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

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(A) Given,$A = \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix}$.The transpose of matrix $A$ is $A^{\prime} = \be

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$I$

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(A) Given,$A = \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix}$.The transpose of matrix $A$ is $A^{\prime} = \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix}$.Now,calculate the product $A \cdot A^{\prime}$:$A \cdot A^{\prime} = \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix} \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix}$Performing matrix multiplication:$A \cdot A^{\prime} = \begin{bmatrix} (\cos 	heta)(\cos 	heta) + (\sin 	heta)(\sin 	heta) & (\cos 	heta)(-\sin 	heta) + (\sin 	heta)(\cos 	heta) \ (-\sin 	heta)(\cos 	heta) + (\cos 	heta)(\sin 	heta) & (-\sin 	heta)(-\sin 	heta) + (\cos 	heta)(\cos 	heta) \end{bmatrix}$Using the trigonometric identity $\cos^{2} 	heta + \sin^{2} 	heta = 1$:$A \cdot A^{\prime} = \begin{bmatrix} \cos^{2} 	heta + \sin^{2} 	heta & -\sin 	heta \cos 	heta + \sin 	heta \cos 	heta \ -\sin 	heta \cos 	heta + \sin 	heta \cos 	heta & \sin^{2} 	heta + \cos^{2} 	heta \end{bmatrix}$$A \cdot A^{\prime} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I$Thus,$A \cdot A^{\prime} = I$,which is the identity matrix.

If $A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$,then $A \cdot A^{\prime}$ is

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