Problem 2 - Arbeitsbuch Mathematik für Ingenieure

Question

Das in Beispiel (7) behandelte Gleichungssystem liege in folgender Form vor: $$ \left(\begin{array}{cc} 1 & 200 \\ 1 & 1 \end{array}\right) \cdot\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)=\left(\begin{array}{c} 100 \\ 1 \end{array}\right) $$ Lösen Sie dieses mit 2 signifikanten Dezimalstellen und üblicher Rundung a) ohne Pivotsuche, b) mit Spaltenpivotwahl, c) mit Restmatrixpivotwahl. Beurteilen Sie die Oualität der Lösungen

Step-by-Step Solution

Answer

The solutions are the same: \(x_1 = 0.00\), \(x_2 = 0.50\) for all methods. Pivot strategies did not significantly impact this small system's solutions.

Step 1: Identify the given matrices and vectors

Consider the equation given: \[\begin{pmatrix} 1 & 200 \ 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 100 \ 1 \end{pmatrix} \] where the coefficient matrix is \[A = \begin{pmatrix} 1 & 200 \ 1 & 1 \end{pmatrix}\], the variable vector is \(\mathbf{x} = \begin{pmatrix} x_1 \ x_2 \end{pmatrix}\), and the constant vector is \(\mathbf{b} = \begin{pmatrix} 100 \ 1 \end{pmatrix}\).

Step 2: Solve without pivoting

Set up the augmented matrix: \[\begin{pmatrix} 1 & 200 & | & 100 \ 1 & 1 & | & 1 \end{pmatrix}\].Perform Gaussian elimination by subtracting the first row from the second row. New matrix: \[\begin{pmatrix} 1 & 200 & | & 100 \ 0 & -199 & | & -99 \end{pmatrix}\].Solve for \(x_2\): \[x_2 = \frac{-99}{-199} \approx 0.50\].Substitute \(x_2\) into the first equation: \[1 \cdot x_1 + 200 \cdot 0.50 = 100\], \[x_1 + 100 = 100\], \[x_1 = 0.00\].

Step 3: Solve with column pivoting

Identify the pivot element in each column: both 1 and 200. Since no swapping is needed, proceed with Gaussian elimination using the pivot element (1) in each step. Follow similar steps to the one without pivoting, resulting in the same final augmented matrix and solving the same way: \[x_1 = 0.00, x_2 = 0.50\].

Step 4: Solve with rest matrix pivoting

Identify the largest pivot from entire submatrix in each step: 200. Swap rows or columns to position the largest element as pivot. The rest matrix after swapping becomes: \[\begin{pmatrix} 1 & 1 & | & 1 \ 200 & 1 & | & 100 \end{pmatrix}\].Continue with Gaussian elimination to get a similar resulting matrix, hence solving again results as: \[x_1 = 0.00, x_2 = 0.50\].

Step 5: Analyze and compare the quality of solutions

Since the primary matrix operation resulted in the same numbers in all cases, without significant rounding errors or drastic differences, it showcases the stability of this small system. The solutions from all methods are relatively similar: \(x_1 = 0.00\), \(x_2 = 0.50\).

Key Concepts

Matrix Operations

Matrix operations are fundamental in solving linear equation systems such as the one given in our exercise. The given system of equations can be represented as a matrix equation: Consider the coefficient matrix, variable vector, and constant vector: $$ \begin{pmatrix} 1 & 200 \ 1 & 1 \end{pmatrix} \times \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 100 \ 1 \end{pmatrix} $$ By arranging them in an augmented matrix form, we can perform Gaussian elimination to solve for variables x1 and x2. Augmented matrices combine the coefficients and constants from the equations: $$ \begin{pmatrix} 1 & 200 & | & 100 \ 1 & 1 & | & 1 \end{pmatrix}$$

Pivoting Strategies

Pivoting plays a critical role in ensuring the numerical stability and accuracy of Gaussian elimination. Here are the primary pivoting strategies assessed in the exercise:

• Without Pivoting: Operates directly on the augmented matrix without adjusting for any numerical large or small coefficients.
• Column Pivoting: Selects the largest absolute value within each column to act as the pivot, ensuring numerical stability against round-off errors.
• Rest Matrix Pivoting: Considers the largest absolute value from the entire submatrix in each step, prioritizing overall matrix stability over individual columns.

To explain further, the rest matrix pivoting strategy swaps rows or columns to place the largest element as the pivot. In our exercise, selecting the largest pivot (200) ensures better accuracy in solutions: $$ \begin{pmatrix} 1 & 1 & | & 1 \ 200 & 1 & | & 100 \end{pmatrix} $$ Subsequently leading to improved Gaussian elimination for accurate results.

Solution Stability

Stability in solutions is essential for reliable and reproducible results. This exercise demonstrates that regardless of the pivoting strategy used, the final solutions for x1 and x2 were consistent across different methods:

• Without Pivoting: x1 = 0.00, x2 = 0.50
• Column Pivoting: x1 = 0.00, x2 = 0.50
• Rest Matrix Pivoting: x1 = 0.00, x2 = 0.50

The primary reason for this stability lies in the minimal rounding errors due to significant pivots like 200. However, for larger and more complex systems, choosing an appropriate pivoting strategy can greatly influence the numerical stability of solutions. This critical aspect ensures that accurate and stable results are achieved even under varied numerical conditions.