I'm discretizing the following Poisson equation using FVM where the domain $\Omega$ of the solution is a regular hexagon of side $1$ centered about the origin. $$\Delta u =k,\text{ $k$ constant}\\ \partial_n u = 0\\ \Omega \subset \mathbb{R^2}$$
Let $V_i\subset \Omega$, $i=1,\dots,6$, be one of the six equilateral triangles comprising the hexagon; so $\bigcup\limits_{i=1}^6 V_i=\Omega$.
Then we obtain a system of six equations, that is a linear system after the following steps: $$\int_{V_i} \Delta udx = \int_{\partial V_i} \nabla u(x)\cdot n ds =\int_{V_i}kdx = 0$$ $$\int_{\partial V_i} \nabla u(x)\cdot nds=\int_{\partial V_i}\partial_n u ds$$ $$\approx \sum\limits_{j\in n_i}\frac{u_j-u_i}{h_{i,j}}l_{i,j}+\int_{\Gamma_i}0ds=\sum\limits_{j\in n_i} \frac{u_j-u_i}{h_{i,j}}l_{i,j}+\text{constant $C$}$$ $$=|V_i|k=|V_i|\cdot 0 = 0$$
Where $|V_i|$ means the area of $V_i$ and $n_i$ is the set of cells immediately neighboring $V_i$ (two such cells for a hexagon).
Now, we can use some geometry to deduce that
$$|V_i| = \frac{\sqrt{3}}{4}; h_{i,j} = \frac{\sqrt{3}}{2}; l_{i,j}=1$$
So the system becomes:
$$\frac{2}{\sqrt3}\sum_{j\in n_i} (u_j-u_i) + C = \frac{\sqrt3}{4}k$$
Which can be represented in matrix-vector form as
$$B\vec{u}=\vec{c}$$
where $B$ is a square $6\times 6$ matrix with the following components
$$b_{ij}=\frac{2}{\sqrt3}, i\ne j; b_{ii}=-\frac{2\sqrt2}{3}$$ $$c_i =-C$$
The matrix $B$ is, explicitly, this:
$$\begin{bmatrix} -2\sqrt2/3 & 2/\sqrt3 & & & &2/\sqrt3 \\ 2/\sqrt3 & -2\sqrt2/3 & 2/\sqrt3 & & & \\ & 2/\sqrt3 &-2\sqrt2/3 & 2/\sqrt3 & & \\ & & 2/\sqrt3 & -2\sqrt2/3 & 2/\sqrt3 & \\ & & & 2/\sqrt3 & -2\sqrt2/3 &2/\sqrt3 \\ 2/\sqrt3 & & & & 2/\sqrt3 & -2\sqrt2/3 \end{bmatrix}$$
But this matrix is non-singular so it has only one unique solution, whereas the original non-discrete system has infinitely many solutions $u+c$. Can someone please tell me what my error(s) is(are)? I would really appreciate that.