Question

If $AB=0$ for the matrices
$A=\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right]$ and $B=\left[\begin{array}{cc}{\cos }^2\varphi & \cos \varphi \sin \varphi \\ \cos \varphi \sin \varphi & {\sin }^2\varphi \end{array}\right]$ then $\theta -\varphi$ is

• A. an odd multiple of $\frac{\pi }{2}$
• B. an odd multiple of $\pi$
• C. an odd even of $\frac{\pi }{2}$
• D. $0$

Answer 1

The correct option is A an odd multiple of $\frac{\pi }{2}$

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

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

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

$=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\Rightarrow$ $\left[\begin{array}{cc}{\cos }^2\theta \cos \varphi \cos (\theta -\varphi )& \cos \theta \sin \varphi \cos (\theta -\varphi )\\ \sin \varphi \cos \varphi & {\sin }^2\theta \sin \varphi \cos (\theta -\varphi )\end{array}\right]$ = $\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\Rightarrow$ $\cos (\theta -\varphi )$$\left[\begin{array}{cc}{\cos }^2\theta \cos \varphi & \cos \theta \sin \varphi \\ \sin \varphi \cos \varphi & {\sin }^2\theta \sin \varphi \end{array}\right]$ = $\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\Rightarrow \cos (\theta -\varphi )(\cos \theta \cos \varphi \sin \theta \sin \varphi -\cos \theta \cos \varphi \sin \theta \sin \varphi )=0$

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

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

i.e an odd multiple of $\frac{\pi }{2}$