If A = begin{bmatrix} 1 & 2 3 & -5 end{bmatr | vedclass

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(B) Given $A = \begin{bmatrix} 1 & 2 \ 3 & -5 \end{bmatrix}$.First,calculate the determinant $|A| = (1)(-5) - (2)(3) = -5 - 6 = -11$.Next,find the ad

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$\begin{bmatrix} \frac{5}{11} & \frac{2}{11} \ \frac{3}{11} & -\frac{1}{11} \end{bmatrix}$

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(B) Given $A = \begin{bmatrix} 1 & 2 \ 3 & -5 \end{bmatrix}$.First,calculate the determinant $|A| = (1)(-5) - (2)(3) = -5 - 6 = -11$.Next,find the adjoint of $A$,denoted as $adj(A)$. For a $2 	imes 2$ matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$,the adjoint is $\begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Thus,$adj(A) = \begin{bmatrix} -5 & -2 \ -3 & 1 \end{bmatrix}$.The inverse is given by $A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-11} \begin{bmatrix} -5 & -2 \ -3 & 1 \end{bmatrix} = \begin{bmatrix} \frac{5}{11} & \frac{2}{11} \ \frac{3}{11} & -\frac{1}{11} \end{bmatrix}$.

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

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