Simple Gauss-Jordan Elimination

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Homework Help Overview

The discussion revolves around the application of Gauss-Jordan elimination to solve a system of linear equations represented in matrix form. Participants are analyzing their steps and results while attempting to identify potential errors in their calculations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants describe their attempts to perform row operations on the matrix, questioning the correctness of their results and the potential for consistent mistakes. They discuss specific steps taken, such as adding and scaling rows, and express confusion over discrepancies in their final results.

Discussion Status

Some participants have identified specific errors in their calculations and are reflecting on their approaches. There is an ongoing exploration of the steps involved in the elimination process, with some guidance provided regarding the correct operations to perform.

Contextual Notes

Participants mention the possibility of the book answer being incorrect, indicating a lack of clarity in the expected outcomes. There is also a focus on maintaining integer values during calculations as a strategy to simplify the process.

Saladsamurai
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I have done the following problem over and over and keep getting the same result. So either the book answer is wrong or I am making the same mistake over again:

\left[\begin{array} {cccc}<br /> 5&amp;-2&amp;6&amp;0\\<br /> -2&amp;1&amp;3&amp;1\\<br /> \end{array}\right]

\left[\begin{array} {cccc}<br /> 1&amp;-\frac25&amp;\frac65&amp;0\\<br /> -2&amp;1&amp;3&amp;1\\<br /> \end{array}\right]

\left[\begin{array} {cccc}<br /> 1&amp;-\frac25&amp;\frac65&amp;0\\<br /> 0&amp;\frac15&amp;\frac{17}{5}&amp;1\\<br /> \end{array}\right]

\left[\begin{array} {cccc}<br /> 1&amp;0&amp;8&amp;2\\<br /> 0&amp;1&amp;17&amp;5\\<br /> \end{array}\right]

Am I screwing this up somewhere?
 
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Saladsamurai said:
I have done the following problem over and over and keep getting the same result. So either the book answer is wrong or I am making the same mistake over again:

\left[\begin{array} {cccc}<br /> 5&amp;-2&amp;6&amp;0\\<br /> -2&amp;1&amp;3&amp;1\\<br /> \end{array}\right]

\left[\begin{array} {cccc}<br /> 1&amp;-\frac25&amp;\frac65&amp;0\\<br /> -2&amp;1&amp;3&amp;1\\<br /> \end{array}\right]

\left[\begin{array} {cccc}<br /> 1&amp;-\frac25&amp;\frac65&amp;0\\<br /> 0&amp;\frac15&amp;\frac{17}{5}&amp;1\\<br /> \end{array}\right]

\left[\begin{array} {cccc}<br /> 1&amp;0&amp;8&amp;2\\<br /> 0&amp;1&amp;17&amp;5\\<br /> \end{array}\right]

Am I screwing this up somewhere?
I ended up with different numbers in the 3rd column.
\left[\begin{array} {cccc}<br /> 1&amp;0&amp;12&amp;2\\<br /> 0&amp;1&amp;27&amp;5\\<br /> \end{array}\right]

In my first step, I added 2 times row 1 to -5 times row 2. Then I added 2 times row 2 to row 1. Finally, I replaced row 1 with 1/5 of itself. It is usually simpler to keep integer values for as long as possible.
 
@Saladsamurai, [STRIKE]When you go to eliminate the -2, you want to add row 1 to row 2. You're subtracting it.[/STRIKE]

3 + 12/5 = 15/5 + 12/5 = 27/5
 
Yeah! I just caught that :/ Man.. I must have added 6/5+3=17/5
like ten times! Stupid brain gets sucked into following
a pattern

thanks! :smile:
 

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