This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chap . 20 Numeric Linear Algebra On p. 833 of the book it is stated that this chapter concerns three main areas of numeric linear algebra, namely, linear systems, curve fitting by least squares, and eigenvalue problems for matrices. Sec . 20 . 1 Linear Systems : Gauss Elimination In this section, Gauss elimination is explained independently of Chap. 7 and this time with emphasis on numeric aspects, in particular, the difficulties arising from small pivots and the count of numeric operations. The latter is an indication of the quality of a numeric method (here: the quality of Gauss elimination and of back substitution). Problem Set 20 . 1 . Page 839 3 . System without solution . The left side of the second equation equals minus three times the left side of the first equation. Hence for a solution to exist the right sides should be related in the same fashion; they should equal, for instance, 16 and u 48 (instead of u 48.5). Of course, for most systems with more than two equations, one cannot immediately see whether there will be solutions, but the Gauss elimination (with partial pivoting) will work in each case, giving the solution(s) or indicating that there is none. 7 . System with a unique solution . Pivoting . Workedout examples are given in the text. They show all the details. Review those first because there is little we can do for a better understanding and we shall have to restrict ourselves to a more detailed discussion of Table 20.1, which contains the algorithm for the Gauss elimination, and the addition of a few remarks. Consider Table 20.1. To follow the discussion, control it for Prob. 7 in terms of matrices with paper and pencil. In each case, write down all three rows of a matrix, not just one or two rows, as was done below to save some space and to avoid copying the same numbers several times. At the beginning, k u 1. Since a 11 u 0, you must pivot. Line 2 in Table 20.1 requests to look for the absolutely greatest a j 1 . This is a 31 . According to the algorithm, you have to interchange Equations 1 and 3, that is, Rows 1 and 3 of the augmented matrix. This gives 13 u 8 6 u 8 6 13 178.54 u 85.88 137.86 . (A) Don’t forget to interchange the entries on the right side (that is, in the last column of the augmented matrix). In line 2 of Table 20.1, the phrase ‘the smallest’ j U k is necessary since there may be several entries of the same absolute value (or even of the same size), and the computer needs unique instructions what to do in each operation. To get 0 as the first entry of Row 2, subtract 6/13 times Row 1 from Row 2. The new Row 2 is 3.692308 u 8 u 168.28308 . (B) This was k u 1 and j u 2 in lines 3 and 4 in the table. Now comes k u 1 and j u n u 3 in line 3. The calculation is m 31 u a 31 / a 11 u 0/13 u 0. Hence the operations in line 4 simply have no effect, they merely reproduce Row 3 of the matrix in (A). This was k u 1....
View
Full
Document
This note was uploaded on 04/08/2012 for the course MT 423 taught by Professor M.lee during the Spring '12 term at National Taiwan University.
 Spring '12
 M.lee

Click to edit the document details