$\let\sumnonlimits\sum
\let\prodnonlimits\prod
\renewcommand{\sum}{\sumnonlimits\limits}
\renewcommand{\prod}{\prodnonlimits\limits}
$ I will write $R$ for $R_{q}$, because $q$ is constant. In the following, I am
going to assume that $R$ is an *arbitrary* right inverse of $M_{q}$, rather
than the specific right inverse $M_{q}^{T}\left(  M_{q}M_{q}^{T}\right)
^{-1}$ which you have suggested. This makes the statement a tad more general
and rids us of a red herring.

I shall prove that for any
$i\in\left\{  1,2,...,n\right\}  $, $j\in\left\{  1,2,...,n\right\}  $ and
$k\in\left\{  1,2,...,n-1\right\}  $, the value $C_{qi}R_{jk}-C_{qj}R_{ik}$ is
a homogeneous polynomial of degree $n-2$ in the entries of the matrix $M_{q}$
(independently, and independent, of the choice of right inverse $R$).

For every $\ell\in\left\{  1,2,...,n\right\}  $, we abbreviate $C_{q\ell}$ by
$v_{q}$. Thus,

$v_{\ell}=C_{q\ell}=\left(  -1\right)  ^{q+\ell}\det\left(
\underbrace{M\text{ without row }q}_{=M_{q}}\text{ and column }\ell\right)  $

**(1)** $=\left(  -1\right)  ^{q+\ell}\det\left(  M_{q}\text{ without column
}\ell\right)  $.

From this it easy to obtain

**(2)** $\sum_{\ell\in\left\{  1,2,...,n\right\}  }\left(  M_{q}\right)
_{j\ell}v_{\ell}=0$ for every $j \in \left\{1,2,...,n-1\right\}$.

[*Proof of **(2)**:* Let $j \in \left\{1,2,...,n-1\right\}$. Let $G$ be the result of inserting a copy of row $j$ of the matrix $M_{q}$ between the rows $q-1$ and $q$ of this matrix. Then, $G$ is an $n\times n$-matrix with two equal rows, and hence has determinant $\det G=0$. But Laplace expansion of $\det G$ along the row we have inserted yields

$\det G=\sum_{\ell\in\left\{  1,2,...,n\right\}  }\left(  M_{q}\right)
_{j\ell}\left(  -1\right)  ^{q+\ell}\det\left(  M_{q}\text{ without column
}\ell\right)  $,

since the entries of this row are $\left(  M_{q}\right)  _{j\ell}$ and the
cofactors corresponding to this row are $\left(  -1\right)  ^{q+\ell}
\det\left(  M_{q}\text{ without column }\ell\right)  $. Now, **(1)** yields

$\sum_{\ell\in\left\{  1,2,...,n\right\}  }\left(  M_{q}\right)  _{j\ell
}v_{\ell}$

$=\sum_{\ell\in\left\{  1,2,...,n\right\}  }\left(  M_{q}\right)  _{j\ell
}\left(  -1\right)  ^{q+\ell}\det\left(  M_{q}\text{ without column }
\ell\right)  $

$=\det G=0$,

so that **(2)** is proven.]

Since $M$ is generic, its cofactors $v_{1}$, $v_{2}$, $...$, $v_{n}$ are nonzero.

Fix $i\in\left\{  1,2,...,n\right\}  $. Let $N_{i}$ denote the matrix $M_{q}$
without column $i$. Thus, $\det\left(  N_{i}\right)  =\det\left(  M_{q}\text{
without column }i\right)  =\left(  -1\right)  ^{q+i}v_{i}$ (because **(1)**
yields $v_{i}=\left(  -1\right)  ^{q+i}\det\left(  M_{q}\text{ without column
}i\right)  $).

Let $S_{i}$ denote the matrix $\left(  R_{jk}-\dfrac{v_{j}}{v_{i}}
R_{ik}\right)  _{j\in\left\{  1,2,...,n\right\}  ;\ k\in\left\{
1,2,...,n-1\right\}  }$. The $i$-th row of this matrix $S_{i}$ is zero; let
$S_{i}^{\prime}$ denote the matrix $S_{i}$ without row $i$.

