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Question
Answer the following:
Prove by method of induction
`[(3, -4),(1, -1)]^"n" = [(2"n" + 1, -4"n"),("n", -2"n" + 1)], ∀ "n" ∈ "N"`
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Solution
Let P(n) ≡ `[(3, -4),(1, -1)]^"n" = [(2"n" + 1, -4"n"),("n", -2"n" + 1)]`, for all n ∈ N.
Step 1:
For n = 1,
L.H.S. = `[(3, -4),(1, -1)]`
R.H.S. = `[(2(1) + 1, -4(1)),(1, -2(1) + 1)] = [(3, -4),(1, -1)]`
∴ L.H.S. = R.H.S. for n = 1
∴ P(1) is true
Step 2:
Let us assume that for some k ∈ N, P (k) is true.
i.e., `[(3, -4),(1, -1)]^"k" = [(2"k" + 1, -4"k"),("k", -2"k" + 1)]` ...(1)
Step 3:
To prove that P(k + 1) is true, i.e., to prove that
`[(3, -4),(1, -1)]^("k"+1) = [(2("k" + 1) + 1, -4("k" + 1)),(("k" + 1), -2("k" + 1) + 1)]`
Now, L.H.S. = `[(3, -4),(1, - 1)]^("k"+1) = [(3, -4),(1, -1)]^"k" [(3, -4),(1, -1)]`
= `[(2"k" + 1, -4"k"),("k", -2"k" + 1)] [(3, -4),(1, -1)]` ...[By (1)]
= `[(3(2"k" + 1) - 4"k", -4(2"k" + 1) + 4"k"),(3"k" - 2"k" + 1, -4"k" + 2"k" - 1)]`
= `[(2"k" + 3, -4"k" - 4),("k" + 1, -2"k" - 1)]`
= `[(2("k" + 1) + 1, -4("k" + 1)),("k" + 1, -2("k" + 1) + 1)]`
= R.H.S.
∴ P(k + 1) is true.
Step 4:
From all the above steps and by the principle of mathematical induction P(n) is true for all n ∈ N.
i.e., `[(3, -4),(1, -1)]^"n" = [(2"n" + 1, -4"n"),("n", -2"n" + 1)]`, for all n ∈ N.
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