Q. 124.3( 30 Votes )

If A and B are sq

Answer :

To prove: ABn = BnA

Given A and B are square matrices of same order such that AB = BA.


We have to prove it using mathematical induction.


Steps involved in mathematical induction are-


1. Prove the equation for n=1


2. Assume the equation to be true for n=k, where k ϵ N


3. Finally prove the equation for n=k+1


Let P(n): ABn = BnA


For n=1,


L.H.S: ABn = AB1 = AB


R.H.S: BnA = B1A = BA = AB


So, L.H.S = R.H.S


P(n) is true for n=1.


Now assuming P(n) to be true for n=k, where k ϵ N


P(k): ABk = BkA …… (1)


Now proving for n=k+1, i.e. P(k+1) is also true


L.H.S = ABn


= ABk+1


= (ABk).B


= (BkA).B …… from (1)


= Bk(A.B)


= Bk(BA) (AB = BA)


= Bk+1A


R.H.S = BnA


= Bk+1A


L.H.S = R.H.S


All conditions are proved. Hence P(k+1) is true.


By mathematical induction we have proved that ABn = BnA.


Now, to prove: (AB)n = AnBn for all n ϵ N


For n=1,


L.H.S = (AB)n = (AB)1 = AB


R.H.S = AnBn = A1B1 = AB


L.H.S = R.H.S


It is true for n=1


Assuming it to be true for n=k then,


(AB)k = AkBk ……(2)


Now proving for n=k+1,


L.H.S = (AB)n


= (AB)k+1


= (AB)k(AB)1


= (AkBk)AB


= Ak(Bk.A)B


= Ak(A.Bk)B (ABn = BnA)


= (AkA)(BkB)


= Ak+1Bk+1


R.H.S = AnBn


= Ak+1Bk+1


L.H.S = R.H.S


All conditions are proved. Hence P(k+1) is true.


By mathematical induction we have proved that (AB)n = AnBn for all nϵN


Hence proved.


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