Let A be a 2 x 2 matrix with real entries such that  $\mathrm{A}\text{'}=\mathrm{\alpha A}+\mathrm{I}$, where $\mathrm{\alpha }\in \mathrm{ℝ}-\{-1,1\}$.  If $\det ({\mathrm{A}}^2-\mathrm{A})=4$, then the sum of all possible values of $\mathrm{\alpha }$ is equal to 

$\mathrm{A}\text{'}=\mathrm{\alpha A}+\mathrm{I}$

$\mathrm{A}=\mathrm{\alpha A}\text{'}+\mathrm{I}\Rightarrow \mathrm{A}-\mathrm{I}=\mathrm{\alpha A}\text{'} ...\left(1\right)$

$$\begin{gathered} \mathrm{A}=\mathrm{\alpha }(\mathrm{\alpha A}+\mathrm{I})+\mathrm{I} \\ \Rightarrow (1-{\mathrm{\alpha }}^2)\mathrm{A}=(1+\mathrm{\alpha })\mathrm{I} \\ \mathrm{\alpha }\ne -1\begin{array}{l}\Rightarrow \mathrm{A}=\frac{\mathrm{I}}{1-\mathrm{\alpha }}\\ \end{array} \end{gathered}$$

$|{\mathrm{A}}^2-\mathrm{A}|=4\Rightarrow \left|\mathrm{A}\right||\mathrm{A}-\mathrm{I}|=4$

From  eq. (1)

$\left|\mathrm{A}\right||\mathrm{\alpha A}\text{'}|=4 \Rightarrow \frac{1}{{(1-\mathrm{\alpha })}^2}\times \frac{{\mathrm{\alpha }}^2}{{(1-\mathrm{\alpha })}^2}=4$

$\frac{\mathrm{\alpha }}{{(1-\mathrm{\alpha })}^2}=\pm 2$

$\therefore \mathrm{\alpha }=+2{(1-\mathrm{\alpha })}^2, \mathrm{\alpha }=-2{(1-\mathrm{\alpha })}^2$
  $$\begin{gathered} \therefore \mathrm{\alpha }=-2{(1-\mathrm{\alpha })}^2\Rightarrow 2{\mathrm{\alpha }}^2-5\mathrm{\alpha }+2=0 , 2{\mathrm{\alpha }}^2-3\mathrm{\alpha }+2=0 \\ \Rightarrow \mathrm{\alpha }=2,\frac{1}{2} \end{gathered}$$

$\mathrm{\alpha }=+2{(1-\mathrm{\alpha })}^2,\text{ Imaginary roots }$

$\therefore \mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{possible} \mathrm{values} \mathrm{of} \mathrm{\alpha }=2+\frac{1}{2}=\frac{5}{2}$