Now, we claim that $N_{i}S_{i}^{\prime}=I_{n-1}$ (the $\left(  n-1\right)
\times\left(  n-1\right)  $ identity matrix). Indeed, for every $\left(
j,k\right)  \in\left\{  1,2,...,n-1\right\}  ^{2}$, the $\left(  j,k\right)
$-th entry of $N_{i}S_{i}^{\prime}$ is

$\left(  N_{i}S_{i}^{\prime}\right)  _{jk}=\sum_{u\in\left\{
1,2,...,n-1\right\}  }\left(  N_{i}\right)  _{ju}\left(  S_{i}^{\prime
}\right)  _{uk}$

$=\sum_{\ell\in\left\{  1,2,...,n\right\}  \setminus\left\{  i\right\}
}\left(  M_{q}\right)  _{j\ell}\underbrace{\left(  S_{i}\right)  _{\ell k}
}_{=R_{\ell k}-\dfrac{v_{\ell}}{v_{i}}R_{ik}}$

(since $N_{i}$ is the matrix $M_{q}$ without column $i$, while $S_{i}^{\prime
}$ is the matrix $S_{i}$ without row $i$)

$=\sum_{\ell\in\left\{  1,2,...,n\right\}  \setminus\left\{  i\right\}
}\left(  M_{q}\right)  _{j\ell}\left(  R_{\ell k}-\dfrac{v_{\ell}}{v_{i}
}R_{ik}\right)  $

$=\sum_{\ell\in\left\{  1,2,...,n\right\}  }\left(  M_{q}\right)  _{j\ell
}\left(  R_{\ell k}-\dfrac{v_{\ell}}{v_{i}}R_{ik}\right)  $

(here, we added an $\ell=i$ addend to the sum; this did not change the sum
because this addend is $0$)

$=\underbrace{\sum_{\ell\in\left\{  1,2,...,n\right\}  }\left(  M_{q}\right)
_{j\ell}R_{\ell k}}_{\substack{=\left(  M_{q}R\right)  _{jk}=\delta
_{jk}\\\text{(since }R\text{ is a}\\\text{right inverse of }M_{q}\text{)}
}}-\underbrace{\sum_{\ell\in\left\{  1,2,...,n\right\}  }\left(  M_{q}\right)
_{j\ell}\dfrac{v_{\ell}}{v_{i}}R_{ik}}_{=\dfrac{1}{v_{i}}R_{ik}\sum_{\ell
\in\left\{  1,2,...,n\right\}  }\left(  M_{q}\right)  _{j\ell}v_{\ell}}$

$=\delta_{jk}-\dfrac{1}{v_{i}}R_{ik}\underbrace{\sum_{\ell\in\left\{
1,2,...,n\right\}  }\left(  M_{q}\right)  _{j\ell}v_{\ell}}
_{\substack{=0\\\text{(by \textbf{(2)})}}}=\delta_{jk}$.

This shows that $N_{i}S_{i}^{\prime}=I_{n-1}$, and thus $S_{i}^{\prime}
=N_{i}^{-1}$ (since $N_{i}$ and $S_{i}^{\prime}$ are $\left(  n-1\right)
\times\left(  n-1\right)  $-matrices).

Now, we recall Cramer's rule for the inverse of a matrix. It essentially says
that the inverse of a square matrix is obtained by dividing its adjoint by its
determinant. In other words, if $X$ is an invertible $p\times p$-matrix for
some $p\in\mathbb{N}$, and $j$ and $k$ are elements of $\left\{
1,2,...,p\right\}  $, then

**(3)** $\left(  X^{-1}\right)  _{jk}=\dfrac{1}{\det X}\left(  -1\right)
^{j+k}\det\left(  X\text{ without row }k\text{ and column }j\right)  $.

Now, recall that $S_{i}^{\prime}=N_{i}^{-1}$. Hence,

$\left(  S_{i}^{\prime}\right)  _{jk}=\left(  N_{i}^{-1}\right)  _{jk}
=\dfrac{1}{\det\left(  N_{i}\right)  }\left(  -1\right)  ^{j+k}\det\left(
N_{i}\text{ without row }k\text{ and column }j\right)  $ (by **(3)**)

$=\dfrac{1}{\left(  -1\right)  ^{q+i}v_{i}}\left(  -1\right)  ^{j+k}
\det\left(  N_{i}\text{ without row }k\text{ and column }j\right)  $ (since
$\det\left(  N_{i}\right)  =\left(  -1\right)  ^{q+i}v_{i}$)

$=\dfrac{1}{v_{i}}\left(  -1\right)  ^{i+j+q+k}\det\left(  N_{i}\text{ without
row }k\text{ and column }j\right)  $

for all $\left(  j,k\right)  \in\left\{  1,2,...,n-1\right\}  ^{2}$. In other words,

**(4)** $v_{i}\left(  S_{i}^{\prime}\right)  _{jk}=\left(  -1\right)
^{i+j+q+k}\det\left(  N_{i}\text{ without row }k\text{ and column }j\right)  $

for all $\left(  j,k\right)  \in\left\{  1,2,...,n-1\right\}  ^{2}$.

