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# 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 $$n ∈ N.$$

Toolbox:
• The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 ≤ i ≤ m and 1 ≤ j ≤ n.
• 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
Let (i) P(n):$AB^n=B^nA$
Given AB=BA
Put n=1 ${AB}^1=B^1A$.
AB=BA
$\Rightarrow$ P(n) is true for n=1
Step2
To prove P(n) is true for n=k
$\Rightarrow AB^k=B^kA$
Multiply both sides by B
$AB^k.B=A(B^k.B)$
$AB^k.B=A.B^{k+1}$[By associative property]
RHS
$AB^k.B$
$\Rightarrow B^k.AB$.
Given AB=BA.
Hence $B^k.BA=(B^k.B)A=B^{k+1}.A$
we have LHS=$A.B^{k+1}$
Hence LHS=RHS
Hence p(k+1) is true for n=k+1.
$\Rightarrow$ By mathematical induction P(n) is true for $n\in N$
Step3
(ii)$P(n):(AB)^n=A^nB^n.$
Let n=1 ${AB}^1=A^1B^1$
Given AB=BA
Hence P(n) is true for n=1.
Step4
LHS:
Let P(n) be true for n=k.
Multiply both side by AB
$\Rightarrow (AB)^k=A^kB^k$
$(AB)^k.AB$
$\Rightarrow A^kB^k.AB.$
$\Rightarrow (AB)^{k+1}$
RHS:
$A^kB^k.AB$
$A^kB^k(BA)$
$A^k(B^k.B)A$
$A^k.(B^{k+1}.A)$
$A^k.AB^{k+1}$$[AB^n=B^nA$]
$A^{k+1}B^{k+1}$
$\Rightarrow$ P(n) is true for n=k+1.
By principle of mathematical induction
P(n) is true for all $n\in N.$