Question

Let \( A = \begin{pmatrix} \sin \theta & \cos \theta \\ \cos \theta & \sin \theta \end{pmatrix} \) and \( A + A^T - 2I = 0 \), where \( A^T \) is the transpose of \( A \) and \( I \) is the identity matrix. The value of \( \theta \) (in degrees) is ............

Hint:

When solving matrix equations, break down each element and solve for the variables involved.

Answer 1

Correct Answer: 90

Step 1: Understanding the given condition. 
We are given the matrix: \[ A = \begin{pmatrix} \sin \theta & \cos \theta \\ \cos \theta & \sin \theta \end{pmatrix} \] and the condition: \[ A + A^T - 2I = 0, \] where \( A^T \) is the transpose of \( A \), and \( I \) is the identity matrix. 

tep 2: Taking the transpose of matrix \( A \). 

The transpose of \( A \) is: \[ A^T = \begin{pmatrix} \sin \theta & \cos \theta \\ \cos \theta & \sin \theta \end{pmatrix}. \] 

Step 3: Substituting into the equation. 
Substituting \( A \) and \( A^T \) into the given equation: \[ \begin{pmatrix} \sin \theta & \cos \theta \\ \cos \theta & \sin \theta \end{pmatrix} + \begin{pmatrix} \sin \theta & \cos \theta \\ \cos \theta & \sin \theta \end{pmatrix} - 2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = 0. \] Simplifying: \[ \begin{pmatrix} 2\sin \theta - 2 & 2\cos \theta \\ 2\cos \theta & 2\sin \theta - 2 \end{pmatrix} = 0. \] 

Step 4: Solving the equation. 
For the matrix to be zero, each element must be zero: \[ 2\sin \theta - 2 = 0 \text{and} 2\cos \theta = 0. \] Solving these gives: \[ \sin \theta = 1 \text{and} \cos \theta = 0. \] 

Step 5: Conclusion.
The value of \( \theta \) that satisfies these equations is \( \theta = 90^\circ \).

Final Answer: \( \boxed{90^\circ} \).