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Context

I have a collection of nine distinct points $P_{i} = [X_{i} : Y_{i} : Z_{i}]$ in the projective plane, with the index $i \in [1, 9]\cap\mathbb{Z}$. I want to show that a cubic curve passes through all nine points.

A cubic curve in this context is a set of points in the projective plane that are solutions of the equation $$ a_{1}X^{3} + a_{2}X^{2}Y + a_{3}XY^{2} + a_{4}Y^{3} + a_{5}X^{2}Z + a_{6}XYZ + a_{7}Y^{2}Z + a_{8}XZ^{2} + a_{9}YZ^{2} + a_{10}Z^{3} = 0 $$ for some coefficients $a_{1}, \ldots, a_{10} \in \mathbb{R}$ not all zero (we do not assume this polynomial to be irreducible).

Ignoring the obvious pairwise collinearity, there are some important collinearity relations between the points. For brevity, the notation $\{i, j, k\}$ will be used to indicate that $P_{i}, P_{j}, P_{k}$ are collinear. The collinear points are the following: $$ \{1, 2, 3\},\quad \{4, 5, 6\},\quad \{7, 8, 9\},\quad \{1, 4, 8\},\quad \{2, 4, 7\},\\ \{1, 5, 9\},\quad \{2, 5, 8\},\quad \{3, 5, 7\},\quad \{2, 6, 9\},\quad \{3, 6, 8\}. $$ These are the only collinearity relations for more than two points, and this is the only information we have regarding the points.

Attempt

I wanted to appeal to the Cayley-Bacharach Theorem, but I am struggling to show that the points arise as the intersection of two cubics. Reading around the subject is little help at the moment since I have no Algebraic Geometry knowledge, so all the papers I've tried to read are inaccessible to me for the time being.

Instead, I have been treating this as a Linear Algebra problem (of sorts) as it is enough to show that the equation $$ \begin{pmatrix} X_{1}^{3} & X_{1}^{2} Y_{1} & X_{1} Y_{1}^{2} & Y_{1}^{3} & X_{1}^{2} Z_{1} & X_{1} Y_{1} Z_{1} & Y_{1}^{2} Z_{1} & X_{1} Z_{1}^{2} & Y_{1} Z_{1}^{2} & Z_{1}^{3} \\ X_{2}^{3} & X_{2}^{2} Y_{2} & X_{2} Y_{2}^{2} & Y_{2}^{3} & X_{2}^{2} Z_{2} & X_{2} Y_{2} Z_{2} & Y_{2}^{2} Z_{2} & X_{2} Z_{2}^{2} & Y_{2} Z_{2}^{2} & Z_{2}^{3} \\ X_{3}^{3} & X_{3}^{2} Y_{3} & X_{3} Y_{3}^{2} & Y_{3}^{3} & X_{3}^{2} Z_{3} & X_{3} Y_{3} Z_{3} & Y_{3}^{2} Z_{3} & X_{3} Z_{3}^{2} & Y_{3} Z_{3}^{2} & Z_{3}^{3} \\ X_{4}^{3} & X_{4}^{2} Y_{4} & X_{4} Y_{4}^{2} & Y_{4}^{3} & X_{4}^{2} Z_{4} & X_{4} Y_{4} Z_{4} & Y_{4}^{2} Z_{4} & X_{4} Z_{4}^{2} & Y_{4} Z_{4}^{2} & Z_{4}^{3} \\ X_{5}^{3} & X_{5}^{2} Y_{5} & X_{5} Y_{5}^{2} & Y_{5}^{3} & X_{5}^{2} Z_{5} & X_{5} Y_{5} Z_{5} & Y_{5}^{2} Z_{5} & X_{5} Z_{5}^{2} & Y_{5} Z_{5}^{2} & Z_{5}^{3} \\ X_{6}^{3} & X_{6}^{2} Y_{6} & X_{6} Y_{6}^{2} & Y_{6}^{3} & X_{6}^{2} Z_{6} & X_{6} Y_{6} Z_{6} & Y_{6}^{2} Z_{6} & X_{6} Z_{6}^{2} & Y_{6} Z_{6}^{2} & Z_{6}^{3} \\ X_{7}^{3} & X_{7}^{2} Y_{7} & X_{7} Y_{7}^{2} & Y_{7}^{3} & X_{7}^{2} Z_{7} & X_{7} Y_{7} Z_{7} & Y_{7}^{2} Z_{7} & X_{7} Z_{7}^{2} & Y_{7} Z_{7}^{2} & Z_{7}^{3} \\ X_{8}^{3} & X_{8}^{2} Y_{8} & X_{8} Y_{8}^{2} & Y_{8}^{3} & X_{8}^{2} Z_{8} & X_{8} Y_{8} Z_{8} & Y_{8}^{2} Z_{8} & X_{8} Z_{8}^{2} & Y_{8} Z_{8}^{2} & Z_{8}^{3} \\ X_{9}^{3} & X_{9}^{2} Y_{9} & X_{9} Y_{9}^{2} & Y_{9}^{3} & X_{9}^{2} Z_{9} & X_{9} Y_{9} Z_{9} & Y_{9}^{2} Z_{9} & X_{9} Z_{9}^{2} & Y_{9} Z_{9}^{2} & Z_{9}^{3} \\ \end{pmatrix}\begin{pmatrix} a_{1} \\ a_{2} \\ a_{3} \\ a_{4} \\ a_{5} \\ a_{6} \\ a_{7} \\ a_{8} \\ a_{9} \\ a_{10} \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix} $$ has a solution for $a_{1}, a_{2}, \ldots, a_{10} \in \mathbb{R}$ not all zero. With an explicit example I have used Maple to solve for the values $a_{i}$, but I cannot figure out how to show that they exist in general (given the collinearity relations).

How can I show the existence of such a cubic can be proved using Cayley-Bacharach, the Linear Algebra approach, or by a whole new method?

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You have a system of nine homogeneous linear equations in ten variables. As $9<10$ then they have a nonzero solution.

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  • $\begingroup$ Ah, so the variables $X_{i}^{d_{1}}Y_{j}^{d_{2}}Z_{k}^{d_{3}}$ form a ten-dimensional vector space, which the vector of $0$'s is an element of? $\endgroup$ – Bill Wallis Aug 9 '18 at 16:15

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