Mathematics
Solving Linear Systems
Solving linear systems involves finding the values of variables that satisfy multiple linear equations simultaneously. This can be done using methods such as substitution, elimination, or matrix operations. The goal is to determine the unique solution, no solution, or infinitely many solutions for the system of equations.
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10 Key excerpts on "Solving Linear Systems"
eBook - PDFFinite Mathematics
An Applied Approach
- Michael Sullivan(Author)
- 2011(Publication Date)
- Wiley(Publisher)
Notice that each equation in the system has been numbered for easy reference. A solution of a system of linear equations consists of values of the variables that are solutions of each equation of the system. To solve a system of linear equations means to find all solutions of the system. For example, is a solution of the system in Example 2(a) because Furthermore, x = 2, y = 1 is the only solution of this system. Do you know why? The lines represented by the equations have different slopes and so they intersect in a single point. (1) (2) b 2(2) + 1 = 4 + 1 = 5 - 4(2) + 6(1) = - 8 + 6 = - 2 (1) (2) b 2x + y = 5 - 4x + 6y = - 2 x = 2, y = 1 (1) (2) Two equations containing four variables, x 1 , x 2 , x 3 , and x 4 b x 1 - 2x 2 + x 3 - x 4 = 5 3x 1 + x 2 - x 3 - 5x 4 = 2 (1) (2) (3) (4) (5) Five equations containing three variables, x, y, and z e x + y + z = 6 2x + z = 4 y + z = 2 x = 4 y = 2 (1) (2) Two equations containing three variables, x, y, and z b x + y + z = 6 x - y = 2 (1) (2) (3) Three equations containing three variables, x, y, and z c x + y + z = 6 3x - 2y + 4z = 9 x - y - z = 0 (1) (2) Two equations containing two variables, x and y b 2x + y = 5 - 4x + 6y = - 2 x + y = 525 8x + 6y = 3580 x + y = 525 ▲ 52 Chapter 2 Systems of Linear Equations A solution of the system in Example 2(b) is because Note that is not a solution of the system in Example 2(b). Although these values satisfy Equations (1) and (3), they do not satisfy Equation (2). Any solution of the system must satisfy each equation of the system. The system of equations given in Example 2(c) has multiple solutions. For example, x = 2, y = 0, z = 4; x = 3, y = 1, z = 2; x = 4, y = 2, z = 0; x = 5, y = 3, z = -2 are each solutions, as you can verify. As it turns out, when a system of equations has multiple solutions, the number of solutions is infinite! The system of equations given in Example 2(d) has no solution. Look at Equations (4) and (5), which tell us x = 4 and y = 2.
No longer available |Learn more- Alfred Basta, Stephan DeLong, Nadine Basta, , Alfred Basta, Stephan DeLong, Nadine Basta(Authors)
- 2013(Publication Date)
- Cengage Learning EMEA(Publisher)
These issues lead us to consideration of methods for solution that are immune from the ambiguity of graphical analysis, methods that are wholly algebraic and computational. Such methods are free from the requirement that a high-quality graph be in our A Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Solving Systems of Linear Equations Algebraically and with Matrices 163 Objectives When you have successfully completed the materials of this chapter, you will be able to: Solve a system of linear equations by the method of substitution. Solve a system of linear equations by the method of elimination. Use substitutions to convert systems of equations into linear systems and solve them. Understand the basic definitions and operations of matrices. Solve a system of linear equations using matrices. possession and are also immune from the problem of identifying noninteger coordinates. We will consider several computational techniques, including the methods of substitution and elimination and the matrix method. We will see that these approaches offer us protection from the vagaries of the graphing method and that they employ mathematical tech-niques already well known to us. 5.1 S OLVING S YSTEMS OF L INEAR E QUATIONS BY THE S UBSTITUTION M ETHOD Let’s consider a system of linear equations whose solution would be difficult to determine using the method of graphing.
