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Prove of some property of square matrix

Asked by 14th March 2009, 6:17 PM
Answered by Expert
Answer:

For square matices A and B

AB = BA and ABn=BnA

Let P(n): (AB)n=A nBn

For n=1,

L.H.S = AB and R.H.S =AB

 P(n) is true for 1.

Let P(n) be true for n = k.

(AB)k = AkBk

Multiply both the side by AB

L.H.S = (AB)k(AB) =(AB)k+1

R.H.S = AkBk (AB) = AkBk (BA)             [ AB = BA]

           = Ak (Bk B) A

           = Ak Bk+1  A              [ABn=BnA ]

           = A (A Bk+1)  

           = Ak+1 Bk+1

⇒ P(n) is true for n = k+1

By principle of mathematical induction

P(n) is true for n N.

Hence, (AB)n = AnBn

Answered by Expert 20th March 2009, 4:05 PM
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