Let `A ,B` be two matrices such that they commute. Show that for any positive integer `n ,` (i) `A B^n=B^n A` (ii) `(A B)^n=A^nB^n`

Explanation

Let `P(n):(AB)^(n)=A^(n)B^(n)`
`therefore P(1):(AB)^(1)=A^(1)B^(1)rArrAB=AB`
So P(1) is true
Now, `P(k):(AB)^(k)rArrA^(k)B^(k) ,K in N`
So, P(K) is true, whenever `P(k+1)` is true
`therefore p(k+1:AB)^(k+1)=A^(k+1)B^(k+1)`
`rArr AB^(k),AB^(1)`
`rArr A^(k)B^(k).BArArrA^(k)B^(k+1)A`
`rArr A^(k).A.B^(k+1)rArrA^(k+1)B^(k+1)`
`rArr (A.B)^(k+1)=A^(k+1)B^(k+1)`
So, P(k+1) is true for all `n in N` , whenever P(k) is true.
By mathematical induction (AB)=`A^(n)B^(n)` is true for all `n in N`.