No solutions to a matrix inequality?

Solution 1:

Your inequality $(A-I)x \ge b$ has no solutions in $x$ as soon as $b>0$. Indeed, any potential solution would have to satisfy $A x \ge x + b$ and, since rows of $A$ are nonnegative and sum to one, each element of vector $Ax$ is a convex combination of the components of $x$, which must be less than $x_{max}$, the largest component of $x$. On the other hand, at least one element of $x+b$ is greater than $x_{max}$, which proves the impossibility.

By the way, applying Farkas' Lemma to this impossible system shows that the following always admits solutions in y $$y^T (A-I) = 0, y \ge 0 \text{ and } y^T b > 0$$ which expresses the fact that $A$ necessarily admits a nonnegative left eigenvector with eigenvalue $1$ (the last inequality ensures that $y$ is nonzero).