Complexity of checking whether linear equations have a positive solution
First, this can be solved by linear programming. Let $x_1$, $x_2$, $\ldots$, $x_n$ be your variables. The linear program that solves your question is then

$\max t$
subject to
$t \leq x_i$, for $i=1 ... n$,
$Ax = 0$.

If the maximum is 0, then there is no positive solution. If the maximum is $\infty$ (i.e., the linear program is unbounded) then there is a positive solution.



Second, using standard transformations on linear programs, the feasibility problem for an arbitrary linear program with strict inequalities can be reduced to your problem. We start with the feasibility problem

Does there exist $x$ such that
$A x &lt; b$ ?

We can now add a new variable $x_{n+1}$ to the right-hand-side of all these equations and an inequality $x_{n+1} &gt; 0$ to make everything homogeneous. So for the $k$th equation, we now have

$a_{k,1} x_1 + \ldots + a_{k,n} x_n - b_k x_{n+1} &lt; 0$.  

This gives an equivalent problem on a new matrix $A&#39;$:

Does there exist $x$ such that
$A&#39;x &lt; 0$ ?

Next, we can turn the inequalities into equations by adding some variables $y_i$ and requiring that $y_i &gt; 0$. The $k$th equation on the new problem turns into

$a&#39;_{k,1} x_1 + \ldots + a&#39;_{k,n} x_n + a&#39;_{k,n+1} x_{n+1} +y_k = 0$.

Finally, we want to allow all the variables to be positive. How we do this? For every variable $x_i$ with an arbitrary sign, we replace $x_i$ by $z_i - w_i$, and require $z_i&gt;0$, $w_i &gt; 0$.

The feasibility problem is essentially as hard as an arbitrary linear programming problem, so in general there won't be any easier way to solve your problem than using linear programming.