$\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