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).