If A = f(x) = begin{bmatrix} cos x & sin x & | vedclass

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(A) Given $A = f(x) = \begin{bmatrix} \cos x & \sin x & 0 \ -\sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix}$.First,calculate the determinant $|A|$:$|

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$f(-x)$

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(A) Given $A = f(x) = \begin{bmatrix} \cos x & \sin x & 0 \ -\sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix}$.First,calculate the determinant $|A|$:$|A| = \cos x(\cos x - 0) - \sin x(-\sin x - 0) + 0 = \cos^2 x + \sin^2 x = 1$.Since $|A| = 1$,$A^{-1} = 	ext{adj}(A)$.The cofactor matrix $C$ is calculated by finding the cofactor of each element.For a matrix of this form (orthogonal matrix),the inverse is equal to its transpose:$A^{-1} = A^T = \begin{bmatrix} \cos x & -\sin x & 0 \ \sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix}$.Now,evaluate $f(-x)$:$f(-x) = \begin{bmatrix} \cos(-x) & \sin(-x) & 0 \ -\sin(-x) & \cos(-x) & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos x & -\sin x & 0 \ \sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix}$.Thus,$A^{-1} = f(-x)$.

If $A = f(x) = \begin{bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$,then $A^{-1}$ is equal to:

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