Direct Methods for Solving Linear Systems. Linear Systems of Equations
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1 Direct Methods for Solving Linear Systems Linear Systems of Equations Numerical Analysis (9th Edition) R L Burden & J D Faires Beamer Presentation Slides prepared by John Carroll Dublin City University c 2011 Brooks/Cole, Cengage Learning
2 Outline 1 Notation & Basic Terminology Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 2 / 43
3 Outline 1 Notation & Basic Terminology 2 3 Operations to Simplify a Linear System of Equations Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 2 / 43
4 Outline 1 Notation & Basic Terminology 2 3 Operations to Simplify a Linear System of Equations 3 Gaussian Elimination Procedure Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 2 / 43
5 Outline 1 Notation & Basic Terminology 2 3 Operations to Simplify a Linear System of Equations 3 Gaussian Elimination Procedure 4 The Gaussian Elimination with Backward Substitution Algorithm Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 2 / 43
6 Outline 1 Notation & Basic Terminology 2 3 Operations to Simplify a Linear System of Equations 3 Gaussian Elimination Procedure 4 The Gaussian Elimination with Backward Substitution Algorithm Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 3 / 43
7 Introduction Linear Systems of Equations We will consider direct methods for solving a linear system of n equations in n variables. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 4 / 43
8 Introduction Linear Systems of Equations We will consider direct methods for solving a linear system of n equations in n variables. Such a system has the form: E 1 : a 11 x 1 + a 12 x a 1n x n = b 1 E 2 : a 21 x 1 + a 22 x a 2n x n = b 2 E n : a n1 x 1 + a n2 x a nn x n = b n. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 4 / 43
9 Introduction Linear Systems of Equations We will consider direct methods for solving a linear system of n equations in n variables. Such a system has the form: E 1 : a 11 x 1 + a 12 x a 1n x n = b 1 E 2 : a 21 x 1 + a 22 x a 2n x n = b 2 E n : a n1 x 1 + a n2 x a nn x n = b n In this system we are given the constants a ij, for each i, j = 1, 2,...,n, and b i, for each i = 1, 2,...,n, and we need to determine the unknowns x 1,...,x n.. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 4 / 43
10 Introduction Direct Methods & Round-off Error Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 5 / 43
11 Introduction Direct Methods & Round-off Error Direct techniques are methods that theoretically give the exact solution to the system in a finite number of steps. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 5 / 43
12 Introduction Direct Methods & Round-off Error Direct techniques are methods that theoretically give the exact solution to the system in a finite number of steps. In practice, of course, the solution obtained will be contaminated by the round-off error that is involved with the arithmetic being used. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 5 / 43
13 Introduction Direct Methods & Round-off Error Direct techniques are methods that theoretically give the exact solution to the system in a finite number of steps. In practice, of course, the solution obtained will be contaminated by the round-off error that is involved with the arithmetic being used. Analyzing the effect of this round-off error and determining ways to keep it under control will be a major component of this presentation. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 5 / 43
14 Introduction Direct Methods & Round-off Error Direct techniques are methods that theoretically give the exact solution to the system in a finite number of steps. In practice, of course, the solution obtained will be contaminated by the round-off error that is involved with the arithmetic being used. Analyzing the effect of this round-off error and determining ways to keep it under control will be a major component of this presentation. We begin, however, by introducing some important terminology and notation. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 5 / 43
15 Matrices & Vectors Definition of a Matrix An n m (n by m) matrix is a rectangular array of elements with n rows and m columns in which not only is the value of an element important, but also its position in the array. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 6 / 43
