If A=begin{bmatrix} cos 2 theta & -sin 2 the | vedclass

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(A) Given that $ A = \begin{bmatrix} \cos 2 	heta & -\sin 2 	heta \ \sin 2 	heta & \cos 2 	heta \end{bmatrix} $. The transpose of $ A $ is $ A^{T

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$ \frac{\pi}{6} $

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(A) Given that $ A = \begin{bmatrix} \cos 2 	heta & -\sin 2 	heta \ \sin 2 	heta & \cos 2 	heta \end{bmatrix} $. The transpose of $ A $ is $ A^{T} = \begin{bmatrix} \cos 2 	heta & \sin 2 	heta \ -\sin 2 	heta & \cos 2 	heta \end{bmatrix} $. Adding $ A $ and $ A^{T} $,we get: $ A+A^{T} = \begin{bmatrix} \cos 2 	heta + \cos 2 	heta & -\sin 2 	heta + \sin 2 	heta \ \sin 2 	heta - \sin 2 	heta & \cos 2 	heta + \cos 2 	heta \end{bmatrix} = \begin{bmatrix} 2 \cos 2 	heta & 0 \ 0 & 2 \cos 2 	heta \end{bmatrix} $. This can be written as $ (2 \cos 2 	heta) I $,where $ I $ is the identity matrix. Given $ A+A^{T} = I $,we have $ (2 \cos 2 	heta) I = I $. Comparing the elements,we get $ 2 \cos 2 	heta = 1 $,which implies $ \cos 2 	heta = \frac{1}{2} $. Since $ \cos \frac{\pi}{3} = \frac{1}{2} $,we have $ 2 	heta = \frac{\pi}{3} $,which gives $ 	heta = \frac{\pi}{6} $.

If $ A=\begin{bmatrix} \cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{bmatrix} $ and $ A+A^{T}=I $,where $ I $ is the $ 2 \times 2 $ identity matrix and $ A^{T} $ is the transpose of $ A $,then the value of $ \theta $ is equal to

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