Row Reduction over indicated field

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Homework Help Overview

The discussion revolves around solving a system of linear equations using Gauss-Jordan elimination in the field Z3. The original poster expresses confusion regarding the application of row reduction techniques within this specific modular arithmetic context.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial steps of row reduction, including swapping rows and performing operations on them. There are questions about interpreting arithmetic in Z3, particularly how negative numbers are represented.

Discussion Status

Participants are actively engaging with the problem, providing suggestions for how to approach the row reduction while emphasizing the importance of understanding arithmetic in Z3. There is a focus on clarifying the representation of numbers in this field, indicating a productive exploration of the topic.

Contextual Notes

The original poster notes that their textbook does not cover the material, which may contribute to their uncertainty. There is an emphasis on understanding the arithmetic rules specific to Z3, which is a key aspect of the discussion.

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Hi-

I am taking linear algebra and the teacher threw some problems at us that I cannot seem to figure out or find examples for online. Also, the textbook doesn't even cover the material.

The problem states:
Using Gauss-Jordan elimination, solve the following system with coefficients in indicated field.

In Z3:

2x + y = 1
x + y = 2

Now I understand completely how to row reduce this system, but am not sure how to proceed in Z3.

Any help appreciated. Thanks!
 
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The best thing to do here is to solve the system as you normally would and then interpret the solution in [itex]\mathbb{Z}_3[/itex].

So, how would you proceed normally?? Just write the first step, don't solve it completely yet.
 
Thanks for the reply.

The first thing I would do is swap row 1 and row 2 to get a leading 1 in row 1.

[1 1 l 2]
[2 1 l 1]

From here would take R2-2R1.
 
OK, so do R2-2R1. But be aware that the arithmetic is in [itex]\mathbb{Z}_3[/itex].

So, for example, the second column gives us 1-2*1=-1=2.
 
or, (to keep it straight as you go) realize that -2 in Z3 is actually 1, so instead of R2-2R1, use R2+R1 (which gives you 3 = 0 in the first column).

that is, instead of aiming to "zero out" entries, you just want to get them to the nearest multiple of 3 (which amounts to the same thing in Z3). with a little practice, you can see how to do this and always keep all the entries positive, which might help a bit.
 
micromass said:
OK, so do R2-2R1. But be aware that the arithmetic is in [itex]\mathbb{Z}_3[/itex].

So, for example, the second column gives us 1-2*1=-1=2.

Deveno said:
or, (to keep it straight as you go) realize that -2 in Z3 is actually 1, so instead of R2-2R1, use R2+R1 (which gives you 3 = 0 in the first column).

that is, instead of aiming to "zero out" entries, you just want to get them to the nearest multiple of 3 (which amounts to the same thing in Z3). with a little practice, you can see how to do this and always keep all the entries positive, which might help a bit.

Thanks for the reply. The part I don't understand is how -1 = 2 and -2 = 1 in Z3. Would you be able to explain this part to me?
 

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