This is a follow up of my previous question. I repeat it below with an additional condition (in italics).
Say $\mathbf{X}$ is a $n\times m$ matrix with no negative entries and such that every row and every column have at least a non zero element.
Let $\mathbf{x}^i$ be the $i$-th row of $\mathbf{X}$ and assume further that $\mathbf{x}^i\not\leq\mathbf{x}^j$ and $\mathbf{x}^j\not\leq\mathbf{x}^i\;\forall i,j$.
Denote $\langle\mathbf{X}\rangle$ the cone spanned by $\mathbf{X}$. Let $\mathbf{q},\mathbf{k}, 1$, and $0$ be vectors of the suitable dimension. In the following LP $$ \begin{array}{ll} \text{maximize} &\quad \mathbf{q}\cdot 1\\ \text{subject to} & \quad \mathbf{X}^{\top}\mathbf{q}\leq\mathbf{k}\\ &\quad\mathbf{q}\geq 0, \end{array}$$
let $\mathbf{q}^*$ be its solution. I see that $\mathbf{k}\not\in\langle\mathbf{X}\rangle\implies \mathbf{X}^{\top}\mathbf{q}^*\leq\mathbf{k}$ (with $\mathbf{X}^{\top}\mathbf{q}^*\ne\mathbf{k}$), because $\mathbf{k}\not\in\langle\mathbf{X}\rangle$ implies that there does not exist a set of positive weights satisfying $\mathbf{X}^{\top}\mathbf{q}=\mathbf{k}$.
However, I believe that $\mathbf{k}\in\langle\mathbf{X}\rangle\implies \mathbf{X}^{\top}\mathbf{q}^*=\mathbf{k}$. Is this true? Why?
