/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Find the LU factorization of (a)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 39

Find the LU factorization of (a) \(A=\left[\begin{array}{rrr}1 & -3 & 5 \\ 2 & -4 & 7 \\ -1 & -2 & 1\end{array}\right]\), (b) \(\quad B=\left[\begin{array}{rrr}1 & 4 & -3 \\ 2 & 8 & 1 \\ -5 & -9 & 7\end{array}\right]\) (a) Reduce \(A\) to triangular form by the following operations: "Replace \(R_{2}\) by \(-2 R_{1}+R_{2}, " \quad\) "Replace \(R_{3}\) by \(R_{1}+R_{3}, " \quad\) and then "Replace \(R_{3}\) by \(\frac{5}{2} R_{2}+R_{3} "\) These operations yield the following, where the triangular form is \(U\) : $$ A \sim\left[\begin{array}{rrr} 1 & -3 & 5 \\ 0 & 2 & -3 \\ 0 & -5 & 6 \end{array}\right] \sim\left[\begin{array}{rrr} 1 & -3 & 5 \\ 0 & 2 & -3 \\ 0 & 0 & -\frac{3}{2} \end{array}\right]=U \quad \text { and } \quad L=\left[\begin{array}{rrr} 1 & 0 & 0 \\ 2 & 1 & 0 \\ -1 & -\frac{5}{2} & 1 \end{array}\right] $$ The entries \(2,-1,-\frac{5}{2}\) in \(L\) are the negatives of the multipliers \(-2,1, \frac{5}{2}\) in the above row operations. (As a check, multiply \(L\) and \(U\) to verify \(A=L U .\) ) (b) Reduce \(B\) to triangular form by first applying the operations "Replace \(R_{2}\) by \(-2 R_{1}+R_{2} "\) and "Replace \(R_{3}\) by \(5 R_{1}+R_{3} . "\) These operations yield $$ B \sim\left[\begin{array}{rrr} 1 & 4 & -3 \\ 0 & 0 & 7 \\ 0 & 11 & -8 \end{array}\right] $$ Observe that the second diagonal entry is 0 . Thus, \(B\) cannot be brought into triangular form without row interchange operations. Accordingly, \(B\) is not \(L U\)-factorable. (There does exist a \(P L U\) factorization of such a matrix \(B\), where \(P\) is a permutation matrix, but such a factorization lies beyond the scope of this text.)

Short Answer

Expert verified
The LU factorizations of the given matrices are: (a) For matrix A, we have \(U = \left[\begin{array}{rrr}1 & -3 & 5 \\0 & 2 & -3 \\0 & 0 & -\frac{3}{2}\end{array}\right]\) and \(L = \left[\begin{array}{rrr}1 & 0 & 0 \\2 & 1 & 0 \\-1 & -\frac{5}{2} & 1\end{array}\right]\). (b) For matrix B, it is not LU-factorable since it cannot be brought into upper triangular form without row interchange operations.

Step by step solution

01

Analyze matrix A

Determine if matrix A can be brought to upper triangular form using row operations without row interchange operations. Matrix A is: \[ A=\left[\begin{array}{rrr} 1 & -3 & 5 \\ 2 & -4 & 7 \\ -1 & -2 & 1 \end{array}\right] \]
02

Apply row operations to find U

Perform the following row operations: Replace \(R_{2}\) by \(-2 R_{1}+R_{2}\), Replace \(R_{3}\) by \(R_{1}+R_{3}\), Replace \(R_{3}\) by \(\frac{5}{2} R_{2}+R_{3}\) The result is: \[ U = \left[\begin{array}{rrr} 1 & -3 & 5 \\ 0 & 2 & -3 \\ 0 & 0 & -\frac{3}{2} \end{array}\right] \]
03

Find matrix L

Determine the matrix L using the multipliers from the row operations performed on matrix A. L will have the same dimensions as A. The entries in L are the negatives of the multipliers used in the row operations. Matrix L is: \[ L = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 2 & 1 & 0 \\ -1 & -\frac{5}{2} & 1 \end{array}\right] \] Check the result by multiplying L and U to verify that \(A = LU\). (b)
04

