Question

if A and B are square matrices of the same order such that AB=BA, then prove by induction that ABn = BnA further prove that (AB )n=AnBn for all natural number ​

Answer 1

Answer:

1.

Given:

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

Let P(n) denote the statement

$"AB^n=B^nA"$

Now,

AB=BA

$AB^1=B^1A$

P(1) is true

Hence the result is true for n=1.

Assume that P(k) is true.

that is

$AB^k=B^kA$ is true

To Prove: P(k+1) is true

That is to prove:

$AB^{k+1}=B^{k+1}A$ is true

Now,

$AB^{k+1}$

$=A(B^kB)$

$=(AB^k)B$

$=(B^kA)B$

$=B^k(AB)$

$=B^k(BA)$ (since AB=BA)

$=(B^kB)A$

$=B^{k+1}A$

Therefore P(k+1) is true.

Hence P(n) is true for all natural numbers.

2.

Let P(n) denote the statement

$"(AB)^n=A^nB^n"$

Put n=1,

$(AB)^1=AB=A^1B^1$

P(1) is true.

Hence the result is true for n=1.

Assume that P(k) is true.

That is, $(AB)^k=A^kB^k$ is true

To prove: P(k+1) is true

That is to prove:

$(AB)^{k+1}=A^{k+1}B^{k+1}$ is true

Now,

$(AB)^{k+1}$

$=(AB)^k(AB)$

$=A^kB^k(AB)$

$=A^k(B^kA)B$

$=A^k(AB^k)B$

$=(A^kA)(B^kB)$

$=A^{k+1}B^{k+1}$

Therefore P(k+1) is true

Hence P(n) is true for all natural numbers.