Chapter 8 Linear Algebraic Equations

Similar documents
Chapter 8: Linear Algebraic Equations

Solutions of Linear system, vector and matrix equation

Lecture 2 Systems of Linear Equations and Matrices, Continued

System of Linear Equations. Slide for MA1203 Business Mathematics II Week 1 & 2

Row Reduction and Echelon Forms

DM559 Linear and Integer Programming. Lecture 2 Systems of Linear Equations. Marco Chiarandini

Number of solutions of a system

Lecture 1 Systems of Linear Equations and Matrices

Find the solution set of 2x 3y = 5. Answer: We solve for x = (5 + 3y)/2. Hence the solution space consists of all vectors of the form

Notes on Row Reduction

Section Gaussian Elimination

MATH.2720 Introduction to Programming with MATLAB Vector and Matrix Algebra

12/1/2015 LINEAR ALGEBRA PRE-MID ASSIGNMENT ASSIGNED BY: PROF. SULEMAN SUBMITTED BY: M. REHAN ASGHAR BSSE 4 ROLL NO: 15126

Linear Algebra. Solving SLEs with Matlab. Matrix Inversion. Solving SLE s by Matlab - Inverse. Solving Simultaneous Linear Equations in MATLAB

Identity Matrix: EDU> eye(3) ans = Matrix of Ones: EDU> ones(2,3) ans =

MTH 102A - Linear Algebra II Semester

Chapter 4. Solving Systems of Equations. Chapter 4

Chapter 5. Linear Algebra. Sections A linear (algebraic) equation in. unknowns, x 1, x 2,..., x n, is. an equation of the form

Lecture 4: Gaussian Elimination and Homogeneous Equations

Name: Section Registered In:

Rank and Nullity. MATH 322, Linear Algebra I. J. Robert Buchanan. Spring Department of Mathematics

Finite Mathematics Chapter 2. where a, b, c, d, h, and k are real numbers and neither a and b nor c and d are both zero.

Chapter 5. Linear Algebra. A linear (algebraic) equation in. unknowns, x 1, x 2,..., x n, is. an equation of the form

18.06 Problem Set 3 - Solutions Due Wednesday, 26 September 2007 at 4 pm in

Lecture 21: 5.6 Rank and Nullity

Numerical Analysis Fall. Gauss Elimination

Linear Algebra I Lecture 10

POLI270 - Linear Algebra

Reduction to the associated homogeneous system via a particular solution

Chapter 1 Linear Equations. 1.1 Systems of Linear Equations

Recall, we solved the system below in a previous section. Here, we learn another method. x + 4y = 14 5x + 3y = 2

Chapter 5. Linear Algebra. Sections A linear (algebraic) equation in. unknowns, x 1, x 2,..., x n, is. an equation of the form

Methods for Solving Linear Systems Part 2

MTH 2032 Semester II

+ MATRIX VARIABLES AND TWO DIMENSIONAL ARRAYS

Statistical methods. Mean value and standard deviations Standard statistical distributions Linear systems Matrix algebra

SECTION 3.3. PROBLEM 22. The null space of a matrix A is: N(A) = {X : AX = 0}. Here are the calculations of AX for X = a,b,c,d, and e. =

MTH Linear Algebra. Study Guide. Dr. Tony Yee Department of Mathematics and Information Technology The Hong Kong Institute of Education

Vector Spaces, Orthogonality, and Linear Least Squares

MAC Learning Objectives. Learning Objectives (Cont.) Module 10 System of Equations and Inequalities II

The Gauss-Jordan Elimination Algorithm

7.6 The Inverse of a Square Matrix

9.1 - Systems of Linear Equations: Two Variables

Lecture Notes in Mathematics. Arkansas Tech University Department of Mathematics. The Basics of Linear Algebra

Solutions to Math 51 Midterm 1 July 6, 2016

Systems of Linear Equations in Two Variables. Break Even. Example. 240x x This is when total cost equals total revenue.

18.06 Spring 2012 Problem Set 3

Lecture 7: Introduction to linear systems

Math 54 HW 4 solutions

The value of a problem is not so much coming up with the answer as in the ideas and attempted ideas it forces on the would be solver I.N.

