Matrix A𝐴 is such that A2=2A−I,𝐴2=2𝐴-𝐼, where I𝐼 is the indentity matrix, then for n≥2,An is equal to
Question
Solution
To solve this problem, we can use the principle of mathematical induction.
Base Case (n=2): We are given that A^2 = 2A - I. So, the base case is true.
Inductive Step: Assume that the statement is true for some k >= 2, i.e., A^k = kA - (k-1)I.
We need to prove that the statement is true for k+1, i.e., A^(k+1) = (k+1)A - kI.
A^(k+1) = A^k * A Using the inductive hypothesis, we can substitute A^k with kA - (k-1)I. So, A^(k+1) = (kA - (k-1)I) * A = kA^2 - (k-1)A Substitute A^2 with 2A - I (given in the problem). So, A^(k+1) = k(2A - I) - (k-1)A = 2kA - kI - (k-1)A = (2k-k+1)A - kI = (k+1)A - kI
So, by the principle of mathematical induction, for n>=2, A^n is equal to nA - (n-1)I.
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