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(D) Since $A$ is an invertible matrix,$A^{-1}$ exists and $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.As matrix $A$ is of order $2$,let $A = \begin{bmatrix

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$\frac{1}{\det(A)}$

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(D) Since $A$ is an invertible matrix,$A^{-1}$ exists and $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.As matrix $A$ is of order $2$,let $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$.Then,$|A| = ad - bc$ and $	ext{adj}(A) = \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Now,$A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \begin{bmatrix} \frac{d}{|A|} & \frac{-b}{|A|} \ \frac{-c}{|A|} & \frac{a}{|A|} \end{bmatrix}$.Therefore,$\det(A^{-1}) = \begin{vmatrix} \frac{d}{|A|} & \frac{-b}{|A|} \ \frac{-c}{|A|} & \frac{a}{|A|} \end{vmatrix}$.$= \frac{1}{|A|^2} \begin{vmatrix} d & -b \ -c & a \end{vmatrix}$.$= \frac{1}{|A|^2} (ad - bc)$.$= \frac{1}{|A|^2} \cdot |A| = \frac{1}{|A|}$.Therefore,$\det(A^{-1}) = \frac{1}{\det(A)}$.Hence,the correct answer is $D$.

If $A$ is an invertible matrix of order $2$,then $\det(A^{-1})$ is equal to

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