Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1 5 Systems and Matrices Copyright © 2013, 2009, 2005 Pearson Education, Inc.

2 5.2 Matrix Solution of Linear Systems The Gauss-Jordan Method Special Systems

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 3 Matrix Solutions Since systems of linear equations occur in so many practical situations, computer methods have been developed for efficiently solving linear systems. Computer solutions of linear systems depend on the idea of a matrix (plural matrices), a rectangular array of numbers enclosed in brackets. Each number is an element of the matrix.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 4 Gauss-Jordan Method In this section, we develop a method for solving linear systems using matrices. We start with a system and write the coefficients of the variables and the constants as an augmented matrix of the system.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 5 Gauss-Jordan Method Linear system of equations

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 6 Gauss-Jordan Method The vertical line, which is optional, separates the coefficients from the constants. Because this matrix has 3 rows (horizontal) and 4 columns (vertical), we say its dimension (also described as order and size) is 3  4 (read “three by four”). The number of rows is always given first. To refer to a number in the matrix, use its row and column numbers.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 7 Matrix Row 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. Interchange any two rows. 2. Multiply or divide the elements of any row by a nonzero real number. 3. Replace any row of the matrix by the sum of the elements of that row and a multiple of the elements of another row.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 8 Gauss-Jordan Method These transformations are restatements in matrix form of the transformations of systems discussed in the previous section. From now on, when referring to the third transformation, we will abbreviate “a multiple of the elements of a row” as “a multiple of a row.”

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 9 Gauss-Jordan Method The Gauss-Jordan method is a systematic technique for applying matrix row transformations in an attempt to reduce a matrix to diagonal form, with 1s along the diagonal from which the solutions are easily obtained. This form is also called reduced-row echelon form.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 10 Using the Gauss-Jordan Method to Put a Matrix into Diagonal Form Step 1 Obtain 1 as the first element of the first column. Step 2 Use the first row to transform the remaining entries in the first column to 0. Step 3 Obtain 1 as the second entry in the second column. Step 4 Use the second row to transform the remaining entries in the second column to 0. Step 5 Continue in this manner as far as possible.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 11 Note The Gauss-Jordan method proceeds column by column, from left to right. In each column, we work to obtain 1 in the appropriate diagonal location, and then use it to transform the remaining elements in that column to 0s. When we are working with a particular column, no row operation should undo the form of a preceding column.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 12 Example 1 USING THE GAUSS-JORDAN METHOD Solve the system. Solution Both equations are in the same form, with variable terms in the same order on the left, and constant terms on the right. Write the augmented matrix.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 13 Example 1 USING THE GAUSS-JORDAN METHOD The goal is to transform the augmented matrix into one in which the value of the variables will be easy to see. That is, since each of the first two columns in the matrix represents the coefficients of one variable, the augmented matrix should be transformed so that it is of the following form. (Here k and j are real numbers.) Once the augmented matrix is in this form, the matrix can be rewritten as a linear system. This form is our goal.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 14 Example 1 USING THE GAUSS-JORDAN METHOD It is best to work in columns beginning in each column with the element that is to become 1. In the augmented matrix 3 is in the first row, first column position. Use transformation 2, multiplying each entry in the first row by 1/3 to get 1 in this position. (This step is abbreviated as.)

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 15 Example 1 USING THE GAUSS-JORDAN METHOD Introduce 0 in the second row, first column by multiplying each element of the first row by – 5 and adding the result to the corresponding element in the second row, using transformation 3.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 16 Example 1 USING THE GAUSS-JORDAN METHOD Obtain 1 in the second row, second column by multiplying each element of the second row by using transformation 2.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 17 Example 1 USING THE GAUSS-JORDAN METHOD Finally, get 0 in the first row, second column by multiplying each element of the second row by and adding the result to the corresponding element in the first row.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 18 Example 1 USING THE GAUSS-JORDAN METHOD This last matrix corresponds to the system which indicates the solution (3, 2). We can read this solution directly from the third column of the final matrix.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 19 Example 1 USING THE GAUSS-JORDAN METHOD CHECK Substitute the solution in both equations of the original system. True Since true statements result, the solution set is {(3,2)}.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 20 Example 2 USING THE GAUSS-JORDAN METHOD Solve the system. Solution

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 21 Example 2 USING THE GAUSS-JORDAN METHOD There is already a 1 in the first row, first column. Introduce 0 in the second row of the first column by multiplying each element in the first row by – 3 and adding the result to the corresponding element in the second row.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 22 Example 2 USING THE GAUSS-JORDAN METHOD To change the third element in the first column to 0, multiply each element of the first row by – 1. Add the result to the corresponding element of the third row.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 23 Example 2 USING THE GAUSS-JORDAN METHOD Use the same procedure to transform the second and third columns. For both of these columns, perform the additional step of getting 1 in the appropriate position of each column. Do this by multiplying the elements of the row by the reciprocal of the number in that position.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 24 Example 2 USING THE GAUSS-JORDAN METHOD The linear system associated with this final matrix is The solution set is {(1, 2, – 1)}. Check the solution in the original system.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 25 Example 3 SOLVING AN INCONSISTENT SYSTEM Use the Gauss-Jordan method to solve the system. Solution

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 26 Example 3 SOLVING AN INCONSISTENT SYSTEM The next step would be to get 1 in the second row, second column. Because of the 0 there, it is impossible to go further. Since the second row corresponds to the equation 0x + 0y = 1, which has no solution, the system is inconsistent and the solution set is Ø.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 27 Example 4 SOLVING A SYSTEM WITH INFINITELY MANY SOLUTIONS Use the Gauss-Jordan method to solve the system. Solution Recall from the previous section that a system with two equations in three variables usually has an infinite number of solutions. We can use the Gauss-Jordan method to give the solution with z arbitrary.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 28 Example 4 SOLVING A SYSTEM WITH INFINITELY MANY SOLUTIONS It is not possible to go further with the Gauss-Jordan method.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 29 Example 4 SOLVING A SYSTEM WITH INFINITELY MANY SOLUTIONS The equations that correspond to the final matrix are Solve these equations for x and y, respectively. The solution set, written with z arbitrary, is

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 30 Summary of Possible Cases When matrix methods are used to solve a system of linear equations and the resulting matrix is written in diagonal form: 1. If the number of rows with nonzero elements to the left of the vertical line is equal to the number of variables in the system, then the system has a single solution. See Examples 1 and 2.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 31 Summary of Possible Cases 2. If one of the rows has the form [0 0    0  a] with a ≠ 0, then the system has no solution. See Example If there are fewer rows in the matrix containing nonzero elements than the number of variables, then the system has either no solution or infinitely many solutions. If there are infinitely many solutions, give the solutions in terms of one or more arbitrary variables. See Example 4.