Have been having problems with this equation system for a while, \begin{array}{l} x - y - az = 1\\ ax + y + az = a\\ ax + 3y + 3z = -1 \end{array} where I need to find all the values of $a\in \mathbb{R}$.

I have tried to solve the system with elimination, by subtracting the first line multiplied by $(-a)$ with the second and third line and so on. I have found $z$ to be $\frac{-1 -a}{3+3a}$ but after integrating it and solving for $y$ I'm lost.

Any help is appreciated, thanks in advance!

  • $\begingroup$ Welcome to Maths.SX! Are you sure of the value found for $z$? I obtain $\frac{-1-a}{3\color{red}-3a}$. $\endgroup$ – Bernard Feb 25 '19 at 20:31
  • $\begingroup$ See this en.wikipedia.org/wiki/Rouch%C3%A9%E2%80%93Capelli_theorem $\endgroup$ – mfl Feb 25 '19 at 20:36
  • $\begingroup$ Sorry for a late response. @Bernad Yes your value of z is correct. I missed a minus there thanks :) $\endgroup$ – Irrumasti Feb 26 '19 at 14:45
  • $\begingroup$ @mfl I usually use matrixes when solving on paper but decided not to use it here as I already have the code used written in a latex document. Thanks anyways though. $\endgroup$ – Irrumasti Feb 26 '19 at 14:48

Hint: Adding equation 1 and 2, we get $$x(a+1)=a+1$$ so $$(a+1)(x-1)=0$$ Can you proceed?

  • $\begingroup$ Hi, sorry for a late response. I did manage to solve the equation finally. As Bernard also stated I had my z incorrect and after doing it right I managed to find y and could deduct that when a = 1 there are no solutions and when a = -1 there should be an infinite amount of solutions. The main problem was that I would rather think I had done something wrong than for it to be possible that a = 1 did not have any solutions. Thanks for the quick response! $\endgroup$ – Irrumasti Feb 26 '19 at 14:59
  • $\begingroup$ If $$x=1$$ so you will get $$z(a-1)=\frac{1}{3}$$ and so $$z=\frac{1}{3(a-1)}$$ for $$a\neq 1$$ $\endgroup$ – Dr. Sonnhard Graubner Feb 26 '19 at 15:04
  • $\begingroup$ I $$a=-1$$ then there are infinity many solutions. $\endgroup$ – Dr. Sonnhard Graubner Feb 26 '19 at 15:06

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