If A and B are square matrices of the same order such that AB = BA, then prove by induction that.

A and B are square matrices of the same order such that AB = BA.

To prove: $\quad \mathrm{P}(n): A B^{n}=B^{n} A, n \in \mathbf{N}$

For n = 1, we have:

$\begin{aligned} \mathrm{P}(1): & A B=B A \\ & \Rightarrow A B^{1}=B^{1} A \end{aligned}$

Therefore, the result is true for n = 1.

Let the result be true for n = k.

$\mathrm{P}(k): A B^{k}=B^{k} A$    .....(1)

Now, we prove that the result is true for $n=k+1$.

$\begin{aligned} A B^{k+1} &=A B^{k} \cdot B & & \\ &=\left(B^{k} A\right) B & &[\text { By }(1)] \\ &=B^{k}(A B) & &[\text { Associative law }] \\ &=B^{k}(B A) & &[A B=B A \text { (Given) }] \\ &=\left(B^{k} B\right) A & &[\text { Associative law }] \\ &=B^{k+1} A & & \end{aligned}$

Therefore, the result is true for $n=k+1$.

Thus, by the principle of mathematical induction, we have $A B^{n}=B^{n} A, n \in \mathbf{N}$.

Now, we prove that $(A B)^{n}=A^{n} B^{n}$ for all $n \in \mathbf{N}$

For n = 1, we have:

$(A B)^{1}=A^{1} B^{1}=A B$

Therefore, the result is true for n = 1.

Let the result be true for n = k.

$(A B)^{k}=A^{k} B^{k}$     ....(2)

Now, we prove that the result is true for n = k + 1.

$\begin{aligned}(A B)^{k+1} &=(A B)^{k} \cdot(A B) & & \\ &=\left(A^{k} B^{k}\right) \cdot(A B) & &[\text { By }(2)] \\ &=A^{k}\left(B^{k} A\right) B & &[\text { Associative law }] \\ &=A^{k}\left(A B^{k}\right) B & &\left[A B^{n}=B^{n} A \text { for all } n \in \mathbf{N}\right] \\ &=\left(A^{k} A\right) \cdot\left(B^{k} B\right) & &[\text { Associative law }] \\ &=A^{k+1} B^{k+1} & & \end{aligned}$

Therefore, the result is true for n = k + 1.

Thus, by the principle of mathematical induction, we have $(A B)^{n}=A^{n} B^{n}$, for all natural numbers.