Let $X^{n-2} \subset \mathbb{P}^n$ a complex projective $n-2$-dimensional variety of degree $3$. Assume that $\mathbb{P}^n$ is the smallest linear space containing $X$. That means it's not possible to find a proper linear subspace $L^k \subset \mathbb{P}^n$ containing $X$.
How it can be shown that $X$ is defined by the vanishing set of the three $2 \times 2$ minors of a linear $3 \times 2$ matrix
$$\begin{pmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ \end{pmatrix}$$
$a_i, b_i$ are linear forms in $X_0,..., X_n$.
Could somebody explain why is this the case?
Source: David Mumford's Algebraic geometry I, Complex Projective Varieties page 80.
What I know. Degree $3$ assumption imply that there exist a general $2$-dimensional $\mathbb{P}^2 \cong L^2 \subset \mathbb{P}^n$ with $X \cap L = \{y^1, y^2, y^3\}$. After a well chosen $PGL(n)$ transformation of $\mathbb{P}^n$ we can assume $L^2= V(X_3, X_4,..., X_n)$. Additionally assume that the points $y^1, y^2, y^3$ are contained in affine part $D(X_2) \cong \mathbb{A}^2 \subset L^2$; equivalently for $ V(X_2, X_3,..., X_n) = L^2 \cap V(X_2) \cong \mathbb{P}^1 $ the intersection with $X$ is empty.
If we denote $y^i= (y^i_0: y^i_1: 1: y^i_2,..., y^i_n)$ (that's possible since $y_i \not \in V(X_2)$) then in $\mathbb{A}^2$ we can write down the intersection $X \cap L^2$ as algebraic set
$$V( (X_0- y_0^{a}) \cdot (X_1 - y_1^{b} \ \vert \ a, b \in \{1,2,3\}, a \neq b )$$
Can we derive from that the structure of vanishing polynomials for $X$.