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

Question

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

Prepp Team · Accepted Answer

Matrix Equation Solution: $A^T + A = I$

This problem requires us to find the value of $\theta$ by solving a matrix equation involving the transpose of a given matrix $A$ and the identity matrix $I$. We need to determine which of the given options for $\theta$ satisfies the condition $A^T + A = I$.

Given Matrices and Condition

We are provided with the following matrices:

$A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$ $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

The condition that needs to be satisfied is:

$A^T + A = I$

Calculating the Transpose of Matrix A

The transpose of a matrix, denoted as $A^T$, is obtained by interchanging its rows and columns. Applying this to matrix $A$:

$A^T = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$

Performing Matrix Addition: $A^T + A$

Next, we add the matrix $A^T$ to matrix $A$. Matrix addition involves adding the corresponding elements of the matrices:

$A^T + A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} + \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$

Performing the element-wise addition:

$A^T + A = \begin{bmatrix} \cos\theta + \cos\theta & \sin\theta + (-\sin\theta) \\ -\sin\theta + \sin\theta & \cos\theta + \cos\theta \end{bmatrix}$

Simplifying the elements:

$A^T + A = \begin{bmatrix} 2\cos\theta & 0 \\ 0 & 2\cos\theta \end{bmatrix}$

Equating with the Identity Matrix

The problem states that $A^T + A = I$. By substituting the result from the previous step, we get:

$\begin{bmatrix} 2\cos\theta & 0 \\ 0 & 2\cos\theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

For two matrices to be equal, all their corresponding elements must be equal. This leads to the following equation involving $\cos\theta$:

$2\cos\theta = 1$

Solving the Trigonometric Equation for $\theta$

We need to solve the equation $2\cos\theta = 1$ for $\theta$. Dividing both sides by 2 gives:

$\cos\theta = \frac{1}{2}$

The principal value of $\theta$ for which $\cos\theta = \frac{1}{2}$ is $\frac{\pi}{3}$. The general solution for the equation $\cos\theta = \cos\alpha$ is given by $\theta = 2n\pi \pm \alpha$, where $n$ is any integer ($n \in Z$).

Substituting $\alpha = \frac{\pi}{3}$, the general solution for $\cos\theta = \frac{1}{2}$ is:

$\theta = 2n\pi \pm \frac{\pi}{3}$, where $n \in Z$.

Analyzing the Options

Now, we compare our derived general solution with the given options:

Option 1: $\theta = 2n\pi + \frac{\pi}{3}$, $n \in Z$. This option represents one set of solutions within the general solution $\theta = 2n\pi \pm \frac{\pi}{3}$. For any integer $n$, this value of $\theta$ yields $\cos\theta = \frac{1}{2}$.
Option 2: $\theta = n\pi$, $n \in Z$. For these values, $\cos\theta$ is either $1$ or $-1$, which does not satisfy $\cos\theta = \frac{1}{2}$.
Option 3: $\theta = (2n + 1)\frac{\pi}{3}$, $n \in Z$. This option includes angles like $\pi$ (when $n=1$), for which $\cos\pi = -1$. Therefore, it does not consistently satisfy the condition $\cos\theta = \frac{1}{2}$.
Option 4: $\theta = 2n\pi + \frac{\pi}{2}$, $n \in Z$. For these values, $\cos\theta = 0$, which does not satisfy $\cos\theta = \frac{1}{2}$.

Option 1 correctly identifies a subset of the angles that satisfy the equation $A^T + A = I$. The calculation confirms that for $\theta = 2n\pi + \frac{\pi}{3}$, the condition holds true.

Prepp Team · Answer

$	heta = n\pi$, $n \in Z$

Prepp Team · Answer

$	heta = (2n + 1)\frac{\pi}{3}$, $n \in Z$

Prepp Team · Answer

$	heta = 2n\pi + \frac{\pi}{2}$, $n \in Z$