Given a matrix $A_{ij}$, can you find whether or not there exists a non-negative linear combination of its columns $p_j \ge 0$ such that for all $i$ $$\sum_j A_{ij}p_j > 0 \quad?$$
This subproblem came up in my attempt to remove dominated strategies from a strategic form game.
for example: $$\pmatrix{-1&0\\0&-1}$$ has no such linear combination, while $$\pmatrix{2&-1\\-1&2}$$ has $p = \pmatrix{1&1}$ as a viable linear combination $$ Ap = \pmatrix{1&1} > \pmatrix{0&0}$$
You might be interested in a result known as the Motzkin Transposition Theorem. It's not going to provide a specific answer for you, but it will give you an equivalent system of linear inequalities to test to the one you're after that might be easier for the specifics of your problem (or might not). In particular the characterization in the link of system (f) is a general version of what you're after: let $A$ be your matrix, and $B = \textrm{Id}$, $C = 0$, then the solvability of the system: $$ Ax > 0, \; \; \; Bx \ge 0, \;\;\;Cx = 0 $$ is precisely what you're after. An $x$ solving the above system exists if and only if there do not exist vectors $y_1, y_2$ such that: $$ A^Ty_1 + y_2 = 0, \; \; \; y_1 \gneq 0, \; \; \; y_2 \ge 0. $$ In other words, these outcomes are 'alternatives' or a complete set of mutually exclusive possibilities. Intuitively, you're asking whether or not the convex hull of the columns of your matrix $A$ (note without loss of generality you can take the sum of your $p_j$ to be unity) intersects the positive orthant (equivalently, can be separated from the strictly positive orthant by a hyperplane); the above gives a characterization in terms of the dual formulation.
In the examples you gave, this is really easy to see geometrically: the convex hull of the columns of your first matrix is the line segment connecting $(-1,0)$ and $(0,-1)$ and hence clearly can be separated from $\mathbb{R}^2_{++}$. As you noted, the convex hull of the columns of the second matrix contains a point in the positive orthant, given by weights $(.5 , .5)$.