Whenever I try to represent a relation \(F(u,v,w)=0\) as a matrix in standard nomographic form, I temporarily ignore numerical coefficients. I figure that if I can get the linearly independent terms to come out correctly, it'll be easy to add in the necessary coefficients. For example, if the equation is \(F(u,v,w) = 3u - 2vw + w^2 = 0\), I will simply focus on the three core terms \(u\), \(vw\), and \(w^2\), then add in the constants 3, -2, and 1 afterwards.
Recently, it occured to me to ask whether my optimism about coefficients is warranted. To phrase it as a question:
Question: Given six numerical coefficients \(x_1, \ldots, x_6\), is it always possible to pick entries of a 3x3 matrix $$\begin{bmatrix}a&b&c\\ d&e&f \\ g & h& i \end{bmatrix}$$ such that the six diagonal terms are the six coefficients? (In other words, so that \(aei = x_1, bfg = x_2, cdh=x_3, afh=-x_4, bdi = -x_5, ceg=-x_6\)?)And here's the (negative, or rather provisionally positive) answer:
Theorem: Given six nonzero constants, you'll be able to find a 3x3 matrix whose terms are those constants if and only if $$x_1x_2x_3 + x_4x_5x_6 = 0.$$
Proof: (\(\Rightarrow\)): If you've found a 3x3 matrix whose terms are those constants, then $$x_1x_2x_3 = (aei)(bfg)(cdh) = abcdefghi = (afh)(bdi)(ceg) = -x_4x_5x_6$$ and so \(x_1x_2x_3 + x_4x_5x_6 = 0\), as required.
(\(\Leftarrow\)). If the constants satisfy \(x_1,x_2,x_3 + x_4x_5x_6 = 0\), then it's possible to construct a 3x3 matrix whose six determinant terms are the six constants. (In fact, there are infinitely many such matrices.) For example, $$\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix} = \begin{bmatrix}x_1 & x_2 & x_3 \\ \frac{-1}{x_4} & \frac{x_2}{x_4x_5} & \frac{1}{x_1} \\ x_1 & -x_4 & \frac{x_4x_5}{x_2}.\end{bmatrix}$$ is such a matrix. ■
