AME404LectureNotes

AME404LectureNotes - AME 404 Lecture Notes by M Dravinski...

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AME 404 Lecture Notes by M. Dravinski Department of Aerospace and Mechanical Engineering University of Southern California 1 Upper Triangular Linear Systems Consider the following system of linear algebraic equations of the type Ux = b u 11 x 1 + u 12 x 2 + u 13 x 3 = b 1 (1) u 22 x 2 + u 23 x 3 = b 2 u 33 x 3 = b 3 or in the so called augmented form u 11 u 12 u 13 b 1 0 u 22 u 23 b 2 00 u 33 b 3 (2) then this system is called to be an upper triangular of order n =3 . De f nition 1 An n by n matrix U =[ u ij ]; i, j =1: n is called upper trian- gular provided that the elements satisfy u ij =0 for i>j. In general, an augmented matrix for an upper triangular system is u 11 u 12 u 13 ··· u 1 n b 1 0 u 22 u 23 u 2 n b 2 . . . 000 u n 1 ,n 1 u n 1 ,n b n 1 0 u nn b n (3) Theorem 2 (Back Substitution) Suppose Ux = b is an upper triangular system of form (3). If u i,i 6 ; i n (4) then there exist a unique solution to (3). Proof. We start with the last equation to get x n = b n u n,n (5) Then from the next two equations it follows that x n 1 = 1 u n 1 ,n 1 ( b n 1 u n 1 x n ) x n 2 = 1 u n 2 ,n 2 ( b n 2 u n 2 ,n 1 x n 1 u n 2 ,n x n ) 1
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or in general x k = 1 u kk ( b k n X j = k +1 u k,j x j ); k = n 1:1 (6) Since x n is unique, by f nite induction so are x n 1 , ..., x 1 . Example 3 Consider the following upper triangular system 4 12 3 2 0 0 27 4 7 0065 4 0003 6 Then according to Eqns.(5 and ?? ) we get x 4 = 6 3 =2 x 3 = 1 6 (4 5 2) = 1 x 2 = 1 2 ( 7 7 ( 1) + 4 2) = 4 x 1 = 1 4 (20 + ( 4) 2 ( 1) 3 2) = 3 Note, the condition u ii 6 =0 is essential. If u ii , there are two possibilities: 1. Nor solution exits, or 2. in f nitely many solutions exist. These two cases are illustrated in the following examples. Example 4 Consider the following system 4 0 007 4 7 Then we get x 4 = 6 3 x 3 = 1 6 (4 5 2) = 1 7 x 3 4 x 4 = 7 ⇒− 15 6 = 7 and the solution does not exist. Example 5 Now for the system 4 1232 0 0070 7 00654 00036 2
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we get x 4 = 6 3 =2 x 3 = 1 6 (4 5 2) = 1 7 x 3 = 7 7( 1) = 7 4 x 1 x 2 +2 ( 1) + 3 (2) = 20 x 2 =4 x 1 16 and the solution is given by x = x 1 4 x 1 16 1 2 where x 1 is an arbitrary parameter. Therefore, we have an in f nitely many solutions. Theorem 6 If the n by n matrix A is either upper or lower triangular, then det A = a 11 ...a nn Algorithm. The algorithm is based on Eqns.(5 and ?? )andi tcanb e realized using the following f owchart. Then, the corresponding MATLAB code may be written as follows: –––––––––––––––––––––––- function x = MyBackSub(U,b) % This program solves a triangular system of n equations Ux=b.
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AME404LectureNotes - AME 404 Lecture Notes by M Dravinski...

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