Consider a generic $n \times n$ matrix $M$.

Define the $(n-1) \times n$ matrix $M_q$ to be $M$ with the $q$th row omitted, and assume that $M_q$ possesses a right inverse, $R_q$:

$$R_q = M_q^T (M_q M_q^T)^{-1}$$

The components of $R_q$ will be rational functions of the components of $M$, with the determinant $\det(M_q M_q^T)$ as their denominator.

Now, consider the $n \times n$ matrix $C$ of cofactors of $M$. That is, $C_{i j}$ is $(-1)^{i+j}$ times the determinant of the submatrix of $M$ obtained by omitting the $i$th row and the $j$th column.

I have found by direct computation for low values of $n$ that the antisymmetric product of the $q$th row of $C$ and the vector formed by summing the components in each row of $R_q$:

$$(A_q)_{i j} = \sum_{k=1}^{n-1} (C_{q i} (R_q)_{j k} - C_{q j} (R_q)_{i k})$$

has components that are polynomials of degree $n-2$ in the components of $M$, with the determinant that one might expect to be present in the denominator, inherited from $R_q$, factoring out.

To give an example, for $n=3$:

$$M_1= \left( \begin{array}{ccc} m_{2,1} & m_{2,2} & m_{2,3} \\ m_{3,1} & m_{3,2} & m_{3,3} \\ \end{array} \right)$$

$$C= \left( \begin{array}{ccc} m_{2,2} m_{3,3}-m_{2,3} m_{3,2} & m_{2,3} m_{3,1}-m_{2,1} m_{3,3} & m_{2,1} m_{3,2}-m_{2,2} m_{3,1} \\ m_{1,3} m_{3,2}-m_{1,2} m_{3,3} & m_{1,1} m_{3,3}-m_{1,3} m_{3,1} & m_{1,2} m_{3,1}-m_{1,1} m_{3,2} \\ m_{1,2} m_{2,3}-m_{1,3} m_{2,2} & m_{1,3} m_{2,1}-m_{1,1} m_{2,3} & m_{1,1} m_{2,2}-m_{1,2} m_{2,1} \\ \end{array} \right)$$

$$ R_1 = M_1^T (M_1 M_1^T)^{-1} = \\ {\scriptsize \frac{ \left( \begin{array}{cc} m_{2,1} \left(m_{3,2}^2+m_{3,3}^2\right)-m_{3,1} \left(m_{2,2} m_{3,2}+m_{2,3} m_{3,3}\right) & m_{3,1} \left(m_{2,2}^2+m_{2,3}^2\right) -m_{2,1} \left(m_{2,2} m_{3,2}+m_{2,3} m_{3,3}\right) \\ m_{2,2} \left(m_{3,1}^2+m_{3,3}^2\right)-m_{3,2} \left(m_{2,1} m_{3,1}+m_{2,3} m_{3,3}\right) & m_{3,2} \left(m_{2,1}^2+m_{2,3}^2\right) -m_{2,2} \left(m_{2,1} m_{3,1}+m_{2,3} m_{3,3}\right) \\ m_{2,3} \left(m_{3,1}^2+m_{3,2}^2\right)-m_{3,3} \left(m_{2,1} m_{3,1}+m_{2,2} m_{3,2}\right) & m_{3,3} \left(m_{2,1}^2+m_{2,2}^2\right) -m_{2,3} \left(m_{2,1} m_{3,1}+m_{2,2} m_{3,2}\right) \\ \end{array} \right)} {\left(m_{3,2}^2+m_{3,3}^2\right) m_{2,1}^2-2 m_{2,3} m_{3,1} m_{3,3} m_{2,1}+m_{2,3}^2 \left(m_{3,1}^2+m_{3,2}^2\right)-2 m_{2,2} m_{3,2} \left(m_{2,1} m_{3,1}+m_{2,3} m_{3,3}\right)+m_{2,2}^2 \left(m_{3,1}^2+m_{3,3}^2\right)} } $$

$$\sum_{k=1}^{n-1} (R_1)_{i k} = \\ {\scriptsize \frac{ \left( \begin{array}{c} m_{3,1} \left(m_{2,2}^2-m_{3,2} m_{2,2}+m_{2,3}^2-m_{2,3} m_{3,3}\right)+m_{2,1} \left(m_{3,2}^2-m_{2,2} m_{3,2}+m_{3,3}^2-m_{2,3} m_{3,3}\right) \\ m_{3,2} \left(m_{2,1}^2-m_{3,1} m_{2,1}+m_{2,3}^2-m_{2,3} m_{3,3}\right)+m_{2,2} \left(m_{3,1}^2-m_{2,1} m_{3,1}+m_{3,3}^2-m_{2,3} m_{3,3}\right) \\ m_{2,3} \left(m_{3,1}^2-m_{2,1} m_{3,1}+m_{3,2}^2-m_{2,2} m_{3,2}\right)+ m_{3,3} \left(m_{2,1}^2-m_{3,1} m_{2,1}+m_{2,2}^2-m_{2,2} m_{3,2}\right) \\ \end{array} \right)} {\left(m_{3,2}^2+m_{3,3}^2\right) m_{2,1}^2-2 m_{2,3} m_{3,1} m_{3,3} m_{2,1}+m_{2,3}^2 \left(m_{3,1}^2+m_{3,2}^2\right)-2 m_{2,2} m_{3,2} \left(m_{2,1} m_{3,1}+m_{2,3} m_{3,3}\right)+m_{2,2}^2 \left(m_{3,1}^2+m_{3,3}^2\right)} }$$

$$A_1= \sum_{k=1}^{n-1} (C_{1 i} (R_1)_{j k} - C_{1 j} (R_1)_{i k}) = \left( \begin{array}{ccc} 0 & m_{3,3}-m_{2,3} & m_{2,2}-m_{3,2} \\ m_{2,3}-m_{3,3} & 0 & m_{3,1}-m_{2,1} \\ m_{3,2}-m_{2,2} & m_{2,1}-m_{3,1} & 0 \\ \end{array} \right)$$

My question is: why are the components of $A_q$ polynomials rather than rational functions? Can this be proved for general $n$? And can $A_q$ be reduced, for general $n$, to a simpler expression that makes no reference to $R_q$?