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I am trying to solve a problem that seems like a standard idea from linear algebra but with a the notion of wedge product and exterior algebra added it gets more complicated for someone who isn't very comfortable using the wedge product notation. Any explanation of the following problem would be greatly appreciated.

Let $\sum_{j=1}^{n} a_{ij} v_j = b_i$ for $1 \leq i \leq m$ where the system of linear equations is given over an arbitrary commutative ring $R$. Let $c_j$ be columns of the matrix $(a_{ij}) $.

Suppose that the all minors of the matrix $(a_{ij})$ of order greater than $p$ are zero but that $c_1 \wedge c_2 \wedge \ldots \wedge c_p \neq 0$.

If we let $b = ( b_i)$ then why is it true that for $\sum_{j=1}^{n} a_{ij} v_j = b_i$ to have a solution for $1\leq i \leq m$ then it is necessary that $c_1 \wedge c_2 \wedge \ldots \wedge c_p \wedge b = 0$?

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up vote 4 down vote accepted

The key point, I think, is the translation of your problem into a geometric situation.
Let $f:R^n\to R^m$ be the $R$-linear map whose matrix in the canonical bases of source and target is $A=(a_{ij})$.
The system of equations $\sum_{j=1}^{n} a_{ij} v_j = b_i$ says exactly that $b$ is in the image $Im(f)$ of $f$ and , as is well known, $Im(f)$ is spanned by the columns $c_1,...,c_n$ of $A$.

If $R$ is a field
Then you get a better than required result.

The hypothesis $c_1 \wedge c_2 \wedge \ldots \wedge c_p \neq 0$ says that the the columns $c_1,...,c_p$ are linearly independant and the conditions on the minors say that these $p$ vectors are a basis of $Im(f)$.
It follows that the vector $b$ is in the span of $c_1,...,c_p$ if and only if $c_1 \wedge c_2 \wedge \ldots \wedge c_p \wedge b = 0$, and we saw that this condition is equivalent to the original system being solvable. So we have more than you asked for: we get a necessary and sufficient condition.

If $R$ is not a field
Then sufficiency may be false (see Edit below), which is why you were not asked to prove it!
The necessity, which is what you asked about, is still true. Here is why:
The hypothesis that minors of size $p+1$ of $A$ are zero says exactly that $\Lambda^{p+1} u:\Lambda^{p+1}R^n \to \Lambda^{p+1}R^m$ is zero.
Now suppose $b=u(a) \; (a\in R^n)$ is in the image of $f$.
Then $c_1 \wedge c_2 \wedge \ldots \wedge c_p \wedge b =u(e_1) \wedge u(e_2) \wedge \ldots \wedge u(e_p) \wedge u(a) =(\Lambda^{p+1} u)(e_1 \wedge e_2 \wedge \ldots \wedge e_p \wedge a)=0$
as required.
(By the way, the hypothesis $c_1 \wedge c_2 \wedge \ldots \wedge c_p \neq 0$ is not used )

Edit: a warning and the falsity of sufficiency
a) It is not true that $c_1 \wedge c_2 \wedge \ldots \wedge c_p \neq 0$ implies that the $c_i$'s are linearly independent.
Here is a counterexample:
Take $n=2$, $R=\mathbb Z/(4)$ and $c_1=2e_1, c_2=e_1+e_2 \in R^2$.
Then $c_1\wedge c_2\neq o \in \Lambda^2R^2$, because the isomorphism $det:\Lambda^2R^2 \stackrel {\simeq}{\to} R$ sends $c_1\wedge c_2$ to $det(c_1\wedge c_2)=det\begin{pmatrix} 2 & 1 \\ 0 & 1 \\ \end{pmatrix} = 2\neq 0 \in R$.
However the vectors $c_1, c_2$ are not linearly independent since $2e_1+0e_2=0$.

b) As a consequence, we see that the the system $\begin{pmatrix} 2 & 1 \\ 0 & 1 \\ \end{pmatrix} \: \begin{pmatrix} x \\ y \\ \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix}=b $
has no solution although $c_1\wedge c_2\neq0,c_1 \wedge c_2 \wedge b= 0$ and the condition of nullity of all $3(=p+1)$-minors is (vacuously) satisfied.
So the conditions mentioned in the question are not always sufficient for solvability when $R$ is not a field.

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Let me denote your $v_j$ by $x_j$. Assume $$ \sum_{j=1}^n\ a_{ij}\ x_j=b_i $$ for all $i$. We can rewrite this as $$ \sum_{j=1}^n\ x_j\ c_j=b. $$ We have $$ c_1\wedge\cdots\wedge c_p\wedge b=\sum_{j=p+1}^n\ x_j\ c_1\wedge\cdots\wedge c_p\wedge c_j=0, $$ the last equality following from the vanishing of the minors of order greater than $p$.

EDIT. Here is a complement to the Edit in Georges's nice answer. This is Proposition 12, Section III.7.9, of Bourbaki's Algèbre:

Let $M$ be a projective $R$-module, and let $x_1,\dots,x_n$ be in $M$. Then the $x_i$ are linearly dependent if and only if there is a nonzero $r$ in $R$ such that $$ r\ x_1\wedge\cdots\wedge x_n=0. $$

As Georges points out, the condition $$ x_1\wedge\cdots\wedge x_n\neq0 $$ does not imply that the $x_i$ are linearly independent.

Georges gives an example. Here is a minor variation: take $R:=\mathbb Z/4\mathbb Z$, $n=1$, $x_1=2$.

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