1. Write the matrix and use the given relation

Given

$$A=\begin{bmatrix}1&5\\ \lambda&10\end{bmatrix}, \qquad A^{-1}=\alpha A+\beta I$$

and

$$\alpha+\beta=-2.$$

We use the identity

$$AA^{-1}=I.$$

So,

$$A(\alpha A+\beta I)=I \implies \alpha A^2+\beta A=I.$$

2. Compute $A^2$

$$A^2=\begin{bmatrix}1&5\\ \lambda&10\end{bmatrix} \begin{bmatrix}1&5\\ \lambda&10\end{bmatrix} = \begin{bmatrix} 1+5\lambda & 5+50\\ \lambda+10\lambda & 5\lambda+100 \end{bmatrix} = \begin{bmatrix} 1+5\lambda & 55\\ 11\lambda & 5\lambda+100 \end{bmatrix}.$$

Hence,

$$\alpha A^2+\beta A = \begin{bmatrix} \alpha(1+5\lambda)+\beta & 55\alpha+5\beta\\ 11\alpha\lambda+\beta\lambda & \alpha(5\lambda+100)+10\beta \end{bmatrix}.$$

Since this equals $I=\begin{bmatrix}1&0\\0&1\end{bmatrix}$, compare entries.

3. Use off-diagonal entries

From the $(1,2)$ entry:

$$55\alpha+5\beta=0 \implies 11\alpha+\beta=0.$$

Given also

$$\alpha+\beta=-2.$$

Subtracting,

$$(11\alpha+\beta)-(\alpha+\beta)=0-(-2)$$ $$10\alpha=2 \implies \alpha=\frac15.$$

Then

$$\beta=-2-\frac15=-\frac{11}{5}.$$

4. Find $\lambda$

From the $(2,1)$ entry:

$$11\alpha\lambda+\beta\lambda=0 \implies \lambda(11\alpha+\beta)=0.$$

But already $11\alpha+\beta=0$, so this gives no new information.

Now use the $(1,1)$ entry:

$$\alpha(1+5\lambda)+\beta=1.$$

Substitute $\alpha=\frac15$, $\beta=-\frac{11}{5}$:

$$\frac15(1+5\lambda)-\frac{11}{5}=1.$$ $$\frac{1+5\lambda-11}{5}=1 \implies \frac{5\lambda-10}{5}=1$$ $$\lambda-2=1 \implies \lambda=3.$$

5. Compute the required expression

$$4\alpha^2+\beta^2+\lambda^2 =4\left(\frac15\right)^2+\left(-\frac{11}{5}\right)^2+3^2$$ $$=4\cdot \frac1{25}+\frac{121}{25}+9 =\frac{4+121}{25}+9 =\frac{125}{25}+9 =5+9=14.$$

6. Final answer

$$\boxed{14}$$

So the correct option is D.