$A =\alpha A ^{ T }+ I$

$A =\alpha(\alpha A + I )+ I$

$A =\alpha^2 A +(\alpha+1) I$

$A \left(1-\alpha^2\right)=(\alpha+1) I$

$A =\frac{ I }{1-\alpha}$

$| A |=\frac{1}{(1-\alpha)^2}$

$\left| A ^2- A \right|=| A || A - I |$

$A - I =\frac{ I }{1-\alpha}- I =\frac{\alpha}{1-\alpha} I$

$| A - I |=\left(\frac{\alpha}{1-\alpha}\right)^2$

$\text { Now }\left| A ^2- A \right|=4$

$| A || A - I |=4$

$\Rightarrow \frac{1}{(1-\alpha)^2} \frac{\alpha^2}{(1-\alpha)^2}=4$

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

$\Rightarrow 2(1-\alpha)^2= \pm \alpha$

Sum of value of $\alpha=\frac{5}{2}$