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mkumar441
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!).
So what we have to do .. ?
as given that A is n by n, rank(A) \(\le\) n and conversely : n = rank(\(I_n\))=rank(AB)\(\le\)rank(A)
got my point @mkumar441 ?
yes i got it thank you for ur quick response
Your welcome dear.. also: \[\huge{\mathbb{WE}\textbf{LC}\mathcal{OME}\space\textit{TO OPENSTUDY}}\]
i need small favour at wat timings do u present inorder post questions.
i didn't get you .. it seems that you are Indian. You can ask in Hindi?
yes i am indian but i dont Hindi.When do you be in online inorder to clarifiy my doubts?
ok i got you now: actually i am present here from 3:00 PM to 9:00 PM (Indian Standard Time_ or there may be breaks but yes: There are gr8 experts present you don't need to care for your clarification of doubts. In case, if your doubt is not answered then you can post the link of your question in the chat. or tag some people having good smart score....
lemme introduce you to some experts for your future help: experimentx mukushla Vaidehi09 amistre64 vishweshshrimali5 unklerhaukus and many more
just put @ sign before there names and they will be looking on your question as soon as possible.
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