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

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(D) Given $A = \begin{bmatrix} 2 & 3 \ 5 & -2 \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = (2)(-2) - (3)(5) = -4 - 15 = -19$.Next,fin

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$\frac{1}{19}$

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(D) Given $A = \begin{bmatrix} 2 & 3 \ 5 & -2 \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = (2)(-2) - (3)(5) = -4 - 15 = -19$.Next,find the adjoint of $A$:$	ext{adj}(A) = \begin{bmatrix} -2 & -3 \ -5 & 2 \end{bmatrix}$.We know that $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$:$A^{-1} = \frac{1}{-19} \begin{bmatrix} -2 & -3 \ -5 & 2 \end{bmatrix} = \frac{1}{19} \begin{bmatrix} 2 & 3 \ 5 & -2 \end{bmatrix}$.Since $A^{-1} = KA$ and $A = \begin{bmatrix} 2 & 3 \ 5 & -2 \end{bmatrix}$,we have:$KA = \frac{1}{19} A$.Therefore,$K = \frac{1}{19}$.

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

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