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2009501436_ì� ê´�ì� _hw#3

Course: ME master, Spring 2009
School: Hanyang University
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Word Count: 787

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Work. Home #3 P roblem Set 1.7 1.7.6) Verify Sol) 1 1 max ( AT A) = (3 + 5) a nd its sqeare root A = (1 + 5) . 2 2 1 1 1 0 T A= , A = 1 1 0 1 e igen value 1 1 A AT = 1 2 1 A AT = 1 1 T det( A A ) = (1 )(2 ) 1 = 0 2 1 1 max = (3 + 5) , min = (3 5) 2 2 . 1 = (3 5) 2 1.7.18) Compute AT A a nd AAT and their eigenvalue and unit eigenvector for 1 1 0 A= 0 1 1 Sol) 1 1 0 2 1 A A = 1 2 1 a nd AAT = 1 2 0 1 1 T AT A 1 1 1 eigenvalue = 3,1, 0 , eigenvector = 2 0 1 1 1 1 2009-03-25 Home Work. #3 unit eigenvector = 1 6 1 6 1 6 1 2 0 1 2 1 3 1 3 1 3 1 1 AAT A eigenvalue = 3,1 , eigenvector = 1 1 1 1 2 2 unit eigenvector = 1 1 2 2 1 2 3 0 0 1 0 1 0 2 1 6 1 6 1 6 1 2 0 1 2 1 3 1 1 1 0 = 3 0 1 1 1 3 T U V = 1 2 1 2 1.7.28) What condition number do you compute K 9 a nd T9 , using the eig command to f ind max / min ? Compute wi th the estimate in equation(19) Sol) R = 0.0979 0 0 0 0 0 0 0 0 0 0.3820 0 0 0 0 0 0 0 0 0 0.8244 0 0 0 0 0 0 0 0 0 1.3820 0 0 0 0 0 0 0 0 0 2.0000 0 0 0 0 0 0 0 0 0 2.6180 0 0 0 0 0 0 0 0 0 3.1756 0 0 0 0 0 0 0 0 0 3.6180 0 0 0 0 0 0 0 0 0 3.9021 2009-03-25 Home Work. #3 max / min = 39.8635 , equation = 4 (9 + 1) 2 = 40.5284 2 R= 0.0273 0 0 0 0 0 0 0 0 0 0.2411 0 0 0 0 0 0 0 0 0 0.6454 0 0 0 0 0 0 0 0 0 1.1966 0 0 0 0 0 0 0 0 0 1.8348 0 0 0 0 0 0 0 0 0 2.4910 0 0 0 0 0 0 0 0 0 3.0939 0 0 0 0 0 0 0 0 0 3.5783 0 0 0 0 0 0 0 0 0 3.8916 4 (9 + 1) 2 = 40.5284 2 max / min = 142.6689 , equation = 1.7.29) How do you know that 50 < max ( AT A) < 51? 1 7 AT A = 7 50 1 7 A= 0 1 Sol) 7 1 T 2 AT A = det( A A) = (1 )(50 ) 49 = 51 + 1 = 0 7 50 X 1 = 0.0196 2 = 50.9804 . 2009-03-25 Home Work. #3 P roblem Set 2.1 2.1.4) Find solution! AT CAu = f , f1 + f 2 + f3 = 0 f = (1, 0,1) Sol) C2 A CAu = C2 0 T C2 C2 + C3 C3 0 u1 f1 C3 u2 = f 2 C3 u3 f 3 f1 + f 2 + f 3 = 0 ? f1 + f 2 + f 3 = C2u1 C2u2 C2u1 + (C2 + C3 )u2 C3u3 C3u2 + C3u3 = o f = (1, 0,1) ? r u = (1,1,1) ? 2.1.8) 1 1 K = c = e lement mat rix 1 1 c K2 = 2 c2 c2 c3 a nd K 3 = c c2 3 c3 c3 (a) Assemble K c2 K = c2 0 c2 0 c2 + c3 c3 c3 c3 K fixed free i n (8) (b) N ow include K for spring1( top fixed ) to reach 2009-03-25 Home Work. #3 c2 0 c2 + c1 c c2 + c3 c3 , c1 = K fixed free K = 2 0 c3 c3 spring1 e nt ry (c) Now place K for spring 4 (bottom fixed ) to reach K fixed fixed i n (7) c2 0 c2 + c1 c c2 + c3 c3 , c4 = K fixed fixed K = 2 0 c3 c3 + c4 spring4 e nt ry 2009-03-25

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