Analyze matrix B

Determine if matrix B can be brought to upper triangular form using row operations without row interchange operations. Matrix B is: \[ B=\left[\begin{array}{rrr} 1 & 4 & -3 \\ 2 & 8 & 1 \\ -5 & -9 & 7 \end{array}\right] \]
05

Apply row operations to find U

Perform the following row operations: Replace \(R_{2}\) by \(-2 R_{1}+R_{2}\), Replace \(R_{3}\) by \(5 R_{1}+R_{3}\) The result is: \[ U = \left[\begin{array}{rrr} 1 & 4 & -3 \\ 0 & 0 & 7 \\ 0 & 11 & -8 \end{array}\right] \] Observe that the second diagonal entry is 0. We conclude that matrix B cannot be brought into triangular form without row interchange operations and therefore is not LU-factorable.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Matrix Row Operations
Understanding matrix row operations is crucial to grasping the process of LU Factorization. These operations involve manipulating the rows of a matrix to simplify it or put it in a more useful form. Such transformations are key in solving systems of linear equations, a fundamental concept in linear algebra.

When we talk about matrix row operations, we refer to three basic actions. First, we can swap two rows, giving us the flexibility to rearrange the equations in a system. Second, we can multiply a row by a non-zero scalar, which essentially scales the equation up or down without changing its solution. Third, we can add or subtract a scalar multiple of one row from another, allowing us to eliminate variables systematically.

These row operations are used to achieve an upper triangular matrix in LU Factorization, which is essential in finding the unique L (lower triangular) and U (upper triangular) matrices. They conform to the same elementary principles followed when solving linear equations; thus, ensuring every step is reversible and the original matrix's properties are preserved.
Upper Triangular Matrix
In linear algebra, an upper triangular matrix plays a pivotal role, especially when it comes to matrix decomposition techniques like LU Factorization. An upper triangular matrix is a square matrix where all the elements below the main diagonal are zero.

In the context of LU Factorization, after applying a series of matrix row operations, we aim to transform the original matrix into an upper triangular form, designated as the matrix U. This helps simplify complex matrix equations. The upper triangular matrix is important because it enables straightforward backward substitution to solve for unknown variables.

For instance, if we consider the matrix A from the exercise, after applying the respective row operations, we obtain an upper triangular matrix U. This U matrix contains all the information needed to solve the system using backward substitution, where we start from the bottom-right element and work upward. However, it is crucial to ensure that no zeros appear on the main diagonal during the factorization process, as those would imply that the system may not have a unique solution or the matrix is not LU-factorizable without further modifications.
Linear Algebra
Linear algebra is an extensive field that deals with vectors, vector spaces, linear mappings, and systems of linear equations. At its core, it is about understanding spaces and how different transformations affect them.

One of the fundamental tools of linear algebra is the concept of matrix operations, which includes addition, subtraction, multiplication, and inversion of matrices. These operations help in representing and solving linear equations efficiently. LU Factorization falls under this umbrella, being a method that decomposes a matrix into a product of a lower and an upper triangular matrix. This decomposition is particularly useful in solving systems of linear equations, determining the invertibility of a matrix, and facilitating matrix inversion and determinant calculation.

While the LU Factorization can appear complex, it's essentially a structured way to do what we naturally do when solving linear equations: simplify and then solve. By breaking down matrices into lower and upper components, linear algebra allows us to understand the structure and solution space of linear systems more deeply.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the following systems in unknowns \(x\) and \(y:\) (a) \(a x+b y=1\) \(c x+d y=0\) (b) \(a x+b y=0\) \(c x+d y=1\)

