Consider the 1D poisson equation
$$ \frac{d^2 u}{dx^2} = -\rho $$
with Dirichlet boundary conditions $u(0) = u(l) = g$. Using a finite difference scheme, with a 5-point grid $u_1,u_2,u_3,u_4,u_5$ (excluding boundary points $u_0$ and $u_l$), we get the set of linear equations
$$ \left( \begin{array}{ccc} 2 & -1 & & & \\ -1 & 2 & -1 & & \\ & -1 & 2 & -1 & \\ & & -1 & 2 & -1 \\ & & & -1 & 2 \\ \end{array} \right)\left( \begin{array}{c} u_1 \\ u_2 \\ u_3 \\ u_4 \\ u_5\end{array} \right) = \left( \begin{array}{c} \rho_1+g \\ \rho_2 \\ \rho_3 \\ \rho_4 \\ \rho_5+g\end{array} \right) $$
My question is: What would the matrix look like if I made $u_3$ a boundary point too?
Would it look like
$$ \left( \begin{array}{ccc} 2 & -1 & & & \\ -1 & 2 & 0 & & \\ & 0 & 1 & 0 & \\ & & 0 & 2 & -1 \\ & & & -1 & 2 \\ \end{array} \right)\left( \begin{array}{c} u_1 \\ u_2 \\ u_3 \\ u_4 \\ u_5\end{array} \right) = \left( \begin{array}{c} \rho_1+g \\ \rho_2+g \\ g \\ \rho_4+g \\ \rho_5+g\end{array} \right) $$
I ask because I have a 3d system with small but irregular internal boundary regions, and it would be currently more convenient for my purposes to leave them in the matrix (even if it means extra computational cost).