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): AB^k = B^kA …… (1)

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

L.H.S = ABⁿ

= AB^k+1

= (AB^k).B

= (B^kA).B …… from (1)

= B^k(A.B)

= B^k(BA) (∵AB = BA)

= B^k+1A

R.H.S = BⁿA

= B^k+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 = A^kB^k ……(2)

Now proving for n=k+1,

L.H.S = (AB)ⁿ

= (AB)^k+1

= (AB)^k(AB)¹

= (A^kB^k)AB

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

= A^k(A.B^k)B (ABⁿ = BⁿA)

= (A^kA)(B^kB)

= A^k+1B^k+1

R.H.S = AⁿBⁿ

= A^k+1B^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 = AⁿBⁿ for all nϵN

Hence proved.