If A and B are square matrices of the same order such that `AB=BA`, then prove by induction that `AB^(n)=B^(n)A`. Further prove that `(AB)^(n)=A^(n)B^(n)` for all `n in N`.

Explanation

We are given, `AB = BA`

We have to prove, `AB^n = B^nA`

When `n = 1`,

`AB^1 = B^1A => AB = BA`

So, given equation is true for `n =1`

Let this equation is true for `n=k`.

Then, `AB^k = B^kA`

Now, we have to prove it is true for `n = k+1` that is `AB^(k+1) = B^(k+1)A`

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