Let $\mathbf{B}$ be the (unique) increasing bijection from $\left\{
1,2,...,n\right\}  \setminus\left\{  i\right\}  $ to $\left\{
1,2,...,n-1\right\}  $. Then, for all $j\in\left\{  1,2,...,n\right\}
\setminus\left\{  i\right\}  $ and $k\in\left\{  1,2,...,n-1\right\}  $, we have

**(5)** $v_{i}\left(  S_{i}\right)  _{jk}=\left(  -1\right)  ^{i+\mathbf{B}
\left(  j\right)  +q+k}\det\left(  M_{q}\text{ without row }k\text{ and
columns }i\text{ and }j\right)  $.

(Indeed, this follows from **(4)**, applied to $\mathbf{B}\left(  j\right)  $
instead of $j$, because $\left(  S_{i}^{\prime}\right)  _{\mathbf{B}\left(
j\right)  ,k}=\left(  S_{i}\right)  _{jk}$ (since $S_{i}^{\prime}$ is the
matrix $S_{i}$ without row $i$) and because column $j$ of $M_{q}$ is column
$\mathbf{B}\left(  j\right)  $ of $N_{i}$ (since $N_{i}$ is the matrix $M_{q}$
without column $i$).)

Now, fix $j\in\left\{  1,2,...,n\right\}  $ and $k\in\left\{
1,2,...,n-1\right\}  $. We need to show that $C_{qi}R_{jk}-C_{qj}R_{ik}$ is a
homogeneous polynomial of degree $n-2$ in the entries of the matrix $M_{q}$.
We WLOG assume that $j\neq i$ (since otherwise, $C_{qi}R_{jk}-C_{qj}R_{ik}
=0$). Thus, $j\in\left\{  1,2,...,n\right\}  \setminus\left\{  i\right\}  $.
Since $C_{qi}=v_{i}$ and $C_{qj}=v_{j}$ (by the definitions of $v_{i}$ and
$v_{j}$), we have

$\underbrace{C_{qi}}_{=v_{i}}R_{jk}-\underbrace{C_{qj}}_{=v_{j}}R_{ik}
=v_{i}R_{jk}-v_{j}R_{ik}=v_{i}\underbrace{\left(  R_{jk}-\dfrac{v_{j}}{v_{i}
}R_{ik}\right)  }_{\substack{=\left(  S_{i}\right)  _{jk}\\\text{(by the
definition of }S_{i}\text{)}}}$

**(6)** $=v_{i}\left(  S_{i}\right)  _{jk}=\left(  -1\right)  ^{i+\mathbf{B}\left(
j\right)  +q+k}\det\left(  M_{q}\text{ without row }k\text{ and columns
}i\text{ and }j\right)  $

(by **(5)**),

which is obviously a homogeneous polynomial of degree $n-2$ in the entries of
the matrix $M_{q}$ (and independent of the choice of $R$), qed.

Thanks for a very nice question, and sorry for this mess of an answer...

**PS.** I believe the genericity of $M$ is not required for **(6)** to hold. Does anyone see a nice proof of this? I don't.

**PPS.** Here is a more general statement which, I think, is true. Let $A$ be a commutative ring. Let $N$ be an $\left(n-1\right) \times n$-matrix over $A$. Let $s \in A^n$ be a vector. For every $\ell \in \left\{1,2,...,n\right\}$, let $p_\ell$ denote the scalar $\left(  -1\right)  ^{\ell}\det\left( N \text{ without column
}\ell\right)$. If $w$ is a vector and $i$ is an integer, we denote by $w_i$ the $i$-th coordinate of $w$ (whenever this makes sense). Then, every two distinct $i \in \left\{1,2,...,n\right\}$ and $j \in \left\{1,2,...,n\right\}$ satisfy

**(11)** $p_i s_j - p_j s_i = \sum_{k=1}^{n-1} \pm \left(Ns\right)_k \det\left(N \text{ without row } k \text{ and columns } i \text{ and } j\right)$,

where $\pm$ is something like $\left(-1\right)^{i+j+\left[i<j\right]+k}$ using the Iverson bracket.

