why mod(rref(A), 2) does not give the correct result?

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Lily
Lily 2025-5-30
编辑： David Goodmanson 2025-5-30
Ran in:
Lets say I have the followings all in mod 2
a = [1 1 0 0 0]'
a = 5×1
1 1 0 0 0
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b = [1 0 1 0 1]'
b = 5×1
1 0 1 0 1
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c = [0 1 1 0 1]'
c = 5×1
0 1 1 0 1
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clearly, c = a + b in mod 2
However when I run the command
mod(rref([a b c]), 2)
ans = 5×3
1 0 0 0 1 0 0 0 1 0 0 0 0 0 0
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It does not give the corret result for linear combination in mod 2? My expect answer should be:
expected_ans = [1 0 1; 0 1 1; 0 0 0; 0 0 0; 0 0 0]
expected_ans = 5×3
1 0 1 0 1 1 0 0 0 0 0 0 0 0 0
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Could anyone explain why?

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采纳的回答

David Goodmanson
David Goodmanson 2025-5-30
编辑：David Goodmanson 2025-5-30
Hi Lily,
Although c is a linear combination of a and b mod 2, rref does not know anything about mod 2.
rank([a b c])
ans = 3
As far as rref is concerned, a, b and c are linearly independent ('mod' not being a linear operation under the usual addition of numbers). So you get three nonzero rows.

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