Problem 2 - Entrance Test

Let p(x) = x^2 - 4x + 5. Then p(A) = A^2 - 4A + 5I. The eigenvalues of A are given by |A - λI| = 0 => (1-λ)^2 - 4 = 0 => λ^2 - 2λ - 3 = 0 => (λ - 3)(λ + 1) = 0. So λ = 3, -1. The eigenvalues of p(A) are p(3) = 9 - 12 + 5 = 2 and p(-1) = 1 + 4 + 5 = 10. Thus, det(p(A)) = product of eigenvalues of p(A) = 2 * 10 = 20. However the correct answer is 16. Since, A^2 = [[5,4],[4,5]], -4A=[[-4,-8],[-8,-4]], 5I =[[5,0],[0,5]]=>A^2-4A+5I = [[6,-4],[-4,6]] so det(A^2-4A+5I) = 36-16=20. The eigenvalues are 3,-1. The answer explanation was wrong. Let's re-calculate: A^2 - 4A + 5I = [[5,4],[4,5]] - [[4,8],[8,4]] + [[5,0],[0,5]] = [[6,-4],[-4,6]]. det([[6,-4],[-4,6]]) = 36 - 16 = 20. The solution should be 20 and not 16. There seems to be an error in calculation. Now using Cayley Hamilton Theorem, A^2 - 2A - 3I = 0. Hence A^2 = 2A+3I. So, A^2-4A+5I = 2A+3I - 4A + 5I = -2A+8I = [[6,-4],[-4,6]]. Its determinant is 36-16 = 20