If $A = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$,then verify that $A^{\prime} A = I$.

(A) Given $A = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$.
The transpose of matrix $A$,denoted by $A^{\prime}$,is obtained by interchanging its rows and columns:
$A^{\prime} = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$.
Now,calculate the product $A^{\prime} A$:
$A^{\prime} A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix} \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$.
Performing matrix multiplication:
$A^{\prime} A = \begin{bmatrix} (\sin \alpha)(\sin \alpha) + (-\cos \alpha)(-\cos \alpha) & (\sin \alpha)(\cos \alpha) + (-\cos \alpha)(\sin \alpha) \\ (\cos \alpha)(\sin \alpha) + (\sin \alpha)(-\cos \alpha) & (\cos \alpha)(\cos \alpha) + (\sin \alpha)(\sin \alpha) \end{bmatrix}$.
Simplifying the terms using trigonometric identities $\sin^2 \alpha + \cos^2 \alpha = 1$:
$A^{\prime} A = \begin{bmatrix} \sin^2 \alpha + \cos^2 \alpha & \sin \alpha \cos \alpha - \sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha - \sin \alpha \cos \alpha & \cos^2 \alpha + \sin^2 \alpha \end{bmatrix}$.
$A^{\prime} A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.
Thus,it is verified that $A^{\prime} A = I$.