Question

If the value of \( \cos \alpha \) is \( \frac{\sqrt{3}}{2} \), then \( A + A = I \), where \[ A = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}. \]

Hint:

For matrix equations, use properties of transpose and identity matrices to simplify calculations.

Answer 1

Step 1: Compute the transpose of matrix \( A \). The transpose of \( A \), denoted as \( A \), is obtained by swapping the off-diagonal elements: \[ A = \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}. \] Step 2: Add \( A \) and \( A \). \[ A + A = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix} + \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}. \] Adding corresponding elements: \[ A + A = \begin{bmatrix} \sin\alpha + \sin\alpha & -\cos\alpha + \cos\alpha \\ \cos\alpha - \cos\alpha & \sin\alpha + \sin\alpha \end{bmatrix}. \] Simplifying: \[ A + A = \begin{bmatrix} 2\sin\alpha & 0 \\ 0 & 2\sin\alpha \end{bmatrix}. \] Step 3: Equating to the identity matrix. The identity matrix is: \[ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \] Thus, equating: \[ \begin{bmatrix} 2\sin\alpha & 0 \\ 0 & 2\sin\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \] From which: \[ 2\sin\alpha = 1 \Rightarrow \sin\alpha = \frac{1}{2}. \] Step 4: Finding \( \cos\alpha \). Using the Pythagorean identity: \[ \sin^2\alpha + \cos^2\alpha = 1. \] Substituting \( \sin\alpha = \frac{1}{2} \): \[ \left( \frac{1}{2} \right)^2 + \cos^2\alpha = 1. \] \[ \frac{1}{4} + \cos^2\alpha = 1. \] \[ \cos^2\alpha = \frac{3}{4}. \] \[ \cos\alpha = \frac{\sqrt{3}}{2}. \] Conclusion: The correct answer is \( \mathbf{\frac{\sqrt{3}}{2}} \).