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

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

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$\alpha - \beta$

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(C) Given $A = \begin{bmatrix} \cos^2 \alpha & \sin \alpha \cos \alpha \ \sin \alpha \cos \alpha & \sin^2 \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \cos^2 \beta & \sin \beta \cos \beta \ \sin \beta \cos \beta & \sin^2 \beta \end{bmatrix}$.Calculating the product $AB$:$AB = \begin{bmatrix} \cos^2 \alpha & \sin \alpha \cos \alpha \ \sin \alpha \cos \alpha & \sin^2 \alpha \end{bmatrix} \begin{bmatrix} \cos^2 \beta & \sin \beta \cos \beta \ \sin \beta \cos \beta & \sin^2 \beta \end{bmatrix}$$AB = \begin{bmatrix} \cos \alpha \cos \beta (\cos \alpha \cos \beta + \sin \alpha \sin \beta) & \cos \alpha \sin \beta (\cos \alpha \cos \beta + \sin \alpha \sin \beta) \ \sin \alpha \cos \beta (\cos \alpha \cos \beta + \sin \alpha \sin \beta) & \sin \alpha \sin \beta (\cos \alpha \cos \beta + \sin \alpha \sin \beta) \end{bmatrix}$Using the identity $\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$:$AB = \cos(\alpha - \beta) \begin{bmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta \ \sin \alpha \cos \beta & \sin \alpha \sin \beta \end{bmatrix}$For $AB$ to be a null matrix,we must have $\cos(\alpha - \beta) = 0$.Therefore,$\alpha - \beta$ must be an odd integral multiple of $\frac{\pi}{2}$.

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