I have this linear system:
$$ \left\{ \begin{array}{r} x+2y=4 \\ y-z=0 \\ x+2z=4 \end{array} \right. $$ This Linear system can be converted to echelon form as follows:


$$ \left[ \begin{array}{ccc|c} 1&2&0&4\\ 0&1&-1&0\\ 0&-2&2&0 \end{array} \right] $$


$$ \left[ \begin{array}{ccc|c} 1&2&0&4\\ 0&1&-1&0\\ 0&0&0&0 \end{array} \right] $$

The solution set here is $\{(4-2z,z,z) | z \in\Bbb R \}$

My question is:
I really don't understand why "$(4-2z,z,z)$" is the solution set?
What is the meaning of two single $z$ in this solution set?
Why not $x+2y=4$, $y-z=0$ is the solution set?


1 Answer 1


The last tableau exhibits the equations

$$x+2y=4$$ and $$y-z=0$$

$z$ can be chosen arbitarily.

It follows $y=z$ and $x=4-2y=4-2z$, so the general solution is $(4-2z/z/z)$

In general, if you have an echelon form which has not diagonal shape, the variables right from the diagonal part can be chosen arbitarily. The rest of the variables can be calculated easily depending on the chosen variables. There are infinite many solutions in this case because there are infinite many possible choices.

  • $\begingroup$ Thank you @Peter please correct me if I'm wrong, I can choose Any variable from "the free variables which is z here" to find the solution set according to it is this true? also, what is the meaning of | sign in the solution set? $\endgroup$ Commented Aug 22, 2015 at 18:10
  • $\begingroup$ Please forgive my ignorance: y-z=0 therefore y=z, x = 4 - 2y, therefore x=4-2z, So, From where does the other single z come from? I mean it should be (4-2z,z) not (4-2z,z,z) $\endgroup$ Commented Aug 22, 2015 at 18:17
  • $\begingroup$ The meaning of the $|$ is : All the triples $(4-2z/z/z)$, for which z is a real number. So, you can choose $z$ to be any real number and $x,y$ depend on this choice. $\endgroup$
    – Peter
    Commented Aug 22, 2015 at 18:22
  • $\begingroup$ You could also say : For every real number there is a corresponding triple solving the equation. $\endgroup$
    – Peter
    Commented Aug 22, 2015 at 18:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .