Let $\mathrm{A}=\left[\begin{array}{cc}1& 2\\ -1& 4\end{array}\right]$ If ${\mathrm{A}}^{-1}+\mathrm{\alpha I}+\mathrm{\beta A}=0, \mathrm{\alpha }, \mathrm{\beta }\in \mathrm{R},$, I is 2 × 2 identity matrix, then  $4(\mathrm{\alpha }-\mathrm{\beta })$ is equal to

$$\begin{gathered} \mathrm{A}=\left[\begin{array}{cc}1& 2\\ -1& 4\end{array}\right] \\ {\mathrm{A}}^2=(1+4)\mathrm{A}+(4+2)\mathrm{I}=0 \\ {\mathrm{A}}^2-5\mathrm{A}+6\mathrm{I}=0 \\ {\mathrm{A}}^{-1}\left({\mathrm{A}}^2-5\mathrm{A}+6\mathrm{I}\right)=0 \\ {\mathrm{A}}^{-1} \mathrm{A} \mathrm{A}-5{\mathrm{A}}^{-1}\mathrm{A}+6{\mathrm{A}}^{-1}\mathrm{I}=0 \\ \mathrm{A}-5\mathrm{I}+6{\mathrm{A}}^{-1}=0 \\ {\mathrm{A}}^{-1}-\frac{5}{6}\mathrm{I}+\frac{1}{6}\mathrm{A}=0 \\ {\mathrm{A}}^{-1}\mathrm{\alpha } \mathrm{I}+\mathrm{\beta } \mathrm{A}=0 \\ \mathrm{\alpha }=\frac{-5}{6} \mathrm{\beta }=\frac{1}{6} \\ 4(\mathrm{\alpha }-\mathrm{\beta })=4 \left(\frac{-5}{6}-\frac{1}{6}\right) \\ =4(-1) \\ =-4 \end{gathered}$$