Let $M$ be some non-homogeneous linear system of $m$ equations and $n$ unknowns where $m=n+1$. Is it true that if the row echelon form of the augmented matrix (extended coefficient matrix) of $M$ has exactly one row with zeros then $M$ has one solution?
According to my book, the answer is yes. However, I can't understand why would it be true. Regardless of the criteria $m=n+1$, the row echelon form of the augmented matrix might have some free variables which might lead to infinite number of solutions. Why would a one zero-row and $m=n+1$ imply exactly one solution?
If $m=n+1$ and there is exactly one zero row, then we have a $n \times n$ coefficients matrix that might have or might not have a solution. How can we be sure that the augmented matrix doesn't have a row like this $(0,0,0,0,....,a)$ where $a \neq 0$? How can we be sure that the system doesn't have free variables which lead to infinite number of solutions?