If this is proven (and this shouldn't be too hard -- it's a polynomial identity, so you can assume as much genericity as you wish), the original result is obtained by setting $N = M_q$ and $s = \left(k\text{-th column of } R\right)$.

**PPPS.** Indeed, **(11)** is not hard to prove. Let $e_1, e_2, ..., e_n$ be the $n$ standard basis vectors of $A^n$. Let $P$ be the $n \times \left(n-1\right)$-matrix whose columns (from left to right) are $e_1, e_2, ..., \widehat{e_i}, ..., \widehat{e_j}, ..., e_n, s$ (where $\widehat{\text{something}}$ means omission, and the order of $i$ and $j$ is not necessarily the one we have shown). Then, $NP$ is the $\left(n-1\right)\times \left(n-1\right)$-matrix whose columns (from left to right) are the columns $1, 2, ..., \widehat{i}, ..., \widehat{j}, ..., n$ of $N$ and the column-vector $Ns$. The right hand side of **(11)** is the determinant of this matrix $NP$, computed by Laplace expansion along its last column. The left hand side of **(11)** is the same determinant, computed using the [Cauchy-Binet formula][1] (which takes a rather simple form here because if we remove the $\ell$-th row from $P$ for some $\ell \notin \left\{i, j\right\}$, then the resulting matrix has two rows with only their rightmost entries nonzero, and so has determinant $0$).

**PPPPS.** Here is a different way to formulate the above proof of **(11)**,
using exterior algebra instead of the Cauchy-Binet formula and giving some
more detail.

Let $e_{1}$, $e_{2}$, $...$, $e_{n}$ be the $n$ standard basis vectors of
$A^{n}$. Let $f_{1}$, $f_{2}$, $...$, $f_{n-1}$ be the $n-1$ standard basis
vectors of $A^{n-1}$. We identify the matrix $N\in A^{\left(  n-1\right)
\times n}$ with the $A$-linear map $A^{n}\rightarrow A^{n-1}$ it represents.
As usual, we use the notation ''$\widehat{\text{some term}}$'' for omission of a term in a product or list (for example,
$\left(  1,2,...,\widehat{5},...,8\right)  =\left(  1,2,3,4,6,7,8\right)  $).
We also use the Iverson bracket notation, i.e., whenever $\mathcal{A}$ is a
logical statement, we write $\left[  \mathcal{A}\right]  $ for the integer
$\left\{
\begin{array}
[c]{c}
1,\text{ if }\mathcal{A}\text{ is true;}\\
0,\text{ if }\mathcal{A}\text{ is false}
\end{array}
\right.  $.

Let $\mathbf{f}$ be the element $f_{1}\wedge f_{2}\wedge...\wedge f_{n-1}$ of
$\wedge^{n-1}\left(  A^{n-1}\right)  $. It is known that $\left( \mathbf{f}\right)  $
is an $A$-module basis of $\wedge^{n-1}\left(  A^{n-1}\right)  $.

Let $i$ and $j$ be two distinct elements of $\left\{  1,2,...,n\right\}  $. We
have $s=\sum_{k=1}^{n}s_{k}e_{k}$, so that

$e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i}}\wedge...\wedge\widehat{e_{j}
}\wedge...\wedge e_{n}\wedge s$

$=e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i}}\wedge...\wedge
\widehat{e_{j}}\wedge...\wedge e_{n}\wedge\left(  \sum_{k=1}^{n}s_{k}
e_{k}\right)  $

**(12)** $=\sum_{k=1}^{n}s_{k}e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i}
}\wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}\wedge e_{k}$

(where the notation $e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i}}
\wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}$ means that both $e_{i}$
and $e_{j}$ are omitted, but does not necessarily imply that $i<j$). Of all
the addends of the sum on the right hand side of **(12)**, only those for
$k=i$ and for $k=j$ have a chance to be nonzero (every other summand contains
a wedge product with two equal factors, and thus vanishes). Hence, **(12)**
simplifies to

$e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i}}\wedge...\wedge\widehat{e_{j}
}\wedge...\wedge e_{n}\wedge s$

$=s_{i}\underbrace{e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i}}
\wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}\wedge e_{i}}_{=\left(
-1\right)  ^{i+\left[  i<j\right]  }e_{1}\wedge e_{2}\wedge...\wedge
\widehat{e_{j}}\wedge...\wedge e_{n}}+s_{j}\underbrace{e_{1}\wedge e_{2}
\wedge...\wedge\widehat{e_{i}}\wedge...\wedge\widehat{e_{j}}\wedge...\wedge
e_{n}\wedge e_{j}}_{=\left(  -1\right)  ^{j+1-\left[  i<j\right]  }e_{1}\wedge
e_{2}\wedge...\wedge\widehat{e_{i}}\wedge...\wedge e_{n}}$

