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Question

If A and B are two matrices such that AB=BA, then nN

A
AnB=BAn
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B
(AB)n=AnBn
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C
(A+B)n=nC0An+nC1An1B+nC2An2B2+...+nCnBn
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D
A2nB2n=(AnBn)(An+Bn)
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Solution

The correct options are
A AnB=BAn
B (AB)n=AnBn
C (A+B)n=nC0An+nC1An1B+nC2An2B2+...+nCnBn
D A2nB2n=(AnBn)(An+Bn)
Given AB=BA
We check all the options
A. The result AnB=BAn is true for n=2 since
A2B=A(AB)=A(BA)=(AB)A=(BA)A=BA2
Let the result be true for n=k
AkB=BAk
For n=k+1, we have
Ak+1B=A(AkB)=A(BAk)=(AB)Ak=(BA)Ak=BAk+1
Thus the result is true for n=k+1
Hence by principle of Mathematical induction, the result is true for all nN
Options B and C : since A and B commute, both these option hold
Option D: (AnBn)(An+Bn)=A2n+AnBnBnAnB2n=A2nB2n, A and B commute.

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