homework_4_solutions - MA 265 HOMEWORK ASSIGNMENT #4...

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Unformatted text preview: MA 265 HOMEWORK ASSIGNMENT #4 SOLUTIONS #1. Page 114; Exercise 6. Repeat Exercice 5 for the following linear systems: (a) x + y + 2 z + 3 w = 13 x- 2 y + z + w = 8 3 x + y + z- w = 1 (b) x + y + z = 1 x + y- 2 z = 3 2 x + y + z = 2 (c) 2 x + y + z- 2 w = 1 3 x- 2 y + z- 6 w =- 2 x + y- z- w =- 1 6 x + z- 9 w =- 2 5 x- y + 2 z- 8 w = 3 That is, (i) Find all solutions, if any exist, by using the Gaussian elimination method. (ii) Find all solutions, if any exist, by using the Gauss-Jordan reduction method. Solution: (a) The augmented matrix for the system is 1 1 2 3 13 1- 2 1 1 8 3 1 1- 1 1 To use the Gaussian elimination method to solve the system, we compute the row echelon form for this matrix: (- 1) r 1 + r 2 r 2 : (- 3) r 1 + r 3 r 3 : 1 1 2 3 13- 3- 1- 2- 5- 2- 5- 10- 38 (- 1 / 3) r 2 r 2 : 1 1 2 3 13 1 1 3 2 3 5 3- 2- 5- 10- 38 2 r 2 + r 3 r 3 : 1 1 2 3 13 0 1 1 3 2 3 5 3 0 0- 13 3- 26 3- 104 3 (- 3 / 13) r 3 r 3 : 1 1 2 3 13 0 1 1 3 2 3 5 3 0 0 1 2 8 1 2 MA 265 HOMEWORK ASSIGNMENT #4 SOLUTIONS Since the pivots correspond to x , y , and z , we see that w is an arbitrary variable. Denote w = r , so that by back substitution we find x y z w = - 2 + r- 1 8- 2 r r for any real number r . To use the Gauss-Jordan reduction method to solve the system, we compute the reduced row echelon form for this matrix in row echelon form: (- 1) r 2 + r 1 r 1 : 1 0 5 3 7 3 34 3 0 1 1 3 2 3 5 3 0 0 1 2 8 (- 5 / 3) r 3 + r 1 r 1 : (- 1 / 3) r 3 + r 2 r 2 1 0 0- 1- 2 0 1 0- 1 0 0 1 2 8 Since the pivots correspond to x , y , and z , we see that w is an arbitrary variable. Denote w = r , so that we find x y z w = - 2 + r- 1 8- 2 r r for any real number r . (b) The augmented matrix for the system is 1 1 1 1 1 1- 2 3 2 1 1 2 To use the Gaussian elimination method to solve the system, we compute the row echelon form for this matrix: (- 1) r 1 + r 2 r 2 : (- 1) r 1 + r 3 r 3 : 1 1 1 1- 3 2- 1- 1 r 2 r 3 : 1 1 1 1- 1- 1- 3 2 (- 1) r 2 r 2 : (- 1 / 3) r 3 r 3 : 1 1 1 1 0 1 1 0 0 1- 2 3 Using back substitution, we find x y z = 1 2 3- 2 3 MA 265 HOMEWORK ASSIGNMENT #4 SOLUTIONS 3 To use the Gauss-Jordan reduction method to solve the system, we compute the reduced row echelon form for this matrix in row echelon form: (- 1) r 2 + r 1 r 1 1 0 0 1 0 1 1 0 0 1- 2 3 (- 1) r 3 + r 3 r 3 1 0 0 1 0 1 0 2 3 0 0 1- 2 3 Using this, we find x y z = 1 2 3- 2 3 (c) The augmented matrix for the system is 2 1 1- 2 1 3- 2 1- 6- 2 1 1- 1- 1- 1 6 1- 9...
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This note was uploaded on 03/11/2010 for the course MA 261A 0026100 taught by Professor ... during the Spring '10 term at Purdue University Calumet.

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homework_4_solutions - MA 265 HOMEWORK ASSIGNMENT #4...

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