Your question is a bit unclear to me. However, as far I can understand...
if the generalized linear model is $y(x_1,x_2)=b_0+b_1x_1+b_2x_2+b_3x_1x_2+b_4x^2_1+b_5x^2_2$, and there are, say, four data points, then
...then your model matrix $X$ is a $4\times 6$ matrix: $$\begin{bmatrix} 1 & x_{11} & x_{21} & x_{11}x_{21} & x_{11}^2 & x_{21}^2 \\ 1 & x_{12} & x_{22} & x_{12}x_{22} & x_{12}^2 & x_{22}^2 \\ 1 & x_{13} & x_{23} & x_{13}x_{23} & x_{13}^2 & x_{23}^2 \\ 1 & x_{14} & x_{24} & x_{14}x_{24} & x_{14}^2 & x_{24}^2 \end{bmatrix}$$
If $\text{rank}(X)=4$, then $X^TX$ is a $6\times 6$ singular matrix, while $XX^T$ is a $4\times 4$ non singular matrix.
You should need $(X^TX)^{-1}$ to estimate $\beta$, but since $X^TX$ is singular you have to use a right inverse, i.e. a matrix $X_R$ such that $XX_R=I$: $$ X\beta=y,\qquad X\beta=XX_Ry,\qquad \beta=X_Ry$$ A right inverse that always guarantees a solution is: $X_R=X^T(XX^T)^{-1}$. See Wikipedia and Cherkassky & Mulier, Learning From Data: Concepts, Theory, and Methods, John Wiley & Sons, 2007, Appendix B.
