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I am trying to find a real parameter $m$ so that the following system is incompatible $$\begin{cases} x+y+mz = 1 \\ x-2y+z=m \\ mx+y+z=0 \end{cases}$$ I did the determinant of the system $$\Delta=2+m+m+2m^2-1-1=2m^2+2m-4$$ By putting $\Delta =0$ $$m^2+m-2=0\rightarrow(m-1)(m+2)=0$$ So I have either $m=1$ or $m=-2$, but how I proceed now? And what are the conditions for a system to be incompatible?

Harry Peter
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  • Insert the values of $m$ ($1$ and $-2$) at the system and use Gauss elimination to see what kind of system you get! – Arnaldo Jul 16 '18 at 15:58
  • Can I use Crammer Rule? Because I don't know Gauss elimination.. –  Jul 16 '18 at 15:59
  • Cramer just works for a system with an unique solution – Arnaldo Jul 16 '18 at 16:00
  • The point is that if the determinant is nonzero then there is a possibility that the system is incompatible, because row reduction will create a row of zeros in the echelon form. But it is possible that the right hand side is chosen "just right" so that in fact the system is still compatible anyway (in this case reducing the augmented matrix results in a full row of zeros). So you have to attempt to solve the system with determinant zero by hand to see if it has a solution or not. – Ian Jul 16 '18 at 16:08
  • @Sonkun: If you undelete your question https://math.stackexchange.com/questions/2882832/polynomial-problem-with-two-conditions, I'll post an answer. – quasi Aug 14 '18 at 23:08

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You have to replace the values of $m$ at the system and use Gauss elimination. For example, for $m=1$ you get \begin{cases} x+y+z = 1 \\ x-2y+z=1 \\ x+y+z=0 \end{cases}

See that the first and the third equations are incompatible.

Do the same for $m=-2$.

Can you finish?

Arnaldo
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  • $$\begin{cases} x+y-2z = 1 \ x-2y+z=-2 \ -2x+y+z=0 \end{cases}$$ I think I see something. If I add all three of them I get $0=-1$ Is that it? –  Jul 16 '18 at 16:05
  • Thank you! But Is there a general conditions? Maybe in future I will find harder systems. –  Jul 16 '18 at 16:08
  • @Sonkun: the general condition is use Gauss Elimination: https://en.wikipedia.org/wiki/Gaussian_elimination – Arnaldo Jul 16 '18 at 17:51