Let A = begin{bmatrix} 1 & 2 -1 & 4 end{bmat | vedclass

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(B) Given $A = \begin{bmatrix} 1 & 2 \ -1 & 4 \end{bmatrix}$. The determinant $|A| = (1)(4) - (2)(-1) = 4 + 2 = 6 
eq 0$. Since $A^{-1} = \frac{1}{|

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$4$

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(B) Given $A = \begin{bmatrix} 1 & 2 \ -1 & 4 \end{bmatrix}$. The determinant $|A| = (1)(4) - (2)(-1) = 4 + 2 = 6 
eq 0$. Since $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$,we have $A^{-1} = \frac{1}{6} \begin{bmatrix} 4 & -2 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} & -\frac{1}{3} \ \frac{1}{6} & \frac{1}{6} \end{bmatrix}$. Given $A^{-1} = \alpha I + \beta A$,we have: $\begin{bmatrix} \frac{2}{3} & -\frac{1}{3} \ \frac{1}{6} & \frac{1}{6} \end{bmatrix} = \alpha \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 1 & 2 \ -1 & 4 \end{bmatrix} = \begin{bmatrix} \alpha + \beta & 2\beta \ -\beta & \alpha + 4\beta \end{bmatrix}$. Comparing the elements,we get: $-\beta = \frac{1}{6} \implies \beta = -\frac{1}{6}$. $\alpha + \beta = \frac{2}{3} \implies \alpha - \frac{1}{6} = \frac{2}{3} \implies \alpha = \frac{2}{3} + \frac{1}{6} = \frac{5}{6}$. Therefore,$4(\alpha - \beta) = 4(\frac{5}{6} - (-\frac{1}{6})) = 4(\frac{6}{6}) = 4(1) = 4$.

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

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