Find the dimension and a basis for the general solution \(W\) of the following homogeneous system using matrix notation: $$ \begin{array}{r} x_{1}+2 x_{2}+3 x_{3}-2 x_{4}+4 x_{5}=0 \\ 2 x_{1}+4 x_{2}+8 x_{3}+x_{4}+9 x_{5}=0 \\ 3 x_{1}+6 x_{2}+13 x_{3}+4 x_{4}+14 x_{5}=0 \end{array} $$ Show how the basis gives the parametric form of the general solution of the system. When a system is homogeneous, we represent the system by its coefficient matrix \(A\) rather than by its augmented matrix \(M\), because the last column of the augmented matrix \(M\) is a zero column, and it will remain a zero column during any row-reduction process. Reduce the coefficient matrix \(A\) to echelon form, obtaining $$ A=\left[\begin{array}{rrrrr} 1 & 2 & 3 & -2 & 4 \\ 2 & 4 & 8 & 1 & 9 \\ 3 & 6 & 13 & 4 & 14 \end{array}\right] \sim\left[\begin{array}{rrrrr} 1 & 2 & 3 & -2 & 4 \\ 0 & 0 & 2 & 5 & 1 \\ 0 & 0 & 4 & 10 & 2 \end{array}\right] \sim\left[\begin{array}{rrrrr} 1 & 2 & 3 & -2 & 4 \\ 0 & 0 & 2 & 5 & 1 \end{array}\right] $$ (The third row of the second matrix is deleted, because it is a multiple of the second row and will result in a zero row.) We can now proceed in one of two ways. (a) Write down the corresponding homogeneous system in echelon form: $$ \begin{array}{r} x_{1}+2 x_{2}+3 x_{3}-2 x_{4}+4 x_{5}=0 \\ 2 x_{3}+5 x_{4}+x_{5}=0 \end{array} $$ The system in echelon form has three free variables, \(x_{2}, x_{4}, x_{5}\), so \(\operatorname{dim} W=3\). A basis \(\left[u_{1}, u_{2}, u_{3}\right]\) for \(W\) may be obtained as follows: (1) Set \(x_{2}=1, x_{4}=0, x_{5}=0\). Back-substitution yields \(x_{3}=0\), and then \(x_{1}=-2\). Thus, \(u_{1}=(-2,1,0,0,0)\) (2) Set \(x_{2}=0, x_{4}=1, x_{5}=0\). Back-substitution yields \(x_{3}=-\frac{5}{2}\), and then \(x_{1}=\frac{19}{2}\). Thus, \(u_{2}=\left(\frac{19}{2}, 0,-\frac{5}{2}, 1,0\right)\) (3) Set \(x_{2}=0, x_{4}=0, x_{5}=1\). Back-substitution yields \(x_{3}=-\frac{1}{2}\), and then \(x_{1}=-\frac{5}{2}\). Thus, \(u_{3}=\left(-\frac{5}{2}, 0,-\frac{1}{2}, 0,1\right)\) [One could avoid fractions in the basis by choosing \(x_{4}=2\) in (2) and \(x_{5}=2\) in (3), which yields multiples of \(u_{2}\) and \(u_{3}\).] The parametric form of the general solution is obtained from the following linear combination of the basis vectors using parameters \(a, b, c\) : $$ a u_{1}+b u_{2}+c u_{3}=\left(-2 a+\frac{19}{2} b-\frac{5}{2} c, a,-\frac{5}{2} b-\frac{1}{2} c, b, c\right) $$ (b) Reduce the echelon form of \(A\) to row canonical form: $$ A \sim\left[\begin{array}{ccccc} 1 & 2 & 3 & -2 & 4 \\ 0 & 0 & 1 & \frac{5}{2} & \frac{1}{2} \end{array}\right] \sim\left[\begin{array}{rrrrr} 1 & 2 & 3 & -\frac{19}{2} & \frac{5}{2} \\ 0 & 0 & 1 & \frac{5}{2} & \frac{1}{2} \end{array}\right] $$ Write down the corresponding free-variable solution: $$ \begin{aligned} &x_{1}=-2 x_{2}+\frac{19}{2} x_{4}-\frac{5}{2} x_{5} \\ &x_{3}=-\frac{5}{2} x_{4}-\frac{1}{2} x_{5} \end{aligned} $$ Using these equations for the pivot variables \(x_{1}\) and \(x_{3}\), repeat the above process to obtain a basis \(\left[u_{1}, u_{2}, u_{3}\right]\) for \(W\). That is, set \(x_{2}=1, x_{4}=0, x_{5}=0\) to get \(u_{1} ;\) set \(x_{2}=0, x_{4}=1, x_{5}=0\) to get \(u_{2}\); and set \(x_{2}=0\), \(x_{4}=0, x_{5}=1\) to get \(u_{3}\).

