Existence of solution for matrix equation (I−αA)bar x= bar b This is my first question in here and I would be really thankful if someone could help me with understanding the matter. I am solving a matrix equation (I−alphaA) bar x= bar b for a positive vector bar x. I don't know anything about the signs of the scalar alpha and vector bar b (but I can always split the solution into several cases). Matrix A is a symmetric matrix with tr(A)=0 (basically it is an adjacent matrix of an undirected graph so it is built only of 0 and 1 with 0 on its diagonal).

Existence of solution for matrix equation $(I-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}A)\stackrel{¯<!--\; ¯\; -->}{x}=\stackrel{¯<!--\; ¯\; -->}{b}$
This is my first question in here and I would be really thankful if someone could help me with understanding the matter.
I am solving a matrix equation $(I-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}A)\stackrel{¯<!--\; ¯\; -->}{x}=\stackrel{¯<!--\; ¯\; -->}{b}$for a positive vector $\stackrel{¯<!--\; ¯\; -->}{x}$. I don't know anything about the signs of the scalar $\mathrm{\alpha <!--\; \alpha \; -->}$ and vector $\stackrel{¯<!--\; ¯\; -->}{b}$ (but I can always split the solution into several cases). Matrix A is a symmetric matrix with tr(A)=0 (basically it is an adjacent matrix of an undirected graph so it is built only of 0 and 1 with 0 on its diagonal).
My approach would be just to write it down as $\stackrel{¯<!--\; ¯\; -->}{x}=(I-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}A{)}^{-<!--\; -\; -->1}\stackrel{¯<!--\; ¯\; -->}{b}$ whenever $(I-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}A)$ is nonsingular (as far as I understand that happens for a finite number of α and thus Lebesgue measure of this set of "inappropriate" $\mathrm{\alpha <!--\; \alpha \; -->}$ is 0).
In most of the literature, however, it is required that $1>\mathrm{\alpha <!--\; \alpha \; -->}\cdot <!--\; \cdot \; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_{max}(A)$ for the convergence of $I+\mathrm{\alpha <!--\; \alpha \; -->}A+{\mathrm{\alpha <!--\; \alpha \; -->}}^2{A}^2+\dots <!--\; \dots \; -->$ and for the solution $\stackrel{¯<!--\; ¯\; -->}{x}$ to exist (where ${\mathrm{\lambda <!--\; \lambda \; -->}}_{max}(A)$ is a maximum eigenvalue of A) . Moreover, then $det(I-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}A)=1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}\cdot <!--\; \cdot \; -->det(A)=1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}\cdot <!--\; \cdot \; -->\underset{i}{\prod <!--\; \prod \; -->}{\mathrm{\lambda <!--\; \lambda \; -->}}_i>0$ and hence it is also invertible.
So I have two questions here:
Q1: Why do I need convergence of $I+\mathrm{\alpha <!--\; \alpha \; -->}A+{\mathrm{\alpha <!--\; \alpha \; -->}}^2{A}^2+\dots <!--\; \dots \; -->$ in here? I understand that if it converges then $\underset{k=0}{\overset{+\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\mathrm{\alpha <!--\; \alpha \; -->}}^k{A}^k=(I-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}A{)}^{-<!--\; -\; -->1}$, but why do I need that for the existence of a solution x¯? For instance, if we take a simple example:
$A=\left|\begin{array}{cc}0& 1\\ 1& 0\end{array}\right|\mathrm{\alpha <!--\; \alpha \; -->}=30$
${\mathrm{\lambda <!--\; \lambda \; -->}}_{max}(A)=1$ and hence, convergence condition is violated: 30>1. However, I can still calculate $\stackrel{¯<!--\; ¯\; -->}{x}$. Inverse of $(I-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}A)$ exists and is equal to $-<!--\; -\; -->\frac{1}{899}\left|\begin{array}{cc}1& 30\\ 30& 1\end{array}\right|$. If, for instance, $\stackrel{¯<!--\; ¯\; -->}{b}$ was negative, then it would give me a positive $\stackrel{¯<!--\; ¯\; -->}{x}$. Or is the condition $1>\mathrm{\alpha <!--\; \alpha \; -->}\cdot <!--\; \cdot \; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_{max}(A)$ needed to guarantee that inverse will give a positive solution with both positive α and $\stackrel{¯<!--\; ¯\; -->}{b}$?
Q2: What will happen if $\mathrm{\alpha <!--\; \alpha \; -->}<0$ (and potentially $\stackrel{¯<!--\; ¯\; -->}{b}<0$)? What are the conditions for existence of solution $\stackrel{¯<!--\; ¯\; -->}{x}$? What are the conditions for it to be positive?