Answer :

To prove: AB^{n} = B^{n}A

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): AB^{n} = B^{n}A

For n=1,

L.H.S: AB^{n} = AB^{1} = AB

R.H.S: B^{n}A = B^{1}A = 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): AB^{k} = B^{k}A …… (1)

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

L.H.S = AB^{n}

= AB^{k+1}

= (AB^{k}).B

= (B^{k}A).B …… from (1)

= B^{k}(A.B)

= B^{k}(BA) (∵AB = BA)

= B^{k+1}A

R.H.S = B^{n}A

= B^{k+1}A

∴ 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} = B^{n}A**.

Now, to prove: (AB)^{n} = A^{n}B^{n} for all n ϵ N

For n=1,

L.H.S = (AB)^{n} = (AB)^{1} = AB

R.H.S = A^{n}B^{n} = A^{1}B^{1} = AB

∴ L.H.S = R.H.S

∴ It is true for n=1

Assuming it to be true for n=k then,

(AB)^{k} = A^{k}B^{k} ……(2)

Now proving for n=k+1,

L.H.S = (AB)^{n}

= (AB)^{k+1}

= (AB)^{k}(AB)^{1}

= (A^{k}B^{k})AB

= A^{k}(B^{k}.A)B

= A^{k}(A.B^{k})B (AB^{n} = B^{n}A)

= (A^{k}A)(B^{k}B)

= A^{k+1}B^{k+1}

R.H.S = A^{n}B^{n}

= A^{k+1}B^{k+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} = A^{n}B^{n}** for

**all n**

**ϵ**

**N**

**Hence proved.**

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