It means that there is a "free variable", and infinitely many solutions to the system, because that "free variable" can be assigned any value. The other variables can often be expressed in terms of the free unknown variable; that is, their values will often depend on the the unknown value, depending on how "reduced" your matrix is.
E.g. If we have the augmented coefficient matrix reduced to the following, representing a system of three equations in the three unknowns $x, y, z$:
$$
\begin{pmatrix}
1 & 1 & 1 & 1 \\
0 & 1 & 0 & 2 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}
$$
Then $z$ is the free variable. $y = 2$, and $x = 1 - y - z = 1 - 2 - z = -(1 + z)$.
So if we let $z = \alpha$, we'd have the an infinite "family" of solutions which can be represented as
$$ \begin{pmatrix}\\ \\
-(1 + \alpha) \\ \\
2 \\ \\
\alpha \\ \\
\end{pmatrix}$$
which depends on the value of $\alpha$, which can be any of infinitely many values.
Hence, there are an infinite number of solutions.
But this still allows us to impose some constraints on the solutions: specifically, $y$ must be $2$, and whatever the value of $\alpha$, $x = -(1+ \alpha)$.
If the augmented matrix has a pivot in its last column, then the system corresponding to that augmented matrix is inconsistent. This is because in the reduced echolon form of the augmented matrix there will be a row of the form $[0\cdots 0\mid c]$ where $c$ is a nonzero number, namely the pivot. This means that $0=c$, implying the inconsistency of the system.
In fact, the same argument sketched above shows that a linear system is consistent if and only if its augmented matrix does not have a pivot in its last column. If no such pivot exists, you can solve the system by back-substitution.
Best Answer
If the first column of the matrix consists of zeros, this means that the coefficient of the first variable is zero in every equation in the system (how did this happen, anyway?). This means that if there is any solution at all, then there must be an infinite number of them, since you can set that first variable to any value at all and still satisfy all of the equations.