$=s_{i}\left(  -1\right)  ^{i+\left[  i<j\right]  }e_{1}\wedge e_{2}
\wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}+s_{j}\left(  -1\right)
^{j+1-\left[  i<j\right]  }e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i}
}\wedge...\wedge e_{n}$

$=\left(  -1\right)  ^{\left[  i<j\right]  +i+j-1}\left(  s_{j}\left(
-1\right)  ^{i}e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i}}\wedge...\wedge
e_{n}-s_{i}\left(  -1\right)  ^{j}e_{1}\wedge e_{2}\wedge...\wedge
\widehat{e_{j}}\wedge...\wedge e_{n}\right)  $.

Applying the linear map $\wedge^{n-1}N$ to both sides of this equality results in

$\left(  \wedge^{n-1}N\right)  \left(  e_{1}\wedge e_{2}\wedge...\wedge
\widehat{e_{i}}\wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}\wedge
s\right)  $

$=\left(  -1\right)  ^{\left[  i<j\right]  +i+j-1}\left(  s_{j}\left(
-1\right)  ^{i}\underbrace{\left(  \wedge^{n-1}N\right)  \left(  e_{1}\wedge
e_{2}\wedge...\wedge\widehat{e_{i}}\wedge...\wedge e_{n}\right)  }
_{=\det\left(  N\text{ without column }i\right)  \mathbf{f}}\right.  $

$\left.  -s_{i}\left(  -1\right)  ^{j}\underbrace{\left(  \wedge
^{n-1}N\right)  \left(  e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{j}}
\wedge...\wedge e_{n}\right)  }_{=\det\left(  N\text{ without column
}j\right)  \mathbf{f}}\right)  $

$=\left(  -1\right)  ^{\left[  i<j\right]  +i+j-1}\left(  s_{j}
\underbrace{\left(  -1\right)  ^{i}\det\left(  N\text{ without column
}i\right)  }_{=p_{i}} \mathbf{f} -s_{i}\underbrace{\left(  -1\right)  ^{j}
\det\left(  N\text{ without column }j\right)  }_{=p_{j}} \mathbf{f} \right) $

$=\left(  -1\right)  ^{\left[  i<j\right]  +i+j-1}\left(  s_{j}p_{i}
\mathbf{f}-s_{i}p_{j}\mathbf{f}\right)  =\left(  -1\right)  ^{\left[
i<j\right]  +i+j-1}\left(  p_{i}s_{j}-p_{j}s_{i}\right)  \mathbf{f}$,

so that

$\left(  p_{i}s_{j}-p_{j}s_{i}\right)  \mathbf{f}$

**(13)** $=\left(  -1\right)  ^{\left[  i<j\right]  +i+j-1}\left(
\wedge^{n-1}N\right)  \left(  e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i}
}\wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}\wedge s\right)  $.

On the other hand, using $\wedge$ to denote the multiplication in the exterior
algebra $\wedge\left(  A^{n-1}\right)  $, we see

$\left(  \wedge^{n-1}N\right)  \left(  e_{1}\wedge e_{2}\wedge...\wedge
\widehat{e_{i}}\wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}\wedge
s\right)  $

$=\underbrace{\left(  \wedge^{n-2}N\right)  \left(  e_{1}\wedge e_{2}
\wedge...\wedge\widehat{e_{i}}\wedge...\wedge\widehat{e_{j}}\wedge...\wedge
e_{n}\right)  }_{=\sum_{\ell=1}^{n-1}\det\left(  N\text{ without columns
}i\text{ and }j\text{ and row }\ell\right)  f_{1}\wedge f_{2}\wedge
...\wedge\widehat{f_{\ell}}\wedge...\wedge f_{n-1}}\wedge\underbrace{\left(
Ns\right)  }_{=\sum_{k=1}^{n-1}\left(  Ns\right)  _{k}f_{k}}$

$=\left(  \sum_{\ell=1}^{n-1}\det\left(  N\text{ without columns }i\text{ and
}j\text{ and row }\ell\right)  f_{1}\wedge f_{2}\wedge...\wedge
\widehat{f_{\ell}}\wedge...\wedge f_{n-1}\right)  \wedge\left(  \sum
_{k=1}^{n-1}\left(  Ns\right)  _{k}f_{k}\right)  $

**(14)** $=\sum_{\ell=1}^{n-1}\sum_{k=1}^{n-1}\left(  Ns\right)  _{k}
\det\left(  N\text{ without columns }i\text{ and }j\text{ and row }
\ell\right)  f_{1}\wedge f_{2}\wedge...\wedge\widehat{f_{\ell}}\wedge...\wedge
f_{n-1}\wedge f_{k}$.

