If A and B are two square marices of the same order such that `AB=BA`, then `(AB)^(n)` is equal to
where ` n epsilonN`

Options

A. `AB`
B. `A^(n)B`
C. `B^(n)A`
D. `A^(n)B^(n)`

Answer

D

Explanation

Here given that `AB=BA`…………i
We want to prove that `(AB)^(n)=A^(n)B^(n)`………..ii
For `n=1`, Eq. (ii) is obviously true [From eq. i ]
Let Eq. (ii) be true for a positive integer `n=m`
i.e. `(AB)^(m)=A^(m)B^(m)`…………….iii
then of `n=m+1`
`(AB)^(m+1)=(AB)^(m)(AB)=(A^(m)B^(m))(AB)` [From Eq. iii]
`=A^(m)(B^(m)A)B=A^(m)(AB^(m))B`
`[ :'AB^(n)=B^(n)A` for all `n epsilonN`, whenever `AB=BA]`
`=(A^(m)A)(B^(m)B)=A^(m+1)B^(m+1)`
Hence by induction Eq. ii is true for all `n epsilonN`.