• # question_answer If $AB=0$, then for the matrices                 $A=\left[ \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\ \end{matrix} \right]$and                 $B=\left[ \begin{matrix} {{\cos }^{2}}\phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & {{\sin }^{2}}\phi \\ \end{matrix} \right],\,\,\,\theta -\phi$is A)  an odd number of$\frac{\pi }{2}$B)  an odd multiple of$\pi$C)  an even multiple of$\frac{\pi }{2}$D) $0$

We have, $AB=\left[ \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {{\sin }^{2}}\theta \\ \end{matrix} \right]\left[ \begin{matrix} {{\cos }^{2}}\phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & {{\sin }^{2}}\phi \\ \end{matrix} \right]$$=\left[ \begin{matrix} {{\cos }^{2}}\theta {{\cos }^{2}}\phi +\cos \theta \cos \phi \sin \theta \sin \phi \\ \cos \theta \sin \theta {{\cos }^{2}}\phi +{{\sin }^{2}}\theta \cos \phi \sin \phi \\ \end{matrix} \right.$$=\left. \begin{matrix} {{\cos }^{2}}\theta \cos \phi \sin \phi +\cos \theta \sin \theta {{\sin }^{2}}\phi \\ \cos \theta \cos \phi \sin \theta \sin \phi +{{\sin }^{2}}\theta {{\sin }^{2}}\phi \\ \end{matrix} \right]$ $=\cos (\theta -\phi )=\left[ \begin{matrix} \cos \theta \cos \phi & \cos \theta \sin \phi \\ \sin \theta \cos \phi & \sin \theta \sin \phi \\ \end{matrix} \right]$ Since,$AB=0,\,\,\,\,\therefore \,\,\cos (\theta -\phi )=0$ $\therefore \,\,\theta -\phi$is an odd multiple of$\frac{\pi }{2}$