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

$$ \begin{aligned} & \mathrm{A}=\left[\begin{array}{cc} 1 & 5 \\ \lambda & 10 \end{array}\right] \\\\ & \Rightarrow|\mathrm{A}-x \mathrm{I}|=0 \\\\ & \Rightarrow\left|\begin{array}{cc} 1-x & 5 \\ \lambda & 10-x \end{array}\right|=0 \\\\ & \Rightarrow(1-x)(10-x)-5 \lambda=0 \\\\ & \Rightarrow 10-11 x+x^2-5 \lambda=0 \end{aligned} $$ <br/><br/>Also, $\Rightarrow \mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}$ <br/><br/>$\Rightarrow \alpha A^2+\beta A-I=0$ <br/><br/>and $A^2-11 A+(10-5 \lambda) I=0$ <br/><br/>On solving, we get <br/><br/>$\alpha=\frac{1}{5}, \beta=-\frac{11}{5}$ <br/><br/>$$ \begin{aligned} & \text { So, } 5 \lambda-10=5 \Rightarrow \lambda=3 \\\\ & \therefore 4 \alpha^2+\beta^2+\lambda^2 \\\\ & =4\left(\frac{1}{25}\right)+\left(\frac{121}{25}\right)+9 \\\\ & =\frac{125}{25}+9=14 \end{aligned} $$