If A = begin{bmatrix} 2 & -3 5 & 4 end{bmatr | vedclass

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(B) To find the inverse of a $2 	imes 2$ matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$,we use the formula $A^{-1} = \frac{1}{|A|} 	ext{a

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$\begin{bmatrix} \frac{4}{23} & \frac{3}{23} \ -\frac{5}{23} & \frac{2}{23} \end{bmatrix}$

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(B) To find the inverse of a $2 	imes 2$ matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$,we use the formula $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$,where $|A| = ad - bc$ and $	ext{adj}(A) = \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Given $A = \begin{bmatrix} 2 & -3 \ 5 & 4 \end{bmatrix}$.First,calculate the determinant $|A| = (2)(4) - (-3)(5) = 8 + 15 = 23$.Next,find the adjoint of $A$: $	ext{adj}(A) = \begin{bmatrix} 4 & 3 \ -5 & 2 \end{bmatrix}$.Therefore,$A^{-1} = \frac{1}{23} \begin{bmatrix} 4 & 3 \ -5 & 2 \end{bmatrix} = \begin{bmatrix} \frac{4}{23} & \frac{3}{23} \ -\frac{5}{23} & \frac{2}{23} \end{bmatrix}$.Thus,the correct option is $B$.

If $A = \begin{bmatrix} 2 & -3 \\ 5 & 4 \end{bmatrix}$,then $A^{-1} = $ . . . . . . .

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