Linear Equations in Linear Algebra

1. Solve each linear system using Gaussian elimination or Gauss-Jordan reduction. The augmented matrix of this linear system is

Math Week in Review #7

Math 314H EXAM I. 1. (28 points) The row reduced echelon form of the augmented matrix for the system. is the matrix

MATH10212 Linear Algebra B Homework Week 3. Be prepared to answer the following oral questions if asked in the supervision class

Linear Systems of n equations for n unknowns

Chapter 16 Numerical Linear Algebra

Handout 1 EXAMPLES OF SOLVING SYSTEMS OF LINEAR EQUATIONS

Matrices and systems of linear equations

1. (7pts) Find the points of intersection, if any, of the following planes. 3x + 9y + 6z = 3 2x 6y 4z = 2 x + 3y + 2z = 1

Linear Systems. Carlo Tomasi. June 12, r = rank(a) b range(a) n r solutions

(Linear equations) Applied Linear Algebra in Geoscience Using MATLAB

22A-2 SUMMER 2014 LECTURE 5

Math 1021, Linear Algebra 1. Section: A at 10am, B at 2:30pm

x 1 2x 2 +x 3 = 0 2x 2 8x 3 = 8 4x 1 +5x 2 +9x 3 = 9

MAC1105-College Algebra. Chapter 5-Systems of Equations & Matrices

Linear Equation: a 1 x 1 + a 2 x a n x n = b. x 1, x 2,..., x n : variables or unknowns

Linear Algebra Highlights

Elementary operation matrices: row addition

MAC Module 1 Systems of Linear Equations and Matrices I

x y + z = 3 2y z = 1 4x + y = 0

1 Last time: linear systems and row operations

Systems of Equations

Math 414 Lecture 1. Reading assignment: Text: Pages 1-29 and Scilab Users Guide: Sections 1-3.

3.4 Elementary Matrices and Matrix Inverse

(Mathematical Operations with Arrays) Applied Linear Algebra in Geoscience Using MATLAB

Solutions to Exam I MATH 304, section 6

MATH 3321 Sample Questions for Exam 3. 3y y, C = Perform the indicated operations, if possible: (a) AC (b) AB (c) B + AC (d) CBA

1 - Systems of Linear Equations

Chapter 9: Systems of Equations and Inequalities

Ack: 1. LD Garcia, MTH 199, Sam Houston State University 2. Linear Algebra and Its Applications - Gilbert Strang

Fundamentals of Linear Algebra. Marcel B. Finan Arkansas Tech University c All Rights Reserved

Midterm 1 Review. Written by Victoria Kala SH 6432u Office Hours: R 12:30 1:30 pm Last updated 10/10/2015

If A is a 4 6 matrix and B is a 6 3 matrix then the dimension of AB is A. 4 6 B. 6 6 C. 4 3 D. 3 4 E. Undefined

Introduction to Matrices

Linear Algebra. Chapter Linear Equations

System of Linear Equations

THE NULLSPACE OF A: SOLVING AX = 0 3.2

CITS2401 Computer Analysis & Visualisation

Chapter 9: Gaussian Elimination

Chapter 5. Linear Algebra. A linear (algebraic) equation in. unknowns, x 1, x 2,..., x n, is. an equation of the form

0.0.1 Section 1.2: Row Reduction and Echelon Forms Echelon form (or row echelon form): 1. All nonzero rows are above any rows of all zeros.

Chapter 2: Systems of Linear Equations and Matrices

Unit 4 Systems of Equations Systems of Two Linear Equations in Two Variables

Chapter 1. Vectors, Matrices, and Linear Spaces

MATH10212 Linear Algebra B Homework Week 4

Chapter 1: Systems of linear equations and matrices. Section 1.1: Introduction to systems of linear equations

Elementary Linear Algebra Review for Exam 2 Exam is Monday, November 16th.

