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Chapter 3: Problem 3

Solve the system \(A \mathbf{x}=\mathbf{b}\) using the given LU factorization of \(A\). $$\begin{aligned}A=\left[\begin{array}{rrr}2 & 1 & -2 \\\\-2 & 3 & -4 \\\4 & -3 & 0\end{array}\right]=\left[\begin{array}{rrr}1 & 0 & 0 \\ -1 & 1 & 0 \\\2 & -\frac{5}{4} & 1\end{array}\right] \\\\\times\left[\begin{array}{rrr}2 & 1 & -2 \\\0 & 4 & -6 \\\0 & 0 & -\frac{7}{2} \end{array}\right], \mathbf{b}=\left[\begin{array}{r}-3 \\\1 \\\0\end{array}\right]\end{aligned}$$

Short Answer

Expert verified
The solution is \( \begin{bmatrix} \frac{-39}{14} \\ \frac{-26}{7} \\ \frac{-22}{7} \end{bmatrix} \).

Step by step solution

01

Understand the problem

We need to solve the system of linear equations \( A \mathbf{x} = \mathbf{b} \) where \( A \) is already factored into \( LU \). We'll use the fact that \( LU \mathbf{x} = \mathbf{b} \). The steps involve solving two simpler systems: \( L\mathbf{y} = \mathbf{b} \) and then \( U \mathbf{x} = \mathbf{y} \).
02

Solve for \(\mathbf{y}\) in \(L\mathbf{y} = \mathbf{b}\)

The matrix \( L \) is lower triangular:\[L = \begin{bmatrix} 1 & 0 & 0 \ -1 & 1 & 0 \ 2 & -\frac{5}{4} & 1 \end{bmatrix} \]We need to solve:\[\begin{bmatrix} 1 & 0 & 0 \ -1 & 1 & 0 \ 2 & -\frac{5}{4} & 1 \end{bmatrix} \begin{bmatrix} y_1 \ y_2 \ y_3 \end{bmatrix} = \begin{bmatrix} -3 \ 1 \ 0 \end{bmatrix}\]By forward substitution, starting from the top row:- From the first row: \( y_1 = -3 \).- From the second row: \( -y_1 + y_2 = 1 \), so \( 1 + 3 = y_2 \Rightarrow y_2 = 4.\)- From the third row: \( 2y_1 - \frac{5}{4}y_2 + y_3 = 0 \), substitute \( y_1 \) and \( y_2 \):\[ 2(-3) - \frac{5}{4}(4) + y_3 = 0 \]\[ -6 - 5 + y_3 = 0 \]\[ y_3 = 11 \].
03

Solve for \(\mathbf{x}\) in \(U\mathbf{x} = \mathbf{y}\)

The matrix \( U \) is upper triangular:\[U = \begin{bmatrix} 2 & 1 & -2 \ 0 & 4 & -6 \ 0 & 0 & -\frac{7}{2} \end{bmatrix} \]We need to solve:\[\begin{bmatrix} 2 & 1 & -2 \ 0 & 4 & -6 \ 0 & 0 & -\frac{7}{2} \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = \begin{bmatrix} -3 \ 4 \ 11 \end{bmatrix}\]By back substitution, starting from the bottom row:- From the third row: \( -\frac{7}{2} x_3 = 11 \), solving gives \( x_3 = -\frac{22}{7} \).- Second row: \( 4x_2 - 6x_3 = 4 \), substituting \( x_3 \):\[ 4x_2 - 6(-\frac{22}{7}) = 4 \]\[ 4x_2 + \frac{132}{7} = 4 \]\[ 4x_2 = 4 - \frac{132}{7} \]\[ 4x_2 = \frac{28}{7} - \frac{132}{7} = \frac{-104}{7} \]\[ x_2 = \frac{-26}{7} \].- First row: \( 2x_1 + x_2 - 2x_3 = -3 \), substituting \( x_2 \) and \( x_3 \):\[ 2x_1 + \left(-\frac{26}{7}\right) - 2\left(-\frac{22}{7}\right) = -3 \]\[ 2x_1 - \frac{26}{7} + \frac{44}{7} = -3 \]\[ 2x_1 + \frac{18}{7} = -3 \]\[ 2x_1 = -3 - \frac{18}{7} \]\[ 2x_1 = \frac{-21}{7} - \frac{18}{7} = \frac{-39}{7} \]\[ x_1 = \frac{-39}{14} \].
04

Conclusion: Solution vector \(\mathbf{x}\)

Thus, the solution is the vector \( \mathbf{x} = \begin{bmatrix} \frac{-39}{14} \ \frac{-26}{7} \ \frac{-22}{7} \end{bmatrix} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Forward Substitution
Forward substitution is a method used to solve a set of equations, particularly those arranged in a lower triangular matrix form. This technique is useful when you're working with matrices as part of LU factorization. In this approach, you solve for the unknowns from the top row to the bottom row of the matrix.

