Least squares fit for an underdetermined linear system

Question

I want fitting my data using bicubic interpolation: $$f(x,y)=\sum_{i=0}^{3}\sum_{j=0}^{3}a_{ij}x^iy^j$$ Let known $$f(0, 0)=1; f(2, 0)=1;f(1, 1)=0;f(0, 2) = 1; f(2, 2)=1$$ I used least squares method, $$min\sum_{k=1}^{5}(f(x_k, y_k)-\sum_{i=0}^{3}\sum_{j=0}^{3}a_{ij}(x_k)^i(y_k)^j)^2$$ Receiving this system: $$\forall t, s \in [0, 3]: \sum_{k=1}^{5}\sum_{i=0}^{3}\sum_{j=0}^{3}a_{ij}(x_k)^{i+t}(y_k)^{j+s} = \sum_{k=1}^{5}(x_k)^{t}(y_k)^{s}f(x_k, y_k)$$ If present as $Ax = b$: $$A = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ y_1 & y_2 & y_3 & y_4 & y_5 \\ . & . & . & . & .\\ (y_1)^3 & (y_2)^3 & (y_3)^3 & (y_4 )^3 & (y_5)^3 \\ x_1 & x_2 & x_3 & x_4 & x_5 \\ x_1 y_1 & x_2 y_2 & x_3 y_3 & x_4 y_4 & x_5 y_5 \\ x_1(y_1)^3 & x_2(y_2)^3 & x_3(y_3)^3 & x_4(y_4 )^3 & x_5(y_5)^3 \\ . & . & . & . & . \\ (x_1y_1)^3 & (x_2y_2)^3 & (x_3y_3)^3 & (x_4y_4 )^3 & (x_5y_5)^3 \end{pmatrix} \cdot \begin{pmatrix} 1 & y_1 & . & (y_1)^3 & x_1 & x_1y_1 & . & x_1(y_1)^3 & (x_1y_1)^3 \\ 1 & y_2 & . & (y_2)^3 & x_2 & x_2y_2 & . & x_2(y_2)^3 & (x_2y_2)^3 \\1 & y_3 & . & (y_3)^3 & x_3 & x_3y_3 & . & x_3(y_3)^3 & (x_3y_3)^3 \\ 1 & y_4 & . & (y_4)^3 & x_4 & x_4y_4 & . & x_4(y_4)^3 & (x_4y_4)^3 \\1 & y_5 & . & (y_5)^3 & x_5 & x_5y_5 & . & x_5(y_5)^3 & (x_5y_5)^3\end{pmatrix}$$ In my example, $$A = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 2 & 2 \\ 0 & 0 & 1 & 4 & 4 \\ 0 & 0 & 1 & 8 & 8 \\ 0 & 2 & 1 & 0 & 2 \\ 0 & 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & 0 & 16 \\ 0 & 4 & 1 & 0 & 4 \\ 0 & 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & 0 & 16 \\ 0 & 0 & 1 & 0 & 32 \\ 0 & 8 & 1 & 0 & 8 \\ 0 & 0 & 1 & 0 & 16 \\ 0 & 0 & 1 & 0 & 32 \\ 0 & 0 & 1 & 0 & 64 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 4 & 0 & 0 & 0 & 8 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 2 & 4 & 8 & 2 & 4 & 8 & 16 & 4 & 8 & 16 & 32 & 8 & 16 & 32 & 64\end{pmatrix}$$ But resulting matrix has null determinant. Please prompt me, in least squares method, coefficients matrix can be not inverted or need find mistake? Thanks!