Question

The product of matrices $A\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right]$ and $B=\left[\begin{array}{cc}{\cos }^2\varphi & \cos \varphi \sin \varphi \\ cos\varphi \sin \varphi & {\sin }^2\varphi \end{array}\right]$ is null matrix if $\theta -\varphi =$

• A. $2n\pi ,n\in Z$
• B. $n\frac{\pi }{2},n\in Z$
• C. $(2n+1)\frac{\pi }{2},n\in Z$
• D. $n\pi ,n\in Z$

Answer 1

The correct option is C $(2n+1)\frac{\pi }{2},n\in Z$

Given, $AB=0_{2\times 2}$
$\Rightarrow \left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right]\left[\begin{array}{cc}{\cos }^2\varphi & \cos \varphi \sin \varphi \\ \cos \varphi \sin \varphi & {\sin }^2\varphi \end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

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

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