Mathematics
Solving Simultaneous Equations Using Matrices
Solving simultaneous equations using matrices involves representing the coefficients of the equations in a matrix, and the variables in a column matrix. By performing matrix operations such as multiplication and inversion, the values of the variables can be determined. This method provides a systematic approach to solving systems of linear equations and is commonly used in various fields including engineering and physics.
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8 Key excerpts on "Solving Simultaneous Equations Using Matrices"
eBook - PDF- John Gilbert, Camilla Jordan, David A Towers(Authors)
- 2017(Publication Date)
- Red Globe Press(Publisher)
The fifth example shows a non-homogeneous set of equations with an infinite number of solutions. Thus a system of linear equations can have no solution, one (unique) solution, or an infinite number of solutions. Solving a system of equations means finding all possible solutions. 168 Guide to Mathematical Methods Exercises: Section 8.1 Solve the following sets of simultaneous equations. (i) x − y = 1 x + y = 3 ; (ii) x + 2 y = 0 3 x + 6 y = 0 ; (iii) x + 2 y = 1 3 x + 6 y = 2 ; (iv) x + 2 y = 1 3 x + 6 y = 3 . 8.2 Matrices and Gaussian elimination Solving large systems of linear equations requires a systematic approach. This is provided by the method known as Gaussian elimination. To reduce the amount of writing involved we first introduce the idea of a matrix. Definition 8.2 A matrix is a rectangular array of numbers: ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ a 11 a 12 · · · a 1 n a 21 a 22 · · · a 2 n . . . . . . . . . . . . a m 1 a m 2 · · · a mn ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ . This matrix has m rows and n columns and is described as an m × n matrix. The ij th element of the matrix is a ij , the element occurring in the i th row and j th column. A 1 × n matrix is sometimes called a row vector and a m × 1 matrix is sometimes called a column vector . Matrices occur in many contexts in mathematics. Here our concern is their role in the solution of equations. Given the system a 11 x 1 + a 12 x 2 + · · · + a 1 n x n = b 1 a 21 x 1 + a 22 x 2 + · · · + a 2 n x n = b 2 . . . . . . . . . a m 1 x 1 + a m 2 x 2 + · · · + a mn x n = b m . Linear equations and matrices 169 of m equations in n unknowns, the matrix ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ a 11 a 12 · · · a 1 n a 21 a 22 · · · a 2 n . . . . . . . . . . . . a m 1 a m 2 · · · a mn ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ is called the coefficient matrix of the system and the matrix ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ a 11 a 12 · · · a 1 n b 1 a 21 a 22 · · · a 2 n b 2 . . . . . . . . . . . . . . . a m 1 a m 2 · · · a mn b m ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ is called the augmented matrix of the system.
No longer available |Learn more- Rizwan Butt(Author)
- 2011(Publication Date)
- Mercury Learning and Information(Publisher)
Chapter 1 Matrices and Linear Systems 1.1 Introduction When engineering systems are modeled, the mathematical description is frequently developed in terms of a set of algebraic simultaneous equations. Sometimes these equations are nonlinear and sometimes linear. In this chapter, we discuss systems of simultaneous linear equations and describe the numerical methods for the approximate solutions of such systems. The solution of a system of simultaneous linear algebraic equations is proba- bly one of the most important topics in engineering computation. Prob- lems involving simultaneous linear equations arise in the areas of elasticity, electric-circuit analysis, heat transfer, vibrations, and so on. Also, the numerical integration of some types of ordinary and partial differential equations may be reduced to the solution of such a system of equations. It has been estimated, for example, that about 75% of all scientific problems require the solution of a system of linear equations at one stage or another. It is therefore important to be able to solve linear problems efficiently and accurately. 1 2 Applied Linear Algebra and Optimization using MATLAB Definition 1.1 (Linear Equation) It is an equation in which the highest exponent in a variable term is no more than one. The graph of such an equation is a straight line. • A linear equation in two variables x 1 and x 2 is an equation that can be written in the form a 1 x 1 + a 2 x 2 = b, where a 1 , a 2 , and b are real numbers. Note that this is the equation of a straight line in the plane. For example, the equations 5x 1 + 2x 2 = 2, 4 5 x 1 + 2x 2 = 1, 2x 1 - 4x 2 = π are all linear equations in two variables. A linear equation in n variables x 1 , x 2 , . . . , x n is an equation that can be written as a 1 x 1 + a 2 x 2 + · · · + a n x n = b, where a 1 , a 2 , . . . , a n are real numbers and called the coefficients of unknown variables x 1 , x 2 , .
