Complexity of $Ax\geq 0$
Your problem is in polynomial time, as the following claim implies.

Claim Let $A$ be an $m\times n$ matrix with rational entries. There exists an integer $x &gt; 0$ such that $Ax \geq 0$ if and only if there exists a real $x &gt; 0$ such that $Ax \geq 0$.

Proof The forward implication is trivial. For the backward implication, suppose that there exists a real $x &gt; 0$ such that $Ax \geq 0$. In particular, the polytope $P = \{ x \geq 0 : Ax \geq 0 \text{ and } \sum x = 1 \}$ is non-empty. Pick some vertex $x$ of $P$. Since all entries of $A$ are rational, all coordinates of $x$ are also rational. Therefore for some integer $N&gt;0$, all entries of $Nx$ are integral. Furthermore $Nx &gt; 0$ and $A(Nx) = N(Ax) \geq 0$. $\square$

Given a matrix $A$, in order to determine whether there exists a real $x &gt; 0$ such that $Ax \geq 0$, check whether the polytope $P = \{ x \geq 0 : Ax \geq 0 \text{ and } \sum x = 1 \}$ is non-empty using linear programming. This can be done in polynomial time.