16 Matrices & Vectors Definition of a Matrix An n m (n by m) matrix is a rectangular array of elements with n rows and m columns in which not only is the value of an element important, but also its position in the array. Notation The notation for an n m matrix will be a capital letter such as A for the matrix and lowercase letters with double subscripts, such as a ij, to refer to the entry at the intersection of the ith row and jth column; that is: a 11 a 12 a 1m a 21 a 22 a 2m A = [a ij ] =... a n1 a n2 a nm Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 6 / 43
17 Matrices & Vectors A Vector is a special case The 1 n matrix A = [a 11 a 12 a 1n ] is called an n-dimensional row vector, and an n 1 matrix a 11 a 21 A =. a n1 is called an n-dimensional column vector. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 7 / 43
18 Matrices & Vectors A Vector is a special case (Cont d) Usually the unnecessary subscripts are omitted for vectors, and a boldface lowercase letter is used for notation. Thus x 1 x 2 x =. x n denotes a column vector, and a row vector. y = [y 1 y 2... y n ] Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 8 / 43
19 Matrices & Vectors: Augmented Matrix The Augmented Matrix (1/2) An n (n + 1) matrix can be used to represent the linear system a 11 x 1 + a 12 x a 1n x n = b 1, a 21 x 1 + a 22 x a 2n x n = b 2,.. a n1 x 1 + a n2 x a nn x n = b n, Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 9 / 43
20 Matrices & Vectors: Augmented Matrix The Augmented Matrix (1/2) An n (n + 1) matrix can be used to represent the linear system by first constructing A = [a ij ] = a 11 x 1 + a 12 x a 1n x n = b 1, a 21 x 1 + a 22 x a 2n x n = b 2,. a n1 x 1 + a n2 x a nn x n = b n, a 11 a 12 a 1n a 21 a 22 a 2n... a n1 a n2 a nn. and b = b 1 b 2. b n Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 9 / 43
21 Matrices & Vectors: Augmented Matrix The Augmented Matrix (2/2) and then forming the new array [A, b]: a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2 [A, b] =.... a n1 a n2 a nn b n Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 10 / 43
22 Matrices & Vectors: Augmented Matrix The Augmented Matrix (2/2) and then forming the new array [A, b]: a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2 [A, b] =.... a n1 a n2 a nn b n where the vertical line is used to separate the coefficients of the unknowns from the values on the right-hand side of the equations. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 10 / 43
23 Matrices & Vectors: Augmented Matrix The Augmented Matrix (2/2) and then forming the new array [A, b]: a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2 [A, b] =.... a n1 a n2 a nn b n where the vertical line is used to separate the coefficients of the unknowns from the values on the right-hand side of the equations. The array [A, b] is called an augmented matrix. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 10 / 43
24 Matrices & Vectors: Augmented Matrix Representing the Linear System In what follows, the n (n + 1) matrix a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2 [A, b] =.... a n1 a n2 a nn b n Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 11 / 43
25 Matrices & Vectors: Augmented Matrix Representing the Linear System In what follows, the n (n + 1) matrix a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2 [A, b] =.... a n1 a n2 a nn b n will used to represent the linear system a 11 x 1 + a 12 x a 1n x n = b 1 a 21 x 1 + a 22 x a 2n x n = b 2. a n1 x 1 + a n2 x a nn x n. = b n Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 11 / 43
26 Outline 1 Notation & Basic Terminology 2 3 Operations to Simplify a Linear System of Equations 3 Gaussian Elimination Procedure 4 The Gaussian Elimination with Backward Substitution Algorithm Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 12 / 43
27 Simplifying a Linear Systems of Equations The Linear System Returning to the linear system of n equations in n variables: E 1 : a 11 x 1 + a 12 x a 1n x n = b 1 E 2 : a 21 x 1 + a 22 x a 2n x n = b 2 E n : a n1 x 1 + a n2 x a nn x n = b n where we are given the constants a ij, for each i, j = 1, 2,...,n, and b i, for each i = 1, 2,...,n,. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 13 / 43
28 Simplifying a Linear Systems of Equations The Linear System Returning to the linear system of n equations in n variables: E 1 : a 11 x 1 + a 12 x a 1n x n = b 1 E 2 : a 21 x 1 + a 22 x a 2n x n = b 2 E n : a n1 x 1 + a n2 x a nn x n = b n where we are given the constants a ij, for each i, j = 1, 2,...,n, and b i, for each i = 1, 2,...,n, we need to determine the unknowns x 1,..., x n.. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 13 / 43
