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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?

Expert verified
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.
See the step by step solution

## 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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