This is a question about how the determinantal rank of an augmented matrix relates to linear independence. It came up for me when I was trying to follow the proof of Bertini's Theorem in Vakil's Foundations of Algebraic Geometry, but I think it's just a question of linear algebra.
Suppose I have a field $k$ and an $n \times r$ matrix $$ J = \begin{pmatrix} a_{00} & \dotsm & a_{0r} \\ \vdots & \ddots & \vdots \\ a_{n0} & \dotsm & a_{nr} \end{pmatrix} $$ with entries in $k$, and that $J$ has nonzero corank $d$. Let $A = k[x_{0}, \dotsc, x_{n}]$ and consider the augmented matrix $$ \overline{J} = \begin{pmatrix} a_{00} & \dotsm & a_{0r} & x_{0} \\ \vdots & \ddots & \vdots & \vdots \\ a_{n0} & \dotsm & a_{nr} & x_{n} \end{pmatrix} $$ with entries in $A$. The condition $\operatorname{corank} \overline{J} \geq d$ amounts to the vanishing of all $m \times m$ minors of $\overline{J}$ when $m > n - d$. Any such minor not meeting the last column is a minor of $J$, hence will vanish because $\operatorname{corank} J = d$. Any minor meeting the last column will be a linear combination of the $x_{i}$ whose coefficients are $(m-1) \times (m-1)$ minors of $J$, and so will also vanish unless $m = n + 1 - d$. So the condition $\operatorname{corank} \overline{J} \geq d$ amounts to the vanishing of a family of linear polynomials $f_{1}, \dotsc, f_{s} \in A$, where $s = \binom{n}{n+1-d}$ and each $f_{i}$ is the determinant of an $(n+1-d) \times (n+1-d)$ minor of $\overline{J}$ that meets the last column.
My question is: how many of the $f_{1}, \dotsc, f_{s}$ are linearly independent elements of $A$? Vakil's proof of Bertini's Theorem suggests that the answer is $d$ or $d+1$. I'm not really sure how to approach the problem. I thought about trying to use the connection between vanishing minors and linear independence of columns, but that concerns linear independence in $k^{n}$, not linear independence in $A$.