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

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(A) Given $A = \begin{bmatrix} 2 & -1 \ -1 & 3 \end{bmatrix}$.First,calculate $A^2 = A 	imes A = \begin{bmatrix} 2 & -1 \ -1 & 3 \end{bmatrix} \beg

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$\frac{1}{95} \begin{bmatrix} 7 & 3 \ 3 & 4 \end{bmatrix}$

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(A) Given $A = \begin{bmatrix} 2 & -1 \ -1 & 3 \end{bmatrix}$.First,calculate $A^2 = A 	imes A = \begin{bmatrix} 2 & -1 \ -1 & 3 \end{bmatrix} \begin{bmatrix} 2 & -1 \ -1 & 3 \end{bmatrix} = \begin{bmatrix} 4+1 & -2-3 \ -2-3 & 1+9 \end{bmatrix} = \begin{bmatrix} 5 & -5 \ -5 & 10 \end{bmatrix}$.Now,calculate $2A^2 + 5A = 2 \begin{bmatrix} 5 & -5 \ -5 & 10 \end{bmatrix} + 5 \begin{bmatrix} 2 & -1 \ -1 & 3 \end{bmatrix} = \begin{bmatrix} 10 & -10 \ -10 & 20 \end{bmatrix} + \begin{bmatrix} 10 & -5 \ -5 & 15 \end{bmatrix} = \begin{bmatrix} 20 & -15 \ -15 & 35 \end{bmatrix}$.Let $M = 2A^2 + 5A = \begin{bmatrix} 20 & -15 \ -15 & 35 \end{bmatrix}$.The determinant $|M| = (20)(35) - (-15)(-15) = 700 - 225 = 475$.The inverse $M^{-1} = \frac{1}{|M|} 	ext{adj}(M) = \frac{1}{475} \begin{bmatrix} 35 & 15 \ 15 & 20 \end{bmatrix}$.Dividing by $5$,we get $M^{-1} = \frac{1}{95} \begin{bmatrix} 7 & 3 \ 3 & 4 \end{bmatrix}$.

If $A = \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix}$,then the inverse of $(2A^2 + 5A)$ is

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