Solve a Linear System

Question

Trying to solve the linear system below for $x,y$ and $z$ in terms of $a,b,c$ and $\zeta_3$:

$x+y+z=a$

$x+{\zeta_3}y+{\zeta_3}z=b$

$x+{\zeta_3}^2y+{\zeta_3}^2z=c$

where $\zeta_3=-1/2+-i(\frac{\sqrt 3}{2})$ i.e. ${\zeta_3}^3=1$

From a book by the Russian author Smirnoff (translated by Silverman) - the author says the answer is $x=(a+b+c)/3$; $y=(a+b{\zeta_3}^2+c{\zeta_3})/3$; $z=(a+b{\zeta_3}+c{\zeta_3}^2)/3$

Got part of the way with Gaussian elimination, but couldn’t figure out how to use the information about e to finish the solution. Any thoughts?

Let me write $\zeta_3$ for $e$, because $e$ is Euler's number. You only need that $1+\zeta_3+\zeta_3^2=0$ and $\zeta_3^3=1$. — Dietrich Burde, Dec 17 '20 at 09:30
If that is the solution, then the equations are wrong. You need both $\zeta_3$ and $\zeta_3^2$ appear as coefficients in the two latter equations. In which case it is an instance of inverting a discrete Fourier transform. And the solution can be verified using Dietrich's hint. — Jyrki Lahtonen, Dec 17 '20 at 11:40
See this old answer of mine for the correct triple of equations. — Jyrki Lahtonen, Dec 17 '20 at 11:41

score 0 · Answer 1 · answered Dec 17 '20 at 12:15

If you make a few steps of Gauss elimination you can get $$\left[\begin{array}{ccc|c} 1 & 1 & 1 & a\\ 1 & \zeta_3 & \zeta_3 & b\\ 1 & \zeta_3^2 & \zeta_3^2 & c \end{array}\right]\begin{array}{l} \\ F_2\leftarrow F_2-F_1\\ F_3\leftarrow F_3-F_1\\ \end{array} \left[\begin{array}{ccc|c} 1 & 1 & 1 & a\\ 0 & \zeta_3-1 & \zeta_3-1 & b-a\\ 0 & \zeta_3^2-1 & \zeta_3^2-1 & c-a \end{array}\right]$$ $$\begin{array}{ll} \ \\ \ \\ F_3\leftarrow F_3-(\zeta_3+1)F_2 \end{array}\left[\begin{array}{ccc|c} 1 & 1 & 1 & a\\ 0 & \zeta_3-1 & \zeta_3-1 & b-a\\ 0 & 0 & 0 & c-a -(\zeta_3+1)(b-a) \end{array}\right]$$ So, Rouché tells you that the system only has solution if \begin{equation} c-a-(\zeta_3+1)(b-a)=0,\tag{1} \label{eq:abc} \end{equation} and in that case there are infinite solutions. Then, call $z=\lambda$, and the solutions are: $$x=a-\frac{b-a}{\zeta_3-1},\quad y=\frac{b-a}{\zeta_3-1}-\lambda,\quad z=\lambda,\quad\forall\lambda\in\mathbb{C}.$$ If you use the relation \eqref{eq:abc} I think that it can be shown that one of the solutions is exactly the answer you have, but obviously there are (infinitely) many more.

I just read in a comment from Jyrki Lahtonen that the equations may have an error. That is likely to be the case.

I got to the point shown in Gaussian elimination above. You were right ... book had mistakes in equations 2 and 3 that I should’ve caught. Thx so much. — gary, Dec 17 '20 at 15:53

Solve a Linear System

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