Conditions so that a system of equation is incompatible

Question

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?

Answer 1

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?