MCQ
If $\text{A}=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix},$ then $A^{\top}+A=I_2$, if :
  • A
    $\theta=\text{n}\pi,\text{n}\in\text{Z}$
  • B
    $\theta=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
  • $\theta=2\text{n}\pi+\frac{\pi}{3},\text{n}\in\text{Z}$
  • D
    None of these

Answer

Correct option: C.
$\theta=2\text{n}\pi+\frac{\pi}{3},\text{n}\in\text{Z}$
Here,
$\text{A}=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}$
Now,
$\text{A}^\text{T}+\text{A}=\text{I}_2$
$\Rightarrow\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}+\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\cos\theta&0\\0&2\cos\theta\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow2\cos\theta=1$
$\Rightarrow\cos\theta=\frac{1}{2}$
$\Rightarrow\cos\theta=\cos\frac{\pi}{3}$
$\Rightarrow\theta=2\text{n}\pi\pm\frac{\pi}{3}$ $(\text{n}\in\text{Z})$

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