If A = begin{bmatrix} 2 & -2 4 & 3 end{bmatr | vedclass

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(D) The inverse of a $2 	imes 2$ matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ is given by $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$,where $

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$\frac{1}{14} \begin{bmatrix} 3 & 2 \ -4 & 2 \end{bmatrix}$

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(D) The inverse of a $2 	imes 2$ matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ is given by $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$,where $|A| = ad - bc$ and $	ext{adj}(A) = \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Given $A = \begin{bmatrix} 2 & -2 \ 4 & 3 \end{bmatrix}$.First,calculate the determinant $|A| = (2)(3) - (-2)(4) = 6 - (-8) = 6 + 8 = 14$.Next,find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:$	ext{adj}(A) = \begin{bmatrix} 3 & 2 \ -4 & 2 \end{bmatrix}$.Therefore,$A^{-1} = \frac{1}{14} \begin{bmatrix} 3 & 2 \ -4 & 2 \end{bmatrix}$.This matches option $D$.

If $A = \begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}$,then $A^{-1} = $ . . . . . . .

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