Direct Methods for Solving Linear Systems Linear Systems of - PowerPoint PPT Presentation

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Matrices & Vectors: Augmented Matrix The Augmented Matrix (2/2) and then forming the new array [ A , b ] : a 11 a 12 a 1 n b 1   · · · a 21 a 22 a 2 n b 2 · · ·   [ A , b ] =  . . . .  . . . .   . . . .   a n 1 a n 2 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Matrices & Vectors: Augmented Matrix The Augmented Matrix (2/2) and then forming the new array [ A , b ] : a 11 a 12 a 1 n b 1   · · · a 21 a 22 a 2 n b 2 · · ·   [ A , b ] =  . . . .  . . . .   . . . .   a n 1 a n 2 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Matrices & Vectors: Augmented Matrix Representing the Linear System In what follows, the n × ( n + 1 ) matrix a 11 a 12 a 1 n b 1   · · · a 21 a 22 a 2 n b 2 · · ·   [ A , b ] =  . . . .  . . . .   . . . .   a n 1 a n 2 a nn b n · · · Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 11 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Matrices & Vectors: Augmented Matrix Representing the Linear System In what follows, the n × ( n + 1 ) matrix a 11 a 12 a 1 n b 1   · · · a 21 a 22 a 2 n b 2 · · ·   [ A , b ] =  . . . .  . . . .   . . . .   a n 1 a n 2 a nn b n · · · will used to represent the linear system a 11 x 1 + a 12 x 2 + · · · + a 1 n x n b 1 = a 21 x 1 + a 22 x 2 + · · · + a 2 n x n b 2 = . . . . . . a n 1 x 1 + a n 2 x 2 + · · · + a nn x n b n = Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 11 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Outline Notation & Basic Terminology 1 3 Operations to Simplify a Linear System of Equations 2 Gaussian Elimination Procedure 3 The Gaussian Elimination with Backward Substitution Algorithm 4 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 12 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 2 + · · · + a 1 n x n = b 1 E 2 : a 21 x 1 + a 22 x 2 + · · · + a 2 n x n = b 2 . . . E n : a n 1 x 1 + a n 2 x 2 + · · · + 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 2 + · · · + a 1 n x n = b 1 E 2 : a 21 x 1 + a 22 x 2 + · · · + a 2 n x n = b 2 . . . E n : a n 1 x 1 + a n 2 x 2 + · · · + 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Simplifying a Linear Systems of Equations Permissible Operations We will use 3 operations to simplify the linear system: Equation E i can be multiplied by any nonzero constant λ with the 1 resulting equation used in place of E i . This operation is denoted ( λ E i ) → ( E i ) . Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 14 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Simplifying a Linear Systems of Equations Permissible Operations We will use 3 operations to simplify the linear system: Equation E i can be multiplied by any nonzero constant λ with the 1 resulting equation used in place of E i . This operation is denoted ( λ E i ) → ( E i ) . Equation E j can be multiplied by any constant λ and added to 2 equation E i with the resulting equation used in place of E i . This operation is denoted ( E i + λ E j ) → ( E i ) . Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 14 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Simplifying a Linear Systems of Equations Permissible Operations We will use 3 operations to simplify the linear system: Equation E i can be multiplied by any nonzero constant λ with the 1 resulting equation used in place of E i . This operation is denoted ( λ E i ) → ( E i ) . Equation E j can be multiplied by any constant λ and added to 2 equation E i with the resulting equation used in place of E i . This operation is denoted ( E i + λ E j ) → ( E i ) . Equations E i and E j can be transposed in order. This operation is 3 denoted ( E i ) ↔ ( E j ) . Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 14 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Simplifying a Linear Systems of Equations Permissible Operations We will use 3 operations to simplify the linear system: Equation E i can be multiplied by any nonzero constant λ with the 1 resulting equation used in place of E i . This operation is denoted ( λ E i ) → ( E i ) . Equation E j can be multiplied by any constant λ and added to 2 equation E i with the resulting equation used in place of E i . This operation is denoted ( E i + λ E j ) → ( E i ) . Equations E i and E j can be transposed in order. This operation is 3 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Simplifying a Linear Systems of Equations Illustration The four equations E 1 : x 1 + x 2 + 3 x 4 = 4 E 2 : 2 x 1 + x 2 − x 3 + x 4 = 1 E 3 : 3 x 1 − x 2 − x 3 + 2 x 4 = − 3 E 4 : − x 1 + 2 x 2 + 3 x 3 − x 4 = 4 will be solved for x 1 , x 2 , x 3 , and x 4 . Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 15 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Simplifying a Linear Systems of Equations Illustration The four equations E 1 : x 1 + x 2 + 3 x 4 = 4 E 2 : 2 x 1 + x 2 − x 3 + x 4 = 1 E 3 : 3 x 1 − x 2 − x 3 + 2 x 4 = − 3 E 4 : − x 1 + 2 x 2 + 3 x 3 − x 4 = 4 will be solved for x 1 , x 2 , x 3 , and x 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 − 2 E 1 ) ( E 2 ) → ( E 3 − 3 E 1 ) ( E 3 ) → ( E 4 + E 1 ) ( E 4 ) and → Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 15 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Simplifying a Linear Systems of Equations E 1 : x 1 + x 2 + 3 x 4 = 4 E 2 : 2 x 1 + x 2 − x 3 + x 4 = 1 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 16 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Simplifying a Linear Systems of Equations E 1 : x 1 + x 2 + 3 x 4 = 4 E 2 : 2 x 1 + x 2 − x 3 + x 4 = 1 Illustration Cont’d (2/5) For example, in the second equation ( E 2 − 2 E 1 ) → ( E 2 ) produces ( 2 x 1 + x 2 − x 3 + x 4 ) − 2 ( x 1 + x 2 + 3 x 4 ) = 1 − 2 ( 4 ) Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 16 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Simplifying a Linear Systems of Equations E 1 : x 1 + x 2 + 3 x 4 = 4 E 2 : 2 x 1 + x 2 − x 3 + x 4 = 1 Illustration Cont’d (2/5) For example, in the second equation ( E 2 − 2 E 1 ) → ( E 2 ) produces ( 2 x 1 + x 2 − x 3 + x 4 ) − 2 ( x 1 + x 2 + 3 x 4 ) = 1 − 2 ( 4 ) which simplifies to the result shown as E 2 in E 1 : x 1 + x 2 + 3 x 4 = 4 E 2 : − x 2 − x 3 − 5 x 4 = − 7 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 16 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 + 3 x 4 = 4 E 2 : − x 2 − x 3 − 5 x 4 = − 7 E 3 : − 4 x 2 − x 3 − 7 x 4 = − 15 E 4 : 3 x 2 + 3 x 3 + 2 x 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 − 4 E 2 ) → ( E 3 ) and ( E 4 + 3 E 2 ) → ( E 4 ) . Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 18 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 − 4 E 2 ) → ( E 3 ) and ( E 4 + 3 E 2 ) → ( E 4 ) . This results in E 1 : x 1 + x 2 + 3 x 4 = 4 , E 2 : − x 2 − x 3 − 5 x 4 = − 7 , E 3 : 3 x 3 + 13 x 4 = 13 , E 4 : − 13 x 4 = − 13 . Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 18 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 − 4 E 2 ) → ( E 3 ) and ( E 4 + 3 E 2 ) → ( E 4 ) . This results in E 1 : x 1 + x 2 + 3 x 4 = 4 , E 2 : − x 2 − x 3 − 5 x 4 = − 7 , E 3 : 3 x 3 + 13 x 4 = 13 , E 4 : − 13 x 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 − 13 x 4 ) = 1 3 ( 13 − 13 ) = 0 . Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 19 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 − 13 x 4 ) = 1 3 ( 13 − 13 ) = 0 . Continuing, E 2 gives x 2 = − ( − 7 + 5 x 4 + x 3 ) = − ( − 7 + 5 + 0 ) = 2 , Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 19 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 − 13 x 4 ) = 1 3 ( 13 − 13 ) = 0 . Continuing, E 2 gives x 2 = − ( − 7 + 5 x 4 + x 3 ) = − ( − 7 + 5 + 0 ) = 2 , and E 1 gives x 1 = 4 − 3 x 4 − x 2 = 4 − 3 − 2 = − 1 . Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 19 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 − 13 x 4 ) = 1 3 ( 13 − 13 ) = 0 . Continuing, E 2 gives x 2 = − ( − 7 + 5 x 4 + x 3 ) = − ( − 7 + 5 + 0 ) = 2 , and E 1 gives x 1 = 4 − 3 x 4 − x 2 = 4 − 3 − 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Outline Notation & Basic Terminology 1 3 Operations to Simplify a Linear System of Equations 2 Gaussian Elimination Procedure 3 The Gaussian Elimination with Backward Substitution Algorithm 4 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 20 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Constructing an Algorithm to Solve the Linear System E 1 : x 1 + x 2 + 3 x 4 = 4 E 2 : 2 x 1 + x 2 − x 3 + x 4 = 1 E 3 : 3 x 1 − x 2 − x 3 + 2 x 4 = − 3 E 4 : − x 1 + 2 x 2 + 3 x 3 − x 4 = 4 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 21 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Constructing an Algorithm to Solve the Linear System E 1 : x 1 + x 2 + 3 x 4 = 4 E 2 : 2 x 1 + x 2 − x 3 + x 4 = 1 E 3 : 3 x 1 − x 2 − x 3 + 2 x 4 = − 3 E 4 : − x 1 + 2 x 2 + 3 x 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:   1 1 0 3 4 2 1 − 1 1 1     3 − 1 − 1 2 − 3   − 1 2 3 − 1 4 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 21 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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:  1 1 0 3 4   1 1 0 3 4  0 − 1 − 1 − 5 − 7 0 − 1 − 1 − 5 − 7      and     0 − 4 − 1 − 7 − 15 0 0 3 13 13    0 3 3 2 8 0 0 0 − 13 − 13 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 22 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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:  1 1 0 3 4   1 1 0 3 4  0 − 1 − 1 − 5 − 7 0 − 1 − 1 − 5 − 7      and     0 − 4 − 1 − 7 − 15 0 0 3 13 13    0 3 3 2 8 0 0 0 − 13 − 13 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 2 + · · · + a 1 n x n = b 1 E 2 : a 21 x 1 + a 22 x 2 + · · · + a 2 n x n = b 2 . . . . . . E n : a n 1 x 1 + a n 2 x 2 + · · · + 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Gaussian Elimination with Backward Substitution Basic Steps in the Procedure (Cont’d) First form the augmented matrix ˜ A : a 11 a 12 a 1 n a 1 , n + 1   · · · a 21 a 22 a 2 n a 2 , n + 1 · · ·   A = [ A , b ] = ˜  . . . .  . . . .   . . . .   a n 1 a n 2 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Gaussian Elimination with Backward Substitution Basic Steps in the Procedure (Cont’d) First form the augmented matrix ˜ A : a 11 a 12 a 1 n a 1 , n + 1   · · · a 21 a 22 a 2 n a 2 , n + 1 · · ·   A = [ A , b ] = ˜  . . . .  . . . .   . . . .   a n 1 a n 2 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 j 1 / 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 j 1 / 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 i th row and the j th column by a ij . Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 25 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 j 1 / 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 i th row and the j th 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 i th 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 i th for all values of i = 1 , 2 , . . . , n − 1. The resulting matrix has the form: a 11 a 12 a 1 n a 1 , n + 1   · · · a 22 a 2 n a 2 , n + 1 0 · · · ˜   A = ˜   . . . ... ... . . .   . . .   a nn a n , n + 1 0 · · · 0 where, except in the first row, the values of a ij are not expected to A . agree with those in the original matrix ˜ Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 26 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 i th for all values of i = 1 , 2 , . . . , n − 1. The resulting matrix has the form: a 11 a 12 a 1 n a 1 , n + 1   · · · a 22 a 2 n a 2 , n + 1 0 · · · ˜   A = ˜   . . . ... ... . . .   . . .   a nn a n , n + 1 0 · · · 0 where, except in the first row, the values of a ij are not expected to A . agree with those in the original matrix ˜ The matrix ˜ A 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 2 a 1 n x n a 1 , n + 1 + + + = · · · a 22 x 2 a 2 n x n a 2 , n + 1 + · · · + = . . ... . . . . . . ... . . . . . . ... . . . . 