Question

If $A=\left[\begin{array}{cc}\cos \mathrm{\theta }& -\sin \mathrm{\theta }\\ \sin \mathrm{\theta }& \cos \mathrm{\theta }\end{array}\right]$, then A^T + A = I₂, if
(a) θ = n π, n ∈ Z
(b) θ = (2n + 1)$\frac{\mathrm{\pi }}{2}$, n ∈ Z
(c) θ = 2n π + $\frac{\mathrm{\pi }}{3}$, n ∈ Z
(d) none of these

Answer 1

(c) θ = 2nπ + $\frac{\mathrm{\pi }}{3}$, n ∈ Z


$$\begin{gathered} \mathrm{Here}, \\ A=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \\ \Rightarrow A^T=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \\ \mathrm{Now}, \\ {A}^T+A=I_2 \\ \Rightarrow \left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]+\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}2\cos \theta & 0\\ 0& 2\cos \theta \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow 2\cos \theta =1 \\ \Rightarrow \cos \theta =\frac{1}{2} \\ \Rightarrow \cos \theta =\cos \frac{\mathrm{\pi }}{3} \\ \Rightarrow \theta =2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3}\left(n\in Z\right) \end{gathered}$$