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Material Type: Exam; Professor: Yeung; Class: Numerical Analysis; Subject: Mathematics; University: University of Wyoming; Term: Spring 2008;
Typology: Exams
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University of Wyoming Student Name: Department of Mathematics Spring 2008 Instructor: Man-Chung Yeung
2 x 1 +x 2 − 3 x 3 +4x 4 = 2 − 3 x 1 +2x 2 +5x 3 − 3 x 4 = 1 8 x 1 +x 2 − 3 x 3 +2x 4 = 4 − 4 x 1 +2x 2 +3x 3 −x 4 = 5.
a) Solve the linear system by Gaussian elimination with pivoting. b) Solve the linear system by (a) finding A−^1 with Gauss-Jordan method with pivoting, (b) computing A−^1 b, where A is the coefficient matrix and b the right hand side of the linear system. c) Which method requires more operation? Why? d) Find the condition number K(A).
Solution.
a)
⇒ (r 1 ↔ r 3 )
⇒ (r 2 ← r 2 + 38 r 1 , r 3 ← r 3 + − 82 r 1 , r 4 ← r 4 + 48 r 1 )
⇒ (r 2 ↔ r 4 )
⇒ (r 3 ← r 3 + − 20 .. 575 r 2 , r 4 ← r 4 + −^22 ..^3755 r 2 )
⇒ (r 4 ← r 4 + − −^22.^45. 7 r 3 )
Back-substitution, then x 1 = − 1. 52 x 2 = 6. 88 x 3 = − 6. 8 x 4 = − 5. 56
b) (a)
⇒ (r 1 ↔ 12 r 1 )
⇒ (r 2 ↔ r 2 + 3r 1 , r 3 ↔ r 3 − 8 r 1 , r 4 ↔ r 4 + 4r 1 )
⇒ (r 2 ↔ (^31). 5 r 2 )
⇒ (r 1 ↔ r 1 − 0. 5 r 2 , r 3 ↔ r 3 + 3r 2 , r 4 ↔ r 4 − 4 r 2 )
The number of floating-point multiplications and divisions needed in the elimination phase, that is, Steps #1 - #3 of Alg. 2.3.1, is 2(n − 1). The back-substitution phase, that is, Step #4 in Alg. 2.3.1, is 3(n − 1) FLOPs. So in total,
5(n − 1)