Eigenvalue of a $3\times 3$ complex matrix

Question

In the following question I got that $0$ is not an eigenvalue, then $M$ is invertible and using Cayley-Hamilton then I got the last option correctly, then why given that $\alpha+\beta \neq 0$? What is the importance of that? Question is

Suppose that the characteristic equation of $M \in \mathbb{C}^{3 \times 3}$ is $$ \lambda^{3}+\alpha \lambda^{2}+\beta \lambda-1=0 $$ where $\alpha, \beta \in \mathbb{C}$ with $\alpha+\beta \neq 0$. Which of the following statements is TRUE? $$ M(I-\beta M)=M^{-1}(M+\alpha I) $$ $$ M(I+\beta M)=M^{-1}(M-\alpha I) $$ $$ M^{-1}\left(M^{-1}+\beta I\right)=M-\alpha I $$ $$ M^{-1}\left(M^{-1}-\beta I\right)=M+\alpha I $$

Answer 1

I believe that this condition was used to ensure that the fourth equation was the unique solution.

If $\alpha=-1, \beta=1$ then the first equation is true.

If $\alpha=i, \beta=-i$ then the second equation is true.

If $\alpha=0, \beta=0$ then the third equation is true.

Thus $\alpha +\beta \ne 0$ prevents all of these possibilities.

Answer 2

$\lambda^{3}+\alpha \lambda^{2}+\beta \lambda-1=0$

The characteristic equation implies:

$M^3 + \alpha M^2 + \beta M - I = 0$

At which point is it basic algebra.

$M^3 + \alpha M^2 = I - \beta M\\ M^{-2}(M^3 + \alpha M^2) = M^{-2}(I - \beta M)\\ M + \alpha = M^{-1}(M^{-1} - \beta)$