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(A) 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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$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(A) 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 the condition $A+A^{\prime} = I$,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 this to the identity matrix $I$:$\begin{bmatrix} 2\sin \alpha & 0 \ 0 & 2\sin \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.This implies $2\sin \alpha = 1$,so $\sin \alpha = \frac{1}{2}$.Since $\sin^2 \alpha + \cos^2 \alpha = 1$,we have $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - (\frac{1}{2})^2 = 1 - \frac{1}{4} = \frac{3}{4}$.Therefore,$\cos \alpha = \pm \frac{\sqrt{3}}{2}$.Looking at the options,$\frac{\sqrt{3}}{2}$ is provided.

The value of $\cos \alpha$ is . . . . . . if $A+A^{\prime}=I$,where $A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$.

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