$\begin{aligned} & A^{-1}=\alpha I+\beta A \\ & \Rightarrow\left[\begin{array}{ll}1 & 2 \\ 3 & 5\end{array}\right]^{-1}=\alpha\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]+\beta\left[\begin{array}{ll}1 & 2 \\ 3 & 5\end{array}\right] \\ & \Rightarrow\left[\begin{array}{cc}-5 & 2 \\ 3 & -1\end{array}\right]=\left[\begin{array}{cc}\alpha+\beta & 2 \beta \\ 3 \beta & \alpha+5 \beta\end{array}\right] \\ & \Rightarrow \alpha=-6, \beta=1\end{aligned}$
Now $\alpha+\beta+\alpha \beta=-6+1+(-6) \times 1=-11$