eBook - PDF- Sheldon Axler(Author)
- 2011(Publication Date)
- Wiley(Publisher)
Now we turn to another way to express a system of linear equations using matrices. We will make the coefficients of each equation into one row of a matrix, but now we will not include the constant term. Instead, the constant terms will form their own matrix with one column. The variables will also be put into a matrix with one column. The next example shows how these ideas are used to rewrite a system of linear equations into a matrix equation. Instead of calling the variables x, y , and z, the variables here have been called x 1 , x 2 , and x 3 so that you will become comfortable with the subscript notation that is used when dealing with a large number of variables. example 14 Consider the following system of equations: x 1 + 2x 2 + 3x 3 = 9 5x 1 - x 2 + 6x 3 = -4 7x 1 - 2x 2 + 8x 3 = 6. (a) Rewrite the system of equations above as a matrix equation. (b) Use an inverse matrix to solve the matrix equation and find the values of x 1 , x 2 , and x 3 . solution (a) Form a matrix A whose rows consist of the coefficients on the left side of the equations above. Form another matrix X with one column consisting of the variables. Finally, form another matrix B with one column consisting of the constant terms from the right side of the equations above. In each of these matrices, keep everything in the proper order as listed above. In other words, we have A = 1 2 3 5 -1 6 7 -2 8 , X = x 1 x 2 x 3 , and B = 9 -4 6 . The system of linear equations above can now be expressed as the single This matrix equation provides another rea- son why the definition of matrix multiplica- tion, which at first seem strange, turns out to be so useful. matrix equation AX = B. To verify that the matrix equation above is the same as the original system of equations, use the definition of matrix multiplication. For example, the entry in section 7.4 Matrix Algebra 437 row 1, column 1 of AX equals row 1 of A times column 1 of X (which has only one column).
eBook - PDF- Sheldon Axler(Author)
- 2011(Publication Date)
- Wiley(Publisher)
chapter 7 Systems of Equations The standard method of solving systems of linear equations is called Gaussian elimi- nation. This technique is named in honor of Carl Friedrich Gauss, pictured here on an old German 10-mark bill. Gauss used this method in a book pub- lished in 1809. How- ever, the method was also used in a Chinese book published over 1600 years earlier. This chapter provides only a taste of some topics in systems of equations. Proper treatment of these subjects requires a full book devoted just to them. The first course in linear algebra focuses on systems of linear equations and matrices. In the first section of this chapter we look at equations and systems of equations. We will use graphical methods to find approximate solutions to a system of equations with two variables. We will also learn how to solve some systems of equations by substitution. In the second section of this chapter we specialize to systems of linear equations. Unlike the situation with more general systems of equations, good methods exist for finding the solutions to systems of linear equations. We will learn about Gaussian elimination, the most efficient and most commonly used of these methods. The third section of this chapter recasts Gaussian elimination in the context of matrices. This helps us understand how computers deal with large systems of linear equations. We delve more deeply into matrix algebra in the last section of this chapter. Key topics here include matrix multiplication and the inverse of a matrix. 387 388 chapter 7 Systems of Equations 7.1 Equations and Systems of Equations learning objectives By the end of this section you should be able to find solutions of equations, when possible; graphically solve a system of equations with two variables; solve a system of equations by substitution. Solving an Equation In this chapter we will discuss solutions to systems of equations.