Here's how forward substitution works in simple terms:
  • Start with the first equation (first row) and solve for the first unknown.
  • Move down to the next row, plug in the known values (those you've already solved) into the equation and solve for the next unknown.
  • Continue this process until all unknowns are solved.
The beauty of forward substitution is that it's straightforward, as each equation depends only on the previously solved unknowns. It鈥檚 like solving a puzzle one piece at a time from top to bottom. This method is critical for effectively solving the system of equations, especially when utilizing LU factorization.
Back Substitution
Back substitution is another efficient technique employed to solve equations, but this time with an upper triangular matrix. It's often the step that follows forward substitution when dealing with LU factorization.

This method involves:
  • Starting from the last equation (bottom row) and solving for the last unknown.
  • Taking that solved value and substituting it into the equation above it to solve for the next unknown.
  • Repeating the process, moving upwards through the matrix until all unknowns are determined.
Back substitution works backwards through the rows, hence its name. Just like forward substitution, it turns a complex problem into a series of simpler ones, but in reverse order. It ensures that you can efficiently determine all the unknowns in your system of equations when working with an upper triangular matrix.
System of Linear Equations
A system of linear equations is a collection of two or more linear equations involving the same set of variables. These systems can be quite common in mathematical models and real-life applications, where you need to find the values of variables that satisfy all equations simultaneously.

Key features of a system of linear equations include:
  • Each equation represents a line in a multidimensional space.
  • The solution of the system is the point where all lines intersect, which satisfies all equations simultaneously.
  • The number of equations usually equals the number of unknowns in order to find a unique solution.
In the context of matrices, a system of linear equations is represented as \( A \mathbf{x} = \mathbf{b} \), where \( A \) is a matrix containing coefficients, \( \mathbf{x} \) is the vector of unknowns, and \( \mathbf{b} \) is the constants vector. Solving such systems, especially using techniques like LU factorization, is crucial in many fields, from engineering to economics.
Triangular Matrices
Triangular matrices are a special type of matrix, crucial in simplifying and solving systems of linear equations. These matrices come in two forms: lower triangular and upper triangular.

Features of triangular matrices include:
  • **Lower Triangular Matrix**: All entries above the diagonal are zero, making it easy to solve with forward substitution.
  • **Upper Triangular Matrix**: All entries below the diagonal are zero, suitable for solving using back substitution.
  • Both forms are pivotal in simplifying many matrix operations.
In LU factorization, you factor the matrix \( A \) into a product of a lower triangular matrix \( L \) and an upper triangular matrix \( U \). The triangular structure is what makes forward and back substitution possible, streamlining the process of finding solutions to systems of linear equations. Thus, understanding triangular matrices is integral to mastering these solving techniques.

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Most popular questions from this chapter

Give bases for \(\operatorname{row}(A), \operatorname{col}(A),\) and \(\operatorname{mull}(A)\). \(A=\left[\begin{array}{rrrr}1 & 1 & 0 & 1 \\ 0 & 1 & -1 & 1 \\ 0 & 1 & -1 & -1\end{array}\right]\)

Give a counterexample to show that the given transformation is not a linear transformation. $$T\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} y \\ x^{2} \end{array}\right]$$

Solve the system \(A \mathbf{x}=\mathbf{b}\) using the given factorization \(A=P^{T} L U .\) Because \(P P^{T}=I, P^{T} L U x=\mathbf{b}\) can be rewritten as \(L U x=P\) b. This system can then be solved using the method of Example 3.34. $$\begin{array}{l}A=\left[\begin{array}{lll}8 & 3 & 5 \\\4 & 1 & 2 \\\4 & 0 & 3\end{array}\right]=\left[\begin{array}{lll}0 & 1 & 0 \\\0 & 0 & 1 \\\1 & 0 & 0\end{array}\right]\left[\begin{array}{rrr}1 & 0 & 0 \\\1 & 1 & 0 \\\2 & -1 & 1\end{array}\right] \\\\\times\left[\begin{array}{rrr} 4 & 1 & 2 \\\0 & -1 & 1 \\\0 & 0 & 2\end{array}\right]=P^{T} L U, \mathbf{b}=\left[\begin{array}{r}16 \\\\-4 \\\4\end{array}\right]\end{array}$$

Find the standard matrix of the given linear transformation from \(\mathbb{R}^{2}\) to \(\mathbb{R}^{2}\) Reflection in the line \(y=x\)

A population with four age classes has a Leslie matrix \(L-\left[\begin{array}{cccc}0 & 1 & 2 & 5 \\ 0.5 & 0 & 0 & 0 \\ 0 & 0.7 & 0 & 0 \\ 0 & 0 & 0.3 & 0\end{array}\right] .\) If the initial population vector is \(\mathbf{x}_{0}=\left[\begin{array}{l}10 \\ 10 \\ 10 \\\ 10\end{array}\right],\) compute \(\mathbf{x}_{1}, \mathbf{x}_{2},\) and \(\mathbf{x}_{3}\)
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