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GO Classes DPP | GATE CS, DA | Linear Algebra | Matrix

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4 4 votes

If\[
A=\begin{bmatrix}
\cos\alpha & \sin\alpha\\
-\sin\alpha & \cos\alpha
\end{bmatrix}
\]
and \(A+A^T=I\), where \(I\) is the identity matrix, then for \(0\le \alpha<2\pi\), the value of \(\alpha\) is

  1. \(\dfrac{\pi}{6},\ \dfrac{11\pi}{6}\)
     
  2. \(\dfrac{\pi}{3},\ \dfrac{5\pi}{3}\)
     
  3. \(\dfrac{\pi}{2},\ \dfrac{3\pi}{2}\)
     
  4. \(\dfrac{2\pi}{3},\ \dfrac{4\pi}{3}\)

3 Answers

0 0 votes
Given $A=\begin{bmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{bmatrix}$

First, find $A^T$:

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

Now,
$A+A^T=\begin{bmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{bmatrix}+\begin{bmatrix}\cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha\end{bmatrix}=\begin{bmatrix}2\cos\alpha & 0\\ 0 & 2\cos\alpha\end{bmatrix}$

Since $A+A^T=I$, we compare with $I=\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$

Therefore, $2\cos\alpha=1$

So, $\cos\alpha=\frac{1}{2}$

For $0\le \alpha<2\pi$, $\alpha=\frac{\pi}{3},\ \frac{5\pi}{3}$

Hence, the correct option is $\boxed{(B)\ \frac{\pi}{3},\ \frac{5\pi}{3}}$
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