Explanation Formula used: The procedure to solve the system of liner equation using Gauss –elimination, Step-1: The first unknown x 1 is eliminated from the last remaining equations by subtracting multiples of the first equation. Step-2: The second unknown x 2 is eliminated from the last remaining equations by subtracting multiples of the reduced second equation. Step-3: The third unknown x 3 is eliminated from the last remaining equations by subtracting multiples of the reduced third equation. Step-4: The nth unknown x n is eliminated from the last remaining equations by subtracting multiples of the reduced nth equation. Step-5: Continue above step until all system becomes uncoupled system. Calculation: Consider the system of equation: x 1 − 3 x 2 + x 3 = 0 …… (1) − 2 x 1 + 2 x 2 − x 3 = 0 …… (2) 4 x 1 − 4 x 2 + 2 x 3 = 0 …… (3) To eliminate x 1 multiply the equation (1) by − 2 and the divide the equation (3) by − 2 : − 2 x 1 + 6 x 2 − 2 x 3 = 0 …… (4) − 2 x 1 + 2 x 2 − x 3 = 0 …… (2) − 2 x 1 + 2 x 2 − x 3 = 0 …… (5) Subtract the equation (2) and (5) from (4): − 2 x 1 + 6 x 2 − 2 x 3 = 0 …… (4) 4 x 2 − x 3 = 0 …… (2)

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 6E

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To solve: The system of equation

−2x1+2x2−x3=0, x1−3x2+x3=0 and 4x1−4x2+2x3=0 using the Gauss-Jordan elimination algorithm.

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Chapter 9.2, Problem 5E

Chapter 9.2, Problem 7E

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Chapter 9 Solutions

Fundamentals of Differential Equations and Boundary Value Problems

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The system of equation − 2 x 1 + 2 x 2 − x 3 = 0 , x 1 − 3 x 2 + x 3 = 0 and 4 x 1 − 4 x 2 + 2 x 3 = 0 using the Gauss-Jordan elimination algorithm.

Solution Summary: The author explains the procedure to solve the system of liner equation using Gauss-Jordan elimination.

Videos

To solve: The system of equation

−2x1+2x2−x3=0, x1−3x2+x3=0 and 4x1−4x2+2x3=0 using the Gauss-Jordan elimination algorithm.

Want to see the full answer?

Chapter 9 Solutions

The system of equation − 2 x 1 + 2 x 2 − x 3 = 0 , x 1 − 3 x 2 + x 3 = 0 and 4 x 1 − 4 x 2 + 2 x 3 = 0 using the Gauss-Jordan elimination algorithm.

Solution Summary: The author explains the procedure to solve the system of liner equation using Gauss-Jordan elimination.

Videos

To solve: The system of equation −2x1+2x2−x3=0, x1−3x2+x3=0 and 4x1−4x2+2x3=0 using the Gauss-Jordan elimination algorithm.

Want to see the full answer?

Chapter 9 Solutions

To solve: The system of equation

−2x1+2x2−x3=0, x1−3x2+x3=0 and 4x1−4x2+2x3=0 using the Gauss-Jordan elimination algorithm.