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Is there any difference regarding row operations and such with matrices when they are over fields? For instance, I have the following matrix over GF(3):

0 0 2 2 0 2
2 2 0 2 1 2
1 1 2 0 2 2
1 1 0 1 2 1

I did a bunch of row operations and put it into reduced echelon form which gave me:

1 1 0 0 0 2/3
0 0 1 0 0 2/3
0 0 0 1 0 1/3
0 0 0 0 1  0

Is this correct or is something done differently with operations when it's a field?

EDIT: Used all the suggestions listed below. Could someone confirm my answer and let me know if there's any quicker steps to the answer? Listing all my steps below.

R1 <-> R4

1 1 0 1 2 1
2 2 0 2 1 2
1 1 2 0 2 2
0 0 1 1 0 1

R2 = R2 + R3

1 1 0 1 2 1
0 0 2 2 0 1
1 1 2 0 2 2
0 0 2 2 0 2

R3 = R3 + 2*R1

1 1 0 1 2 1
0 0 2 2 0 1
0 0 2 2 0 1
0 0 2 2 0 2

R2 = R2 + R3

1 1 0 1 2 1
0 0 1 1 0 2
0 0 2 2 0 1
0 0 2 2 0 2

R3 = R2 + R3, R4 = R2 + R4

1 1 0 1 2 1
0 0 1 1 0 2
0 0 0 0 0 0
0 0 0 0 0 1

R4 <-> R3

1 1 0 1 2 1
0 0 1 1 0 2
0 0 0 0 0 1
0 0 0 0 0 0

R2 = R2 + R3

1 1 0 1 2 1
0 0 1 1 0 0 
0 0 0 0 0 1
0 0 0 0 0 0

R1 = R1 + 2*R3

1 1 0 1 2 0
0 0 1 1 0 0
0 0 0 0 0 1
0 0 0 0 0 0
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    $\begingroup$ What the others said. Everything is the same, and you must not divide by zero, in this case you must not divide by $3$ either, because $3=0$. See this old answer for an example of doing row reduction over $GF(29)$. +1 to all $\endgroup$ Oct 16, 2015 at 15:11

3 Answers 3

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Everything is done the same, with the adjustment that all the matrix entries and the scalars you use come from the field.

So as mentioned, if you are working in $\mathbb{F}_{3}$ then you can't divide a row by $3=0$. Any time you perform a row operation you can reduce the entries mod $3$.

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  • $\begingroup$ One quick question. Can I still do division normally with a single row? For instance, if I have a row of all 2's, can I just do 1/2 times that row to make them 1's? Or is it different with GF(3)? $\endgroup$
    – pfinferno
    Oct 17, 2015 at 4:51
  • $\begingroup$ wait, the inverse of 2 aka 1/2, is equal to 2? That part doesn't make sense to me. If I had a row of 2's and did 1/2 times that row, it should give me all 1's right? $\endgroup$
    – pfinferno
    Oct 17, 2015 at 13:57
  • $\begingroup$ Okay I think I understand what you mean now. But if I do 1/2 or a row of 2's, that gives me 1, and 1 mod 3 = 1 right? So a 1 would go in the spot in the matrix. Could you take a look at my edit on the original post and see if my calculations are correct? $\endgroup$
    – pfinferno
    Oct 17, 2015 at 20:40
  • $\begingroup$ Awesome. Think I get it now, thanks! $\endgroup$
    – pfinferno
    Oct 17, 2015 at 20:51
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Hint. Does $2/3$ make sense over GF(3)?

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  • $\begingroup$ Nope, co-efficients can be 0,1, or 2 I remember now. Is the addition = subtraction here? I remember working with the fields before 2 + 2 = 0. $\endgroup$
    – pfinferno
    Oct 16, 2015 at 5:30
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    $\begingroup$ @pfinferno: In characteristic 2, addition is subtraction, because $2=0$ implies $1=-1$. In characteristic 3, that doesn't work: you get, e.g., $-1 = 2$. I suppose if you wanted, you could say that subtraction and doubling are the same. Or that doubling is the same as halving, because $4=1$ ,and thus $2 = \frac{1}{2}$. $\endgroup$
    – user14972
    Oct 16, 2015 at 5:49
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Yes, everything here is done the same way — including the fact you shouldn't divide by zero like you did in your calculations.

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