29 Simplifying a Linear Systems of Equations Permissible Operations We will use 3 operations to simplify the linear system: Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 14 / 43
30 Simplifying a Linear Systems of Equations Permissible Operations We will use 3 operations to simplify the linear system: 1 Equation E i can be multiplied by any nonzero constant λ with the resulting equation used in place of E i. This operation is denoted (λe i ) (E i ). Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 14 / 43
31 Simplifying a Linear Systems of Equations Permissible Operations We will use 3 operations to simplify the linear system: 1 Equation E i can be multiplied by any nonzero constant λ with the resulting equation used in place of E i. This operation is denoted (λe i ) (E i ). 2 Equation E j can be multiplied by any constant λ and added to equation E i with the resulting equation used in place of E i. This operation is denoted (E i + λe j ) (E i ). Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 14 / 43
32 Simplifying a Linear Systems of Equations Permissible Operations We will use 3 operations to simplify the linear system: 1 Equation E i can be multiplied by any nonzero constant λ with the resulting equation used in place of E i. This operation is denoted (λe i ) (E i ). 2 Equation E j can be multiplied by any constant λ and added to equation E i with the resulting equation used in place of E i. This operation is denoted (E i + λe j ) (E i ). 3 Equations E i and E j can be transposed in order. This operation is denoted (E i ) (E j ). Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 14 / 43
33 Simplifying a Linear Systems of Equations Permissible Operations We will use 3 operations to simplify the linear system: 1 Equation E i can be multiplied by any nonzero constant λ with the resulting equation used in place of E i. This operation is denoted (λe i ) (E i ). 2 Equation E j can be multiplied by any constant λ and added to equation E i with the resulting equation used in place of E i. This operation is denoted (E i + λe j ) (E i ). 3 Equations E i and E j can be transposed in order. This operation is denoted (E i ) (E j ). By a sequence of these operations, a linear system will be systematically transformed into to a new linear system that is more easily solved and has the same solutions. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 14 / 43
34 Simplifying a Linear Systems of Equations Illustration The four equations E 1 : x 1 + x 2 + 3x 4 = 4 will be solved for x 1, x 2, x 3, and x 4. E 2 : 2x 1 + x 2 x 3 + x 4 = 1 E 3 : 3x 1 x 2 x 3 + 2x 4 = 3 E 4 : x 1 + 2x 2 + 3x 3 x 4 = 4 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 15 / 43
35 Simplifying a Linear Systems of Equations Illustration The four equations E 1 : x 1 + x 2 + 3x 4 = 4 will be solved for x 1, x 2, x 3, and x 4. E 2 : 2x 1 + x 2 x 3 + x 4 = 1 E 3 : 3x 1 x 2 x 3 + 2x 4 = 3 E 4 : x 1 + 2x 2 + 3x 3 x 4 = 4 We first use equation E 1 to eliminate the unknown x 1 from equations E 2, E 3, and E 4 by performing: (E 2 2E 1 ) (E 2 ) (E 3 3E 1 ) (E 3 ) and (E 4 + E 1 ) (E 4 ) Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 15 / 43
36 Simplifying a Linear Systems of Equations E 1 : x 1 +x 2 +3x 4 =4 E 2 : 2x 1 +x 2 x 3 + x 4 =1 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 16 / 43
37 Simplifying a Linear Systems of Equations Illustration Cont d (2/5) E 1 : x 1 +x 2 +3x 4 =4 E 2 : 2x 1 +x 2 x 3 + x 4 =1 For example, in the second equation (E 2 2E 1 ) (E 2 ) produces (2x 1 + x 2 x 3 + x 4 ) 2(x 1 + x 2 + 3x 4 ) = 1 2(4) Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 16 / 43
38 Simplifying a Linear Systems of Equations Illustration Cont d (2/5) E 1 : x 1 +x 2 +3x 4 =4 E 2 : 2x 1 +x 2 x 3 + x 4 =1 For example, in the second equation (E 2 2E 1 ) (E 2 ) produces (2x 1 + x 2 x 3 + x 4 ) 2(x 1 + x 2 + 3x 4 ) = 1 2(4) which simplifies to the result shown as E 2 in E 1 : x 1 + x 2 + 3x 4 = 4 E 2 : x 2 x 3 5x 4 = 7 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 16 / 43
39 Simplifying a Linear Systems of Equations Illustration Cont d (3/5) Similarly for equations E 3 and E 4 so that we obtain the new system: E 1 : x 1 + x 2 + 3x 4 = 4 E 2 : x 2 x 3 5x 4 = 7 E 3 : 4x 2 x 3 7x 4 = 15 E 4 : 3x 2 + 3x 3 + 2x 4 = 8 For simplicity, the new equations are again labeled E 1, E 2, E 3, and E 4. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 17 / 43
