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If A and B are square matrices of the same order such that AB = BA, then prove by induction that
. Further, prove that 
for all n ∈ N
Asked by Shilpa Last Modified
Answer 
Step1
Let (i) P(n):ABn=BnA
Given AB=BA
Put n=1 AB1=B1A.
AB=BA
⇒ P(n) is true for n=1
Step2
To prove P(n) is true for n=k
⇒ABk=BkA
Multiply both sides by B
ABk.B=A(Bk.B)
ABk.B=A.Bk+1[By associative property]
RHS
ABk.B
⇒Bk.AB.
Given AB=BA.
Hence Bk.BA=(Bk.B)A=Bk+1.A
we have LHS=A.Bk+1
Hence LHS=RHS
Hence p(k+1) is true for n=k+1.
⇒ By mathematical induction P(n) is true for n∈N
Step3
(ii)P(n):(AB)n=AnBn.
Let n=1 AB1=A1B1
Given AB=BA
Hence P(n) is true for n=1.
Step4
LHS:
Let P(n) be true for n=k.
Multiply both side by AB
⇒(AB)k=AkBk
(AB)k.AB
⇒AkBk.AB.
⇒(AB)k+1
RHS:
AkBk.AB
AkBk(BA)
Ak(Bk.B)A
Ak.(Bk+1.A)
Ak.ABk+1[ABn=BnA]
Ak+1Bk+1
⇒ P(n) is true for n=k+1.
By principle of mathematical induction
P(n) is true for all n∈N.
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