# Gauß-Jordan algorithm - 'reading' the solution

Disclaimer: I'm not really sure how to do a proper coefficient-matrix in latex, if someone could edit it to look properly I'd be really thankful ;)

Given the following system of linear equations, determine the solution set using the Gauß-Jordan-Algorithm

$$(I):3x_1 +2x_2-x_3-2x_4=0$$ $$(II):2x_1+3x_2-4x_3+2x_4=0$$ $$(III):x_1+3x_2-5x_3+4x_4=0$$ $$(IV):x_1+4x_2-7x_3+6x_4=0$$

So to solve this I used the Gauß-Jordan-Algorithm as asked by the task and ended up with these last two steps:

$$\begin{pmatrix} 1 & 4 & -7 & 6 &|&0 \\ 0 & 1 & -2 &2 &|&0 \\ 0 & 1 & -2 & 2 &|&0 \\ 0 & 1 & -2 & 2 &|&0 \\ \end{pmatrix} \text{ (III & IV-II)}= \begin{pmatrix} 1 & 4 & -7 & 6 &|&0 \\ 0 & 1 & -2 &2 &|&0 \\ 0 & 0 & 0 & 0 &|&0 \\ 0 & 0 & 0 & 0 &|&0 \\ \end{pmatrix}$$

Would someone of you mind explaining how exactly to continue from this point? How exactly do I 'read/see' the solution in this last matrix?

P.S.: I'm from Germany and therefore I'm only familiar with the german terminology, please bear with me if something was lost in translation

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You have reduced it to row echelon form. You can go further to reduced row echelon form as follows: $$\begin{pmatrix} 1 & 0 & 1 & -2 &|&0 \\ 0 & 1 & -2 &2 &|&0 \\ 0 & 0 & 0 & 0 &|&0 \\ 0 & 0 & 0 & 0 &|&0 \\ \end{pmatrix}$$

You can read this matrix as:

$$x_1+x_3-2x_4=0\\x_2-2x_3+2x_4=0\\x_3~\mathrm{and}~x_4~\text{are free}$$

This system has an infinite number of solutions. You can parameterize:

$$x_4=t, ~x_3=s, ~x_2=2s-2t,~x_1=2t-s$$

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Maybe it's me, but I don't see how you reduced that further. Mind telling me? ;) – Rickyfox Jun 19 '13 at 17:11
(Row 1) - 4*(Row 2) – yousufmsoliman Jun 19 '13 at 17:29
Bloody hell, the heat and the long day have left their marks on me, thanks tho ;) – Rickyfox Jun 19 '13 at 17:32
No worries haha. Hope I helped out! – yousufmsoliman Jun 19 '13 at 17:36