Given,
A=[1−124]⇒∣A∣=6
So, A−1=∣A∣adjA=61[41−21]=[3261−3161]
Here, [3261−3161]=[α00α]+[β−β2β4β]
\begin{matrix}\alpha +\beta =\frac{2}{3} \\ \beta =-\frac{1}{6}\end{matrix}}\Rightarrow \alpha =\frac{2}{3}+\frac{1}{6}=\frac{5}{6}
Hence, 4(α−β)=4(1)=4