Solve Guass Jordan Problem: Step-by-Step Guide

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Homework Help Overview

The discussion revolves around solving a system of linear equations using Gauss-Jordan elimination. The original poster presents a set of equations and their corresponding augmented matrix, expressing frustration with the reduction process and seeking assistance in identifying errors.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss redoing the work from scratch to avoid repeating mistakes, with one suggesting a specific method to eliminate coefficients greater than one. There is also mention of arithmetic errors in the original poster's calculations.

Discussion Status

The discussion is ongoing, with participants providing suggestions for reformatting the problem and re-evaluating the steps taken. There is no explicit consensus on the best approach yet, but several ideas are being explored.

Contextual Notes

Participants note the complexity of the numerical representation and suggest using LaTeX for clearer formatting. The original poster's calculations are acknowledged as potentially erroneous, but specific details of the errors are not fully resolved.

paulchem
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Use Gauss jordan reduction


Hi, sorry for introducing myself. I was just frustrated because I've been trying to do this problem for awhile.

4x+y-3z=11
2x-3y+2z=9
x+y+z=-3



4 1 -3 11
0 -1 0 3
0 3 7 -23

R2+2R3-->R2
4R3+-R1-->R3


3R1+-R3-->R1
4 0 2 56
0 -1 0 3
0 3 7 -23

I'm stuck here.

The calculator gave me

1 0 0 7/18
0 1 0 -3
0 0 1 -7/18

Please tell me where I went wrong thanks
 
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One thing I often do with these type of problems is that I redo the work from scratch -- I usually won't make the same mistake both times. (swapping rows around decreases the odds of remaking the mistake too)


That being said, you made an arithmetic error computing R2+2R3-->R2.
 
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I think a good idea when you start one of these questions is to first eliminate any coefficients of x which are greater than 1. So in the case of your question, starting with the given system, I would do:

R1' = R1 - 4R3
R2' = R2 - 2R3
 
Hello PaulChem. You know, I'm not trying to be critical, really I'm not, but those numbers look a little "busy", hard to follow. I don't expect you to know now how to make some really nice math format using LaTex which you can learn all about by jumping to the Physics Forum and checking out "Introducing LaTex". But here is what it would look like with a little formatting. Just double click on it and a small window will pop up showing the LaTex commands:

[tex] \left[ <br /> \begin{array}{cccc}<br /> 4 & 1 & -3 & 11 \\<br /> 0 & -1 & 0 & 3 \\<br /> 0 & 3 & 7 & -23 <br /> \end{array}<br /> \right][/tex]

Edit: alright single click, whatever. I get confussed.
 
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And if you want to draw a partitioned array...

[tex] <br /> \left(<br /> <br /> \begin{array}{ccc|c}<br /> <br /> 4 & 1 & -3 & 11 \\<br /> <br /> 0 & -1 & 0 & 3 \\<br /> <br /> 0 & 3 & 7 & -23 <br /> <br /> \end{array}<br /> <br /> \right)<br /> [/tex]
 
Hurkyl said:
And if you want to draw a partitioned array...

[tex] <br /> \left(<br /> <br /> \begin{array}{ccc|c}<br /> <br /> 4 & 1 & -3 & 11 \\<br /> <br /> 0 & -1 & 0 & 3 \\<br /> <br /> 0 & 3 & 7 & -23 <br /> <br /> \end{array}<br /> <br /> \right)<br /> [/tex]

Oh that is so much nicer. Mine is crummy.

Edit: Yep, PaulChem, I say do the whole problem again using Hurkly's format, step by step, nice "partitioned arrays", cut and past his commands into your post. With a final line saying: x, y, and z are: :smile:
 
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