Question

Let A = $\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$ and I = $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$. If AT + A = I, then

• A. $\theta = 2n\pi + \frac{\pi}{3}, n \in Z$
• B. $\theta = n\pi, n \in Z$
• C. $\theta = (2n + 1)\frac{\pi}{2}, n \in Z$
• D. $\theta = 2n\pi + \frac{\pi}{6}, n \in Z$

Hint:

The matrix A is a standard rotation matrix. Knowing its properties can sometimes be helpful. In this case, direct calculation is straightforward. Remember the general solutions for trigonometric equations: for $\cos x = \cos \alpha$, $x = 2n\pi \pm \alpha$. Always check if the provided options are a subset of your general solution.

Answer 1

The Correct Option is A

Step 1: Concept Identification:

The problem requires solving a matrix equation involving matrix \( A \), its transpose \( A^T \), and the identity matrix \( I \), to determine the general value of the angle \( \theta \).

Step 2: Methodological Outline:

1. Compute \( A^T \).
2. Compute the sum \( A^T + A \).
3. Equate \( A^T + A \) to \( I \).
4. Solve the resulting trigonometric equation for \( \theta \).

Step 3: Detailed Derivation:

Given matrix \( A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \).

The transpose of \( A \) is:

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

The sum \( A^T + A \) is:

\[ A^T + A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} + \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \] \[ A^T + A = \begin{bmatrix} 2\cos\theta & 0 \\ 0 & 2\cos\theta \end{bmatrix}. \]

The given condition is \( A^T + A = I \):

\[ \begin{bmatrix} 2\cos\theta & 0 \\ 0 & 2\cos\theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \]

Equating corresponding elements yields:

\[ 2\cos\theta = 1 \] \[ \cos\theta = \frac{1}{2} \]

The principal value of \( \theta \) is \( \frac{\pi}{3} \).

The general solution for \( \cos\theta = \cos\alpha \) is \( \theta = 2n\pi \pm \alpha \), where \( n \) is an integer.

Therefore, the general solution for \( \cos\theta = \frac{1}{2} \) is \( \theta = 2n\pi \pm \frac{\pi}{3} \), with \( n \in \mathbb{Z} \).

From the provided options, \( \theta = 2n\pi + \frac{\pi}{3} \) represents a subset of the general solution and is the most suitable choice.

Step 4: Conclusion:

The general solution for \( \theta \) is \( \theta = 2n\pi + \frac{\pi}{3} \), where \( n \in \mathbb{Z} \).