Question

If \[A = \begin{bmatrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta\end{bmatrix}\]  then A^T + A = I₂, if

Options

• θ = n π, n ∈ Z

• θ     = (2n + 1) \[\frac{\pi}{2}\] n ∈ Z 

• θ = 2n π +\[\frac{\pi}{3}\] n ∈ Z

• none of these

Answer 1

θ = 2nπ + \[\frac{\pi}{3}\]n ∈ Z

\[Here, \]

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

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

\[Now, \]

\[ A^T + A = I_2 \]

\[ \Rightarrow \begin{bmatrix}\cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{bmatrix} + \begin{bmatrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]

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

\[ \Rightarrow 2\cos \theta = 1\]

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

\[ \Rightarrow \cos \theta = \cos\frac{\pi}{3}\]

\[ \Rightarrow \theta = 2n\pi \pm \frac{\pi}{3} \left( n \in Z \right)\]