eBook - PDFLinear Algebra: Gateway to Mathematics
Second Edition
- Robert Messer(Author)
- 2021(Publication Date)
- American Mathematical Society(Publisher)
2 Systems of Linear Equations Before continuing the study of vector spaces, we want to develop a system-atic computational technique for solving certain systems of simultaneous equations. This chapter presents a simple but powerful algorithm for deter-mining all solutions of systems of linear equations. Matrices will provide a convenient notation for conceptualizing the massive arrays of coefficients. Watch for the vector space operations of addition and scalar multiplication lurking in the background at all times. 2.1 Notation and Terminology You are undoubtedly familiar with methods for solving systems of two simultaneous equations in two unknowns, such as 2? + 3? = 7 6? − ? = 2 You may also have considered systems of three or more equations in three or more unknowns, such as ? + ? − ? = 1 2? + ? = 2 ? − 2? + 4? = 0 The technique we will develop is based on the familiar method of using one equation to eliminate the occurrence of a variable in another of the equations and gradually accumulating values of all the variables. 67 68 Chapter 2. Systems of Linear Equations Mathematical Strategy Session: There are many advantages to pro-ceeding systematically: We will be able to deal with any number of equations in any number of unknowns. If there is a solution, the method will find it. If there is more than one solution, the method will find all of them. If there are no solutions, the method will let us know. We will be able to adapt the method to solve related problems. We will understand how a computer program can assist with the routine arithmetic. Let us begin with an example of three linear equations in three unknowns: ? + ? − ? = −1 2? + 4? = −4 −? + 3? + 2? = −6 The term linear refers to the fact that the only operations applied to the variables are multiplication by various constants and addition of the resulting terms. The analogy with the two vector space operations gives a hint as to why linear equations are so useful in linear algebra.
eBook - PDF- Sheldon Axler(Author)
- 2011(Publication Date)
- Wiley(Publisher)
How can we find the solutions to a system of linear equations such as the one above? One method that works for any system of linear equations is substitution, which was discussed in Section 7.1. example 4 Explain how substitution could be used to solve the following system of linear equations: Here we have a sys- tem of three linear equations with three variables (x, y , and z). x + 2y + 3z = 4 3x - 3y + 4z = 1 2x + y - z = 7. solution To solve this system of equations, we could solve the first equation for x in terms of y and z (getting x = 4 - 2y - 3z) and then substitute this value for x into the last two equations. The last two equations would then become a system of two linear equations with two variables (y and z). We could solve one of these equations for y in terms of z and substitute that value for y into the other equation. This procedure would give us a single linear equation with one variable z. That equation could easily be solved for z. The value of z would then give us the value of y , and the values of y and z would then give us the value of x (from x = 4 - 2y - 3z). 404 chapter 7 Systems of Equations Systems of linear equations arise in numerous contexts, usually with many more variables than the two or three variables used in examples in textbooks. Mathematical models of a sector of the economy or of an industrial production process typically involve dozens, hundreds, or thousands of variables. The substitution technique discussed in the example above could be used Google computers solve a very large sys- tem of linear equa- tions to rank the re- sults of web searches. to solve any system of linear equations. However, substitution is not as efficient for solving typical large systems of linear equations as another tech- nique called Gaussian elimination that we will soon learn. Thus substitution is not used much in practice. Another method for solving systems of linear equations is called Cramer’s rule.
- Kevin D. Dorfman, Prodromos Daoutidis(Authors)
- 2017(Publication Date)
- Cambridge University Press(Publisher)
2 Linear Algebraic Equations 2.1 Definition of a Linear System of Algebraic Equations In this chapter, we will discuss a myriad of methods for solving a system of linear algebraic equations on the computer. The general form of a system of linear algebraic equations is a 1,1 x 1 + a 1,2 x 2 + . . . + a 1, n x n = b 1 a 2,1 x 1 + a 2,2 x 2 + . . . + a 2, n x n = b 2 . . . . . . a n ,1 x 1 + a n ,2 x 2 + . . . + a n , n x n = b n (2.1.1) where we are trying to solve for the various unknowns x i . As shown in the above descrip-tion, we will assume that the number of equations is equal to the number of unknowns. In the context of our previous definition of mathematical models, let’s write this equation system in matrix form Ax = b (2.1.2) The coefficient matrix A = ⎡ ⎢ ⎢ ⎢ ⎣ a 1,1 a 1,2 · · · a 1, n a 2,1 a 2,2 · · · a 2, n . . . . . . . . . a n ,1 a n ,2 · · · a n , n ⎤ ⎥ ⎥ ⎥ ⎦ (2.1.3) is the operator for this problem, the unknown vector is x = ⎡ ⎢ ⎢ ⎢ ⎣ x 1 x 2 . . . x n ⎤ ⎥ ⎥ ⎥ ⎦ (2.1.4) 46 Linear Algebraic Equations and the forcing function is the constant vector b = ⎡ ⎢ ⎢ ⎢ ⎣ b 1 b 2 . . . b n ⎤ ⎥ ⎥ ⎥ ⎦ (2.1.5) You may wonder why we are devoting so much time in this book to the “simple” problem of solving this system of equations. In particular, you may remember solving numerous mass and energy balance problems that ultimately reduced to a system of linear algebraic equations and hardly felt the need to resort to a computer to get an answer. This is because the problems you were trying to solve are too small, not because linear algebraic systems are easy to solve! Consider the two problems in Fig. 2.1 . In most introductory chemical engineering classes, the problems are similar to Fig. 2.1 a. These are small problems, and the unknowns are selected such that the equations are only weakly coupled. In these small problems, you can proceed easily through the equations, solving for one variable at a time and then proceeding to the next equation.
