Q:

Does this system of equations has a solution?

Let $R$ be an integral domain with identity. Let $A$ be an $n\times n$ matrix with entries from $R$. Assume that $A$ has the following properties:
(1) Every row of $A$ has exactly one $1$ and one $-1$.
(2) Every column of $A$ has exactly one $1$ and one $-1$.
(3) All other elements are $0$.
(4) $A_{ii}=1$ (all the diagonal entries are 1)
Let $B$ be an $n\times 1$. The $i$-th element of $B$ is $b_i$. Assume that $\sum_{i=1}^{n}b_i=0$. 
Does $AX=B$ has a solution (where $X$ is an $n\times 1$ matrix) ?

A:

No: Consider the following system of equations (we work in $R = \mathbb{Z}$):
$$\begin{pmatrix}
1 & -1 & 0 & 0\\
-1 & 1 & 0 & 0\\
0 &  0 & 1& -1\\
0 & 0 & -1 & 1
\end{pmatrix}\cdot 
\begin{pmatrix}
x_1 \\
x_2\\
x_3\\
x_4
\end{pmatrix}
=
\begin{pmatrix}
1\\
2\\
-1\\
-2
\end{pmatrix}$$
Then we would have that $1 = x_1 - x_2 = -(x_1 - x_2) = - 2$.

