A=left[begin{array}{rr}cos theta & -sin theta | vedclass

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(B) Given,$A=\left[\begin{array}{rr}\cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta\end{array}ight]$.Since $AB=BA=I$,$B$ is the inverse of $

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$\left[\begin{array}{rr}\cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta\end{array}ight]$

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(B) Given,$A=\left[\begin{array}{rr}\cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta\end{array}ight]$.Since $AB=BA=I$,$B$ is the inverse of $A$,i.e.,$B=A^{-1}$.The determinant of $A$ is $|A| = \cos^2 	heta - (-\sin^2 	heta) = \cos^2 	heta + \sin^2 	heta = 1$.The inverse of a $2 	imes 2$ matrix $\left[\begin{array}{rr}a & b \ c & d\end{array}ight]$ is $\frac{1}{ad-bc} \left[\begin{array}{rr}d & -b \ -c & a\end{array}ight]$.Applying this to $A$,we get $B = A^{-1} = \frac{1}{1} \left[\begin{array}{rr}\cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta\end{array}ight]$.Thus,$B = \left[\begin{array}{rr}\cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta\end{array}ight]$.

$A=\left[\begin{array}{rr}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$ and $AB=BA=I$,then $B$ is equal to

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