eBook - PDF- Mark D. Turner, Charles P. McKeague(Authors)
- 2016(Publication Date)
- XYZ Textbooks(Publisher)
I persevered and found myself in the Proofs in Mathematics course, where I learned to truly appreciate math again. The class utilized a different way of thinking than I was used to. It was a challenging class; I struggled throughout most of it, but I was eager to learn the new material despite my grades not being what I desired. It was at that moment that I did not care how my grades ended up. Enjoying and understanding what I was learning was my main priority. 287 4.3 Linear Equations Learning Objectives Learning Objectives 4.3 In this section, we will learn how to: 1. Solve a system of linear equations using the substitution method. The Substitution Method Introduction There is a third method of solving systems of equations. It is the substitution method, and, like the elimination method, it can be used on any system of linear equations. Some systems, however, lend themselves more to the substitution method than others do. Solve the following system. x + y = 2 y = 2 x − 1 SOLUTION If we were to solve this system by the methods used in the previous section, we would have to rearrange the terms of the second equation so that similar terms would be in the same column. There is no need to do this, however, because the second equation tells us that y is 2 x − 1. We can replace the variable y in the first equation with the expression 2 x − 1 from the second equation; that is, we substitute 2 x − 1 from the second equation for y in the first equation. Here is what it looks like: x + (2 x − 1) = 2 The equation we end up with contains only the variable x . The variable y has been eliminated by substitution. Solving the resulting equation, we have x + (2 x − 1) = 2 3 x − 1 = 2 3 x = 3 x = 1 This is the x -coordinate of the solution to our system. To find the y -coordinate, we can substitute x = 1 into the second equation of our system. y = 2(1) − 1 y = 2 − 1 y = 1 The solution to our system is the ordered pair (1, 1).
eBook - PDF- Stefan Waner, Steven Costenoble(Authors)
- 2017(Publication Date)
- Cengage Learning EMEA(Publisher)
The only solution turns out to be x 5 2 and y 5 1 , a solution you might easily guess. But how do we know that this is the only solution, and how do we find solutions systematically? When we restrict ourselves to systems of linear equations, there is a very elegant method for determining the number of solutions and finding them all. Moreover, as we will see, many real-world applications give rise to just such systems of linear equations. We begin in Section 3.1 with systems of two linear equations in two unknowns and some of their applications. In Section 3.2 we study a powerful matrix method, called row reduction , for solving systems of linear equations in any number of unknowns. In Section 3.3 we look at more applications. Computers have been used for many years to solve the large systems of equations that arise in the real world. You probably already have access to devices that will do the row operations that are used in row reduction. Many graphing calculators can do them, as can spreadsheets and various special-purpose applications, including utilities avail-able at the Website. Using such a device or program makes the calculations quicker and helps to avoid arithmetic mistakes. Then there are programs (and calculators) into which you simply feed the system of equations and out pop the solutions. We can think of what we do in this chapter as looking inside the “black box” of such a program. More impor-tant, we talk about how, starting from a real-world problem, to get the system of equa-tions to solve in the first place. No computer will do this conversion for us yet. 3.1 Systems of Two Equations in Two Unknowns Linear Equations and Solutions Suppose you have $3 in your pocket to spend on snacks and a drink. If x represents the amount you’ll spend on snacks and y represents the amount you’ll spend on a drink, you can say that x 1 y 5 3 .
