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

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(D) Given $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \ \sin \alpha & \cos \alpha \end{bmatrix}$.The transpose of matrix $A$ is $A^{\prime} = \be

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$\frac{\pi}{3}$

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(D) Given $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \ \sin \alpha & \cos \alpha \end{bmatrix}$.The transpose of matrix $A$ is $A^{\prime} = \begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix}$.Given the condition $A + A^{\prime} = I$,where $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$ is the identity matrix.Adding $A$ and $A^{\prime}$:$\begin{bmatrix} \cos \alpha & -\sin \alpha \ \sin \alpha & \cos \alpha \end{bmatrix} + \begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$$\begin{bmatrix} 2 \cos \alpha & 0 \ 0 & 2 \cos \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$Comparing the corresponding elements,we get:$2 \cos \alpha = 1$$\cos \alpha = \frac{1}{2}$Since $\cos \alpha = \frac{1}{2}$,the value of $\alpha$ is $\frac{\pi}{3}$.

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

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