PowerPoints organized by Dr. Michael R. Gustafson II, Duke University

MODULE 8 Topics: Null space, range, column space, row space and rank of a matrix

Transcription:

PowerPoint to accompany Introduction to MATLAB for Engineers, Third Edition William J. Palm III Chapter 8 Linear Algebraic Equations Copyright 2010. The McGraw-Hill Companies, Inc. This work is only for non-profit use by instructors in courses for which this textbook has been adopted. Any other use without publisher s consent is unlawful.

A singular problem refers to a set of equations having either no unique solution or no solution at all. For example, the set 3x - 4y = 5 6x - 8y = 10 is singular and has no unique solution because the second equation is identical to the first equation, multiplied by 2. The graphs of these two equations are identical. All we can say is that the solution must satisfy y = (3x - 5)/4, which describes an infinite number of solutions. 8-2

The equations 6x 10y = 2 3x 4y = 5 have graphs that intersect at the solution y = 4, x = 7. On the other hand, the set 3x - 4y = 5 6x - 8y = 3 8-3 is singular but has no solution. The graphs of these two equations are distinct but parallel (see the next slide). Because they do not intersect, no solution exists.

The graphs of two equations that intersect at a solution. 8-4

Parallel graphs indicate that no solution exists. 8-5

Consider the following set of homogeneous equations (which means that their right sides are all zero) 6x + ay = 0 2x + 4y = 0 where a is a parameter. Multiply the second equation by 3 and subtract the result from the first equation to obtain (a - 12)y = 0 The solution is y = 0 only if a 12; if a = 12, there is an infinite number of solutions for x and y, where x = -2y. 8-6

Matrix notation enables us to represent multiple equations as a single matrix equation. For example, consider the following set: 2x 1 + 9x 2 = 5 3x 1-4x 2 = 7 This set can be expressed in vector-matrix form as 2 9 3-4 x 1 x 2 = 5 7 which can be represented in the following compact form 8-7 Ax = b (see pages 332-333)

For the equation set Ax = b, if A = 0, then there is no unique solution. Depending on the values in the vector b, there may be no solution at all, or an infinite number of solutions. 8-8

The MATLAB command inv(a) computes the inverse of the matrix A. The following MATLAB session solves the following equations using MATLAB. 2x + 9y = 5 3x - 4y = 7 >>A = [2,9;3,-4];b = [5;7] >>x = inv(a)*b x = 2.3714 0.0286 If you attempt to solve a singular problem using the inv command, MATLAB displays an error message. 8-9

8-10 Existence and uniqueness of solutions (pages 334-335) The set Ax = b with m equations and n unknowns has solutions if and only if Let r = rank[a]. rank[a] = rank[a b] (1) If condition (1) is satisfied and if r = n, then the solution is unique. If condition (1) is satisfied but r < n, an infinite number of solutions exists and r unknown variables can be expressed as linear combinations of the other n - r unknown variables, whose values are arbitrary.

Homogeneous case. The homogeneous set Ax = 0 is a special case in which b = 0. For this case rank[a] = rank[a b] always, and thus the set always has the trivial solution x = 0. A nonzero solution, in which at least one unknown is nonzero, exists if and only if rank[a] < n. If m < n, the homogeneous set always has a nonzero solution. 8-11

If the number of equations equals the number of unknowns and if A 0, then the equation set has a solution and it is unique. If A = 0 or if the number of equations does not equal the number of unknowns, then you must use the methods presented in Sections 8.3 or 8.4. 8-12

MATLAB provides the left-division method for solving the equation set Ax = b. The left-division method is based on Gauss elimination. (Section 8.2 on page 335) To use the left-division method to solve for x, type x = A\b. For example, >> A = [6, -10; 3, -4]; b = [2; 5]; >> x = A\b x = 7 4 This method also works in some cases where the number of unknowns does not equal the number of equations. 8-13 For more examples, see pages 335-341.

