Question

If A and B are square matrices of the same order such that $AB=BA$,then prove by induction that $AB^n=B^nA$.Further, prove that $(AB)^n=A^nB^n$ for all $n∈N$

Answer 1

A and B are square matrices of the same order such that $AB = BA$. To prove $P(n):AB^n=B^nA\,\,\, n∈N$
For $n = 1$, we have:
$P(1):AB^1=B^1A\implies AB=BA$
Therefore, the result is true for $n = 1$.
Let the result be true for $n = k$.
$P(k):AB^k=B^kA .....(1)$
Now, we prove that the result is true for $n = k + 1$.
$AB^k+1=AB^k.B$
$=(B^kA)B\,\,\,\,\,\, [by (1)]$
$=B^k(AB) [Associative law]$
$=B^k(BA) [AB=BA\,\, given]$
$=(B^kB)A [Associative\,\, law]$
$=B^k+1A$
Therefore, the result is true for $n = k + 1.$
Thus, by the principle of mathematical induction, we have
$AB^n=B^nA,n∈N$
Now, we prove that $(AB)^n=A^nB^n$ for all $n∈N$
For $n = 1$, we have:
$(AB)^1=A^1B^1=AB$
Therefore, the result is true for $n = 1$.
Let the result be true for $n = k$.
$(AB)^k=A^kB^k ....(2)$
Now, we prove that the result is true for $n = k + 1$.
$(AB)^{k+1}=(AB)^k.(AB)$
$=(A^kB^k).(AB) [By(2)]$
$= A^k(B^kA)B [Associative\,\, law]$
$=A^k(AB^k)B [AB^n=B^nA\,\, for\,\, all\,\, n∈N]$
$=(A^kA).(B^kB) [Associative\,\, law]$
$=A^{k+1}B^{k+1}$
Therefore, the result is true for $n = k + 1$
Thus, by the principle of mathematical induction, we have $(AB)^n=A^nB^n$ , for all natural numbers.