Any difference working with matrices over fields?

Question

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

Answer 1

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 $3$.

Answer 2

Hint. Does $2/3$ make sense over GF(3)?

Answer 3

Yes, everything here is done the same way — including the fact you shouldn't divide by zero like you did in your calculations.