Question

If the product of matrices $A=\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right],B=\left[\begin{array}{cc}{\cos }^2\varphi & \cos \varphi \sin \varphi \\ \cos \varphi \sin \varphi & {\sin }^2\varphi \end{array}\right]$ is a null matrix, then $\theta -\varphi$ is equal to

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

Answer 1

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

$AB=\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right]\left[\begin{array}{cc}{\cos }^2\varphi & \cos \varphi \sin \varphi \\ \cos \varphi \sin \varphi & {\sin }^2\varphi \end{array}\right]=\left[\begin{array}{cc}{\cos }^2\theta {\cos }^2\varphi +\cos \theta \sin \theta \cos \varphi \sin \varphi & {\cos }^2\theta \cos \varphi \sin \varphi +{\sin }^2\varphi \sin \theta \cos \theta \\ \cos \theta \sin \theta {\cos }^2\varphi +{\sin }^2\theta \sin \varphi \cos \varphi & \cos \theta \sin \theta \sin \varphi \cos \varphi +{\sin }^2\varphi {\sin }^2\theta \end{array}\right]=\left[\begin{array}{cc}\cos (\theta -\varphi )\cos \theta \cos \varphi & \cos \theta \sin \varphi \cos (\theta -\varphi )\\ \cos (\theta -\varphi )\sin \theta \cos \varphi & \sin \theta \sin \varphi \cos (\theta -\varphi )\end{array}\right]=\cos (\theta -\varphi )\left[\begin{array}{cc}\cos \theta \cos \varphi & \cos \theta \sin \varphi \\ \sin \theta \cos \varphi & \sin \theta \sin \varphi \end{array}\right]$ Now, $AB=O$
As $\left[\begin{array}{cc}\cos \theta \cos \varphi & \cos \theta \sin \varphi \\ \sin \theta \cos \varphi & \sin \theta \sin \varphi \end{array}\right]\ne O$ for any values of $(\theta ,\varphi )$
$\therefore \cos (\theta -\varphi )=0\Rightarrow \theta -\varphi =(2n+1)\frac{\pi }{2},n\in Z$