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

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

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$\begin{bmatrix} 25 & -15 \ 25 & -20 \end{bmatrix}$

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(C) Given $A = \begin{bmatrix} 2 & -3 \ 5 & -7 \end{bmatrix}$.First,we find the determinant $|A| = (2)(-7) - (-3)(5) = -14 + 15 = 1$.Since $|A| 
eq 0$,$A^{-1}$ exists.The inverse is given by $A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{1} \begin{bmatrix} -7 & 3 \ -5 & 2 \end{bmatrix} = \begin{bmatrix} -7 & 3 \ -5 & 2 \end{bmatrix}$.Now,calculate $2A = 2 \begin{bmatrix} 2 & -3 \ 5 & -7 \end{bmatrix} = \begin{bmatrix} 4 & -6 \ 10 & -14 \end{bmatrix}$.Next,calculate $3A^{-1} = 3 \begin{bmatrix} -7 & 3 \ -5 & 2 \end{bmatrix} = \begin{bmatrix} -21 & 9 \ -15 & 6 \end{bmatrix}$.Finally,$2A - 3A^{-1} = \begin{bmatrix} 4 & -6 \ 10 & -14 \end{bmatrix} - \begin{bmatrix} -21 & 9 \ -15 & 6 \end{bmatrix} = \begin{bmatrix} 4 - (-21) & -6 - 9 \ 10 - (-15) & -14 - 6 \end{bmatrix} = \begin{bmatrix} 25 & -15 \ 25 & -20 \end{bmatrix}$.

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

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