Of all the addends of the inner sum on the right hand side of **(14)**, only
those for $k=\ell$ have a chance to be nonzero (because all other addends have
the form $f_{1}\wedge f_{2}\wedge...\wedge\widehat{f_{\ell}}\wedge...\wedge
f_{n-1}\wedge f_{k}$ for $k\neq\ell$, which is a wedge product with two equal
factors and thus equals $0$). Hence, **(14)** simplifies to

$\left(  \wedge^{n-1}N\right)  \left(  e_{1}\wedge e_{2}\wedge...\wedge
\widehat{e_{i}}\wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}\wedge
s\right)  $

$=\sum_{\ell=1}^{n-1}\left(  Ns\right)  _{\ell}\det\left(  N\text{ without
columns }i\text{ and }j\text{ and row }\ell\right)  \underbrace{f_{1}\wedge
f_{2}\wedge...\wedge\widehat{f_{\ell}}\wedge...\wedge f_{n-1}\wedge f_{\ell}
}_{=\left(  -1\right)  ^{n-1-\ell}f_{1}\wedge f_{2}\wedge...\wedge f_{n-1}}$

$=\sum_{\ell=1}^{n-1}\left(  Ns\right)  _{\ell}\det\left(  N\text{ without
columns }i\text{ and }j\text{ and row }\ell\right)  \left(  -1\right)
^{n-1-\ell}\underbrace{f_{1}\wedge f_{2}\wedge...\wedge f_{n-1}}_{=\mathbf{f}
}$

$=\sum_{\ell=1}^{n-1}\left(  Ns\right)  _{\ell}\det\left(  N\text{ without
columns }i\text{ and }j\text{ and row }\ell\right)  \left(  -1\right)
^{n-1-\ell}\mathbf{f}$.

Thus, **(13)** becomes

$\left(  p_{i}s_{j}-p_{j}s_{i}\right)  \mathbf{f}$

$=\left(  -1\right)  ^{\left[  i<j\right]  +i+j-1}\underbrace{\left(
\wedge^{n-1}N\right)  \left(  e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i}
}\wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}\wedge s\right)  }
_{=\sum_{\ell=1}^{n-1}\left(  Ns\right)  _{\ell}\det\left(  N\text{ without
columns }i\text{ and }j\text{ and row }\ell\right)  \left(  -1\right)
^{n-1-\ell}\mathbf{f}}$

$=\left(  -1\right)  ^{\left[  i<j\right]  +i+j-1}\sum_{\ell=1}^{n-1}\left(
Ns\right)  _{\ell}\det\left(  N\text{ without columns }i\text{ and }j\text{
and row }\ell\right)  \left(  -1\right)  ^{n-1-\ell}\mathbf{f}$.

Since $\left(  \mathbf{f}\right)  $ is a basis of $\wedge^{n-1}\left(
A^{n-1}\right)  $, we can compare coefficients before $\mathbf{f}$ in this
equality, and obtain

$p_{i}s_{j}-p_{j}s_{i}$

$=\left(  -1\right)  ^{\left[  i<j\right]  +i+j-1}\sum_{\ell=1}^{n-1}\left(
Ns\right)  _{\ell}\det\left(  N\text{ without columns }i\text{ and }j\text{
and row }\ell\right)  \left(  -1\right)  ^{n-1-\ell}$

$=\sum_{\ell=1}^{n-1}\left(  -1\right)  ^{\left[  i<j\right]  +i+j+n-\ell
}\left(  Ns\right)  _{\ell}\det\left(  N\text{ without columns }i\text{ and
}j\text{ and row }\ell\right)  $

$=\sum_{k=1}^{n-1}\left(  -1\right)  ^{\left[  i<j\right]  +i+j+n-k}\left(
Ns\right)  _{k}\det\left(  N\text{ without columns }i\text{ and }j\text{ and
row }k\right)  $.

This proves **(11)**, up to all the sign errors I surely have made.

As a consequence, we obtain a new proof of your statement, and it does no longer rely on genericity of $M$.

  [1]: https://en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula