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

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(B) Given $A = \begin{bmatrix} 2 & -2 \ 4 & 3 \end{bmatrix}$.First,we find the determinant of $A$:$|A| = (2)(3) - (-2)(4) = 6 + 8 = 14$.Since $|A|

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

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(B) Given $A = \begin{bmatrix} 2 & -2 \ 4 & 3 \end{bmatrix}$.First,we find the determinant of $A$:$|A| = (2)(3) - (-2)(4) = 6 + 8 = 14$.Since $|A| 
eq 0$,$A^{-1}$ exists.The adjoint of $A$ is obtained 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}$.Using the formula $A^{-1} = \frac{1}{|A|} 	ext{Adj } A$:$A^{-1} = \frac{1}{14} \begin{bmatrix} 3 & 2 \ -4 & 2 \end{bmatrix}$.

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

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