Let A = 1 2 - 2 - 5 . Let α , β ∈ ℝ be such that α A 2 + β A = 2 I . Then…

We know that characteristic equation of matrix $A$ is given by, 

$\left|A-\lambda I\right|=0$

$\Rightarrow \left|\begin{array}{cc}1-\lambda & 2\\ -2& -5-\lambda \end{array}\right|=0$

$\Rightarrow {\lambda }^2+4\lambda =1$

$\Rightarrow A^2+4A=I$

Multiplying by $2$ both side we get,

$\Rightarrow 2A^2+8A=2I \cdots \left(1\right)$

Given that $\alpha A^2+\beta A=2I \cdots \left(2\right)$

Now on Comparing equation $\left(1\right) \& \left(2\right)$ we get

$\alpha =2, \beta =8$

$\therefore \alpha +\beta =10$