Let $$A = \begin{bmatrix} 1 & 5\\ \lambda & 10 \end{bmatrix}$$ and $A^{-1} = \alpha A + \beta I$ and $\alpha + \beta = -2$. Find $4\alpha^{2} + \beta^{2} + \lambda^{2}$.
I am interested in knowing how to solve this problem by using eigenvalues and characteristic equation. I created the characteristic equation but could not get to the right answer. My progress is as follows. Let
$$A^{-1} = \alpha A + \beta I \tag{1}$$
We multiply $(1)$ by $A$
$$\alpha A^{2}+\beta A-I=0 \tag{2}$$
The characteristic equation
$$A^{2}-11A+(10-5\lambda)I=0 \tag{3}$$
We can compare $(2)$ and $(3)$ to get $\alpha=1$, $\beta=-11$ and $\lambda = \frac{11}{5}$ but that does not make sense considering $\alpha+\beta=-2$. What am I doing wrong?
Note that the problem setter did not expect me to solve the problem by using eigenvalues and characteristic equations. I'm just experimenting. Infact we aren't even taught that topic at my level.
Your approach is quite perfect. The mistake appears at the step, when you equalize the coefficients of (2) and (3).
Before doing so, the equation (3) needs to be multiplied by $\alpha.$
The system of equations becomes $$\begin{cases} \alpha&=\alpha\\ \beta&=-11\alpha\\ -1&=\alpha(10-5\lambda)\\ \alpha+\beta&=-2\end{cases}$$ I bet you can finish now.