Express each of the following matrices as a product of elementary matrices: $$ A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right], \quad B=\left[\begin{array}{rr} 3 & -6 \\ -2 & 4 \end{array}\right], \quad C=\left[\begin{array}{rr} 2 & 6 \\ -3 & -7 \end{array}\right], \quad D=\left[\begin{array}{lll} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 3 & 8 & 7 \end{array}\right] $$

Describe the pivoting row-reduction algorithm. Also describe the advantages, if any, of using this pivoting algorithm.

Write each of the following matrices as a product of elementary matrices: (a) \(A=\left[\begin{array}{rr}1 & -3 \\ -2 & 4\end{array}\right]\) (b) \(B=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1\end{array}\right]\) (c) \(C=\left[\begin{array}{rrr}1 & 1 & 2 \\ 2 & 3 & 8 \\ -3 & -1 & 2\end{array}\right]\) The following three steps write a matrix \(M\) as a product of elementary matrices: Step 1. Row reduce \(M\) to the identity matrix \(I\), keeping track of the elementary row operations. Step 2. Write down the inverse row operations. Step 3. Write \(M\) as the product of the elementary matrices corresponding to the inverse operations. This gives the desired result. If a zero row appears in Step 1 , then \(M\) is not row equivalent to the identity matrix \(I\), and \(M\) cannot be written as a product of elementary matrices. (a) (1) We have $$ A=\left[\begin{array}{rr} 1 & -3 \\ -2 & 4 \end{array}\right] \sim\left[\begin{array}{ll} 1 & -3 \\ 0 & -2 \end{array}\right] \sim\left[\begin{array}{rr} 1 & -3 \\ 0 & 1 \end{array}\right] \sim\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=I $$ where the row operations are, respectively, "Replace \(R_{2}\) by \(2 R_{1}+R_{2}, " \quad\) "Replace \(R_{2}\) by \(-\frac{1}{2} R_{2}, " \quad\) "Replace \(R_{1}\) by \(3 R_{2}+R_{1} "\) (2) Inverse operations: "Replace \(R_{2}\) by \(-2 R_{1}+R_{2}, " \quad\) "Replace \(R_{2}\) by \(-2 R_{2}, " \quad\) "Replace \(R_{1}\) by \(-3 R_{2}+R_{1} "\) (3) \(A=\left[\begin{array}{rr}1 & 0 \\ -2 & 1\end{array}\right]\left[\begin{array}{rr}1 & 0 \\ 0 & -2\end{array}\right]\left[\begin{array}{rr}1 & -3 \\ 0 & 1\end{array}\right]\) (b) (1) We have $$ B=\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{array}\right] \sim\left[\begin{array}{lll} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \sim\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=I $$ where the row operations are, respectively, "Replace \(R_{2}\) by \(-4 R_{3}+R_{2}, " \quad\) "Replace \(R_{1}\) by \(-3 R_{3}+R_{1}, " \quad\) "Replace \(R_{1}\) by \(-2 R_{2}+R_{1} "\) (2) Inverse operations: "Replace \(R_{2}\) by \(4 R_{3}+R_{2}, " \quad\) "Replace \(R_{1}\) by \(3 R_{3}+R_{1}, " \quad\) "Replace \(R_{1}\) by \(2 R_{2}+R_{1} "\) (3) \(B=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 4 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\) (c) (1) First row reduce \(C\) to echelon form. We have $$ C=\left[\begin{array}{rrr} 1 & 1 & 2 \\ 2 & 3 & 8 \\ -3 & -1 & 2 \end{array}\right] \sim\left[\begin{array}{lll} 1 & 1 & 2 \\ 0 & 1 & 4 \\ 0 & 2 & 8 \end{array}\right] \sim\left[\begin{array}{lll} 1 & 1 & 2 \\ 0 & 1 & 4 \\ 0 & 0 & 0 \end{array}\right] $$ In echelon form, \(C\) has a zero row. "STOP." The matrix \(C\) cannot be row reduced to the identity matrix \(I\), and \(C\) cannot be written as a product of elementary matrices. (We note, in particular, that \(C\) has no inverse.)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.