design matrices 
Given a linear model $Y = X\beta + \epsilon$ with three treatments and six subjects where $X$ is the design matrix, suppose $X = \begin{matrix}1 & 1 & 0\\
1 & 1 & 0\\
1 & 0 & 1\\
1 & 0 & 1\\
1 & -1 & -1\\
1 & -1 & -1
\end{matrix}$ and  
X'=
  \begin{matrix}1 & 0 & 0\\
1 & 0 & 0\\
1 & 1 & 0\\
1 & 1 & 0\\
1 & 0 & 1\\
1 & 0 & 1
\end{matrix} 
with response vector $Y=[Y_{11} Y_{12} Y_{21} Y_{22} Y_{31} Y_{32}]^{T}$ 
Do these two matrices give you the same model? 

I've had some trouble trying to figure out how, given that $\beta = (\beta_1 \beta_2 \beta_3)^T$, these two designs can give the equivalent model. Don't they have entirely different values of $\beta$ for each $Y_{ij}$?
EDIT: The constraints are $\sum_{i=1}^{3}\beta_i = 0$

$\beta=X^\dagger Y$ where $X^\dagger=(X^t X)^{-1} X^t$ is the pseudo inverse of X.
If you compute the pseudo inverse of both $X$ and $X'$ you will see that they give different matrix (octave code below). Hence your models will, most likely, be different. Now there might be some $Y$ such that the $\beta$  will be the same (i.e. such that $(X')^\dagger Y = (X)^\dagger Y$) but I'm thinking this is somehow irrelevant here (however, as J.M. points out, this case can happen).
X = [1 1 0; 1 1 0; 1 0 1; 1 0 1; 1 -1 -1; 1 -1 -1];

Xb = [1 0 0; 1 0 0; 1 1 0; 1 1 0; 1 0 1; 1 0 1];
inv(X'*X)*X'
inv(Xb'*Xb)*Xb'
Edit: a counter example shows the same: let Y=[1;2;3;4;5;6], then 
X\Y yields [3.5;-2;0] and Xb\Y yields [1.5;2;4] where A\b operation amounts to applying the pseudo inverse of A to the vector b.
