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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

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We can prove it by mathematical induction.
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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 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)...
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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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