TITLE: Using the notation of wedge product to solve a linear system of equations
QUESTION [5 upvotes]: 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 $c_1 \wedge c_2 \wedge \ldots \wedge c_p \wedge b = 0$?

REPLY [2 votes]: 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 $R:=\mathbb Z/4\mathbb Z$, $n=1$, $x_1=2$.