0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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 at 20:31
  • $\begingroup$ See this en.wikipedia.org/wiki/Rouch%C3%A9%E2%80%93Capelli_theorem $\endgroup$ – mfl Feb 25 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 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 at 14:48
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 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 at 15:04
  • $\begingroup$ I $$a=-1$$ then there are infinity many solutions. $\endgroup$ – Dr. Sonnhard Graubner Feb 26 at 15:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.