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

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(D) Given $A = \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = (\

Vedclass · Accepted Answer

$A$ is not invertible.

Vedclass · Answer

(D) Given $A = \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = (\cos 	heta)(\cos 	heta) - (-\sin 	heta)(\sin 	heta) = \cos^2 	heta + \sin^2 	heta = 1$.Since $|A| = 1 
eq 0$,the matrix $A$ is invertible.Next,check if $A$ is orthogonal:$A A^T = \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix} \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix} = \begin{bmatrix} \cos^2 	heta + \sin^2 	heta & \cos 	heta \sin 	heta - \sin 	heta \cos 	heta \ \sin 	heta \cos 	heta - \cos 	heta \sin 	heta & \sin^2 	heta + \cos^2 	heta \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I$.Since $A A^T = I$,$A$ is an orthogonal matrix. The transpose of an orthogonal matrix is also orthogonal,so $A'$ is also orthogonal.Therefore,the statement '$A$ is not invertible' is incorrect.

If $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$,then which of the following statements is not correct?

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