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Prove by Mathematical Induction that (A’)n = (An)’, where n ∈ N for any square matrix A.

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By principle of mathematical induction we say that if a statement P(n) is true for n = 1 and if we assume P(k) to be true for some random natural number k and usnig it if we prove P(k+1) to be true we can say that P(n) is true for all natural numbers.

We are given to prove that (A’)n = (An)’.

Let P(n) be the statement : (A’)n = (An)’.

Clearly, P(1): (A’)1 = (A1)’

⇒ P(1) : A’ = A’

⇒ P(1) is true

Let P(k) be true.

∴ (A’)k = (Ak)’ …(1)

Let’s take P(k+1) now:

∵ (Ak+1)’ = (AkA)’

We know that by properties of transpose of a matrix:


∴ (AkA)’ = A’(Ak)’ = A’(A’)k = (A’)k+1

Thus, (Ak+1)’ = (A’)k+1

∴ P(k+1) is true.


We can say that: (A’)n = (An)’ is true for all n ∈ N.

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