Question

If $A=\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right]$ and $B=\left[\begin{array}{cc}{\cos }^2\Phi & \cos \Phi \sin \Phi \\ \cos \Phi \sin \Phi & {\sin }^2\Phi \end{array}\right]$, show that AB is a zero matrix if $\theta$ and $\Phi$ differ by an odd multiple of $\frac{\pi }{2}$.

Answer 1

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

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

For $AB$ to be a zero matrix, each element of matrix should be 0.

$\therefore$ from 1st term and 4th term, we have

${\cos }^2\theta {\cos }^2\varphi +\cos \theta \sin \theta \cos \varphi \sin \varphi +\cos \theta \sin \theta \cos \varphi \sin \varphi +{\sin }^2\theta {\sin }^2\varphi =0$

$\therefore {\cos }^2\theta {\cos }^2\varphi +2\cos \theta \sin \theta \cos \varphi \sin \varphi +{\sin }^2\theta {\sin }^2\varphi =0$

$\therefore (\cos \theta \cos \varphi +\sin \theta \sin \varphi {)}^2=0$

$\therefore \cos \theta \cos \varphi +\sin \theta \sin \varphi =0$

$\therefore \cos (\theta -\varphi )=0$

$\therefore \theta -\varphi =n\pi +\frac{\pi }{2}$

Hence, proved.