74. If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABⁿ = BⁿA. Further, prove that (AB)ⁿ = AⁿBⁿ for all n Î N.

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Vishal Baghel | Contributor-Level 10

8 months ago

We have, AB = BA. (given)

(E) P (n):AB' = B'A.

P (i):AB¹ = B¹A. Þ AB = BA

so, the result is true for n = 1.

Let the result be true for n = k.

P (k):AB^k = B^kA

Then,

P (k + 1) : AB^{k + 1} = A. B^k. B = B^kA.B = B^k.BA ${?} AB=BA$

= Bk + 1.A .

So, AB^{k + 1}= B^{k + 1}A.

The result also holds for n = k + 1.

Hence, AB^n = B^n A^n ho

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