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 on the preceding page, 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 by Maclaurin [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 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

\[\alpha x^3 + \beta x^2y + \gamma xy^2 + \delta y^3 + \varepsilon x^2 + \zeta xy + \eta y^2 + \theta x + \iota y + \kappa = 0.\]

If we now substitute \((x_1,y_1),\) \((x_2,y_2),\) …, \((x_9,y_9),\) we have a system of 9 equations in the 10 unknowns \(\alpha,\) \(\beta,\) …, \(\kappa,\) represented by the following augmented matrix:\[\left[\begin{array}{rrrrrrrrrrr} x_1^3 & x_1^2y_1 & x_1y_1^2 & y_1^3 & x_1^2 & x_1y_1 & y_1^2 & x_1 & y_1 & 1 & 0 \\ x_2^3 & x_2^2y_2 & x_2y_2^2 & y_2^3 & x_2^2 & x_2y_2 & y_2^2 & x_2 & y_2 & 1 & 0 \\ x_3^3 & x_3^2y_3 & x_3y_3^2 & y_3^3 & x_3^2 & x_3y_3 & y_3^2 & x_3 & y_3 & 1 & 0 \\ x_4^3 & x_4^2y_4 & x_4y_4^2 & y_4^3 & x_4^2 & x_4y_4 & y_4^2 & x_4 & y_4 & 1 & 0 \\ x_5^3 & x_5^2y_5 & x_5y_5^2 & y_5^3 & x_5^2 & x_5y_5 & y_5^2 & x_5 & y_5 & 1 & 0 \\ x_6^3 & x_6^2y_6 & x_6y_6^2 & y_6^3 & x_6^2 & x_6y_6 & y_6^2 & x_6 & y_6 & 1 & 0 \\ x_7^3 & x_7^2y_7 & x_7y_7^2 & y_7^3 & x_7^2 & x_7y_7 & y_7^2 & x_7 & y_7 & 1 & 0 \\ x_8^3 & x_8^2y_8 & x_8y_8^2 & y_8^3 & x_8^2 & x_8y_8 & y_8^2 & x_8 & y_8 & 1 & 0 \\ x_9^3 & x_9^2y_9 & x_9y_9^2 & y_9^3 & x_9^2 & x_9y_9 & y_9^2 & x_9 & y_9 & 1 & 0\end{array}\right].\]

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 \[\alpha x^3 + \beta x^2y + \gamma xy^2 + \delta y^3 + \varepsilon x^2 + \zeta xy + \eta y^2 + \theta x + \iota y + \kappa = 0\] 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.