40 Simplifying a Linear Systems of Equations Illustration Cont d (4/5) In the new system, E 2 is used to eliminate the unknown x 2 from E 3 and E 4 by performing (E 3 4E 2 ) (E 3 ) and (E 4 + 3E 2 ) (E 4 ). Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 18 / 43
41 Simplifying a Linear Systems of Equations Illustration Cont d (4/5) In the new system, E 2 is used to eliminate the unknown x 2 from E 3 and E 4 by performing (E 3 4E 2 ) (E 3 ) and (E 4 + 3E 2 ) (E 4 ). This results in E 1 : x 1 + x 2 + 3x 4 = 4, E 2 : x 2 x 3 5x 4 = 7, E 3 : 3x x 4 = 13, E 4 : 13x 4 = 13. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 18 / 43
42 Simplifying a Linear Systems of Equations Illustration Cont d (4/5) In the new system, E 2 is used to eliminate the unknown x 2 from E 3 and E 4 by performing (E 3 4E 2 ) (E 3 ) and (E 4 + 3E 2 ) (E 4 ). This results in E 1 : x 1 + x 2 + 3x 4 = 4, E 2 : x 2 x 3 5x 4 = 7, E 3 : 3x x 4 = 13, E 4 : 13x 4 = 13. This latter system of equations is now in triangular (or reduced) form and can be solved for the unknowns by a backward-substitution process. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 18 / 43
43 Simplifying a Linear Systems of Equations Illustration Cont d (5/5) Since E 4 implies x 4 = 1, we can solve E 3 for x 3 to give x 3 = 1 3 (13 13x 4) = 1 (13 13) = 0. 3 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 19 / 43
44 Simplifying a Linear Systems of Equations Illustration Cont d (5/5) Since E 4 implies x 4 = 1, we can solve E 3 for x 3 to give Continuing, E 2 gives x 3 = 1 3 (13 13x 4) = 1 (13 13) = 0. 3 x 2 = ( 7 + 5x 4 + x 3 ) = ( ) = 2, Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 19 / 43
45 Simplifying a Linear Systems of Equations Illustration Cont d (5/5) Since E 4 implies x 4 = 1, we can solve E 3 for x 3 to give Continuing, E 2 gives and E 1 gives x 3 = 1 3 (13 13x 4) = 1 (13 13) = 0. 3 x 2 = ( 7 + 5x 4 + x 3 ) = ( ) = 2, x 1 = 4 3x 4 x 2 = = 1. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 19 / 43
46 Simplifying a Linear Systems of Equations Illustration Cont d (5/5) Since E 4 implies x 4 = 1, we can solve E 3 for x 3 to give Continuing, E 2 gives and E 1 gives x 3 = 1 3 (13 13x 4) = 1 (13 13) = 0. 3 x 2 = ( 7 + 5x 4 + x 3 ) = ( ) = 2, x 1 = 4 3x 4 x 2 = = 1. The solution is therefore, x 1 = 1, x 2 = 2, x 3 = 0, and x 4 = 1. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 19 / 43
47 Outline 1 Notation & Basic Terminology 2 3 Operations to Simplify a Linear System of Equations 3 Gaussian Elimination Procedure 4 The Gaussian Elimination with Backward Substitution Algorithm Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 20 / 43
48 Constructing an Algorithm to Solve the Linear System E 1 : x 1 + x 2 + 3x 4 = 4 E 2 : 2x 1 + x 2 x 3 + x 4 = 1 E 3 : 3x 1 x 2 x 3 + 2x 4 = 3 E 4 : x 1 + 2x 2 + 3x 3 x 4 = 4 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 21 / 43
49 Constructing an Algorithm to Solve the Linear System E 1 : x 1 + x 2 + 3x 4 = 4 E 2 : 2x 1 + x 2 x 3 + x 4 = 1 E 3 : 3x 1 x 2 x 3 + 2x 4 = 3 E 4 : x 1 + 2x 2 + 3x 3 x 4 = 4 Converting to Augmented Form Repeating the operations involved in the previous illustration with the matrix notation results in first considering the augmented matrix: Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 21 / 43
50 Constructing an Algorithm to Solve the Linear System Reducing to Triangular Form Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 22 / 43
51 Constructing an Algorithm to Solve the Linear System Reducing to Triangular Form Performing the operations as described in the earlier example produces the augmented matrices: and Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 22 / 43
52 Constructing an Algorithm to Solve the Linear System Reducing to Triangular Form Performing the operations as described in the earlier example produces the augmented matrices: and The final matrix can now be transformed into its corresponding linear system, and solutions for x 1, x 2, x 3, and x 4, can be obtained. The procedure is called Gaussian elimination with backward substitution. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 22 / 43
53 Gaussian Elimination with Backward Substitution Basic Steps in the Procedure The general Gaussian elimination procedure applied to the linear system E 1 : a 11 x 1 + a 12 x a 1n x n = b 1 E 2 : a 21 x 1 + a 22 x a 2n x n = b 2. E n : a n1 x 1 + a n2 x a nn x n = b n will be handled in a similar manner.. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 23 / 43