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)
The solution to the system is found by computation of the matrix product X A B = = ----- -1 1 1 2 1 2 6 7 2 1 5 2 3 2 1 2 3 10 / / / / / / 5 9 2 22 10 -- = uni00A0 / . As in the previous example, this indicates that the solution to the system of equations is the ordered triple -- 9 2 22 10 , , . Exercises Solve each of the following systems of linear equations, using the matrix method. 1. 4 3 1 3 9 x y x y -= + = 3. -+ = -= -8 6 6 4 6 5 x y x y 2. 8 14 46 2 5 3 x y x y -= -+ = -4. 6 5 1 8 3 11 x y x y -= + = 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 203 5. 2 9 5 3 2 4 x y x y + = + = -10. 5 6 4 47 7 8 2 2 7 3 4 14 x y z x y z x y z --= -+ -= + + = 6. 2 2 6 4 22 x y x y -= -+ = 11. 3 2 1 5 3 13 2 1 3 x y z x y x y z -+ = -+ = + -= 7. 3 4 7 2 1 3 2 4 x y x y + = + = 12. 4 2 9 3 3 19 2 3 3 x y z x y z x y z -+ = -+ = -+ = 8. x y x y -= + = 1 2 7 1 3 1 2 5 13. x y z x y z x x z + -= + -= + -= 7 2 30 3 2 4 2 5 2 4 18 9. 2 3 11 3 1 2 2 5 5 x y z x y z x y z -+ = -+ = --= -14. x y z x y z x y z -+ = -+ -= -+ = -2 8 3 5 2 4 2 1 6 Summary In this chapter, you learned about: Solving systems of linear equations by the substitution method and the method of elimination. Conversions into linear equations by substitution. Matrices, their attributes, and their operations; equality, addition, sub-traction, multiplication, and multiplicative inverse; and solving systems of linear equations.
eBook - PDF- David W Lewis(Author)
- 1991(Publication Date)
- WSPC(Publisher)
1 Chapter 1 MATRICES AND LINEAR EQUATIONS A familiarity with matrices is necessary nowadays in many areas of mathematics and in a wide variety of other disciplines. Areas of mathematics where matrices occur include algebra, differential equations, calculus of several variables, probability and statistics, optimization, and graph theory. Other disciplines using matrix theory include engineering, physical sciences, biological sciences, economics and management science. In this first chapter we give the fundamentals of matrix algebra, determinants, and systems of linear equations. At the end of the chapter we give some examples of situations in mathematics and other disciplines where matrices arise. 1.1 Matrices and matrix algebra A matrix is a rectangular array of symbols. In this book the symbols will usually be either real or complex numbers. The separate elements of the array are known as the entries of the matrix. Let m and n be positive integers. An rrun matrix A consists of m rows and n columns of numbers written in the following manner. f a ,, a n . ■ . a, ) 11 12 In a a . . . a 21 22 2n A = * a a . . . a I ml m2 mnl 2 We often write A = (a..) for short. The entry a lies in the i-th row and ij ij the j-th column of the matrix A. Two mxn matrices A = (a.) and B = (b.) are equal if and only if all ij u the corresponding entries of A and B are equal. i.e. a.. = b.. for each i and j . ij ij The sum of the mxn matrices A = (a.) and B = (b.) is the mxn matrix denoted A + B which has entry a.. + b.. in the (ij)-place for each i,j. Let X be a scalar (i.e. a real or complex number) and let A = (a.) be an mxn matrix. The scalar multiple of A by X is the nun matrix denoted XA which has entry Xajn the (ij)-place for each i,j. 1.1.2 Proposition The following properties hold. (i) A + B = B + A for all mxn matrices A and B, i.e. addition of matrices is commutative. (ii) (A + B) + C = A + (B + C) for all nun matrices A,B, and C, i.e.addition of matrices is associative.
