Question

If $A = \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix}$ and $A + A' = I$, then the value of $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is:

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{3}$
• C. 0
• D. $-\frac{\pi}{2}$

Hint:

For matrix equations like $A+A'=I$, you only need to calculate a single diagonal entry to set up your equation: $a_{11} + a_{11} = 1 \Rightarrow 2\cos x = 1$. This saves time!

Answer 1

The Correct Option is B

Concept: If \[ A+A^T=I, \] where \(A^T\) denotes the transpose of \(A\) and \[ I= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}, \] then we first find \(A^T\), add it to \(A\), and compare the result with the identity matrix to determine the value of \(x\).

Step 1: Find the transpose of \(A\).
Given, \[ A= \begin{bmatrix} \cos x & \sin x\\ -\sin x & \cos x \end{bmatrix}. \] The transpose of \(A\) is \[ A^T= \begin{bmatrix} \cos x & -\sin x\\ \sin x & \cos x \end{bmatrix}. \]

Step 2: Compute \(A+A^T\).
\[ A+A^T= \begin{bmatrix} \cos x & \sin x\\ -\sin x & \cos x \end{bmatrix} + \begin{bmatrix} \cos x & -\sin x\\ \sin x & \cos x \end{bmatrix}. \] Adding corresponding entries, \[ A+A^T= \begin{bmatrix} 2\cos x & 0\\ 0 & 2\cos x \end{bmatrix}. \]

Step 3: Compare with the identity matrix.
Since \[ A+A^T=I, \] we have \[ \begin{bmatrix} 2\cos x & 0\\ 0 & 2\cos x \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}. \] Therefore, \[ 2\cos x=1, \] which gives \[ \cos x=\frac{1}{2}. \]

Step 4: Find the value of \(x\).
Since \[ x\in\left[-\frac{\pi}{2},\,\frac{\pi}{2}\right], \] the only value satisfying \[ \cos x=\frac{1}{2} \] is \[ x=\frac{\pi}{3}. \]

Hence,

\[ \boxed{x=\frac{\pi}{3}.} \]