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Mathematics 18 Online
OpenStudy (trojanpoem):

If two matrices A, B verify the relationship AB = BA Prove that: \[AB^{n} = B^{n}A\] for any integer n.

OpenStudy (jango_in_dtown):

The proposition is ture for n=1, by the given condition. Let the proposition be true for some integer k>1. Then AB^k=B^kA Now AB^(k+1)=AB^kB=(AB^k)B=B^kAB=B^kBA=B^(k+1)A/ Hence the proposition is true for k+1 whenever it is true for k. Hence by the method of induction....

OpenStudy (trojanpoem):

Thanks, nevertheless, My professor said : "Using induction is forbidden.",

OpenStudy (jango_in_dtown):

So let me rethink

OpenStudy (jango_in_dtown):

AB^n=AB...B=(BA)B...B=B(AB)B...B=B(BA)B...B=BB(AB)B...B=BB(BA)B...B=...=BB...B(AB)=BB...B(BA)=B^nA

OpenStudy (trojanpoem):

What did you do ? the so many B are confusing.

OpenStudy (jango_in_dtown):

just used the property BB...B (n times)=B^n

OpenStudy (jango_in_dtown):

and the given property AB=BA multiples times

OpenStudy (trojanpoem):

Nice.

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