Chapter 16: Chapter 16

Problem 358

Use a partial pivoting strategy to solve the system $$ \begin{aligned} &(0.100) 10^{-3} \mathrm{x}_{1}+(0.100) 10^{1} \mathrm{x}_{2}=(0.200) 10^{1} \\\ &(0.100) 10^{1} \mathrm{x}_{1}+(0.100) 10^{1} \mathrm{x}_{2}=(0.300) 10^{1} \end{aligned} $$ Is the answer markedly superior to merely using Gauss elimination?

Short Answer

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In this specific problem, using a partial pivoting strategy resulted in the correct solutions $x_1 \approx 1$ and $x_2 \approx 2$, while merely using Gauss elimination without row swapping led to an incorrect result. Partial pivoting reduced the chance of round-off errors and provided a markedly superior solution by reordering the rows.

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TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: 1. Rewrite the system in a more convenient form

Rewrite the system of linear equations in the following form to work with it more comfortably: $$ \begin{aligned} 0.001 x_{1}+10 x_{2} &= 20 \\ 10 x_{1}+10 x_{2} &= 30 \end{aligned} $$

Step 2: 2. Swap rows

Swap the two equations to move the largest coefficients to the top row. This action should reduce the chance of round-off errors. $$ \begin{aligned} 10 x_{1}+10 x_{2} &=30 \\ 0.001 x_{1}+10 x_{2} &= 20 \end{aligned} $$

Step 3: 3. Perform Gaussian elimination

Perform the Gaussian elimination on the rearranged system of linear equations. Divide the first row by the coefficient of x1, 10, in this case, to make its leading coefficient equal to one. $$ \begin{aligned} x_{1}+x_{2} &=3 \\ 0.001 x_{1}+10 x_{2}&=20 \end{aligned} $$

Step 4: 4. Eliminate x1 from the second equation

Multiply the first row by -0.001 and add it to the second equation to eliminate $x_1$ from the second row: $$ \begin{aligned} x_{1}+x_{2}&=3 \\ 0+9.997 x_{2}&=19.997 \end{aligned} $$

Step 5: 5. Solve for x2

Solve for $x_2$ by dividing the coefficient of x2 in the second equation, 9.997, and obtain: $$ x_{2} = \frac{19.997}{9.997} \approx 2 $$

Step 6: 6. Solve for x1

Substituting the value of $x_2$ into the first equation, we get the value of $x_1$: $$ x_{1}+2=3 $$ So, $x_1 \approx 1$

Step 7: 7. Compare with classical Gaussian elimination

Let's perform the classical Gaussian elimination on the initial system without reordering the rows: $$ \begin{aligned} 0.001 x_{1}+10 x_{2}&=20 \\ 10 x_{1}+10 x_{2}&= 30 \end{aligned} $$ Upon eliminating the first variable $x_1$ from the second equation by subtracting 10,000 times the first equation from the second equation, we get: $$ \begin{aligned} 0.001 x_{1} + 10 x_{2}&= 20 \\ 0 x_{2}&=10 \end{aligned} $$ The classical Gaussian elimination without row swapping led to a result with an incorrect second equation, which indicates that partial pivoting provided a markedly superior solution.

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Chapter 16: Chapter 16

Short Answer

Step by step solution

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Step 1: 1. Rewrite the system in a more convenient form

Step 2: 2. Swap rows

Step 3: 3. Perform Gaussian elimination

Step 4: 4. Eliminate x1 from the second equation

Step 5: 5. Solve for x2

Step 6: 6. Solve for x1

Step 7: 7. Compare with classical Gaussian elimination

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Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Chapter 9

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 14

Chapter 15

Chapter 17

Chapter 18

Chapter 19

Chapter 20

Chapter 21

Chapter 22

Chapter 23

Chapter 24

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