Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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:

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

share|improve this answer
    
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) –  SyntacticSugar 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! –  SyntacticSugar Jun 19 '13 at 17:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.