Explanation Proof: Consider the system of equation x 1 + 2 x 2 − x 3 = r x 1 x 1 + x 3 = r x 2 4 x 1 − 4 x 2 + 5 x 3 = r x 3 Rearrange the system of equation; ( r − 1 ) x 1 + 2 x 2 − x 3 = 0 …… (1) x 1 − r x 2 + x 3 = 0 …… (2) 4 x 1 − 4 x 2 + ( 5 − r ) x 3 = 0 …… (3) Find the solution of system of equation corresponding to r = − 1 Substitute − 1 for r 1 in the system of equation: x 1 + x 2 + x 3 = 0 …… (4) − 2 x 1 + 2 x 2 − x 3 = 0 …… (5) 4 x 1 − 4 x 2 + 6 x 3 = 0 …… (6) To eliminate x 1 multiply the equation (4) by 4 and the equation (5) by − 2 : 4 x 1 + 4 x 2 + 4 x 3 = 0 …… (7) 4 x 1 − 4 x 2 + 2 x 3 = 0 …… (8) 4 x 1 − 4 x 2 + 6 x 3 = 0 …… (6) Subtract the equation (8) and (6) from (7): 4 x 1 + 4 x 2 + 4 x 3 = 0 …… (7) 8 x 2 + 2 x 3 = 0 …… (8) 8 x 2 − 2 x 3 = 0 …… (6) To eliminate x 2 subtract the equation (6) from (8): 4 x 1 + 4 x 2 + 4 x 3 = 0 …… (7) 8 x 2 + 2 x 3 = 0 …… (8) 4 x 3 = 0 …… (6) Solve above uncoupled equations separately: x 3 = 0 Substitute 0 for x 3 in the equation (8): 8 x 2 + 2 ( 0 ) = 0 8 x 2 = 0 x 2 = 0 Substitute 0 for x 3 and 0 for x 2 in the equation (4):

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 9.2, Problem 14E

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To prove: The system of equation x1+2x2−x3=rx1, 4x1−4x2+5x3=rx3 and x1+x3=rx2 using the Gauss–Jordan elimination algorithm to has a unique solution for r1=−1 but an infinite number of solutions for r1=2.

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Chapter 9.2, Problem 13E

Chapter 9.3, Problem 1E

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Fundamentals of Differential Equations and Boundary Value Problems

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To prove: The system of equation x 1 + 2 x 2 − x 3 = r x 1 , 4 x 1 − 4 x 2 + 5 x 3 = r x 3 and x 1 + x 3 = r x 2 using the Gauss–Jordan elimination algorithm to has a unique solution for r 1 = − 1 but an infinite number of solutions for r 1 = 2 .

Solution Summary: The author analyzes the Gauss–Jordan elimination algorithm to have a unique solution for r_1=-1, but an infinite number of solutions.

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To prove: The system of equation x1+2x2−x3=rx1, 4x1−4x2+5x3=rx3 and x1+x3=rx2 using the Gauss–Jordan elimination algorithm to has a unique solution for r1=−1 but an infinite number of solutions for r1=2.

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Chapter 9 Solutions