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: Understanding the Concept:

We are given a matrix equation involving a matrix \( A \), its transpose \( A^T \), and the identity matrix \( I \). We need to solve this equation to find the general value of the angle \( \theta \).

Step 2: Key Formula or Approach:

1. Find the transpose of matrix \( A \), \( A^T \). 
2. Calculate the sum \( A^T + A \). 
3. Set this sum equal to the identity matrix \( I \). 
4. Solve the resulting trigonometric equation for \( \theta \).

Step 3: Detailed Explanation:

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

First, find the transpose of \( A \), which is obtained by interchanging rows and columns:

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

Now, add \( A^T \) and \( A \):

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

We are given that \( A^T + A = I \):

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

By equating the corresponding elements of the matrices, we get the equation:

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

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

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

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

Looking at the options provided, the option \( \theta = 2n\pi + \frac{\pi}{3} \) is one part of the general solution. It is the most appropriate choice among the given options.

Step 4: Final Answer:

The value of \( \theta \) is given by \( \theta = 2n\pi + \frac{\pi}{3} \), where \( n \in \mathbb{Z} \).