I just finished a course about systems of linear équations and I'm trying to do some exercises. Here is the system to solve using Gaussian elimination:
$$ \begin{cases} -5x-2y+z=a \\ -4x-4y-4z=b \\ x-2y-5z=c \end{cases} $$
Here is what I did: $$\Leftrightarrow \begin{cases} x-2y-5z=c \\ -4x-4y-4z=b \\ -5x-2y+z=a \end{cases} \Leftrightarrow \begin{cases} x-2y-5z=c \\ -12y-24z=b+4c \\ -12y-24z=a+5c \end{cases} \Leftrightarrow \begin{cases} x-2y-5z=c \\ y+2z=\frac{-b-4c}{12} \\ y+2z=\frac{-a-5c}{12} \end{cases} $$ But what to do now? I don't know how to proceed because I can't use y nor z as a pivot element. Maple tells me that there isn't any solution, but I don't know how to prove that...
Thank you in advance.
You ended up with the following:
$\begin{cases} x-2y-5z=c, \\ y+z=\frac{-b-4c}{12}, \\ y+z=\frac{-a-5c}{12}. \end{cases}$
Notice the two last equations. They each say that $y+z$ is equal to two things, so these two things better not be different, because otherwise there are no solutions.
So a necessary condition for there to be any solutions is that $\frac{-b-4c}{12}=\frac{-a-5c}{12}$ (and let's say that their common value is $d$). Thus, you have the two equations
$\begin{cases} x-2y-5z=c, \\ y+z=d. \end{cases}$
You may notice that you are not quite done with your Gaussian elimination. It is possible for you to eliminate the $y$ in the top equation by using the bottom equation.