If A=begin{bmatrix} 2 & -3 5 & -7 end{bmatri | vedclass

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(C) Given $A = \begin{bmatrix} 2 & -3 \ 5 & -7 \end{bmatrix}$.First,find the determinant $|A| = (2)(-7) - (-3)(5) = -14 + 15 = 1$.The inverse of a $2

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$3\begin{bmatrix} 3 & -2 \ \frac{10}{3} & -3 \end{bmatrix}$

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(C) Given $A = \begin{bmatrix} 2 & -3 \ 5 & -7 \end{bmatrix}$.First,find the determinant $|A| = (2)(-7) - (-3)(5) = -14 + 15 = 1$.The inverse of a $2 	imes 2$ matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ is $A^{-1} = \frac{1}{|A|} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.So,$A^{-1} = \frac{1}{1} \begin{bmatrix} -7 & 3 \ -5 & 2 \end{bmatrix} = \begin{bmatrix} -7 & 3 \ -5 & 2 \end{bmatrix}$.Now,calculate $A - A^{-1} = \begin{bmatrix} 2 & -3 \ 5 & -7 \end{bmatrix} - \begin{bmatrix} -7 & 3 \ -5 & 2 \end{bmatrix}$.$A - A^{-1} = \begin{bmatrix} 2 - (-7) & -3 - 3 \ 5 - (-5) & -7 - 2 \end{bmatrix} = \begin{bmatrix} 9 & -6 \ 10 & -9 \end{bmatrix}$.Factoring out $3$,we get $3 \begin{bmatrix} 3 & -2 \ \frac{10}{3} & -3 \end{bmatrix}$.

If $A=\begin{bmatrix} 2 & -3 \\ 5 & -7 \end{bmatrix}$,then $A-A^{-1}=$

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