Existence of a solution with entries $1,-1$ or $0$ Suppose $A$ is an $m\times n$ matrix and $b$ is an $m\times 1$ vector (where $m,n\geq 3$) such that each of the column vectors of $A$ and the vector $b$ has one entry equal to $1$, another entry equal to $-1$ and rest of the entries are zero.
Now if the system $Ax=b$ has a solution ( where $x$ is $n\times 1$) then does there always exist a solution $x$ such that each of the entries of $x$ are $1,-1$ or $0$ ? If not in general, is it true when rank of $A$ is $m-1$ ?
 A: Any such $A$ can be seen as incidence matrix of some directed graph. If the rank is less than $m-1$, then the corresponding graph will just have more than one connected components. 
The system $Ax=b$ can be seen as $b$ amount of flow being injected at the nodes and $x$ is then a valid flow along the edges satisfying flow conservation at each node. You can show that $Ax=b$ has a solution iff the values in $b$ corresponding to each connected component add up to zero. 
Your requirement that there must exist a solution with only $1$, $-1$ and $0$ is not always true for arbitrary $b$. But for your particular case, you can see that there exists a solution for $Ax=b$ iff the nodes corresponding to $1$ and $-1$ in $b$ are in the same connected component. And if they are in the same connected component, we can find a path (ignoring directions) between those two nodes and send a unit flow along this path. This gives a solution $x$ with just $1$, $-1$ and $0$. 
You can extend this to more general $b$ too, with some conditions on min-cut of the graph.
