Let us consider the short exact sequence of coherent sheaves on $\mathbb{P}^n$ $$0 \to \mathcal{O}_{\mathbb P^n}(-1)^{r} \stackrel{N}{\longrightarrow} \mathcal{O}_{\mathbb P^n}^{r} \longrightarrow \mathcal F \to 0,$$ where $r \geq 2$ and $N=\{L_{ij}\}$ is a $r \times r$ matrix of linear forms in the variables $x_0, \ldots, x_n$ acting on the left (i.e., $\mathbf{x} \to N \mathbf{x}$).

Then the sheaf $\mathcal{F}$ is supported on the determinantal hypersurface $Y:=\{\det N=0 \} \subset \mathbb{P}^n$, and it is locally free of rank $1$ (i.e., a line bundle) on the open dense locus of $Y$ given by $\textrm{rank } N = r-1$.

Passing to cohomology, we obtain an isomorphism $$\sigma \colon H^0(\mathbb{P}^n, \, \mathcal{O}_{\mathbb P^n}^{r}) \stackrel{\cong}{\longrightarrow} H^0(Y, \, \mathcal F),$$ in particular $h^0(Y, \, \mathcal F)=r.$ In other words, this is a bijective correspondence between $n$-ples $(\alpha_1, \ldots, \alpha_r)$ of complex numbers and global sections of $\mathcal F$.

Question.How can we describe the zero locus in $Y$ of the section $\sigma(\alpha_1, \ldots, \alpha_n)$ in terms of the $\alpha_i$ and the matrix $N$?

**Toy model.** Consider the case $n=3$, $r=2$ and $$N = \pmatrix{x_0 & x_1 \\ x_2 & x_3}.$$
In this situation, $Y$ is the smooth quadric $\{x_0x_3-x_1x_2=0 \} \subset \mathbb{P}^3$, and $\mathcal{F}$ is a line bundle with $2$ global sections, hence a pencil corrisponding to one of the two rulings of $Y$. I made some (unelegant) local computation and it seems to me that in this case $\sigma(\alpha_1, \, \alpha_2)$ should be something like $\alpha_1 \sigma_{02} + \alpha_2 \sigma_{13}$, where $\sigma_{02}$ is the section whose zero locus is the line $\{x_0=x_2=0 \}$ and $\sigma_{13}$ is the section whose zero locus is the line (in the same ruling) $\{x_1=x_3=0 \}.$ In other words, the global sections of $\mathcal{F}$ are expressed in terms of the *columns* of $N$.

On the other hand, dualizing our exact sequence and applying Grothendieck duality we get $$0 \to \mathcal{O}_{\mathbb P^3}(-1)^{2} \stackrel{ { }^t N}{\longrightarrow} \mathcal{O}_{\mathbb P^3}^{2} \longrightarrow \mathcal F^*(1) \to 0,$$ where $\mathcal{F}^*:=\mathscr{Hom}(\mathcal{F}, \, \mathcal{O}_{\mathbb P^3})$.

Now the sections of $\mathcal{F}^*(1)$ should correspond the other ruling of the quadric, because $\mathcal{F} \otimes \mathcal{F}^*(1)= \mathcal{O}_Y(1)$. Therefore the global sections of $\mathcal{F}^*(1)$ are given in terms of the *rows* of $N$, that are the columns of ${}^t N$.

Is this correct? If so, is there any intrinsec or general way to see this?