For a given matrix A = begin{bmatrix} cos the | vedclass

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(C) matrix $A$ is orthogonal if $AA^T = I$. Calculating $A^T = \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix}$.

Vedclass · Accepted Answer

$A$ is an orthogonal matrix for $	heta \in \mathbb{R}$

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(C) matrix $A$ is orthogonal if $AA^T = I$. Calculating $A^T = \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix}$. Then $AA^T = \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix} \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix} = \begin{bmatrix} \cos^2 	heta + \sin^2 	heta & 0 \ 0 & \sin^2 	heta + \cos^2 	heta \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I$. Since $AA^T = I$,the matrix $A$ is an orthogonal matrix for all $	heta \in \mathbb{R}$. Thus,option $C$ is correct.

For a given matrix $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$,which of the following statements holds true?

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