Question

Prove that the product of matrices $\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right]$ and $\left[\begin{array}{cc}{\cos }^2\varphi & \cos \varphi \sin \varphi \\ \cos \varphi \sin \varphi & {\sin }^2\varphi \end{array}\right]$ is the null matrix, when $\theta$ and $\varphi$ differ by an odd multiple of $\frac{\pi }{2}$.

Answer 1

$\cos \theta \cos \varphi +\sin \theta \sin \varphi =\cos (\theta -\varphi )$

$\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right]\left[\begin{array}{cc}{\cos }^2\varphi & \cos \varphi \sin \varphi \\ \cos \varphi \sin \varphi & {\sin }^2\varphi \end{array}\right]$

$=\left[\begin{array}{cc}cos\theta cos\varphi (\cos \theta \cos \varphi +\sin \theta \sin \varphi )& cos\theta sin\varphi (\cos \theta \cos \varphi +\sin \theta \sin \varphi )\\ sin\theta cos\varphi (\cos \theta \cos \varphi +\sin \theta \sin \varphi )& sin\theta sin\varphi (\cos \theta \cos \varphi +\sin \theta \sin \varphi )\end{array}\right]$

$=\left[\begin{array}{cc}cos\theta cos\varphi \cos (\theta -\varphi )& cos\theta sin\varphi \cos (\theta -\varphi )\\ sin\theta cos\varphi \cos (\theta -\varphi )& sin\theta sin\varphi \cos (\theta -\varphi )\end{array}\right]$

Given$\theta -\varphi =(2n+1)\frac{\pi }{2}\Rightarrow \cos (\theta -\varphi )=0$

therefore $\left[\begin{array}{cc}cos\theta cos\varphi \cos (\theta -\varphi )& cos\theta sin\varphi \cos (\theta -\varphi )\\ sin\theta cos\varphi \cos (\theta -\varphi )& sin\theta sin\varphi \cos (\theta -\varphi )\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$