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Math 310: Final Exam (Solutions) Using second row, we see det(A—aI) = —(3-2)[(1—-a)(4—a) —5-2] = (a —3) (2? — 52-6) = (@-3)(e + 1)(w-6) So eigenvalues —1,3,6. (b) For each eigenvalue, find a basis for the corresponding eigenspace of A. Solving (A — Af)« = 0 by Chapter 1 for each A gives the columns of: -1 0 01 0]. 10. (c) Is A diagonalizable? If possible, give X such that X 1AX = D is diagonal. Yes. The eigenvectors form a basis. so can use matrix in (b) for X. Problem 2: I give you that X-1AX = D where Prof. S. Sinith: Monday 5 December 1994 Problem 1: 1 ¢ (a) Find the eigenvalues of A = ( 0 3 0 nN a mo cate -1 2 -1 1 —3 0 . a 2 (a) Give the gencral solution of the differential equation system My ta + 2 Ye = 2 — Using cigenvalucs —3, 1 and cigenvectors from given diagonalization, Hw \_ -1 3t L\ ot Ws = 1 e+e, 1 e 8041 = —ae7* + coet and yo = ee * + coe. (b) Give the particular solution when y,(0) = 3 and yo({0) = 1. -1 3 Putting in t = 0 gives system with augmented matrix 14 ; ) ; Then Chapter 1 methods give solution ¢, = —1, cz = 2; 80 yy =e + Qe! yy = —eW# + 2%. Problem 3: (a) Let M be the Markov matrix 3 lo Find the “steady state” eigenvector for 1. (Components of vector should add to 1). For eigenvalue 1, M -—1f= + ( 7 3 ) so eigenvector is span of (3,1)7. Hence (8, +) is steady-state vector. 