54 Gaussian Elimination with Backward Substitution Basic Steps in the Procedure (Cont d) First form the augmented matrix Ã: a 11 a 12 a 1n a 1,n+1 a 21 a 22 a 2n a 2,n+1 Ã = [A, b] =.... a n1 a n2 a nn a n,n+1 where A denotes the matrix formed by the coefficients. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 24 / 43
55 Gaussian Elimination with Backward Substitution Basic Steps in the Procedure (Cont d) First form the augmented matrix Ã: a 11 a 12 a 1n a 1,n+1 a 21 a 22 a 2n a 2,n+1 Ã = [A, b] =.... a n1 a n2 a nn a n,n+1 where A denotes the matrix formed by the coefficients. The entries in the (n + 1)st column are the values of b; that is, a i,n+1 = b i for each i = 1, 2,...,n. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 24 / 43
56 Gaussian Elimination with Backward Substitution Basic Steps in the Procedure (Cont d) Provided a 11 0, we perform the operations corresponding to (E j (a j1 /a 11 )E 1 ) (E j ) for each j = 2, 3,...,n to eliminate the coefficient of x 1 in each of these rows. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 25 / 43
57 Gaussian Elimination with Backward Substitution Basic Steps in the Procedure (Cont d) Provided a 11 0, we perform the operations corresponding to (E j (a j1 /a 11 )E 1 ) (E j ) for each j = 2, 3,...,n to eliminate the coefficient of x 1 in each of these rows. Although the entries in rows 2, 3,...,n are expected to change, for ease of notation we again denote the entry in the ith row and the jth column by a ij. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 25 / 43
58 Gaussian Elimination with Backward Substitution Basic Steps in the Procedure (Cont d) Provided a 11 0, we perform the operations corresponding to (E j (a j1 /a 11 )E 1 ) (E j ) for each j = 2, 3,...,n to eliminate the coefficient of x 1 in each of these rows. Although the entries in rows 2, 3,...,n are expected to change, for ease of notation we again denote the entry in the ith row and the jth column by a ij. With this in mind, we follow a sequential procedure for i = 2, 3,...,n 1 and perform the operation (E j (a ji /a ii )E i ) (E j ) for each j = i + 1, i + 2,..., n, provided a ii 0. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 25 / 43
59 Gaussian Elimination with Backward Substitution Basic Steps in the Procedure (Cont d) This eliminates (changes the coefficient to zero) x i in each row below the ith for all values of i = 1, 2,...,n 1. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 26 / 43
60 Gaussian Elimination with Backward Substitution Basic Steps in the Procedure (Cont d) This eliminates (changes the coefficient to zero) x i in each row below the ith for all values of i = 1, 2,...,n 1. The resulting matrix has the form: a 11 a 12 a 1n a 1,n+1 0 a 22 a 2n a 2,n+1 Ã = a nn a n,n+1 where, except in the first row, the values of a ij are not expected to agree with those in the original matrix Ã. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 26 / 43
61 Gaussian Elimination with Backward Substitution Basic Steps in the Procedure (Cont d) This eliminates (changes the coefficient to zero) x i in each row below the ith for all values of i = 1, 2,...,n 1. The resulting matrix has the form: a 11 a 12 a 1n a 1,n+1 0 a 22 a 2n a 2,n+1 à = a nn a n,n+1 where, except in the first row, the values of a ij are not expected to agree with those in the original matrix Ã. The matrix à represents a linear system with the same solution set as the original system. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 26 / 43
62 Gaussian Elimination with Backward Substitution Basic Steps in the Procedure (Cont d) The new linear system is triangular, a 11 x 1 + a 12 x a 1n x n = a 1,n+1 a 22 x a 2n x n = a 2,n a nn x n = a n,n+1 so backward substitution can be performed. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 27 / 43
63 Gaussian Elimination with Backward Substitution Basic Steps in the Procedure (Cont d) The new linear system is triangular, a 11 x 1 + a 12 x a 1n x n = a 1,n+1 a 22 x a 2n x n = a 2,n a nn x n = a n,n+1 so backward substitution can be performed. Solving the nth equation for x n gives x n = a n,n+1 a nn Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 27 / 43
64 Gaussian Elimination with Backward Substitution Basic Steps in the Procedure (Cont d) Solving the (n 1)st equation for x n 1 and using the known value for x n yields x n 1 = a n 1,n+1 a n 1,n x n a n 1,n 1 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 28 / 43
