### Determination of Higher Order Curves

Cramer is usually given credit for the result that $$\frac{n^2+3n}{2}$$ points generally determine a curve of degree $$n$$, because he gave a proof of it, using the counting argument above, in Chapter 3 of his book Introduction a l'analyse des lignes courbes algébriques (Introduction to the Analysis of Curved Algebraic Lines) [Cramer 1750], which was widely read in the second half of the 18th century. Euler proved the same thing in a paper published in the same year [Euler 1750a], but the result had actually been published much earlier in [Maclaurin 1720].

To determine a curve of degree $$n$$, it is usually enough to know $$\varphi_n = \frac{n^2+3n}{2}$$ points that lie on the curve. These will give rise to a homogeneous system of $$\varphi_n$$ equations in $$\varphi_n + 1$$ unknowns. As long as the rank of this system is $$\varphi_n$$, then the $$\varphi_n + 1$$ coefficients of the polynomial equation will be determined, up to a scalar multiple. However, as Euler's example indicates, the linear equations may not be linearly independent for every possible set of $$\varphi_n$$ points on the curve. In particular, a degenerate conic should be determined by $$\varphi_2=5$$ of its points, but not when 4 or 5 of the points lie on the same line.

Let's consider the case of the cubic equation, i.e. the polynomial equation of degree 3. In [Euler 1750a], Euler wrote the general cubic equation as $$\alpha x^3 + \beta x^2y + \gamma xy^2 + \delta y^3 + \varepsilon x^2 + \zeta xy + \eta y^2 + \theta x + \iota y + \kappa = 0. \label{GeneralCubic}$$ If we now substitute $$(x_1,y_1)$$, $$(x_2,y_2)$$, $$\ldots$$, $$(x_9,y_9)$$, we have a system of 9 equations in the 10 unknowns $$\alpha$$, $$\beta$$, $$\ldots$$, $$\kappa$$, represented by the following augmented matrix: $\begin{bmatrix} x_1^3 & x_1^2y_1 & x_1y_1^2 & y_1^3 & x_1^2 & x_1y_1 & y_1^2 & x_1 & y_1 & 1 & 0 \\ x_2^3 & x_2^2y_2 & x_2y_2^2 & y_2^3 & x_2^2 & x_2y_2 & y_2^2 & x_2 & y_2 & 1 & 0 \\ x_3^3 & x_3^2y_3 & x_3y_3^2 & y_3^3 & x_3^2 & x_3y_3 & y_3^2 & x_3 & y_3 & 1 & 0 \\ x_4^3 & x_4^2y_4 & x_1y_4^2 & y_4^3 & x_4^2 & x_4y_4 & y_4^2 & x_4 & y_4 & 1 & 0 \\ x_5^3 & x_5^2y_5 & x_1y_5^2 & y_5^3 & x_5^2 & x_5y_5 & y_5^2 & x_5 & y_5 & 1 & 0 \\ x_6^3 & x_6^2y_6 & x_1y_6^2 & y_6^3 & x_6^2 & x_6y_6 & y_6^2 & x_6 & y_6 & 1 & 0 \\ x_7^3 & x_7^2y_7 & x_1y_7^2 & y_7^3 & x_7^2 & x_7y_7 & y_7^2 & x_7 & y_7 & 1 & 0 \\ x_8^3 & x_8^2y_8 & x_1y_8^2 & y_8^3 & x_8^2 & x_8y_8 & y_8^2 & x_8 & y_8 & 1 & 0 \\ x_9^3 & x_9^2y_9 & x_1y_9^2 & y_9^3 & x_9^2 & x_9y_9 & y_9^2 & x_9 & y_9 & 1 & 0 \\ \end{bmatrix}$ For example, the entries in the second column are the numbers $$x_i^2y_i$$, because the second unknown coefficient ( $$\beta$$ ) is the coefficient of $$x^2y$$.

Suppose that we apply elementary operations to put this matrix into a reduced row-echelon form. If the rank of the system is 9, then there will be infinitely many solutions depending on one parameter, say $$t$$. Because the system is homogeneous, we will always be able to solve for the non-zero coefficients as multiples of $$t$$. This will give us a unique solution of the general cubic equation up to an arbitrary factor $$t$$ on the left-hand side. However, if the rank of the system is less than 9, then the general cubic equation will not be determined up to scalar multiplicity.