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

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(B) Given $A = \begin{bmatrix} \sin \alpha & -\cos \alpha \ \cos \alpha & \sin \alpha \end{bmatrix}$.Then the transpose $A^{\prime} = \begin{bmatrix}

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$\frac{\pi}{6}$

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(B) Given $A = \begin{bmatrix} \sin \alpha & -\cos \alpha \ \cos \alpha & \sin \alpha \end{bmatrix}$.Then the transpose $A^{\prime} = \begin{bmatrix} \sin \alpha & \cos \alpha \ -\cos \alpha & \sin \alpha \end{bmatrix}$.Given $A + A^{\prime} = I$,where $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.Adding $A$ and $A^{\prime}$:$A + A^{\prime} = \begin{bmatrix} \sin \alpha + \sin \alpha & -\cos \alpha + \cos \alpha \ \cos \alpha - \cos \alpha & \sin \alpha + \sin \alpha \end{bmatrix} = \begin{bmatrix} 2 \sin \alpha & 0 \ 0 & 2 \sin \alpha \end{bmatrix}$.Equating to $I$:$\begin{bmatrix} 2 \sin \alpha & 0 \ 0 & 2 \sin \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.This implies $2 \sin \alpha = 1$,so $\sin \alpha = \frac{1}{2}$.The principal value of $\alpha$ for $\sin \alpha = \frac{1}{2}$ is $\alpha = \frac{\pi}{6}$.

If $A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$ and $A + A^{\prime} = I$,then the value of $\alpha$ is . . . . . . .

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