If A = begin{bmatrix} 2 & 1 7 & 4 end{bmatri | vedclass

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(B) Given $A = \begin{bmatrix} 2 & 1 \ 7 & 4 \end{bmatrix}$.First,calculate $A^2 = A 	imes A = \begin{bmatrix} 2 & 1 \ 7 & 4 \end{bmatrix} \begin{b

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$\frac{1}{4} \begin{bmatrix} -3 & 1 \ 7 & -1 \end{bmatrix}$

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(B) Given $A = \begin{bmatrix} 2 & 1 \ 7 & 4 \end{bmatrix}$.First,calculate $A^2 = A 	imes A = \begin{bmatrix} 2 & 1 \ 7 & 4 \end{bmatrix} \begin{bmatrix} 2 & 1 \ 7 & 4 \end{bmatrix} = \begin{bmatrix} 4+7 & 2+4 \ 14+28 & 7+16 \end{bmatrix} = \begin{bmatrix} 11 & 6 \ 42 & 23 \end{bmatrix}$.Next,calculate $5A = 5 \begin{bmatrix} 2 & 1 \ 7 & 4 \end{bmatrix} = \begin{bmatrix} 10 & 5 \ 35 & 20 \end{bmatrix}$.Then,$A^2 - 5A = \begin{bmatrix} 11 & 6 \ 42 & 23 \end{bmatrix} - \begin{bmatrix} 10 & 5 \ 35 & 20 \end{bmatrix} = \begin{bmatrix} 1 & 1 \ 7 & 3 \end{bmatrix}$.Let $B = A^2 - 5A = \begin{bmatrix} 1 & 1 \ 7 & 3 \end{bmatrix}$.The determinant $|B| = (1)(3) - (1)(7) = 3 - 7 = -4$.The inverse $B^{-1} = \frac{1}{|B|} 	ext{adj}(B) = \frac{1}{-4} \begin{bmatrix} 3 & -1 \ -7 & 1 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} -3 & 1 \ 7 & -1 \end{bmatrix}$.

If $A = \begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}$,then $(A^2 - 5A)^{-1}$ is

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