If $E( heta)=\left[\begin{array}{cc}\cos ^{2} heta & \cos heta \sin heta \ \cos heta \sin

Question

If $E(	heta)=\left[\begin{array}{cc}\cos ^{2} 	heta & \cos 	heta \sin 	heta \ \cos 	heta \sin 	heta & \sin ^{2} 	heta\end{array}ight]$ and $	heta$ and $\phi$ differ by an odd multiple of $\frac{\pi}{2}$, then $E(	heta) E(\phi)$ is\

Professor PC Thomas Classes · Answer

(a) $\mathrm{E}(	heta)=\left[\begin{array}{cc}\cos ^{2} 	heta & \cos 	heta \sin 	heta \ \cos 	heta \sin 	heta & \sin ^{2} 	heta\end{array}ight]$ $$ \begin{aligned} & \mathrm{E}(	heta) \cdot \mathrm{E}(\phi)= \ & {\left[\begin{array}{cc} \cos ^{2} 	heta & \cos 	heta \sin 	heta \ \cos 	heta \sin 	heta & \sin ^{2} 	heta \end{array}ight]\left[\begin{array}{cc} \cos ^{2} \phi & \cos \phi \sin \phi \ \cos \phi \sin \phi & \sin ^{2} \phi \end{array}ight]} \end{aligned} $$ $\left[\begin{array}{ll}\cos ^{2} 	heta \cos ^{2} \phi+\cos 	heta \sin 	heta \cos \phi \sin \phi & \cos ^{2} 	heta \cos \phi \sin \phi+\cos 	heta \sin 	heta \sin ^{2} \phi \ \cos 	heta \sin 	heta \cos ^{2} \phi+\sin ^{2} 	heta \cos \phi \sin \phi & \cos 	heta \sin 	heta \cos \phi \sin \phi+\sin ^{2} 	heta \sin ^{2} \phi\end{array}ight]$ $$ =\left[\begin{array}{cc} \cos 	heta \cos \phi \cos (	heta-\phi) & \cos 	heta \sin \phi \cos (	heta-\phi) \ \sin 	heta \cos \phi \cos (	heta-\phi) & \sin 	heta \sin \phi \cos (	heta-\phi) \end{array}ight] $$ $=\left[\begin{array}{ll}0 & 0 \ 0 & 0\end{array}ight]=$ null matrix\