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# If AB=BA for any two square matrices,prove by mathematical induction that $(AB)^n=A^nB^n$.

Toolbox:
• We use the principle of mathematical induction, where we need to prove P(n) is true for n=1, n=k, n=k+1
Step1:
Given
AB=BA
To prove
$(AB)^n=A^nB^n.$
Let $P(n):(AB)^n=A^nB^n.$
P(1) substitute n=1.
LHS:-$(AB)^1=AB$
RHS:-$A^nB^n=A^1B^1=AB.$
$\Rightarrow LHS=RHS.$
Step2:
$\Rightarrow P(n)$ be true for n=k.
$\Rightarrow (AB)^k=A^kB^k$
Multiply both sides by AB
$\Rightarrow (AB)^k.AB=A^kB^k.AB$
$\Rightarrow (AB)^k.(AB)^1=A^kB^k.AB$
$\Rightarrow (AB)^{k+1}=A^kB^k.BA$ [since AB=BA]
$\Rightarrow (AB)^{k+1}=A^k(B^k.BA)$ [By associative law]
$\qquad\qquad=A^k(B^k.B)A$
$\qquad\qquad=A^k(B^{k+1})A$ [AB=BA]
$\qquad\qquad=A^k(AB^{k+1})$ [$AB^n=B^nA$]
$\qquad\qquad=A^{k+1}.B^{k+1}$
$\Rightarrow$ P(n) is true for k+1
$\Rightarrow (AB)^n=A^nB^n.$