Question

If \( A = \begin{bmatrix} \cos \theta & \sin \theta & 0 \\ \sin \theta & -\cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \), then the value of \( AA' \) is:

• A. \( \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \)
• B. \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
• C. \( \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \)
• D. \( \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 1 & 1 & 0 \end{bmatrix} \)

Hint:

If a matrix \(A\) is orthogonal, then \(AA' = I\). Check orthogonality by verifying that rows (or columns) are orthonormal.

Answer 1

The Correct Option is B

Use the property \( AA' = I \) for orthogonal matrices.
We are given: \[ A = \begin{bmatrix} \cos \theta & \sin \theta & 0 \\ \sin \theta & -\cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \] Compute \( A A' \) using matrix multiplication. Since the dot product of each row and corresponding column gives identity: \[ AA' = I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]