A) unit matrix
B) null matrix
C) diagonal matrix
D) None of the above
Correct Answer: B
Solution :
Given, \[E(\theta )=\left[ \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\ \end{matrix} \right]\] \[E(\theta )E(\phi )=\left[ \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\ \end{matrix} \right]\] \[\times \left[ \begin{matrix} {{\cos }^{2}}\phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & {{\sin }^{2}}\phi \\ \end{matrix} \right]\] \[=\left[ \begin{align} & {{\cos }^{2}}{{\cos }^{2}}\phi +\cos \theta \sin \theta \cos \phi \sin \phi \\ & \cos \theta \sin \theta {{\cos }^{2}}\phi +{{\sin }^{2}}\theta \cos \phi \sin \phi \\ \end{align} \right.\] \[\left. \begin{align} & {{\cos }^{2}}\theta \cos \phi \sin \phi +cos\theta \sin \theta si{{n}^{2}}\phi \\ & \cos \theta \sin \theta \cos \phi sin\phi +si{{n}^{2}}\theta {{\sin }^{2}}\phi \\ \end{align} \right]\] \[=\left[ \begin{align} & \cos \theta \cos \phi \cos (\theta -\phi ) \\ & \cos \phi \sin \theta \cos (\theta -\phi ) \\ \end{align} \right.\] \[\left. \begin{align} & \cos \theta \sin \phi \cos (\theta -\phi ) \\ & \sin \theta \sin \phi \cos (\theta -\phi ) \\ \end{align} \right]\] \[=\left[ \begin{align} & \cos \theta \cos \phi \cos (2n+1)\frac{\pi }{2} \\ & \cos \phi \sin \theta \cos (2n+1)\frac{\pi }{2} \\ \end{align} \right.\] \[\left. \begin{align} & \cos \theta \sin \phi \cos (2n+1)\frac{\pi }{2} \\ & \sin \theta \sin \phi \cos (2n+1)\frac{\pi }{2} \\ \end{align} \right]\] \[=\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]\] \[\left[ \therefore \,\,\cos (2n+1)\frac{\pi }{2}=0 \right]\]
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