Statement $S$:
If an $m \times n$ (consistent) linear system $Ax = b$ has a unique solution then $r = n$. $r$ is the number of nonzero rows in the row reduced echelon form (RREF) of $A$.
How to prove this statement $S$?
Is the word consistent necessary in $S$?
I have another related theorem here:
Theorem $T$: Let $[C | d]$ be an $[m×(n+1)]$ matrix in reduced echelon form, where $[C | d]$ represents a consistent system. Let $[C | d]$ have $r$ nonzero rows. Then $r \le n$ and in the solution of the system there are $n − r$ variables that can be assigned arbitrary values.
and its corollary:
By Theorem $T$, the only possibilities for the system represented by $[C | d]$ (and hence for the system represented by $[A | b]$) are:
- The system is inconsistent.
- The system is consistent and, in the notation of statement $S$, $r < n$. In this case there are $n − r$ unconstrained variables, so the system has infinitely many solutions.
- The system is consistent and $r = n$. In this case there are no unconstrained variables, so the system has a unique solution.
Don't know how to use theorem $T$.
Theorem $T$ means: If there exists some solution $x$, it means $b\in\text{col}A$. Further, if this solution is unique, it means $\ker A=\{0\}$. Next, by Rank Theorem,
$$n=\dim \text{col}A+\dim\ker A=r+0=r $$
So the proof is complete. Note the dimension of the column space of $A$ equals the number of nonzero rows in the row reduced echelon form (RREF) of $A$.