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Unformatted text preview: Lecture Note 3: Sept 18  22, 2006 Dr. Jeff ChakFu WONG Department of Mathematics Chinese University of Hong Kong jwong@math.cuhk.edu.hk MAT 2310 Linear Algebra and Its Applications Fall, 2006 Produced by Jeff ChakFu WONG 1 S OLVING THE L INEAR S YSTEMS Gaussian Elimination and LU Factorization (or Decomposition) 1. Doolittle’s method 2. Crout’s method 3. Choleski’s method SOLVING THE LINEAR SYSTEMS 2 Things to remember Gaussian Elimination and LU Factorization (or Decomposition) • Suppose A is nonsingular • Factor A = LU , where L is lower triangular, U is upper triangular – Computed by Gaussian elimination. – There are many techniques like direct decomposition, Doolittle’s method or Crout’s method executed to find L and U . – Rows and columns often must be reordered to avoid division by zero – pivoting – Given LUfactorization (or decomposition), find x by Solve L z = b by forward substitution Solve U x = z by backward (or back) substitution SOLVING THE LINEAR SYSTEMS 3 The method can be used to solve a system of equations or to find the inverse of a matrix. In this method, the matrix A is decomposed or factorized as the product of a lower triangular matrix L and an upper triangular matrix U . We write the matrix A as A = LU. where L = l 11 ... l 21 l 22 ... l 31 l 32 l 33 ... . . . . . . . . . ... l n 1 l n 2 l l 3 ... l nn , U = u 11 u 12 u 13 ... u 1 n u 22 u 23 ... u 2 n u 33 ... u 3 n . . . . . . . . . ... ... u nn . Multiplying the matrices L and U and comparing the elements of the product matrix with the corresponding elements of A , we get l il u 1 j + l i 2 u 2 j + ··· + l in u nj = a ij , i,j = 1 , 2 , ··· ,n where l ij = 0 , j > i and u ij = 0 , i > j . SOLVING THE LINEAR SYSTEMS 4 To obtain a unique solution, we can choose the values for n elements in either L or U arbitrary. The simplest choices are Doolittle’s method l ii = 1 , i = 1 , 2 , ··· ,n Crout’s method u ii = 1 , i = 1 , 2 , ··· ,n where 1 0 0 * 1 * * 1 , 1 * * 1 * 1 Here * and * are unknowns. The method fails when any of diagonal elements (pivots), l ii , in L or u ii in U becomes zero . SOLVING THE LINEAR SYSTEMS 5 Doolittle’s Method We look for 1 0 0 m 21 1 m 31 m 32 1 • We begin with the augmented matrix, and display ( in the column headed m ) the multipliers required for the transformations. m a 11 a 12 a 13 b 1 a 21 a 22 a 23 b 2 a 31 a 32 a 33 b 3 R 1 R 2 R 3 SOLVING THE LINEAR SYSTEMS 6 Step 1. Eliminate the coefficients a 21 and a 31 , using row R 1 : m 21 = a 21 a 11 m 31 = a 31 a 11 a 11 a 12 a 13 b 1 a 22 a 23 b 2 a 32 a 33 b 3 R 1 R 2 R 3 ( R 2 m 21 × R 1 ) ( R 3 m 32 × R 1 ) Step 2.Step 2....
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 Spring '09
 JeffWong
 Linear Algebra, Algebra

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