If A = begin{bmatrix} cos^2 theta & sin theta | vedclass

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(B) Given $A = \begin{bmatrix} \cos^2 	heta & \sin 	heta \cos 	heta \ \sin 	heta \cos 	heta & \sin^2 	heta \end{bmatrix}$ and $B = \begin{bmatr

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$O$

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(B) Given $A = \begin{bmatrix} \cos^2 	heta & \sin 	heta \cos 	heta \ \sin 	heta \cos 	heta & \sin^2 	heta \end{bmatrix}$ and $B = \begin{bmatrix} \cos^2 \phi & \sin \phi \cos \phi \ \sin \phi \cos \phi & \sin^2 \phi \end{bmatrix}$.We are given that $	heta - \phi = \frac{\pi}{2}$ or $\phi - 	heta = \frac{\pi}{2}$,which implies $\cos(	heta - \phi) = \cos(\frac{\pi}{2}) = 0$.Now,calculate the product $AB$:$AB = \begin{bmatrix} \cos^2 	heta & \sin 	heta \cos 	heta \ \sin 	heta \cos 	heta & \sin^2 	heta \end{bmatrix} \begin{bmatrix} \cos^2 \phi & \sin \phi \cos \phi \ \sin \phi \cos \phi & \sin^2 \phi \end{bmatrix}$Performing matrix multiplication:$AB = \begin{bmatrix} \cos^2 	heta \cos^2 \phi + \sin 	heta \cos 	heta \sin \phi \cos \phi & \cos^2 	heta \sin \phi \cos \phi + \sin 	heta \cos 	heta \sin^2 \phi \ \sin 	heta \cos 	heta \cos^2 \phi + \sin^2 	heta \sin \phi \cos \phi & \sin 	heta \cos 	heta \sin \phi \cos \phi + \sin^2 	heta \sin^2 \phi \end{bmatrix}$Factoring out common terms:$AB = \begin{bmatrix} \cos 	heta \cos \phi (\cos 	heta \cos \phi + \sin 	heta \sin \phi) & \cos 	heta \sin \phi (\cos 	heta \cos \phi + \sin 	heta \sin \phi) \ \sin 	heta \cos \phi (\cos 	heta \cos \phi + \sin 	heta \sin \phi) & \sin 	heta \sin \phi (\cos 	heta \cos \phi + \sin 	heta \sin \phi) \end{bmatrix}$Using the identity $\cos(	heta - \phi) = \cos 	heta \cos \phi + \sin 	heta \sin \phi$:$AB = \cos(	heta - \phi) \begin{bmatrix} \cos 	heta \cos \phi & \cos 	heta \sin \phi \ \sin 	heta \cos \phi & \sin 	heta \sin \phi \end{bmatrix}$Since $	heta - \phi = \frac{\pi}{2}$,$\cos(	heta - \phi) = 0$.Therefore,$AB = 0 	imes \begin{bmatrix} \cos 	heta \cos \phi & \cos 	heta \sin \phi \ \sin 	heta \cos \phi & \sin 	heta \sin \phi \end{bmatrix} = O$ (the zero matrix).

If $A = \begin{bmatrix} \cos^2 \theta & \sin \theta \cos \theta \\ \sin \theta \cos \theta & \sin^2 \theta \end{bmatrix}$,$B = \begin{bmatrix} \cos^2 \phi & \sin \phi \cos \phi \\ \sin \phi \cos \phi & \sin^2 \phi \end{bmatrix}$ and $\theta$ and $\phi$ differ by $\frac{\pi}{2}$,then $AB = $

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