Does $Ax=b$ have a solution $a\ge 0$ when $A_{ij} > 0$ for $i\le j$ and $A_{ij} < 0$ for $i> j$?

Question

Write $x = (x_1, \dots, x_n) > 0$ when $x_1 > 0, \dots, x_n > 0,$ and $x \ge 0$ when $x_1 \ge 0, \dots, x_n \ge 0.$

Consider an invertible square matrix $A$ that satisfies $$ A_{ij} > 0 \text{ for } i \le j, \qquad \text{ and } A_{ij} < 0 \text{ for } i > j; $$ and a vector $b > 0.$ Let $x = A^{-1}b$ solve the linear system $Ax = b.$ Is $x \ge 0$?

My thoughts so far:

• Clearly, $x$ has at least one positive element.
• $A$ looks somewhat related to the class of Z-matrices. Do matrices that look like $A$ also have a name?
• Farkas' lemma might be helpful.

Edit: Even though $x\ge 0$ is not true in general (see copper.hat's counterexample below); are there sufficient conditions that one can impose on $A$?

Answer 1

$101 \begin{bmatrix} 1 & 100 \\ -1 & 1 \end{bmatrix}^{-1}\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -99 \\ 2 \end{bmatrix}$.