Application of linear equations: Calculating cable tension for a suspended mass. Example 8.2-2. Figure 8.2-1, page 337. 8-14

Application of linear equations: An electrical-resistance network. Example 8.2-3. Figure 8.2-2 on page 338 8-15

Underdetermined Systems (Section 8.3 on page 341) An underdetermined system does not contain enough information to solve for all of the unknown variables, usually because it has fewer equations than unknowns. Thus an infinite number of solutions can exist, with one or more of the unknowns dependent on the remaining unknowns. 8-16

A simple example of an underdetermined systems is x + 3y = 6 All we can do is solve for one of the unknowns in terms of the other; for example, x = 6-3y. An infinite number of solutions satisfy this equation. 8-17 When there are more equations than unknowns, the left-division method will give a solution with some of the unknowns set equal to zero. For example, >>A = [1, 3]; b = 6; >>solution = A\b solution = 0 2 which corresponds to x = 0 and y = 2.

An infinite number of solutions might exist even when the number of equations equals the number of unknowns. This situation can occur when A = 0. For such systems the matrix inverse method and Cramer s method will not work, and the left-division method generates an error message warning us that the matrix A is singular. In such cases the pseudoinverse method x = pinv(a)*b 8-18 gives one solution, the minimum norm solution.

A Statically indeterminate problem. A light fixture and its free-body diagram. Example 8.3-2. Figure 8.3-1, page 344. 8-19

In cases that have an infinite number of solutions, some of the unknowns can be expressed in terms of the remaining unknowns, whose values are arbitrary. We can use the rref command to find these relations. See slide 8-24. 8-20

Recall that if A = 0, the equation set is singular. If you try to solve a singular set using MATLAB, it prints a message warning that the matrix is singular and does not try to solve the problem. 8-21

An ill-conditioned set of equations is a set that is close to being singular. The ill-conditioned status depends on the accuracy with which the solution calculations are made. When the internal numerical accuracy used by MATLAB is insufficient to obtain a solution, MATLAB prints a message to warn you that the matrix is close to singular and that the results might be inaccurate. 8-22

The pinv command can obtain a solution of an underdetermined set. To solve the equation set Ax = b using the pinv command, type x = pinv(a)*b Underdetermined sets have an infinite number of solutions, and the pinv command produces a solution that gives the minimum value of the Euclidean norm, which is the magnitude of the solution vector x. 8-23 More? See pages 343-345.

The rref function. We can always express some of the unknowns in an underdetermined set as functions of the remaining unknowns. We can obtain such a form by multiplying the set s equations by suitable factors and adding the resulting equations to eliminate an unknown variable. The MATLAB rref function provides a procedure to reduce an equation set to this form, which is called the reduced row echelon form. The syntax is rref([a b]). The output is the augmented matrix [C d] that corresponds to the equation set Cx = d. This set is in reduced row echelon form. 8-24 More? See pages 345-350.

Overdetermined Systems (Section 8.4 on page 350) An overdetermined system is a set of equations that has more independent equations than unknowns. For such a system the matrix inverse method will not work because the A matrix is not square. However, some overdetermined systems have exact solutions, and they can be obtained with the left division method x = A\b. 8-25

For other overdetermined systems, no exact solution exists. In some of these cases, the left-division method does not yield an answer, while in other cases the left-division method gives an answer that satisfies the equation set only in a least squares sense, as explained in Example 8.4 1 on pages 351-353. When MATLAB gives an answer to an overdetermined set, it does not tell us whether the answer is the exact solution. 8-26

Some overdetermined systems have an exact solution. The left-division method sometimes gives an answer for overdetermined systems, but it does not indicate whether the answer is the exact solution. We need to check the ranks of A and [A b] to know whether the answer is the exact solution. 8-27

8-28 To interpret MATLAB answers correctly for an overdetermined system, first check the ranks of A and [A b] to see whether an exact solution exists; if one does not exist, then you know that the left-division answer is a least squares solution.

