sin \theta \left[\begin{array}{cc}\sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{array}\right]$
$=\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ -\sin \theta \cos \theta & {\cos }^2\theta \end{array}\right]+\left[\begin{array}{cc}{\sin }^2\theta & -\sin \theta \cos \theta \\ \sin \theta \cos \theta & {\sin }^2\theta \end{array}\right]$
$=\left[\begin{array}{cc}{\cos }^2\theta +{\sin }^2\theta & \cos \theta \sin \theta -\sin \theta \cos \theta \\ -\sin \theta \cos \theta +\sin \theta \cos \theta & {\cos }^2\theta +{\sin }^2\theta \end{array}\right]$
$=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$