Exam 4 Study Guide Solution fall 2012 on Engineering Mathematics III (Numerical Methods) - M351 Study Guide 4(S Zhang 1(8.8:8 This is enough The other

# Exam 4 Study Guide Solution fall 2012 on Engineering Mathematics III (Numerical Methods)

• Test Prep
• dlbmcd238
• 8

This preview shows page 1 - 3 out of 8 pages.

M351 Study Guide 4 (S. Zhang) 1. (8.8:8) Find eigenvalues and eigenvectors. 2 1 2 1 ans: det( A - λI ) = 2 - λ 1 2 1 - λ , = ( λ 2 - 3 λ + 2) - 2 = ( λ - 3) λ λ = 0 , 3 For λ = 0, solving ( A - λI ) x = 0 , A - λI = 2 1 2 1 2 1 0 Choose the opposite coeffcients with one negative sign: x = C 1 - 2 λ = 3, solving ( A - λI ) v = 0 , A - λI = - 1 1 1 - 1 - 1 1 0 Choose the opposite coeffcients with one negative sign: x = C 1 1 2. (8.8:11) Find eigenvalues and eigenvectors. - 1 2 - 5 1 ans: det( A - λI ) = - 1 - λ 2 - 5 1 - λ , λ = ± 3 i λ = 3 i , solving ( A - λI ) x = 0 , A - λI = - 1 - 3 i 2 - 5 1 - 3 i - 1 - 3 i 2 0 Choose the opposite coeffcients with one negative sign: x 1 = C 2 1 + 3 i = C ( 2 1 + i 0 3 This is enough. The other eigenvector is a conjugate of this one: x 2 = x 1 = C ( 2 1 + i 0 - 3 Anyway, if we compute, λ = - 3 i , solving ( A - λI ) x = 0 , A - λI = - 1 + 3 i 2 - 5 1 + 3 i - 1 + 3 i 2 0 Choose the opposite coeffcients with one negative sign: x = C 2 1 - 3 i = C ( 2 1 + i 0 - 3 Note that the two roots are conjugate, and the two eigen- vectors are conjugate. So we should skip the second step. 3. (8.8:18) Find eigenvalues and eigenvectors. 1 6 0 0 2 1 0 1 2 ans: det( A - λI ) = 1 - λ 6 0 0 2 - λ 1 0 1 2 - λ , = (1 - λ ) 2 - λ 1 1 2 - λ = (1 - λ )[(2 - λ ) 2 - 1] λ = 1 , 1 , 3 λ = 1 , 1 (repeated roots), we find linearly independent so- lutions of ( A - λI ) x = 0 , A - λI = 0 6 0 0 1 1 0 1 1 0 1 0 0 0 1 0 0 0 x 1 is free: (only 1 freedom, not two) x = C 1 0 0 (note, it is called a defficit eigenvalue, and the matrix cannot be diagonized.) λ = 3 A - λI = - 2 6 0 0 - 1 1 0 1 - 1 1 - 3 0 1 - 1 0 1
x 3 is free: x = C 3 1 1 4. (8.8:22) Find eigenvalues and eigenvectors. 0 0 0 0 0 0 0 0 1 ans: det( A - λI ) = 0 - λ 0 0 0 0 - λ 0 0 0 1 - λ , λ = 0 , 0 , 1 λ = 0 , 0 (repeated roots), we find linearly independent so- lutions of ( A - λI ) x = 0 , A - λI = 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 x 1 and x 2 are free: x = C 1 1 0 0 + C 2 0 1 0 λ = 1, solving ( A - λI ) x = 0 , A - λI = - 1 0 0 0 - 1 0 0 0 0 x 3 is free: x = C 0 0 1 5. (10.1:18) The given vectors are solutions of X 0 = AX . Determine if X i form a set of fundamental solutions of X 0 = AX . X i = 1 - 1 e t , 2 6 e t + 8 - 8 te t ans: Method 1. Wronskian. W = X 1 X 2 = e t 2 e t + 8 te t - e t 6 e t - 8 te t = [(6 - 8 t ) + (2 + 8 t )] e 2 t = 8 e 2 t 6 = 0 The two solutions are linearly independent. So they form a fundamental set of solutions. Method 2. Finding nonzero solutions. c 1 X 1 + c 2 X 2 = 0 c 1 1 - 1 e t + c 2 2 6 e t + 8 - 8 te t = 0 0 Choose t = 0, 1 - 1 + c 2 2 6 = 0 0 1 - 1 | 0 2 6 | 0 1 - 1 | 0 8 | 0 , c 1 = c 2 = 0 Only zero solution. The two solutions are linearly indepen- dent. So they form a fundamental set of solutions. 6. (10.2:14) Solve the system x 0 = Ax by eigenfunctions: A = 1 1 4 0 2 0 1 1 1 , x (0) = 1 3 0 . ans: Find eigenvalues, | A - λI | = 1 - λ 1 4 0 2 - λ 0 1 1 1 - λ by row 2 = 0 + (2 - λ ) 1 - λ 4 1 1 - λ + 0 = (2 - λ )[(1 - λ ) 2 - 4] = (2 - λ )[(1 - λ ) - 2][(1 - λ ) + 2] = 0 λ = - 1 , 2 , 3 Find eigenvectors for each eigenvalue: For λ = - 1, find a nonzero, or linearly independent solu- tions of ( A - λI ) x = 0 2 1 4 | 0 0 3 0 | 0 1 1 2 | 0 2 0 4 | 0 0 3 0 | 0 0 | 0 x = - 2 0 1 For λ = 2, solve ( A - λI ) x = 0 - 1 1 4 | 0 0 0 0 | 0 1 1 - 1 | 0
• • • 