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

A and B are square matrices of the same order such that `AB=BA`.
To prove : `P(n) : AB^(n)=B^(n) A, n in N`
For `n=1`, we have
`P(1) : AB=BA` [Given]
`implies AB^(1)=B^(1)A`
Therefore, the result is true for `n=1`.
Let the result br true from `n=k`.
i.e., `P(k) : AB^(k)=B^(k)A` (1)
Now, we have to prove that the result is true for `n=k+1`.
`AB^(k+1)=AB^(k)B`
`=(B^(k)A) B` [Using (1)]
`=B^(k) (AB)` [Associative law]
`=B^(k) (BA)" "[AB=BA ("Given")]`
`=(B^(k)B)A` [Associative law]
`=B^(k+1) A`
Therefore, the result is true for `n=k+1`.
Thus, by the principle of methematical induction, we have `AB^(n)=B^(n)A, n in N`.
Now, we have to prove that `(AB)^(n)=A^(n)B^(n)` for all `n in N`.
For `n=1`, we have
`(AB)^(1)=A^(1)B^(1)=AB`
Therefore, the result is true for `n=1`.
Let the result be true for `n=k`.
i.e., `(AB)^(k)=A^(k)B^(k)` (2)
Now, we have to prove that the result is true for `n=k+1`.
`(AB)^(k+1)=(AB)^(k) (AB)`
`=(A^(k)B^(k)) (AB)` [Using (2)]
`=A^(k) (B^(k) A)B` [Associative law]
`=A^(k) (AB^(k))B" "[AB^(n)=B^(n)" for all "n in N]`
`=(A^(k) A) (B^(k) B)` [Associative law]
`=A^(k+1) B^(k+1)`
Therefore, the result is true for `n=k+1`.
Thus, by the principle of mathematical induction, we have `(AB)^(n)=A^(n)B^(n)`, for all `n in N`.