The multiplicative inverse of A = begin{bmatr | vedclass

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(B) To find the multiplicative inverse $A^{-1}$,we use the formula $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.First,calculate the determinant $|A|$:$|A| =

Vedclass · Accepted Answer

$\begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix}$

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(B) To find the multiplicative inverse $A^{-1}$,we use the formula $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.First,calculate the determinant $|A|$:$|A| = (\cos 	heta)(\cos 	heta) - (-\sin 	heta)(\sin 	heta) = \cos^2 	heta + \sin^2 	heta = 1$.Next,find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:$	ext{adj}(A) = \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix}$.Therefore,$A^{-1} = \frac{1}{1} \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix} = \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix}$.

The multiplicative inverse of $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$ is

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