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Copyright © 2007 Pearson Education, Inc. Slide 7-1

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Copyright © 2007 Pearson Education, Inc. Slide 7-2 Chapter 7: Matrices and Systems of Equations and Inequalities 7.1Systems of Equations 7.2Solution of Linear Systems in Three Variables 7.3Solution of Linear Systems by Row Transformations 7.4Matrix Properties and Operations 7.5Determinants and Cramer’s Rule 7.6Solution of Linear Systems by Matrix Inverses 7.7Systems of Inequalities and Linear Programming 7.8Partial Fractions

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Copyright © 2007 Pearson Education, Inc. Slide 7-3 7.3 Solution of Linear Systems by Row Transformations Matrices and Technology –Matrix methods suitable for calculator and computer solutions of large systems Matrix Row Transformations –Streamlined use of echelon methods

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Copyright © 2007 Pearson Education, Inc. Slide 7-4 7.3 Solution of Linear Systems by Row Transformations –This is called an augmented matrix where each member of the array is called an element or entry. –The rows of an augmented matrix can be treated just like the equations of a linear system.

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Copyright © 2007 Pearson Education, Inc. Slide 7-5 7.3 Matrix Transformations For any augmented matrix of a system of linear equations, the following row transformations will result in the matrix of an equivalent system. 1.Any two rows may be interchanged. 2.The elements of any row may be multiplied by a nonzero real number. 3.Any row may be changed by adding to its elements a multiple of the corresponding elements of another row.

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Copyright © 2007 Pearson Education, Inc. Slide 7-6 7.3 Solving a System by the Row Echelon Method ExampleSolve the system. Analytic Solution The augmented matrix for the system is

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Copyright © 2007 Pearson Education, Inc. Slide 7-7 7.3 Solving a System by the Row Echelon Method The matrix represents the system The solution to this system using back-substitution is {(1, –2)}.

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Copyright © 2007 Pearson Education, Inc. Slide 7-8 7.3 Solving a System by the Row Echelon Method Graphing Calculator Solution Enter the augmented matrix of the system. Using the ref command on the TI-83, we can find the row echelon form of the matrix.

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Copyright © 2007 Pearson Education, Inc. Slide 7-9 The left screen displays decimal entries, but can be converted to fractions as indicated in the right screen. This corresponds to the augmented matrix in the analytic solution. The rest of the solution process is the same. 7.3 Solving a System by the Row Echelon Method

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Copyright © 2007 Pearson Education, Inc. Slide 7-10 Example Solve the system. The augmented matrix of the system is 7.3 Extending the Row Echelon Method to Larger Systems

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Copyright © 2007 Pearson Education, Inc. Slide 7-11 7.3 Extending the Row Echelon Method to Larger Systems There is already a 1 in row 1, column 1. Next, get 0s in the rest of column 1.

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Copyright © 2007 Pearson Education, Inc. Slide 7-12 7.3 Extending the Row Echelon Method to Larger Systems

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Copyright © 2007 Pearson Education, Inc. Slide 7-13 7.3 Extending the Row Echelon Method to Larger Systems The corresponding matrix to the system is Applying back-substitution we get the solution {(1, 2, –1)}.

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Copyright © 2007 Pearson Education, Inc. Slide 7-14 7.3 Reduced Row Echelon Method Matrix methods for solving systems –Row Echelon Method form of a matrix seen earlier 1s along the diagonal and 0s below –Reduced Row Echelon Method form of a matrix 1s along the diagonal and 0s both below and above

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Copyright © 2007 Pearson Education, Inc. Slide 7-15 7.3 Reduced Row Echelon Method ExampleThe augmented form of the system is Using the row transformations, this augmented matrix can be transformed to

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Copyright © 2007 Pearson Education, Inc. Slide 7-16 7.3 Reduced Row Echelon Method with the Graphing Calculator ExampleSolve the system. SolutionThe augmented matrix of the system is shown below.

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Copyright © 2007 Pearson Education, Inc. Slide 7-17 7.3 Reduced Row Echelon Method with the Graphing Calculator Using the rref command we obtain the row reduced echelon form.

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Copyright © 2007 Pearson Education, Inc. Slide 7-18 7.3 Solving a System with No Solutions ExampleShow that the following system is inconsistent. SolutionThe augmented matrix of the system is

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Copyright © 2007 Pearson Education, Inc. Slide 7-19 7.3 Solving a System with No Solutions The final row indicates that the system is inconsistent and has solution set .

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Copyright © 2007 Pearson Education, Inc. Slide 7-20 7.3 Solving a System with Dependent Equations ExampleShow that the system has dependent equations. Express the general solution using an arbitrary variable. SolutionThe augmented matrix is

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Copyright © 2007 Pearson Education, Inc. Slide 7-21 7.3 Solving a System with Dependent Equations The final row of 0s indicates that the system has dependent equations. The first two rows represent the system

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Copyright © 2007 Pearson Education, Inc. Slide 7-22 7.3 Solving a System with Dependent Equations Start with the augmented matrix

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Copyright © 2007 Pearson Education, Inc. Slide 7-23 7.3 Solving a System with Dependent Equations The equations that correspond to the final matrix are

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Copyright © 2007 Pearson Education, Inc. Slide 7-24 7.3 Solving a System with Dependent Equations Solving for y we get Substitute this result into the expression to find x. Solution set written with z arbitrary:

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