65 Gaussian Elimination with Backward Substitution Basic Steps in the Procedure (Cont d) Solving the (n 1)st equation for x n 1 and using the known value for x n yields x n 1 = a n 1,n+1 a n 1,n x n a n 1,n 1 Continuing this process, we obtain x i = a i,n+1 a i,n x n a i,n 1 x n 1 a i,i+1 x i+1 a ii = a i,n+1 n j=i+1 a ijx j a ii for each i = n 1, n 2,...,2, 1. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 28 / 43
66 Gaussian Elimination with Backward Substitution A More Precise Description Gaussian elimination procedure is described more precisely, although more intricately, by forming a sequence of augmented matrices Ã(1), à (2),..., Ã(n), where Ã(1) is the matrix à given earlier and Ã(k), for each k = 2, 3,...,n, has entries a (k) ij, where: a (k) ij = a (k 1) ij when i = 1,2,...,k 1 and j = 1,2,...,n when i = k,k + 1,...,n and j = 1,2,...,k 1 a (k 1) ij a(k 1) i,k 1 a (k 1) k 1,k 1 a (k 1) k 1,j when i = k,k + 1,...,n and j = k,k + 1,...,n + 1 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 29 / 43
67 Gaussian Elimination with Backward Substitution A More Precise Description (Cont d) Thus à (k) = a (1) 11 a (1) 12 a (1) 13 a (1) 1,k 1 a (1) 1k a (1) 1n a (1) 1,n+1 0 a (2) 22 a (2) 23 a (2) 2,k 1 a (2) 2k a (2) 2n a (2) 2,n a (k 1) k 1,k 1 a (k 1) k 1,k a (k 1) k 1,n. 0 a (k) kk a (k) kn a (k) nk a (k) nn a (k 1) k 1,n+1 a (k) k,n+1. a (k) n,n+1 represents the equivalent linear system for which the variable x k 1 has just been eliminated from equations E k, E k+1,...,e n. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 30 / 43
68 Gaussian Elimination with Backward Substitution A More Precise Description (Cont d) The procedure will fail if one of the elements a (1) 11, a(2) 22, a(3) 33,..., a (n 1) n 1,n 1, a(n) nn is zero because the step E i a(k) i,k a (k) kk (E k ) E i either cannot be performed (this occurs if one of a (1) 11,..., a(n 1) is zero), or the backward substitution cannot be accomplished (in the case a (n) nn = 0). n 1,n 1 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 31 / 43
69 Gaussian Elimination with Backward Substitution A More Precise Description (Cont d) The procedure will fail if one of the elements a (1) 11, a(2) 22, a(3) 33,..., a (n 1) n 1,n 1, a(n) nn is zero because the step E i a(k) i,k a (k) kk (E k ) E i either cannot be performed (this occurs if one of a (1) 11,..., a(n 1) n 1,n 1 is zero), or the backward substitution cannot be accomplished (in the case a (n) nn = 0). The system may still have a solution, but the technique for finding it must be altered. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 31 / 43
70 Illustration of the Gaussian Elimination Procedure Example Represent the linear system E 1 : x 1 x 2 + 2x 3 x 4 = 8 E 2 : 2x 1 2x 2 + 3x 3 3x 4 = 20 E 3 : x 1 + x 2 + x 3 = 2 E 4 : x 1 x 2 + 4x 3 + 3x 4 = 4 as an augmented matrix and use Gaussian Elimination to find its solution. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 32 / 43
71 Illustration of the Gaussian Elimination Procedure Solution (1/6) The augmented matrix is à = Ã(1) = Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 33 / 43
72 Illustration of the Gaussian Elimination Procedure Solution (1/6) The augmented matrix is à = Ã(1) = Performing the operations (E 2 2E 1 ) (E 2 ), (E 3 E 1 ) (E 3 ) and (E 4 E 1 ) (E 4 ) Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 33 / 43
73 Illustration of the Gaussian Elimination Procedure Solution (1/6) The augmented matrix is à = Ã(1) = Performing the operations (E 2 2E 1 ) (E 2 ), (E 3 E 1 ) (E 3 ) and (E 4 E 1 ) (E 4 ) gives à (2) = Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 33 / 43
74 Illustration of the Gaussian Elimination Procedure à (2) = Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 34 / 43
75 Illustration of the Gaussian Elimination Procedure Solution (2/6) Ã (2) = The diagonal entry a (2) 22, called the pivot element, is 0, so the procedure cannot continue in its present form. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 34 / 43
76 Illustration of the Gaussian Elimination Procedure Solution (2/6) Ã (2) = The diagonal entry a (2) 22, called the pivot element, is 0, so the procedure cannot continue in its present form. But operations (E i ) (E j ) are permitted, so a search is made of the elements a (2) 32 and a(2) 42 for the first nonzero element. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 34 / 43
