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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?

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If $M$ (echelon form) has exactly one zero row then

1) The rank of matrix is exactly $n$;

2) The rank of $nxn$ submatrix of $M$ after extracting this zero row (and right hand side column) is $n$.

3) Zero row does not restrict solutions anyhow since $0x=0$ for all $x$. Thus it can be deleted.

4) The final system of $n$ equations has the full rank and then it has the unique solution.

EDITION: Actually the author is absolutely correct asking the question, because the statement is not quite true. Not augment matrix also MUST have only one zero row! Then the logic above is correct. Consider the simple example of 3 equations with 2 variables: $$ x+y=1\\ x+y=2\\ x+y=1\\ $$ Echelon form will be $$ x+y=1\\ 0x+0y=1\\ 0x+0y=0\\ $$ $M$ matrix is $$ \matrix{1 & 1 &1 \\0 & 0 & 1\\ 0 & 0 & 0} $$ The system does not have a solution.

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  • $\begingroup$ Thank you. However I can't understand why the system has a unique solution provided that the rank is $n$. Linear system has solutions if there are no rows of type $(0,0,0,...,a)$ where $a \neq 0$. We can't know if there are such rows here, so the system might have or might not have a solution. And even if it has a solution, how do we know for sure that there are no free variables present? Because if we have a free variable, our system has infinite number of solutions. $\endgroup$
    – user186483
    Oct 23, 2014 at 10:30
  • $\begingroup$ @user186483, you are correct pointing out the problem. I edited my answer. Under condition you define system may have or 1 or 0 solution. Bit the infinite number is impossible because $M$ has rank $n$. $\endgroup$ Oct 23, 2014 at 16:16

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