MHB Can You Solve This Challenging System of Real Number Equations?

[anemone]

Here is this week's POTW:

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If $a,\,b,\,c,\,x,\,y,\,z \in \Bbb{R}$ and $a\ne x,\,b\ne y$ and $c \ne z$, solve the following system of equations:

$-a=b+y\\-b=c+z\\-c=a+x\\x=by\\y=cz\\z=ax$

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to Ackbach for his correct solution, which you can find below:(Cool)

Spoiler

The first thing we do is substitute the last equation into the penultimate to get $y=acx,$ and then plug this into the ante-penultimate to get $x=abcx.$ This equation has two possibilities: either $x=0,$ or $abc=1.$ If we take the first, then we wind up with $x=y=z=0,$ and the first three equations simplify down to
\begin{align*}
-a&=b \\
-b&=c \\
-c&=a,
\end{align*}
from which we get that $c=b$ AND $c=-b,$ which in turn implies that $c=0,$ or $a=b=c=0.$ This is a contradiction, since we are not allowed to have $a=x.$ Hence, $x\not=0,$ but we must have that $abc=1$. It follows from this that none of the six variables can be zero.

One solution, inspired by symmetry and obtained by inspection, is that $a=b=c=1,$ and $x=y=z=-2.$ The question is whether this solution is unique. I'm inclined to think that it is. The system of equations is entirely linear in $a,b,c,$ so if we can show that there can't be any other values of $a,b,c$ that work, we will be done, since if $a,b,$ and $c$ are determined, then by the first three equations, $x,y,$ and $z$ are determined. Consider the following rewrite of the first three equations:
\begin{align*}
-a-b&=y \\
-b-c&=z \\
-a-c&=x.
\end{align*}
We calculate the determinant of the coefficient matrix as
$$\det\left[\begin{matrix}-1 &-1 &0\\0 &-1 &-1\\-1 &0 &-1 \end{matrix}\right]=-1(1)+1(-1)=-2\not=0.$$
This implies that the values of $a,b,c$ are unique, and hence the values of $x,y,z$ are unique.