# When Nine Points Are Worth But Eight: Euler’s Resolution of Cramer’s Paradox - Euler's Elegant Example

Author(s):
Robert E. Bradley (Adelphi University) and Lee Stemkoski (Adelphi University)

In the fall of 1747, Euler read a paper to the Berlin Academy titled "On an apparent contradiction in the theory of curved lines." It was published soon afterwards [Euler 1750a], coincidentally in the same year that Cramer published his book on the analysis of curved lines [Cramer 1750]. In this paper, Euler carefully explained Cramer's Paradox and gave examples of various linear systems in which one or more of the given equations is "contained in the others" – in modern terms, where one or more equation is a linear combination of the remaining equations, so that the rank of the system is less that the number of equations. He then described his resolution of Cramer's paradox and closed with examples of sets of nine points that do not uniquely determine a cubic curve. Most of his examples are degenerate cubics whose equation can be factored as the product of an equation of order two and another of order one, or as a product of three linear equations. Cubic curves such as these generally consist either of a straight line and a conic section, or of three straight lines.

Euler's pièce de resistance was the example consisting of 9 points arranged in a $3 \times 3$ square grid. For simplicity, let's assume these are the points $(0,0),$ $(1,0),$ $(0,1),$ $(-1,0),$ $(0,-1),$ $(1,1),$ $(-1,1),$ $(-1,-1)$ and $(1,-1)$ (Euler considered the slightly more general case with $\pm a$ replacing $\pm 1,$ for an arbitrary constant $a>0$). We can try to fit a cubic equation through these points by substituting the $x$- and $y$-coordinate of each point into the general cubic equation

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

and setting up a system of 9 linear equations in the 10 unknown coefficients of this equation.

A modern brute force method of doing this might begin with the homogeneous augmented matrix $\left[\begin{array}{rrrrrrrrrrr} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 \\0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 & 0 \\-1 & 0 & 0 & 0 & 1 & 0 & 0 &-1 & 0 & 1 & 0 \\0 & 0 & 0 &-1 & 0 & 0 & 1 & 0 &-1 & 1 & 0 \\1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \\-1 & 1 &-1 & 1 & 1 &-1 & 1 &-1 & 1 & 1 & 0 \\-1 &-1 &-1 &-1 & 1 & 1 & 1 &-1 &-1 & 1 & 0 \\1 &-1 & 1 &-1 & 1 &-1 & 1 & 1 &-1 & 1 & 0 \\ \end{array}\right].$

We would then apply elementary matrix operations to transform it into the reduced row-echelon form $\left[\begin{array}{rrrrrrrrrrr}1 & 0 & 0 & 0 & 0 & 0 & 0 &1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 &1 & 0 & 0 \\0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\end{array}\right].$

If we let $\theta = s$ and $\iota = -t$ for arbitrary real numbers $s$ and $t,$ then $\alpha = -s,$ $\delta = t,$ and the general cubic equation $\alpha x^3 + \beta x^2y + \gamma xy^2 + \delta y^3 + \varepsilon x^2 + \zeta xy + \eta y^2 + \theta x + \iota y + \kappa = 0$ becomes $-s x^3 + t y^3 + sx - ty = 0,$ or $t(y^3-y) = s(x^3-x).$

Euler simply gave the solution, which he wrote as $my(yy-aa)=nx(xx-aa),$ and left the details to the reader. His solution process was probably something like the following. We first consider $(0,0),$ which gives $\kappa = 0.$ Next, the points $(\pm 1,0)$ give us the equations $\alpha + \varepsilon + \theta = 0$$-\alpha + \varepsilon - \theta = 0,$ which are equivalent to the equations $\varepsilon = 0$ and $\alpha + \theta = 0.$ Similarly, the points $(0,\pm 1)$ give us $\eta = 0$ and $\delta + \iota = 0.$ Therefore, the general cubic equation $\alpha x^3 + \beta x^2y + \gamma xy^2 + \delta y^3 + \varepsilon x^2 + \zeta xy + \eta y^2 + \theta x + \iota y + \kappa = 0$ reduces to $\alpha(x^3-x) + \beta x^2y + \gamma xy^2 + \delta(y^3-y) + \zeta xy = 0.$ If we substitute the points $(1,1),$ $(-1,1)$ and $(1,-1),$ respectively, we have $\begin{array}{rrr}\beta + \gamma + \zeta &=& 0\\ \beta - \gamma - \zeta &=& 0\\-\beta + \gamma - \zeta &=& 0.\end{array}$

From these we deduce $\beta = \gamma = \zeta = 0$ or $\alpha(x^3-x) + \delta(y^3-y) = 0.$

Because the point $(-1,-1),$ which we have not used yet, yields $0=0$ when substituted, the system has rank 8 or, as Euler put it in his letter to Cramer, the nine points “are worth but 8.”

Robert E. Bradley (Adelphi University) and Lee Stemkoski (Adelphi University), "When Nine Points Are Worth But Eight: Euler’s Resolution of Cramer’s Paradox - Euler's Elegant Example," Convergence (February 2014)