Question

If AB = BA for any two square matrices, prove by mathematical induction that (AB)ⁿ = AⁿBⁿ 

Answer 1

Let P(n): (AB)ⁿ = AⁿBⁿ

So, P(1): (AB)¹ = A¹B¹

⇒ AB = AB

So, P(1) is true.

Let P(n) is true for some k ∈ N

So, P(k): (AB)^k = A^kB^k, k ∈ N  .....(i)

Now (AB)^k+1 = (AB)^k(AB)  ....(Using (i))

= A^kB^k(AB)

= A^kB^k–1(BA)B

= A^kB^k–1(AB)B   .....(As given AB = BA)

= A^kB^k–1AB²

= A^kB^k–2(BA)B²

= A^kB^k–2ABB²

= A^kB^k–2AB³

.......

.......

= A^k+1B^k+1

Thus P(1) is true and whenever P(k) is true P(k + 1) is true.

So, P(n) is true for all n ∈ N.