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homework_4_solutions

homework_4_solutions - MA 265 HOMEWORK ASSIGNMENT#4...

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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 0 - 3 - 1 - 2 - 5 0 - 2 - 5 - 10 - 38 ( - 1 / 3) r 2 r 2 : 1 1 2 3 13 0 1 1 3 2 3 5 3 0 - 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

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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 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 0 0 - 3 2 0 - 1 - 1 0 r 2 r 3 : 1 1 1 1 0 - 1 - 1 0 0 0 - 3 2 ( - 1) r 2 r 2 : ( - 1 / 3) r 3 r 3 : 1 1 1 1 0 1 1 0 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 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 0 1 - 9 - 2 5 - 1 2 - 8 3 To use the Gaussian elimination method to solve the system, we compute the row echelon form for this matrix: (1 / 2) r 1 r 1 : 1 1 2 1 2 - 1 1 2 3 - 2 1 - 6 - 2 1 1 - 1 - 1 - 1 6 0 1 - 9 - 2 5 - 1 2 - 8 3 ( - 3) r 1 + r 2 r 2 : ( - 1) r 1 + r 3 r 3 : ( - 6) r 1 + r 4 r 4 : ( - 5) r 1 + r 5 r 5 :

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