## mkumar441 3 years ago 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!).

1. mathslover

So what we have to do .. ?

2. mathslover

ok sorry wait

3. mathslover

as given that A is n by n, rank(A) $$\le$$ n and conversely : n = rank($$I_n$$)=rank(AB)$$\le$$rank(A)

4. mathslover

got my point @mkumar441 ?

5. mkumar441

yes i got it thank you for ur quick response

6. mathslover

Your welcome dear.. also: $\huge{\mathbb{WE}\textbf{LC}\mathcal{OME}\space\textit{TO OPENSTUDY}}$

7. mkumar441

i need small favour at wat timings do u present inorder post questions.

8. mathslover

i didn't get you .. it seems that you are Indian. You can ask in Hindi?

9. mkumar441

yes i am indian but i dont Hindi.When do you be in online inorder to clarifiy my doubts?

10. mathslover

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....

11. mathslover

lemme introduce you to some experts for your future help: experimentx mukushla Vaidehi09 amistre64 vishweshshrimali5 unklerhaukus and many more

12. mathslover

just put @ sign before there names and they will be looking on your question as soon as possible.

13. mkumar441

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).

14. mathslover

Post that as a new question and tag me there.. I will help you there