eBook - PDF- Stefan Waner, Steven Costenoble(Authors)
- 2017(Publication Date)
- Cengage Learning EMEA(Publisher)
EXAMPLE 2 ✱ We can add these equations because when we add equal amounts to both sides of an equation, the results are equal. That is, if A 5 B and C 5 D , then A 1 C 5 B 1 D . Figure 2 y 2 2 x x H11001 y H11005 3 x H11002 y H11005 1 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-300 262 Chapter 4 Systems of Linear Equations and Matrices Method 3: Algebraic: Solving by Intersection Solution by intersection is based on the idea of the geometric approach: that the solution is the point of intersection of the graphs of two linear functions. To find this intersection point algebraically, we reason that the two functions must have the same value at the point of intersection, so we solve both equations for y , ✱ set the resulting linear functions of x equal to each other, and solve for x : y 5 2 x 1 3 Solve x 1 y 5 3 for y . y 5 x 2 1 Solve x 2 y 5 1 for y . 2 x 1 3 5 x 2 1 Equate the resulting linear functions of x . 2 2 x 5 2 4 x 5 2 Solve for x . We can then substitute this value into either equation (or either linear function we found) to obtain y as in Method 2 and find the solution: 1 x , y 2 5 1 2, 1 2 . Method 4: Algebraic: Solving by Substitution In the substitution method we solve one of the equations for one of the unknowns and then substitute the resulting expres-sion in the other to solve for the other unknown: y 5 2 x 1 3 Solve x 1 y 5 3 for y . x 2 1 2 x 1 3 2 5 1 Substitute the resulting expression in the other equation, x 2 y 5 1 . 2 x 2 3 5 1 1 2 x 5 4 x 5 2. Solve for x . We can then substitute this value into either equation to obtain y as in the elimination method to obtain the solution: 1 x , y 2 5 1 2, 1 2 . ➡ Before we go on . . . As we said above, the method of elimination is our pre-ferred algebraic method of solving systems of linear equations and is the algebraic method we will use in most of the examples that follow.
eBook - PDF- Stefan Waner, Steven Costenoble(Authors)
- 2017(Publication Date)
- Cengage Learning EMEA(Publisher)
EXAMPLE 2 ✱ We can add these equations because when we add equal amounts to both sides of an equation, the results are equal. That is, if A 5 B and C 5 D , then A 1 C 5 B 1 D . Figure 2 y 2 2 x x H11001 y H11005 3 x H11002 y H11005 1 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-300 194 Chapter 3 Systems of Linear Equations and Matrices Method 3: Algebraic: Solving by Intersection Solution by intersection is based on the idea of the geometric approach: that the solution is the point of intersection of the graphs of two linear functions. To find this intersection point algebraically, we reason that the two functions must have the same value at the point of intersection, so we solve both equations for y , ✱ set the resulting linear functions of x equal to each other, and solve for x : y 5 2 x 1 3 Solve x 1 y 5 3 for y . y 5 x 2 1 Solve x 2 y 5 1 for y . 2 x 1 3 5 x 2 1 Equate the resulting linear functions of x . 2 2 x 5 2 4 x 5 2 Solve for x . We can then substitute this value into either equation (or either linear function we found) to obtain y as in Method 2 and find the solution: 1 x , y 2 5 1 2, 1 2 . Method 4: Algebraic: Solving by Substitution In the substitution method we solve one of the equations for one of the unknowns and then substitute the resulting expres-sion in the other to solve for the other unknown: y 5 2 x 1 3 Solve x 1 y 5 3 for y . x 2 1 2 x 1 3 2 5 1 Substitute the resulting expression in the other equation, x 2 y 5 1 . 2 x 2 3 5 1 1 2 x 5 4 x 5 2. Solve for x . We can then substitute this value into either equation to obtain y as in the elimination method to obtain the solution: 1 x , y 2 5 1 2, 1 2 . ➡ Before we go on . . . As we said above, the method of elimination is our pre-ferred algebraic method of solving systems of linear equations and is the algebraic method we will use in most of the examples that follow.
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