$A=\left[\begin{array}{ll}1 & 2 \\ 3 & 5\end{array}\right]$
$\alpha A^2-\beta A=2 I$
$\Rightarrow \alpha\left[\begin{array}{ll}1 & 2 \\ 3 & 5\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 3 & 5\end{array}\right]-\beta\left[\begin{array}{ll}1 & 2 \\ 3 & 5\end{array}\right]=2\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\begin{aligned} & \Rightarrow \alpha\left[\begin{array}{cc}7 & 12 \\ 18 & 31\end{array}\right]-\beta\left[\begin{array}{ll}1 & 2 \\ 3 & 5\end{array}\right]=2\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\ & \Rightarrow\left[\begin{array}{cc}7 \alpha-\beta & 12 \alpha-2 \beta \\ 18 \alpha-3 \beta & 31 \alpha-5 \beta\end{array}\right]=\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]\end{aligned}$
$\therefore 7 \alpha-\beta=2$ ...(i)
$12 \alpha-2 \beta=0 \Rightarrow$ or $6 \alpha-\beta=0$ ...(ii)
Eqn. (i) - Eqn. (ii) $\alpha=2$
$\begin{aligned} & \therefore \beta=6 \alpha=12 \\ & \therefore \alpha^2+\beta=4+12=16\end{aligned}$