Problem in showing that this determinant is greater than zero.

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In one class the teacher showed us one problem about determinants that himself could not solve, and I tried solve the problem too, but I didn't had success even in the $3$ by $3$ case, I'm very curious to know the solution of this problem, so that is the reason I'm posting here. The problem is the following:

If I have a real square matrix $A$ of order $n$ in which:

1) Each diagonal member $a_{ii}>0$;

2) If $i \neq j$ then $a_{ij} \leq 0$;

3) For each colum $j$ we have $\sum_{i=1}^{n} a_{ij} > 0$;

Then how to conclude that $det(A)>0$ ?