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If \( A=\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right] \) and \( A+A^{T}=I \),
where \( \mathrm{I} \) is the unit matrix of \( 2 \times 2 \& \mathrm{~A}^{\mathrm{T}} \) is the transpose of \( \mathrm{A} \), then the value of \( \theta \) is equal to
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where \( \mathrm{I} \) is the unit matrix of \( 2 \times 2 \& \mathrm{~A}^{\mathrm{T}} \) is the transpose of \( \mathrm{A} \), then the value of \( \theta \) is equal to
Solution:
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Verified Answer
The correct answer is:
\( \frac{\pi}{6} \)
Given that, $\mathrm{A}=\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]$
So, $\mathrm{A}^{\mathrm{T}}=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$
Now, $\mathrm{A}+\mathrm{A}^{\mathrm{T}}=\left[\begin{array}{ccc}2 \cos 2 \theta & 0 \\ 0 & 2 \cos 2 \theta\end{array}\right]$
$=2 \cos 2 \theta\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
Given that, $\mathrm{A}+\mathrm{A}^{\mathrm{T}}=\mathrm{I}$ so
$\Rightarrow(2 \cos 2 \theta) I=I$
$\Rightarrow 2 \cos 2 \theta=1 \Rightarrow \cos 2 \theta=\frac{1}{2}$
$\Rightarrow 2 \theta=\frac{\Pi}{3} \Rightarrow \theta=\frac{\Pi}{6}$
So, $\mathrm{A}^{\mathrm{T}}=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$
Now, $\mathrm{A}+\mathrm{A}^{\mathrm{T}}=\left[\begin{array}{ccc}2 \cos 2 \theta & 0 \\ 0 & 2 \cos 2 \theta\end{array}\right]$
$=2 \cos 2 \theta\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
Given that, $\mathrm{A}+\mathrm{A}^{\mathrm{T}}=\mathrm{I}$ so
$\Rightarrow(2 \cos 2 \theta) I=I$
$\Rightarrow 2 \cos 2 \theta=1 \Rightarrow \cos 2 \theta=\frac{1}{2}$
$\Rightarrow 2 \theta=\frac{\Pi}{3} \Rightarrow \theta=\frac{\Pi}{6}$
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