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 2 a 1 n x n a 1 , n + 1 + + + = · · · a 22 x 2 a 2 n x n a 2 , n + 1 + · · · + = . . ... . . . . . . ... . . . . . . ... . . . . a nn x n a n , n + 1 = so backward substitution can be performed. Solving the n th 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 a i , n + 1 − a i , n x n − a i , n − 1 x n − 1 − · · · − a i , i + 1 x i + 1 x i = a ii a i , n + 1 − � n j = i + 1 a ij x 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Gaussian Elimination with Backward Substitution A More Precise Description Gaussian elimination procedure is described more precisely, although A ( 1 ) , more intricately, by forming a sequence of augmented matrices ˜ A ( n ) , where ˜ A ( 1 ) is the matrix ˜ A ( k ) , for A ( 2 ) , . . . , ˜ ˜ A given earlier and ˜ each k = 2 , 3 , . . . , n , has entries a ( k ) ij , where:  a ( k − 1) when i = 1 , 2 , . . . , k − 1 and j = 1 , 2 , . . . , n + 1 ij     when i = k, k + 1 , . . . , n and j = 1 , 2 , . . . , k − 1  0  a ( k ) ij = a ( k − 1)  a ( k − 1) i,k − 1 a ( k − 1)  when i = k, k + 1 , . . . , n and j = k, k + 1 , . . . , n + 1 −   ij k − 1 ,j a ( k − 1)   k − 1 ,k − 1 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 29 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Gaussian Elimination with Backward Substitution A More Precise Description (Cont’d) Thus a ( 1 ) a ( 1 ) a ( 1 ) a ( 1 ) a ( 1 ) a ( 1 ) a ( 1 )   · · · · · · 1 , k − 1 1 k 1 n 1 , n + 1 11 12 13 a ( 2 ) a ( 2 ) a ( 2 ) a ( 2 ) a ( 2 ) a ( 2 )  0 · · · · · ·  2 , k − 1 2 k 2 n 2 , n + 1 22 23   . . . . .  ... ...  . . . . .   . . . . .    . . . . .  ... ... . . . . .   . . . . .   A ( k ) = ˜   . ... a ( k − 1 ) a ( k − 1 ) a ( k − 1 ) a ( k − 1 ) .   . · · ·  k − 1 , k − 1 k − 1 , k k − 1 , n k − 1 , n + 1    . a ( k ) a ( k ) a ( k )  .  . 0 · · ·   kk kn k , n + 1   . . . . .   . . . . .  . . . . .    a ( k ) a ( k ) a ( k ) 0 · · · · · · · · · 0 · · · nn nk 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 a ( k )   i , k  E i − ( E k )  → E i a ( k ) kk 11 , . . . , a ( n − 1 ) either cannot be performed (this occurs if one of a ( 1 ) n − 1 , n − 1 is zero), or the backward substitution cannot be accomplished (in the case a ( n ) nn = 0). Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 31 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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 a ( k )   i , k  E i − ( E k )  → E i a ( k ) kk 11 , . . . , a ( n − 1 ) either cannot be performed (this occurs if one of a ( 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Illustration of the Gaussian Elimination Procedure Example Represent the linear system E 1 : x 1 − x 2 + 2 x 3 − x 4 = − 8 E 2 : 2 x 1 − 2 x 2 + 3 x 3 − 3 x 4 = − 20 E 3 : x 1 + x 2 + x 3 = − 2 E 4 : x 1 − x 2 + 4 x 3 + 3 x 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Illustration of the Gaussian Elimination Procedure Solution (1/6) The augmented matrix is  1 − 1 2 − 1 − 8  2 − 2 3 − 3 − 20 A ( 1 ) = A = ˜ ˜     1 1 1 0 − 2   1 − 1 4 3 4 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 33 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Illustration of the Gaussian Elimination Procedure Solution (1/6) The augmented matrix is  1 − 1 2 − 1 − 8  2 − 2 3 − 3 − 20 A ( 1 ) = A = ˜ ˜     1 1 1 0 − 2   1 − 1 4 3 4 Performing the operations ( E 2 − 2 E 1 ) → ( E 2 ) , ( E 3 − E 1 ) → ( E 3 ) ( E 4 − E 1 ) → ( E 4 ) and Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 33 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Illustration of the Gaussian Elimination Procedure Solution (1/6) The augmented matrix is  1 − 1 2 − 1 − 8  2 − 2 3 − 3 − 20 A ( 1 ) = A = ˜ ˜     1 1 1 0 − 2   1 − 1 4 3 4 Performing the operations ( E 2 − 2 E 1 ) → ( E 2 ) , ( E 3 − E 1 ) → ( E 3 ) ( E 4 − E 1 ) → ( E 4 ) and gives  1 − 1 2 − 1 − 8  0 0 − 1 − 1 − 4 A ( 2 ) = ˜     0 2 − 1 1 6   0 0 2 4 12 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 33 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Illustration of the Gaussian Elimination Procedure  1 − 1 2 − 1 − 8  0 0 − 1 − 1 − 4 A ( 2 ) = ˜     0 2 − 1 1 6   0 0 2 4 12 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 34 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Illustration of the Gaussian Elimination Procedure  1 − 1 2 − 1 − 8  0 0 − 1 − 1 − 4 A ( 2 ) = ˜     0 2 − 1 1 6   0 0 2 4 12 Solution (2/6) 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Illustration of the Gaussian Elimination Procedure  1 − 1 2 − 1 − 8  0 0 − 1 − 1 − 4 A ( 2 ) = ˜     0 2 − 1 1 6   0 0 2 4 12 Solution (2/6) 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Illustration of the Gaussian Elimination Procedure  1 − 1 2 − 1 − 8  0 0 − 1 − 1 − 4 A ( 2 ) = ˜     0 2 − 1 1 6   0 0 2 4 12 Solution (2/6) 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Illustration of the Gaussian Elimination Procedure  1 − 1 2 − 1 − 8  0 0 − 1 − 1 − 4 A ( 2 ) = ˜     0 2 − 1 1 6   0 0 2 4 12 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 35 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Illustration of the Gaussian Elimination Procedure  1 − 1 2 − 1 − 8  0 0 − 1 − 1 − 4 A ( 2 ) = ˜     0 2 − 1 1 6   0 0 2 4 12 Solution (3/6) Perform the operation ( E 2 ) ↔ ( E 3 ) to obtain a new matrix:   1 − 1 2 − 1 − 8 A ( 2 ) ′ = 0 2 − 1 1 6 ˜     0 0 − 1 − 1 − 4   0 0 2 4 12 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 35 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Illustration of the Gaussian Elimination Procedure   1 − 1 2 − 1 − 8 A ( 2 ) ′ = 0 2 − 1 1 6 ˜     0 0 − 1 − 1 − 4   0 0 2 4 12 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 36 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Illustration of the Gaussian Elimination Procedure   1 − 1 2 − 1 − 8 A ( 2 ) ′ = 0 2 − 1 1 6 ˜     0 0 − 1 − 1 − 4   0 0 2 4 12 Solution (4/6) A ( 3 ) will be ˜ Since x 2 is already eliminated from E 3 and E 4 , ˜ A ( 2 ) ′ , and the computations continue with the operation ( E 4 + 2 E 3 ) → ( E 4 ) , giving   1 − 1 2 − 1 − 8 0 2 − 1 1 6 A ( 4 ) = ˜     0 0 − 1 − 1 − 4   0 0 0 2 4 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 36 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Illustration of the Gaussian Elimination Procedure  1 − 1 2 − 1 − 8  0 2 − 1 1 6 A ( 4 ) = ˜     0 0 − 1 − 1 − 4   0 0 0 2 4 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 37 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Illustration of the Gaussian Elimination Procedure  1 − 1 2 − 1 − 8  0 2 − 1 1 6 A ( 4 ) = ˜     0 0 − 1 − 1 − 4   0 0 0 2 4 Solution (5/6) The solution may now be found through backward substitution: 4 x 4 = 2 = 2 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 37 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Illustration of the Gaussian Elimination Procedure  1 − 1 2 − 1 − 8  0 2 − 1 1 6 A ( 4 ) = ˜     0 0 − 1 − 1 − 4   0 0 0 2 4 Solution (5/6) The solution may now be found through backward substitution: 4 x 4 = 2 = 2 [ − 4 − ( − 1 ) x 4 ] x 3 = = 2 − 1 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 37 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Illustration of the Gaussian Elimination Procedure  1 − 1 2 − 1 − 8  0 2 − 1 1 6 A ( 4 ) = ˜     0 0 − 1 − 1 − 4   0 0 0 2 4 Solution (5/6) The solution may now be found through backward substitution: 4 x 4 = 2 = 2 [ − 4 − ( − 1 ) x 4 ] x 3 = = 2 − 1 [ 6 − x 4 − ( − 1 ) x 3 ] x 2 = = 3 2 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 37 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Illustration of the Gaussian Elimination Procedure  1 − 1 2 − 1 − 8  0 2 − 1 1 6 A ( 4 ) = ˜     0 0 − 1 − 1 − 4   0 0 0 2 4 Solution (5/6) The solution may now be found through backward substitution: 4 x 4 = 2 = 2 [ − 4 − ( − 1 ) x 4 ] x 3 = = 2 − 1 [ 6 − x 4 − ( − 1 ) x 3 ] x 2 = = 3 2 [ − 8 − ( − 1 ) x 4 − 2 x 3 − ( − 1 ) x 2 ] x 1 = = − 7 1 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 37 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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. A ( k − 1 ) from the k th row to the n th row is The k th column of ˜ searched for the first nonzero entry. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 38 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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. A ( k − 1 ) from the k th row to the n th row is The k th column of ˜ 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 ˜ A ( k − 1 ) ′ . Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 38 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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. A ( k − 1 ) from the k th row to the n th row is The k th column of ˜ 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 ˜ A ( k − 1 ) ′ . A ( k ) , and so on. The procedure can then be continued to form ˜ Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 38 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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. A ( k − 1 ) from the k th row to the n th row is The k th column of ˜ 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 ˜ A ( k − 1 ) ′ . A ( k ) , and so on. The procedure can then be continued to form ˜ 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm 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. A ( k − 1 ) from the k th row to the n th row is The k th column of ˜ 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 ˜ A ( k − 1 ) ′ . A ( k ) , and so on. The procedure can then be continued to form ˜ 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Outline Notation & Basic Terminology 1 3 Operations to Simplify a Linear System of Equations 2 Gaussian Elimination Procedure 3 The Gaussian Elimination with Backward Substitution Algorithm 4 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 39 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Gaussian Elimination with Backward Substitution Algorithm (1/3) To solve the n × n linear system E 1 : a 11 x 1 + a 12 x 2 + · · · + a 1 n x n = a 1 , n + 1 E 2 : a 21 x 1 + a 22 x 2 + · · · + a 2 n x n = a 2 , n + 1 . . . . . . . . . . . . . . . E n : a n 1 x 1 + a n 2 x 2 + · · · + a nn x n = a n , n + 1 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 40 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Gaussian Elimination with Backward Substitution Algorithm (1/3) To solve the n × n linear system E 1 : a 11 x 1 + a 12 x 2 + · · · + a 1 n x n = a 1 , n + 1 E 2 : a 21 x 1 + a 22 x 2 + · · · + a 2 n x n = a 2 , n + 1 . . . . . . . . . . . . . . . E n : a n 1 x 1 + a n 2 x 2 + · · · + a nn x n = a n , n + 1 number of unknowns and equations n ; augmented INPUT 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Gaussian Elimination with Backward Substitution Algorithm (1/3) To solve the n × n linear system E 1 : a 11 x 1 + a 12 x 2 + · · · + a 1 n x n = a 1 , n + 1 E 2 : a 21 x 1 + a 22 x 2 + · · · + a 2 n x n = a 2 , n + 1 . . . . . . . . . . . . . . . E n : a n 1 x 1 + a n 2 x 2 + · · · + a nn x n = a n , n + 1 number of unknowns and equations n ; augmented INPUT 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 OUTPUT has no unique solution. Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 40 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Gaussian Elimination with Backward Substitution Algorithm (2/3) For i = 1 , . . . , n − 1 do Steps 2–4: ( Elimination process ) Step 1 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 41 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Gaussian Elimination with Backward Substitution Algorithm (2/3) For i = 1 , . . . , n − 1 do Steps 2–4: ( Elimination process ) Step 1 Let p be the smallest integer with i ≤ p ≤ n and a pi � = 0 Step 2 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

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Gaussian Elimination with Backward Substitution Algorithm (2/3) For i = 1 , . . . , n − 1 do Steps 2–4: ( Elimination process ) Step 1 Let p be the smallest integer with i ≤ p ≤ n and a pi � = 0 Step 2 If no integer p can be found then OUTPUT (‘no unique solution exists’) STOP If p � = i then perform ( E p ) ↔ ( E i ) Step 3 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 41 / 43

Terminology Permissible Operations Gaussian Elimination GE/BS Algorithm Gaussian Elimination with Backward Substitution Algorithm (2/3) For i = 1 , . . . , n − 1 do Steps 2–4: ( Elimination process ) Step 1 Let p be the smallest integer with i ≤ p ≤ n and a pi � = 0 Step 2 If no integer p can be found then OUTPUT (‘no unique solution exists’) STOP If p � = i then perform ( E p ) ↔ ( E i ) Step 3 For j = i + 1 , . . . , n do Steps 5 and 6: Step 4 Set m ji = a ji / a ii Step 5 Perform ( E j − m ji E i ) → ( E j ) Step 6 Numerical Analysis (Chapter 6) Linear Systems of Equations R L Burden & J D Faires 41 / 43