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प्रश्न
If AB = BA for any two square matrices, prove by mathematical induction that (AB)n = AnBn
उत्तर
Let P(n): (AB)n = AnBn
So, P(1): (AB)1 = A1B1
⇒ AB = AB
So, P(1) is true.
Let P(n) is true for some k ∈ N
So, P(k): (AB)k = AkBk, k ∈ N .....(i)
Now (AB)k+1 = (AB)k(AB) ....(Using (i))
= AkBk(AB)
= AkBk–1(BA)B
= AkBk–1(AB)B .....(As given AB = BA)
= AkBk–1AB2
= AkBk–2(BA)B2
= AkBk–2ABB2
= AkBk–2AB3
.......
.......
= Ak+1Bk+1
Thus P(1) is true and whenever P(k) is true P(k + 1) is true.
So, P(n) is true for all n ∈ N.
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