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

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(C) Given $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \ \sin \alpha & \cos \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \cos \beta & -\sin \bet

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$AB = BA$

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(C) Given $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \ \sin \alpha & \cos \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \cos \beta & -\sin \beta \ \sin \beta & \cos \beta \end{bmatrix}$.Calculating $AB$:$AB = \begin{bmatrix} \cos \alpha & -\sin \alpha \ \sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \beta & -\sin \beta \ \sin \beta & \cos \beta \end{bmatrix}$$= \begin{bmatrix} \cos \alpha \cos \beta - \sin \alpha \sin \beta & -\cos \alpha \sin \beta - \sin \alpha \cos \beta \ \sin \alpha \cos \beta + \cos \alpha \sin \beta & -\sin \alpha \sin \beta + \cos \alpha \cos \beta \end{bmatrix}$$= \begin{bmatrix} \cos(\alpha + \beta) & -\sin(\alpha + \beta) \ \sin(\alpha + \beta) & \cos(\alpha + \beta) \end{bmatrix}$.Similarly,calculating $BA$:$BA = \begin{bmatrix} \cos \beta & -\sin \beta \ \sin \beta & \cos \beta \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha \ \sin \alpha & \cos \alpha \end{bmatrix}$$= \begin{bmatrix} \cos \beta \cos \alpha - \sin \beta \sin \alpha & -\cos \beta \sin \alpha - \sin \beta \cos \alpha \ \sin \beta \cos \alpha + \cos \beta \sin \alpha & -\sin \beta \sin \alpha + \cos \beta \cos \alpha \end{bmatrix}$$= \begin{bmatrix} \cos(\beta + \alpha) & -\sin(\beta + \alpha) \ \sin(\beta + \alpha) & \cos(\beta + \alpha) \end{bmatrix}$.Since $\cos(\alpha + \beta) = \cos(\beta + \alpha)$ and $\sin(\alpha + \beta) = \sin(\beta + \alpha)$,we have $AB = BA$.

If $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \cos \beta & -\sin \beta \\ \sin \beta & \cos \beta \end{bmatrix}$,then the correct relation is

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