If $A=\left(\begin{array}{cc}1& 5\\ \lambda & 10\end{array}\right)$, $A^{-1}=\alpha A+\beta I$ and $\alpha +\beta =-2$ then $4{\alpha }^2+{\beta }^2+{\lambda }^2$ is equal to 

 $$\begin{gathered} A^{-1}=\frac{1}{10-5\lambda }\left(\begin{array}{cc}10& -5\\ -\lambda & 1\end{array}\right) \\ \frac{1}{10-5\lambda }\left(\begin{array}{cc}10& -5\\ -\lambda & 1\end{array}\right)=\left(\begin{array}{cc}\alpha & 5\alpha \\ \lambda \alpha & 10\alpha \end{array}\right)+\left(\begin{array}{cc}\beta & 0\\ 0& \beta \end{array}\right) \end{gathered}$$
     
  $\alpha +\beta =\frac{10}{10-5\lambda }\to$1, $\frac{\cancel{5}}{5\lambda -10}=\cancel{5}\alpha \to$  2, $\frac{\lambda }{5\lambda -10}=\lambda \alpha \to$  3$\frac{1}{10-5\lambda }=10\alpha +\beta \to$  4
$\alpha +\beta$  =-2
$10=2(5\lambda -10)\Rightarrow 5\lambda =15$ ($\lambda$  =3)
$\alpha =\frac{1}{5}$ , $\frac{-1}{5}=2+\beta$
$\beta =\frac{-11}{5}$ 
 $\therefore 4{\alpha }^2+{\beta }^2+{\lambda }^2=\frac{4}{25}+\frac{121}{25}+9$
        = $\frac{125}{25}+9=114$