(Note: I asked a similar question at math.stackexchange but the present one is more precise.)
I have a linear program of the form:
$$\text{minimize} \space\space x_1 \space\space \text{subject to:}$$ $$\mathbf{Ax} \ge \mathbf{b} \tag{1}$$
where $\mathbf{A}$ is a $q \times q$ square matrix with entries $a_{i,j} = 0$ if $i \ge j + 2$, $a_{i,j} = -r = -2/3$ if $i = j + 1$, and $a_{i,j} = \binom{j}{i-1}(-1)^{i+j}$ otherwise, $\mathbf{x}^{\intercal}=(x_1,x_2,\ldots,x_q)$, $\mathbf{b}^{\intercal}=(b_1,b_2,\ldots,b_q)$ with $b_1= \frac{2}{3}\binom{n-1}{2}$ and $b_k=\frac{n+1-2k}{3}\binom{q+1}{k-1}$ for $k \ge 2$. The $\mathbf{A}$ matrix is "nearly" triangular.
Background: this problem can be constructed for a separating union-closed family of $n$ sets and a maximum of $q+1$ elements, using the property that $2/3$ of the couples of sets, both non-empty, have at least an element in common (see here). $x_i = \sum_{S \in \mathcal{P}([q+1]), |S|=i} \binom{f(S)}{2}$, and $f(S)=|\{T \in \mathcal{F}: S \subseteq T\}|$, and we assume that there is no element in at least $n/2$ sets.
My question: Is it possible to solve $(1)$ analytically and get a closed form for the minimum as a function of $n$ and $q$? And in particular, when is it possible to say that $\min x_1 = (1,0,\ldots,0)\mathbf{A}^{-1}\mathbf{b}$?
I read about Fourier–Motzkin elimination but I can't see how to apply it analytically. I think also that a direct elimination starting from the last inequality is made difficult due to the alternating signs.
