Prove by induction that, for n ∈ Z⁺ , [3 , -2 ; 2 , -1]ⁿ = [2n+1 , -2n ; 2n , 1-2n]
STEP 1: Prove that the theorem holds for n = 1. Substitute n = 1 into the equation and show that the LHS = RHS.
STEP 2: Assume that the relation is true when n = k.
STEP 3: Prove that the relation holds for n = k + 1, using the fact that it is true when n = k. This can be done by multiplying the matrix [2k+1 , -2k ; 2k , 1-2k] by the matrix [3 , -2 ; 2 , -1] (equivalent of raising the power on the LHS by 1). Simplifying and rearranging the result will yield: [2(k+1) +1 , -2(k+1) ; 2(k+1) , 1-2(k+1)].
STEP 4: Result shows that the equation holds when n = k + 1. Thus, if true for n = 1, must be true for all positive integers.
FT
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