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please solve this ... suppose A and B are n by n matrices,and AB=I.prove from rank(AB)<=rank(A) that the rank of A is n . So A is invertible and B must be its two-sided inverse. Therefore BA=I(Which is not obvious!).

Mathematics
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So what we have to do .. ?
ok sorry wait
as given that A is n by n, rank(A) \(\le\) n and conversely : n = rank(\(I_n\))=rank(AB)\(\le\)rank(A)

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got my point @mkumar441 ?
yes i got it thank you for ur quick response
Your welcome dear.. also: \[\huge{\mathbb{WE}\textbf{LC}\mathcal{OME}\space\textit{TO OPENSTUDY}}\]
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i didn't get you .. it seems that you are Indian. You can ask in Hindi?
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Another question ..Suppose A is an m by n matrix of rank r.Its reduced echelon form is R. Describe exactly reduced row echelon form of R transpose (not A transpose).
Post that as a new question and tag me there.. I will help you there

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