Verify that A^TA = I₂

Given:

\[ A = \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix} \]

Step 1: Find A^T

\[ A^T = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix} \]

Step 2: Compute A^TA

\[ A^T A = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix} \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix} \]

\[ A^T A = \begin{bmatrix} \sin^2\alpha + \cos^2\alpha & \sin\alpha\cos\alpha – \cos\alpha\sin\alpha \\ \cos\alpha\sin\alpha – \sin\alpha\cos\alpha & \cos^2\alpha + \sin^2\alpha \end{bmatrix} \]

Step 3: Simplify

\[ \sin^2\alpha + \cos^2\alpha = 1,\quad \sin\alpha\cos\alpha – \cos\alpha\sin\alpha = 0 \]

\[ A^T A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I_2 \]

Conclusion:

\[ A^T A = I_2 \]

Hence Verified. A is an orthogonal matrix.