1 / 30

COLLEGE ALGEBRA

COLLEGE ALGEBRA. LIAL HORNSBY SCHNEIDER. 5.2. Matrix Solution of Linear Systems. The Gauss-Jordan Method Special Systems. Matrix Solutions.

lona
Télécharger la présentation

COLLEGE ALGEBRA

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. COLLEGE ALGEBRA LIAL HORNSBY SCHNEIDER

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

  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 iscalled an element of the matrix.

  4. Gauss-Jordan Method Matrices in general are discussed in more detail later in this chapter. In this section, we develop a method for solving linear systems using matrices. As an example, start with a system and write the coefficients of the variables and the constants as a matrix, called the augmented matrix of the system.

  5. Gauss-Jordan Method Linear system of equations

  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 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.

  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.

  8. Gauss-Jordan Method These transformations are just restatements in matrix form of the transformations of systems discussed in the previous section. From now on, when referring to the third transformation, “a multiple of the elements of a row” will be abbreviated as “a multiple of a row.”

  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, such as from which the solutions are easily obtained. This form is also called reduced-row echelon form.

  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.

  11. Note The Gauss-Jordan method proceeds column by column, from left to right. When you are working with a particular column, no row operation should undo the form of a preceding column.

  12. USING THE GAUSS-JORDAN METHOD Example 1 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.

  13. USING THE GAUSS-JORDAN METHOD Example 1 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 column in the matrix represents the coefficients of one variable, the augmented matrix should be transformed so that it is of the form for real numbers k and j. Once the augmented matrix is in this form, the matrix can be rewritten as a linear system to get This form is our goal.

  14. USING THE GAUSS-JORDAN METHOD Example 1 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 to get 1 in this position. (This step is abbreviated as ⅓R1.)

  15. USING THE GAUSS-JORDAN METHOD Example 1 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.

  16. USING THE GAUSS-JORDAN METHOD Example 1 Obtain 1 in the second row, second column by multiplying each element of the second row by using transformation 2.

  17. USING THE GAUSS-JORDAN METHOD Example 1 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.

  18. USING THE GAUSS-JORDAN METHOD Example 1 This last matrix corresponds to the system that has solution set {(3, 2)}. We can read this solution directly from the third column of the final matrix. Check the solution in both equations of the original system.

  19. USING THE GAUSS-JORDAN METHOD Example 2 Solve the system. Solution

  20. USING THE GAUSS-JORDAN METHOD Example 2 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.

  21. USING THE GAUSS-JORDAN METHOD Example 2 To change the third element in the first column to 0, multiply each element of the first row by –1, and add the result to the corresponding element of the third row.

  22. USING THE GAUSS-JORDAN METHOD Example 2 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.

  23. USING THE GAUSS-JORDAN METHOD Example 2 The linear system associated with this final matrix is The solution set is {(1, 2, –1)}. Check the solution in the original system.

  24. SOLVING AN INCONSISTENT SYSTEM Example 3 Use the Gauss-Jordan method to solve the system. Solution

  25. SOLVING AN INCONSISTENT SYSTEM Example 3 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.

  26. SOLVING A SYSTEM WITH INFINITELY MANY SOLUTIONS Example 4 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.

  27. SOLVING A SYSTEM WITH INFINITELY MANY SOLUTIONS Example 4 It is not possible to go further with the Gauss-Jordan method.

  28. SOLVING A SYSTEM WITH INFINITELY MANY SOLUTIONS Example 4 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

  29. 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.

  30. 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 3. 3. 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.

More Related