Suppose an over-constrained system as $$Ax=b$$ where $A\in\mathbb{R}^{m\times n}$, $x\in\mathbb{R}^n$, $b\in\mathbb{R}^m$ and $m\gt n$. Let's consider a special scenario, that coefficient matrix $A$ is rank-deficient with $\text{rank}($A$)=n-1$, but the augmented matrix $\text{rank}([A|b])=n$. So now we have $n$ independent functions and $n$ variables, does this mean we are gonna have a unique solution here? This makes me feel like the system is actually a well-determined system.
But for an example, $$\left[\begin{array}{ll} 1 & 2 \\ 1 & 2 \\ 1 & 2 \end{array}\right]x=\left[\begin{array}{l} 3 \\ 4 \\ 1 \end{array}\right]$$ Here, $\text{rank}(A)=1$,$\text{rank}([A|b])=2$. We now have two independent equations and two variables. However, we can directly see that each equation is contradictory to each other, as we will get $0+0=3$ from these 3 equations. Then we seems to have no solution here.
Meanwhile, there is no way that a vector $b$ that linearly independent from the column vectors from $A$ lies in the column space of $A$. So I must have misunderstood something here. Could anyone help me out of here? How should I classify this linear equations here? It's over-determined or well-determined?
If ${\rm rank} A = n - 1$ and ${\rm rank} (A |b) = n$, that means that ${\rm rref} (A|b)$ has a pivot in the augmented column. This is not hard to show. Because of this, the system is inconsistent and has no solution.