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 $A^T + A = I$, then
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.
Important Questions from Matrix Algebra
- If A = \( \left[\begin{array}{cc}0 & 1 \\ −1 & 0\end{array}\right]\) and (aI 2 + bA)2 = A, then
If A = \(\left[\begin{array}{cc}2 & −3 \\3 & 5\end{array}\right]\), then which of the following statements are correct?
A. A is a square matrix
B. A−1 exists
C. A is a symmetric matrix
D. |A| = 19
E. A is a null matrix
Choose the correct answer from the options given below.
If A is Square Matrix of order 3, then product of A and its transpose is
What is the transformation matrix M that transforms a square in the xy-plane defined by (1, 1) T, (-1, 1) T, (-1, -1) T and (1, -1) T to a parallelogram whose corresponding vertices are (2, 1) T, (0, 1) T, (-2, -1) T and (0, -1) T?
Let \(A = \left[ {\begin{array}{*{20}{c}} 1&1&0\\ 0&1&0\\ 1&1&0\\ 0&0&1 \end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}} 1&0&0&0\\ 0&1&1&0\\ 1&0&1&1\\ \end{array}} \right]\) Find the boolean product A ⊙ B of the two matrices.
