Answer 1

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.