Prove of some property of square matrix
Asked by
| 14th Mar, 2009,
06:17: PM
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 ]
= Ak (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
| 20th Mar, 2009,
04:05: PM
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