77 Illustration of the Gaussian Elimination Procedure Solution (2/6) Ã (2) = The diagonal entry a (2) 22, called the pivot element, is 0, so the procedure cannot continue in its present form. But operations (E i ) (E j ) are permitted, so a search is made of the elements a (2) 32 and a(2) 42 for the first nonzero element. Since a (2) 32 0, the operation (E 2) (E 3 ) can be performed to obtain a new matrix. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 34 / 43
78 Illustration of the Gaussian Elimination Procedure à (2) = Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 35 / 43
79 Illustration of the Gaussian Elimination Procedure à (2) = Solution (3/6) Perform the operation (E 2 ) (E 3 ) to obtain a new matrix: à (2) = Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 35 / 43
80 Illustration of the Gaussian Elimination Procedure à (2) = Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 36 / 43
81 Illustration of the Gaussian Elimination Procedure à (2) = Solution (4/6) Since x 2 is already eliminated from E 3 and E 4, Ã(3) will be Ã(2), and the computations continue with the operation (E 4 + 2E 3 ) (E 4 ), giving à (4) = Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 36 / 43
82 Illustration of the Gaussian Elimination Procedure à (4) = Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 37 / 43
83 Illustration of the Gaussian Elimination Procedure Solution (5/6) Ã (4) = The solution may now be found through backward substitution: x 4 = 4 2 = 2 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 37 / 43
84 Illustration of the Gaussian Elimination Procedure Solution (5/6) Ã (4) = The solution may now be found through backward substitution: x 4 = 4 2 = 2 x 3 = [ 4 ( 1)x 4] 1 = 2 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 37 / 43
85 Illustration of the Gaussian Elimination Procedure Solution (5/6) Ã (4) = The solution may now be found through backward substitution: x 4 = 4 2 = 2 x 3 = [ 4 ( 1)x 4] = 2 1 x 2 = [6 x 4 ( 1)x 3 ] = 3 2 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 37 / 43
86 Illustration of the Gaussian Elimination Procedure Solution (5/6) Ã (4) = The solution may now be found through backward substitution: x 4 = 4 2 = 2 x 3 = [ 4 ( 1)x 4] = 2 1 x 2 = [6 x 4 ( 1)x 3 ] = 3 2 x 1 = [ 8 ( 1)x 4 2x 3 ( 1)x 2 ] 1 = 7 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 37 / 43
87 Illustration of the Gaussian Elimination Procedure Solution (6/6): Some Observations The example illustrates what is done if a (k) kk = 0 for some k = 1, 2,...,n 1. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 38 / 43
88 Illustration of the Gaussian Elimination Procedure Solution (6/6): Some Observations The example illustrates what is done if a (k) kk = 0 for some k = 1, 2,...,n 1. The kth column of Ã(k 1) from the kth row to the nth row is searched for the first nonzero entry. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 38 / 43
89 Illustration of the Gaussian Elimination Procedure Solution (6/6): Some Observations The example illustrates what is done if a (k) kk = 0 for some k = 1, 2,...,n 1. The kth column of Ã(k 1) from the kth row to the nth row is searched for the first nonzero entry. If a (k) pk 0 for some p,with k + 1 p n, then the operation (E k ) (E p ) is performed to obtain Ã(k 1). Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 38 / 43
90 Illustration of the Gaussian Elimination Procedure Solution (6/6): Some Observations The example illustrates what is done if a (k) kk = 0 for some k = 1, 2,...,n 1. The kth column of Ã(k 1) from the kth row to the nth row is searched for the first nonzero entry. If a (k) pk 0 for some p,with k + 1 p n, then the operation (E k ) (E p ) is performed to obtain Ã(k 1). The procedure can then be continued to form Ã(k), and so on. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 38 / 43
91 Illustration of the Gaussian Elimination Procedure Solution (6/6): Some Observations The example illustrates what is done if a (k) kk = 0 for some k = 1, 2,...,n 1. The kth column of Ã(k 1) from the kth row to the nth row is searched for the first nonzero entry. If a (k) pk 0 for some p,with k + 1 p n, then the operation (E k ) (E p ) is performed to obtain Ã(k 1). The procedure can then be continued to form Ã(k), and so on. If a (k) pk = 0 for each p, it can be shown that the linear system does not have a unique solution and the procedure stops. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 38 / 43
