If A = begin{bmatrix} 1 & 2 -1 & 4 end{bmatr | vedclass

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(A) Given $A = \begin{bmatrix} 1 & 2 \ -1 & 4 \end{bmatrix}$.First,we find the characteristic equation of $A$ using $|A - \lambda I| = 0$.$|A - \lamb

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$\frac{8}{3}$

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(A) Given $A = \begin{bmatrix} 1 & 2 \ -1 & 4 \end{bmatrix}$.First,we find the characteristic equation of $A$ using $|A - \lambda I| = 0$.$|A - \lambda I| = \begin{vmatrix} 1 - \lambda & 2 \ -1 & 4 - \lambda \end{vmatrix} = (1 - \lambda)(4 - \lambda) - (-2) = \lambda^2 - 5\lambda + 4 + 2 = \lambda^2 - 5\lambda + 6 = 0$.By the Cayley-Hamilton theorem,$A^2 - 5A + 6I = 0$.Multiplying by $A^{-1}$,we get $A - 5I + 6A^{-1} = 0$.$6A^{-1} = 5I - A$.$A^{-1} = \frac{5}{6}I - \frac{1}{6}A$.Comparing this with $A^{-1} = \alpha I + \beta A$,we get $\alpha = \frac{5}{6}$ and $\beta = -\frac{1}{6}$.Then,$\alpha + \beta = \frac{5}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$.Therefore,$4(\alpha + \beta) = 4 	imes \frac{2}{3} = \frac{8}{3}$.

If $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$ and $A^{-1} = \alpha I + \beta A$,where $\alpha, \beta \in \mathbb{R}$ and $I$ is the identity matrix of order $2$,then $4(\alpha + \beta) = $

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