Determining consistency of a general overdetermined linear system For $m > 2$, consider the $m \times 2$ (overdetermined) linear system
$$A \mathbf{x} = \mathbf{b}$$
with (general) coefficients in a field $\mathbb{F}$; in components we write the system as
$$\left(\begin{array}{cc}a_{11} & a_{12} \\ \vdots & \vdots \\ a_{m1} & a_{m2} \end{array}\right)
\left(\begin{array}{c}x_1 \\ x_2\end{array}\right)
=
\left(\begin{array}{c}b_1 \\ \vdots \\ b_n\end{array}\right),$$
where $m > 2$, so that the system is overdetermined.
If $m = 3$ and the system is consistent, (equivalently, $\mathbf{b}$ is in the column space of $A$), then the columns of the augmented matrix $\pmatrix{ A \mid {\bf b}}$ are linearly dependent, and so
$$\det \pmatrix{ A \mid {\bf b}} = 0.$$
In particular, we have produced a polynomial in the components $(a_{ij}, b_j)$ of the linear system for which vanishing is a necessary condition for system's consistency. I'll call such polynomials polynomial obstructions for the system.
If $m > 3$, then we can produce ${m}\choose{3}$ such polynomials by considering the determinants of the $3 \times 3$ minors of $\pmatrix{ A \mid {\bf b}}$.
Are essentially all polynomials obstructions to the system essentially given by these, or are there others? Put more precisely: By definition the polynomial obstructions comprise an ideal in the polynomial ring $\mathbb{F}[a_{11}, \ldots, a_{m2}, b_1, \ldots b_m]$---do the determinants of the $3 \times 3$ minors generate this ideal? If not, how does one produce a complete set of generators?
More generally, for an $m \times n$ overdetermined linear system (so that $m > n$)
$$A \mathbf{x} = \mathbf{b},$$
we can produce polynomial obstructions by taking the determinants of the ${m}\choose{n+1}$ minors (of size $(n + 1) \times (n + 1)$). What are the answers to the obvious analogues to the above questions in the $n = 2$ case?

We assume that $b\not=0$ and that $\mathbb{F}$ is an infinite field. The required condition is $rank(A)=rank(A|b)$ or equivalently $rank(A|b)\leq rank(A)$. If $A$ is a generic matrix, then $rank(A)=n$ and your conditions are sufficient because they are equivalent to $rank(A|b)\leq n=rank(A)$. Yet, if $A$ is not generic and $rank(A)=r<n$, then it is not sufficient -for instance assume that the first column of $A$ is zero-. You must extract from $A$ a $r\times r$ invertible submatrix $A'$; we can assume that $A'$ is formed using the first $r$ columns and rows of $A$. After you construct, for $r<i\leq m$, the $(r+1)\times (r+1)$ matrices $U_i=\begin{pmatrix}A'&[b_1,\cdots,b_r]^T\\L_i&b_i\end{pmatrix}$ where $L_i=[a_{i,1},\cdots,a_{i,r}]$ and you say that their determinants are zero. Of course, if you know $r$ and not $A'$, then you must consider all the $(r+1)\times (r+1)$ matrices, extracted from $(A|b)$, containing in the last column a part of the vector $b$. If you do not know $r$, that is if $A$ is a formal matrix, then $r=0$ or $r=1$ or ...$r=n$ and you apply that is said above.
