We need to interpret the expression as $A^{-1} = \alpha I + \beta A$ since the question writes “If $A-1 = \alpha I + \beta A$”, which clearly means the inverse of $A$.

Given $A=\begin{pmatrix}1&2\\-1&4\end{pmatrix}$ first find $A^{-1}$.

Compute the determinant: $|A| = (1)(4) - (2)(-1) = 4+2=6$

Hence, $A^{-1} = \frac{1}{6}\begin{pmatrix}4&-2\\1&1\end{pmatrix}$

Now write $\alpha I + \beta A = \alpha \begin{pmatrix}1&0\\0&1\end{pmatrix} + \beta \begin{pmatrix}1&2\\-1&4\end{pmatrix}$ $= \begin{pmatrix}\alpha+\beta & 2\beta \\ -\beta & \alpha+4\beta\end{pmatrix}$

Compare this with $A^{-1}=\begin{pmatrix}\frac{2}{3} & -\frac{1}{3}\\ \frac{1}{6} & \frac{1}{6}\end{pmatrix}$

Equating corresponding entries:

From the $(1,2)$ entry, $2\beta=-\frac{1}{3} \implies \beta=-\frac{1}{6}$

From the $(1,1)$ entry, $\alpha+\beta=\frac{2}{3}$ $\alpha=\frac{2}{3}+\frac{1}{6}=\frac{5}{6}$

Now compute: $\alpha-\beta=\frac{5}{6}-\left(-\frac{1}{6}\right)=1$ Therefore, $4(\alpha-\beta)=4$

So the correct option is $\boxed{\text{D}}$