Problem 33 - Advanced Engineering Mathematics

Question

Use the given LU-factorization $$ \mathbf{A}=\left(\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array}\right)=\left(\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{array}\right)\left(\begin{array}{lll} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right) $$ to solve the linear system \(\mathbf{A X}=\mathbf{B}_{i}\) for the given column matrix \(\mathbf{B}_{i}, i=1,2,3,4\) $$ \mathbf{B}_{3}=\left(\begin{array}{r} \frac{1}{2} \\ \frac{3}{4} \\ -\frac{1}{2} \end{array}\right) $$

Step-by-Step Solution

Answer

\( \mathbf{X} = \begin{pmatrix} \frac{1}{4} \\ \frac{3}{2} \\ -\frac{5}{4} \end{pmatrix} \).‎

Step 1: Understanding LU-Factorization

The goal is to solve the system \( \mathbf{A} \mathbf{X} = \mathbf{B}_3 \) using the provided LU-factorization of \( \mathbf{A} \). The factorization is given by \( \mathbf{A} = \mathbf{L} \mathbf{U} \). \( \mathbf{L} \) is the lower triangular matrix, and \( \mathbf{U} \) is the upper triangular matrix. The solution involves two main steps: forward substitution to solve \( \mathbf{L} \mathbf{Y} = \mathbf{B}_3 \), and backward substitution to solve \( \mathbf{U} \mathbf{X} = \mathbf{Y} \).

Step 2: Forward Substitution

To find \( \mathbf{Y} \) such that \( \mathbf{L} \mathbf{Y} = \mathbf{B}_3 \), we set up the equations using \( \mathbf{L} = \begin{pmatrix} 1 & 0 & 0 \ 1 & 1 & 0 \ 1 & 1 & 1 \end{pmatrix} \) and solve them individually: 1. \( Y_1 = \frac{1}{2} \) 2. \( Y_2 + Y_1 = \frac{3}{4} \Rightarrow Y_2 = \frac{3}{4} - \frac{1}{2} = \frac{1}{4} \) 3. \( Y_3 + Y_2 + Y_1 = -\frac{1}{2} \Rightarrow Y_3 = -\frac{1}{2} - \frac{1}{4} - \frac{1}{2} = -\frac{5}{4} \) Thus, \( \mathbf{Y} = \begin{pmatrix} \frac{1}{2} \ \frac{1}{4} \ -\frac{5}{4} \end{pmatrix} \).

Step 3: Backward Substitution

Now solve \( \mathbf{U} \mathbf{X} = \mathbf{Y} \) with \( \mathbf{U} = \begin{pmatrix} 1 & 1 & 1 \ 0 & 1 & 1 \ 0 & 0 & 1 \end{pmatrix} \):1. \( X_3 = -\frac{5}{4} \) 2. \( X_2 + X_3 = \frac{1}{4} \Rightarrow X_2 = \frac{1}{4} + \frac{5}{4} = \frac{3}{2} \) 3. \( X_1 + X_2 + X_3 = \frac{1}{2} \Rightarrow X_1 = \frac{1}{2} - \frac{3}{2} - \left(-\frac{5}{4}\right) = \frac{1}{4} \) Thus, \( \mathbf{X} = \begin{pmatrix} \frac{1}{4} \ \frac{3}{2} \ -\frac{5}{4} \end{pmatrix} \).

Step 4: Verify the Solution

Double-check by multiplying \( \mathbf{A} \) with \( \mathbf{X} \) to see if it equals \( \mathbf{B}_3 \). Calculate:\[ \mathbf{A} \cdot \mathbf{X} = \begin{pmatrix} 1 & 1 & 1 \ 1 & 2 & 2 \ 1 & 2 & 3 \end{pmatrix} \begin{pmatrix} \frac{1}{4} \ \frac{3}{2} \ -\frac{5}{4} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} \ \frac{3}{4} \ -\frac{1}{2} \end{pmatrix} \] This matches \( \mathbf{B}_3 \), confirming the solution is correct.

Key Concepts

Linear Systems

Linear systems are collections of linear equations involving multiple variables. Each linear equation in the system represents a straight line in a multi-dimensional space. When we solve a linear system, we're looking for the point(s) where these lines intersect. This solution, if it exists, is expressed as a set of values for the variables that satisfy all the equations simultaneously.

Consider the matrix equation \( \mathbf{A} \mathbf{X} = \mathbf{B} \). Here, \( \mathbf{A} \) is the coefficient matrix, \( \mathbf{X} \) is a column vector of variables, and \( \mathbf{B} \) is the column vector of constants. Solving a linear system involves finding the vector \( \mathbf{X} \) that satisfies the equation. Linear systems are fundamental in mathematics and engineering, modeling everything from electrical circuits to economic systems.

Forward Substitution

Forward substitution is a method used to solve linear equations when the coefficient matrix is a lower triangular matrix. In our LU factorization process, we utilized forward substitution to solve \( \mathbf{L} \mathbf{Y} = \mathbf{B}_3 \), where \( \mathbf{L} \) is lower triangular.

To solve for \( \mathbf{Y} \), start from the first equation and work your way down. In a lower triangular matrix, each equation only involves the current and previously solved elements of \( \mathbf{Y} \). This makes forward substitution a straightforward process:

• Solve the first equation for the first variable directly since it doesn't have any previous variables.
• Plug this value into the next equation to solve for the next variable.
• Continue this process sequentially until you solve all variables in \( \mathbf{Y} \).

By the end, you have a vector \( \mathbf{Y} \) that acts as an intermediary solution, crucial for backward substitution.

Backward Substitution

Backward substitution is the counterpart to forward substitution when dealing with upper triangular matrices. After obtaining \( \mathbf{Y} \) through forward substitution, we apply backward substitution to solve \( \mathbf{U} \mathbf{X} = \mathbf{Y} \), where \( \mathbf{U} \) is upper triangular.

In backward substitution, you start from the last equation and work upwards. For each step:

• Solve for the bottom-most variable (the simplest equation with one variable).
• Use this variable to solve the preceding equation, which includes it and one other variable.
• Continue using known values to solve the previous equations until all variables in \( \mathbf{X} \) are found.

This approach is efficient because each equation has decreasing complexity, allowing us to eliminate known values step-by-step.

Triangular Matrices

Triangular matrices are a type of matrix where all the entries above or below the main diagonal are zero. These matrices are particularly easy to work with in solving linear systems.

There are two types of triangular matrices:

• Lower Triangular Matrix: All entries above the main diagonal are zero. It's used in forward substitution. In our problem, the matrix \( \mathbf{L} \) is a lower triangular matrix.
• Upper Triangular Matrix: All entries below the main diagonal are zero. It's used in backward substitution. In the given exercise, \( \mathbf{U} \) serves as the upper triangular matrix.

Triangular matrices make the process of solving linear systems more efficient since they simplify the substitution process, either forward or backward. They are central in methods like LU decomposition, which is used to decompose matrices into triangular matrices, enabling straightforward solutions to otherwise complex systems.