Question

If $E\left(\theta \right)=\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right]$ and $\theta andɸ$ differ by an odd multiple of $\pi /2,thenE\left(\theta \right)E\left(ɸ\right)$ is a

• A. unit matrix
• B. null matrix
• C. diagonal matrix
• D. None of these

Answer 1

The correct option is B

null matrix

Finding the value:

Given that $E\left(\theta \right)=\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right]$

Hence, $E\left(\varphi \right)=\left[\begin{array}{cc}{\cos }^2\varphi & \cos \varphi \sin \varphi \\ \cos \varphi \sin \varphi & {\sin }^2\varphi \end{array}\right]$

$\begin{array}{rcl}E\left(\theta \right)\times E\left(\varphi \right)& =& \left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right]\times \left[\begin{array}{cc}{\cos }^2\varphi & \cos \varphi \sin \varphi \\ \cos \varphi \sin \varphi & {\sin }^2\varphi \end{array}\right]\\ & =& \left[\begin{array}{cc}\cos \theta \sin \varphi \cos (\theta -\varphi )& \cos \theta \sin \varphi \cos (\theta -\varphi )\\ \cos \theta \sin \varphi \cos (\theta -\varphi )& \cos \theta \sin \varphi \cos (\theta -\varphi )\end{array}\right]\\ & =& \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\left(\because (\theta -\varphi )=(2n+1)\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2,and\cos (2$}\right.\mathrm{n}+1)\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=0\right)\\ & & \end{array}$

Hence, the correct answer is option B.