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Is it true that given a matrix $A$, $A$ has L.I columns/rows if and only if $\text {det}(A)=0$?

This is my question I think the statement is something like the above but I'm not 100% sure. If anyone could give me the correct statement and maybe some reasoning (not necessarily proof) as to why this would be the case I would appreciate it.

I'm more than willing to give my input but I'm quite stuck at the moment so have not a lot to offer unfortunately.

Thanks.

Chris
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  • It's supposed to be "$\iff \det(A) \ne 0$". I try to motivate this geometrically in this answer. –  Oct 07 '15 at 13:12
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    Only if $A$ is a square matrix. – Rajat Oct 07 '15 at 13:13
  • ^ Oh right. That too. :) –  Oct 07 '15 at 13:14
  • Yes of course I meant does not equal zero apologies. I will take a look over the answer now. Could you also explain to me if L.I rows is the same as L.I columns I think it would be but I'm not fully confident. – Chris Oct 07 '15 at 13:15
  • There is a theorem which states that the column rank = the row rank of a matrix. This implies that a square matrix with L.I. columns will have L.I. rows. –  Oct 07 '15 at 13:16
  • Unless there's something else you'd like to know, I'm going to vote to close this because asking "Is this statement correct" warrants only a one line answer and I don't think this question will be of use to anyone else. –  Oct 07 '15 at 13:35
  • Also http://math.stackexchange.com/q/585143 –  Oct 11 '15 at 04:36

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