Question

If \( A = \begin{bmatrix} \sin \theta & \cos \theta \\ -\cos \theta & \sin \theta \end{bmatrix} \) and \( A + A' = I \), then \( \theta \) is equal to :

• A. \( 0 \)
• B. \( \frac{\pi}{6} \)
• C. \( \frac{\pi}{3} \)
• D. \( \pi \)

Hint:

When adding a matrix to its transpose, the diagonal elements always double (\( 2a_{ii} \)), while the off-diagonal elements become \( a_{ij} + a_{ji} \). If the matrix is skew-symmetric, the off-diagonal sum cancels out completely to zero.

Answer 1

The Correct Option is C

Concept:
The transpose of a matrix \(A\), denoted by \(A'\) (or \(A^T\)), is obtained by interchanging its rows and columns. The identity matrix of order \(2\times2\) is \[ I= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}. \] Using the given condition \[ A+A'=I, \] we compare the corresponding entries of the matrices to determine the value of \(\theta\). 

Step 1: Write the given matrix and find its transpose. The given matrix is \[ A= \begin{bmatrix} \sin\theta & \cos\theta\\ -\cos\theta & \sin\theta \end{bmatrix}. \] Its transpose is \[ A'= \begin{bmatrix} \sin\theta & -\cos\theta\\ \cos\theta & \sin\theta \end{bmatrix}. \] 

Step 2: Add the matrices \(A\) and \(A'\). \[ A+A' = \begin{bmatrix} \sin\theta & \cos\theta\\ -\cos\theta & \sin\theta \end{bmatrix} + \begin{bmatrix} \sin\theta & -\cos\theta\\ \cos\theta & \sin\theta \end{bmatrix}. \] Adding the corresponding elements, \[ A+A' = \begin{bmatrix} 2\sin\theta & 0\\ 0 & 2\sin\theta \end{bmatrix}. \] 

Step 3: Equate \(A+A'\) to the identity matrix. Since \[ A+A'=I, \] we have \[ \begin{bmatrix} 2\sin\theta & 0\\ 0 & 2\sin\theta \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}. \] Comparing the corresponding entries, \[ 2\sin\theta=1. \] Therefore, \[ \sin\theta=\frac{1}{2}. \] 

Step 4: Find the value of \(\theta\). Since \[ \sin\theta=\frac{1}{2}, \] the principal value of \(\theta\) is \[ \boxed{\theta=\frac{\pi}{6}}. \] Hence, the correct option is \[ \boxed{\text{(B) }\frac{\pi}{6}}. \]