92 Illustration of the Gaussian Elimination Procedure Solution (6/6): Some Observations The example illustrates what is done if a (k) kk = 0 for some k = 1, 2,...,n 1. The kth column of Ã(k 1) from the kth row to the nth row is searched for the first nonzero entry. If a (k) pk 0 for some p,with k + 1 p n, then the operation (E k ) (E p ) is performed to obtain Ã(k 1). The procedure can then be continued to form Ã(k), and so on. If a (k) pk = 0 for each p, it can be shown that the linear system does not have a unique solution and the procedure stops. Finally, if a (n) nn = 0, the linear system does not have a unique solution, and again the procedure stops. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 38 / 43
93 Outline 1 Notation & Basic Terminology 2 3 Operations to Simplify a Linear System of Equations 3 Gaussian Elimination Procedure 4 The Gaussian Elimination with Backward Substitution Algorithm Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 39 / 43
94 Gaussian Elimination with Backward Substitution Algorithm (1/3) To solve the n n linear system E 1 : E 2 :. a 11 x 1 + a 12 x a 1n x n = a 1,n+1 a 21 x 1 + a 22 x a 2n x n = a 2,n E n : a n1 x 1 + a n2 x a nn x n = a n,n+1 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 40 / 43
95 Gaussian Elimination with Backward Substitution Algorithm (1/3) To solve the n n linear system E 1 : E 2 :. a 11 x 1 + a 12 x a 1n x n = a 1,n+1 a 21 x 1 + a 22 x a 2n x n = a 2,n E n : a n1 x 1 + a n2 x a nn x n = a n,n+1 INPUT number of unknowns and equations n; augmented matrix A = [a ij ], where 1 i n and 1 j n + 1. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 40 / 43
96 Gaussian Elimination with Backward Substitution Algorithm (1/3) To solve the n n linear system E 1 : E 2 :. a 11 x 1 + a 12 x a 1n x n = a 1,n+1 a 21 x 1 + a 22 x a 2n x n = a 2,n E n : a n1 x 1 + a n2 x a nn x n = a n,n+1 INPUT OUTPUT number of unknowns and equations n; augmented matrix A = [a ij ], where 1 i n and 1 j n + 1. solution x 1, x 2,..., x n or message that the linear system has no unique solution. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 40 / 43
97 Gaussian Elimination with Backward Substitution Algorithm (2/3) Step 1 For i = 1,...,n 1 do Steps 2 4: (Elimination process) Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 41 / 43
98 Gaussian Elimination with Backward Substitution Algorithm (2/3) Step 1 For i = 1,...,n 1 do Steps 2 4: (Elimination process) Step 2 Let p be the smallest integer with i p n and a pi 0 If no integer p can be found then OUTPUT ( no unique solution exists ) STOP Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 41 / 43
99 Gaussian Elimination with Backward Substitution Algorithm (2/3) Step 1 For i = 1,...,n 1 do Steps 2 4: (Elimination process) Step 2 Let p be the smallest integer with i p n and a pi 0 If no integer p can be found then OUTPUT ( no unique solution exists ) STOP Step 3 If p i then perform (E p ) (E i ) Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 41 / 43
100 Gaussian Elimination with Backward Substitution Algorithm (2/3) Step 1 For i = 1,...,n 1 do Steps 2 4: (Elimination process) Step 2 Let p be the smallest integer with i p n and a pi 0 If no integer p can be found then OUTPUT ( no unique solution exists ) STOP Step 3 If p i then perform (E p ) (E i ) Step 4 For j = i + 1,...,n do Steps 5 and 6: Step 5 Set m ji = a ji /a ii Step 6 Perform (E j m ji E i ) (E j ) Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 41 / 43
101 Gaussian Elimination with Backward Substitution Algorithm (3/3) Step 7 If a nn = 0 then OUTPUT ( no unique solution exists ) Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 42 / 43
102 Gaussian Elimination with Backward Substitution Algorithm (3/3) Step 7 If a nn = 0 then OUTPUT ( no unique solution exists ) Step 8 Set x n = a n,n+1 /a nn (Start backward substitution) Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 42 / 43
103 Gaussian Elimination with Backward Substitution Algorithm (3/3) Step 7 If a nn = 0 then OUTPUT ( no unique solution exists ) Step 8 Set x n = a n,n+1 /a nn (Start backward substitution) Step 9 For i = n 1,...,1 set x i = [ a i,n+1 n j=i+1 a ijx j ] / a ii Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 42 / 43
104 Gaussian Elimination with Backward Substitution Algorithm (3/3) Step 7 If a nn = 0 then OUTPUT ( no unique solution exists ) Step 8 Set x n = a n,n+1 /a nn (Start backward substitution) Step 9 For i = n 1,...,1 set x i = [ a i,n+1 n j=i+1 a ijx j ] / a ii Step 10 OUTPUT (x 1,..., x n ) (Procedure completed successfully) STOP Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 42 / 43
105 Questions?
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