Let $A=\begin{bmatrix}\cos θ&-\sin θ\\\sin θ&\cos θ\end{bmatrix}$ and $I=\begin{bmatrix}1&0\\0&1\end{bmatrix}$. If $A^T+A=I$, then
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $θ=2n\pi+\frac{\pi}{3},n∈Z$
$A=\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix},\quad A^T=\begin{pmatrix}\cos\theta&\sin\theta\\ -\sin\theta&\cos\theta\end{pmatrix}$
$A^T+A=\begin{pmatrix}2\cos\theta&0\\0&2\cos\theta\end{pmatrix}$
$A^T+A=I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$
$2\cos\theta=1$
$\cos\theta=\frac{1}{2}$
$\theta=2n\pi\pm\frac{\pi}{3},\;n\in\mathbb{Z}$
The given condition holds for $\theta=2n\pi\pm\frac{\pi}{3},\;n\in\mathbb{Z}$.