eBook - PDF- Bruce Solomon(Author)
- 2014(Publication Date)
- Chapman and Hall/CRC(Publisher)
Our main 60 2. Solving Linear Systems goal in this chapter will be to discover and understand the Gauss– Jordan algorithm. Before we get there, however, we want to point out that linear systems have not just a dual , but a triple nature. Every linear system can be seen as posing three completely different problems. 1.6. Three ways to interpret systems. We have introduced linear systems as a way of solving the “pre-image” problem: When A is n × m , the solutions of Ax = b are the pre-images (in R m ) of b ∈ R n under the linear map T represented by A , so that for every input x ∈ R m , T ( x ) = Ax Along with this pre-image problem, however, there are two other im-portant ways to interpret the same system. We call them the column and row problems, respectively. Column problem. The column problem is about linear combinations in the range R n . Example 1.7 . Consider the simple 2 × 3 system 3 x -4 y + z = 0 5 x + 3 y -2 z = 1 This has the form Ax = b , where A = 3 -4 1 5 3 -2 , x = x y z , and b = 0 1 As we emphasized in Remark 1.22 , however, matrix/vector multiplica-tion equals linear combination , so we can also read the system as a problem about linear combinations of the columns of A . In this case, for instance, Ax = b becomes x 3 5 + y -4 3 + z 1 -2 = 0 1 or x a 1 + y a 2 + z a 3 = e 2 where a j denotes column j of A . Seen in this light, solving the linear system means finding a linear combination of the a i vectors that adds up to (0 , 1) . 1. THE LINEAR SYSTEM 61 The same reasoning applies to every linear system. Just as in the example above, we can rewrite the matrix/vector equation Ax = b as the vector equation (7) x 1 c 1 ( A ) + x 2 c 2 ( A ) + · · · + x m c m ( A ) = b where the c i ( A ) denotes column i of the coefficient matrix A .
eBook - PDF- Stefan Waner, Steven Costenoble(Authors)
- 2017(Publication Date)
- Cengage Learning EMEA(Publisher)
The only solution turns out to be x 5 2 and y 5 1 , a solution you might easily guess. But how do we know that this is the only solution, and how do we find solutions systematically? When we restrict ourselves to systems of linear equations, there is a very elegant method for determining the number of solutions and finding them all. Moreover, as we will see, many real-world applications give rise to just such systems of linear equations. We begin in Section 4.1 with systems of two linear equations in two unknowns and some of their applications. In Section 4.2 we study a powerful matrix method, called row reduction , for solving systems of linear equations in any number of unknowns. In Section 4.3 we look at more applications. Computers have been used for many years to solve the large systems of equations that arise in the real world. You probably already have access to devices that will do the row operations that are used in row reduction. Many graphing calculators can do them, as can spreadsheets and various special-purpose applications, including utilities avail-able at the Website. Using such a device or program makes the calculations quicker and helps to avoid arithmetic mistakes. Then there are programs (and calculators) into which you simply feed the system of equations and out pop the solutions. We can think of what we do in this chapter as looking inside the “black box” of such a program. More impor-tant, we talk about how, starting from a real-world problem, to get the system of equa-tions to solve in the first place. No computer will do this conversion for us yet. 4.1 Systems of Two Equations in Two Unknowns Linear Equations and Solutions Suppose you have $3 in your pocket to spend on snacks and a drink. If x represents the amount you’ll spend on snacks and y represents the amount you’ll spend on a drink, you can say that x 1 y 5 3 .
eBook - PDFIntermediate Algebra
A Guided Approach
- Rosemary Karr, Marilyn Massey, R. Gustafson, , Rosemary Karr, Marilyn Massey, R. Gustafson(Authors)
- 2014(Publication Date)
- Cengage Learning EMEA(Publisher)
All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 3.4 Solving Systems of Linear Equations Using Matrices 199 In this section, we will discuss an alternative method used for solving systems of linear equations. This method involves the use of matrices . Solve a system of linear equations with the same number of equations as variables using Gaussian elimination. 1 Section Objectives matrix element of a matrix square matrix augmented matrix coefficient matrix Gaussian elimination triangular form of a matrix back substitution Vocabulary Solving Systems of Linear Equations Using Matrices Multiply the first row by 2 and add the result to the second row. 1. 2 3 5 1 2 3 2. 2 1 0 4 2 3 2 1 Multiply the first row by 2 1 and add the result to the second row. 3. 2 3 5 1 2 3 4. 2 1 0 4 2 3 2 1 Getting Ready 3.4 2 3 1 Solve a system of linear equations with the same number of equations as variables using Gaussian elimination. Solve a system with more linear equations than variables using row operations on a matrix. Solve a system with fewer linear equations than variables using row operations on a matrix. MATRIX A matrix is any rectangular array of numbers. Some examples of matrices are A 5 c 1 2 3 4 5 6 d B 5 £ 1 2 3 4 5 6 § C 5 £ 2 4 6 8 10 12 14 16 18 § The numbers in each matrix are called elements . Because matrix A has two rows and three columns, it is called a 2 3 3 matrix (read “2 by 3” matrix). Matrix B is a 3 3 2 matrix, because the matrix has three rows and two columns. Matrix C is a 3 3 3 matrix (three rows and three columns). Copyright 2013 Cengage Learning. All Rights Reserved.
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