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

Prove by mathematical induction that (A')^n = (A^n)', where n ∈ N for any square matrix A.

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By the 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, then 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')\textsuperscript{n} = (A\textsuperscript{n})'.

Let P(n) be the statement :(A')\textsuperscript{n} = (A\textsuperscript{n})'.

Clearly, P(1): (A')\textsuperscript{1} = (A\textsuperscript{1})'

\\ \Rightarrow $ P(1) : A' = A' \\$ \Rightarrow $ $ P(1) is true

Let P(k) be true.

\therefore $ (A')\textsuperscript{k} = (A\textsuperscript{k})' $ \ldots (1)

Let’s take P(k+1) now:

\because (A\textsuperscript{k+1})' = (A\textsuperscript{k}A)'

We know that according to the rule of transpose of a matrix,

(AB)^{T} = B^{T}A^{T} \quad \therefore \quad (A^{k}A)' = A'(A^{k})' = A'(A')^{k} = (A')^{k+1}

Thus,(A\textsuperscript{k+1})' = (A')\textsuperscript{k+1}

\therefore P(k+1) is true.

Hence proved: (A')\textsuperscript{n} = (A\textsuperscript{n})' is true for all n \in N.

 

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