Example of an underdetermined system. A network of oneway streets. Example 8.3-5. Figure 8.3-2, page 349. 8-29

Solving Linear Equations: Summary If the number of equations in the set equals the number of unknown variables, the matrix A is square and MATLAB provides two ways of solving the equation set Ax = b: 1. The matrix inverse method; solve for x by typing x = inv(a)*b. 2. The matrix left-division method; solve for x by typing x = A\b. 8-30 (continued )

Solving Linear Equations: Summary (continued) If A is square and if MATLAB does not generate an error message when you use one of these methods, then the set has a unique solution, which is given by the left-division method. You can always check the solution for x by typing A*x to see if the result is the same as b. 8-31 (continued )

Solving Linear Equations: Summary (continued) If you receive an error message, the set is underdetermined, and either it does not have a solution or it has more than one solution. In such a case, if you need more information, you must use the following procedures. 8-32 (continued )

8-33 Solving Linear Equations: Summary (continued) For underdetermined and overdetermined sets, MATLAB provides three ways of dealing with the equation set Ax = b. (Note that the matrix inverse method will never work with such sets.) 1. The matrix left-division method; solve for x by typing x = A\b. 2. The pseudoinverse method; solve for x by typing x = pinv(a)*b. 3. The reduced row echelon form (RREF) method. This method uses the MATLAB function rref to obtain a solution. (continued )

8-34 Solving Linear Equations: Summary (continued) Underdetermined Systems In an underdetermined system not enough information is given to determine the values of all the unknown variables. An infinite number of solutions might exist in which one or more of the unknowns are dependent on the remaining unknowns. For such systems the matrix inverse method will not work because either A is not square or because A = 0. (continued )

8-35 Solving Linear Equations: Summary (continued) The left-division method will give a solution with some of the unknowns arbitrarily set equal to zero, but this solution is not the general solution. An infinite number of solutions might exist even when the number of equations equals the number of unknowns. The left-division method fails to give a solution in such cases. In cases that have an infinite number of solutions, some of the unknowns can be expressed in terms of the remaining unknowns, whose values are arbitrary. The rref function can be used to find these relations. (continued )

Solving Linear Equations: Summary (continued) Overdetermined Systems An overdetermined system is a set of equations that has more independent equations than unknowns. For such a system Cramer s method and the matrix inverse method will not work because the A matrix is not square. Some overdetermined systems have exact solutions, which can be obtained with the left-division method A\b. (continued ) 8-36

Solving Linear Equations: Summary (continued) For overdetermined systems that have no exact solution, the answer given by the left-division method satisfies the equation set only in a least squares sense. When we use MATLAB to solve an overdetermined set, the program does not tell us whether the solution is exact. We must determine this information ourselves. The first step is to check the ranks of A and [A b] to see whether a solution exists; if no solution exists, then we know that the left-division solution is a least squares answer. 8-37 More? See pages 398-402.

Pseudocode for the linear equation solver. Table 8.5 1, page 354. 8-38 1. If the rank of A equals the rank of [A b], then determine whether the rank of A equals the number of unknowns. If so, there is a unique solution, which can be computed using left division. Display the results and stop. 2. Otherwise, there is an infinite number of solutions, which can be found from the augmented matrix. Display the results and stop. 3. Otherwise (if the rank of A does not equal the rank of [A b]), then there are no solutions. Display this message and stop.

Flowchart of the linear equation solver. Figure 8.5 1 8-39

MATLAB program to solve linear equations. Table 8.5-2 % Script file lineq.m % Solves the set Ax = b, given A and b. % Check the ranks of A and [A b]. if rank(a) == rank([a b]) % The ranks are equal. Size_A = size(a); % Does the rank of A equal the number of unknowns? if rank(a) == size_a(2) % Yes. Rank of A equals the number of unknowns. disp( There is a unique solution, which is: ) x = A\b % Solve using left division. 8-40 (continued )

Linear equation solver (continued) else % Rank of A does not equal the number of unknowns. disp( There is an infinite number of solutions. ) disp( The augmented matrix of the reduced system is: ) rref([a b]) % Compute the augmented matrix. end else % The ranks of A and [A b] are not equal. disp( There are no solutions. ) end 8-41

The following slides contain figures from the chapter s homework problems. 8-42

Figure P4, page 358 8-43

Figure P6, page 360 8-44

Figure P7, page 361 8-45

Figure P8, page 362 8-46

Figure P9, page 363 8-47

Figure P10, page 365 8-48

Figure P13, page 366 8-49