$x_1 + x_2 + x_3 = 6$

$x_1 - x_2 + 2x_3 = 7$

$3x_1 + x_2 + x_3 = 8$

$ 2x_1 + 2x_2 - x_3 = 3$

Find the solutions of the linear equations system.

I tried to find the reduced row echolon form of this linear equations systems and I found,

$$ \begin{bmatrix} 1 & 0 & 0 : 2 \\ 0 & 1 & 0 : 1 \\ 0 & 0 & 1 : 3 \\ 0 & 1 & 0 : 2 \\ \end{bmatrix} $$

So $x_1=2$, $x_2=1$ or $x_2=2$ and $x_3=3$. But I don't know what is the solution of this linear equations system when we have 2 different values for a variable..

  • 1
    $\begingroup$ Are you sure that you haven’t made a mistake in your tow reduction? $\endgroup$ Commented Oct 29, 2019 at 15:00
  • 2
    $\begingroup$ that's not a correct reduction in row-echelon form, since the leading coefficient of the last row is not to the right of the leading coefficient of the second row. I'm going to compute the correct reduction, I'll come back to you in a minute EDIT: the answer by @Ian is correct $\endgroup$
    – sortai
    Commented Oct 29, 2019 at 15:00
  • $\begingroup$ I wrote the third row, I wrote 0 to $a_33$ cell instead of 1. Alsoo, I have checked my row reduction 2 times before I post the question. Thanks! $\endgroup$ Commented Oct 29, 2019 at 15:07
  • 1
    $\begingroup$ Your reduction still cannot be correct because if it were then the first three rows would be a reduction of the first three equations, but $3(2)+1+3 \neq 8$. $\endgroup$
    – Ian
    Commented Oct 29, 2019 at 15:10
  • $\begingroup$ I can't see any mistakes on my row reduction :/ Can you check it please? I checked my row reduction 2 times and 1 more now, honestly. $\endgroup$ Commented Oct 29, 2019 at 15:12

1 Answer 1


The system represented by your augmented matrix doesn't have a solution, because any solution would need to have $x_2=1$ to satisfy the second equation and $x_2=2$ to satisfy the fourth equation.

Incidentally your matrix is technically not in echelon form. To convert to echelon form, you still need to subtract the second row from the fourth row, and then the inconsistent equation would read $0=1$. But your matrix is already in a form where the solution can be read off, so this isn't really a problem unless you were specifically asked to put it in echelon form.

That said when I ask Matlab for the reduced echelon form I also get an inconsistent system. This shouldn't really be a surprise: a linear system of $m$ equations in $n$ variables where $m>n$ and the coefficients are "random" will generally have no solution. Here $m=4,n=3$.

  • $\begingroup$ Continuing the row reduction from where the OP left off to the bitter end results in the identity matrix. Is that what you meant by “a meaningfully different result” from Matlab? $\endgroup$
    – amd
    Commented Oct 29, 2019 at 17:49
  • $\begingroup$ @amd Well, actually now that I think about it what Matlab does is it treats the 1 in the bottom right corner as a pivot which zeroes out the rest of that column, which is equivalent to the OP's version. It didn't occur to me that it would do that in the inconsistent case. $\endgroup$
    – Ian
    Commented Oct 29, 2019 at 18:16

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