A matrix containing a line such as this is invalid, right? $(0 \space0\space 0\space 0\space|\space1)$ A matrix containing a line such as this is invalid, right?
$$
(0 \space0\space 0\space 0\space|\space1)
$$
the matrix in question is this:
$$\left( \begin{array}{rrrr|r}
1&0&1&1&2\\
0&1&-4&0&-6\\
0&0&0&0&1
\end{array} \right)$$
Edit: my original question was bad. I was asked to "solve the system of linear equations, or say there isn't one". My question is if this is a "no answer" case. Thank you.
 A: I wouldn't use the term "invalid".  
The matrix with the row in question - or rather, (as @rschwieb points out below), the associated system of three equations in four unknowns - is inconsistent.
Edit:
Given your clarification, you are correct, there is no solution to the associated system of equations. 
To see why, note that the bottom-most row of your matrix tells you that $$0\cdot x + 0\cdot y + 0\cdot z = 1.$$ There does not exist any solution. Can you see that whatever the values of $x, y, z$, we will never have the left-hand-side $=$ right-hand-side? 
Note also that, in the following example (representing a system of $5$ equations in $4$ unknowns $w,x,y,z$):
$$\left( \begin{array}{rrrr|r}
1&0&0&0&2\\
0&1&0&0&-6\\
0&0&1&0&1\\
0&0&0&1&1\\
0&0&0&0&1\\
\end{array} \right),$$
where it appears that $w=2, x=-6, y=1, z=1$ is a solution, that "pseudo-solution" is incompatible with the equation associated with the $5^{th}$ row: $$0\cdot w +0\cdot x + 0\cdot y + 0\cdot z = 1.$$ Hence the entire associated system of equations is inconsistent (and thus no solution exists).
