DIRECTS METHODS

GAUSSIAN ELIMINATION

0 A system of equations is solved by the Gauss when their solutions are obtained by reducing an equivalent system given in which each equation has one fewer variables than the last.

0 When applying this process, the resulting matrix is known as: “Form staggered”.

EXAMPLE

55

46

117

64

312

95

321

321

321

xxx

xxx

xxx

6

18

117

164

3112

951

0 Solve the following system of linear equations by Gaussian Elimination method.

6

1358

117

164

105590

951

6

18

117

164

3112

95121 12* ff

413

1358

117

35140

105590

951

6

1358

117

164

105590

95131 4* ff

413

591358

117

35140

5910510

951

413

1358

117

35140

105590

951591*1f

595355

591358

117

5959500

5910510

951

413

591358

117

35140

5910510

95132 14* ff

9

591358

117

100

5910510

951

595355

591358

117

5959500

5910510

95159

595*3f

9

591358

117

59105

95

3

32

321

x

xx

xxx

1

7

9

1

2

3

x

x

x

0 The Gauss-Jordan is a variant of the method of Gauss. When you delete an unknown in an equation, Gauss-Jordan eliminates that unknown quantity in the rest of the equations, the basis for disposing the pivot equation. Also all the rows are normalized when taken as pivot equation. The end result of such disposal creates an identity matrix instead of a triangular Gauss as it does, so do not use the back substitution.

GAUSS-JORDAN

0 Solve the following system of linear equations by Gauss-Jordan method.

EXAMPLE

4

3

1

5

423

32

zyx

zyx

zyx

0 Then we write the augmented matrix of the system.

0 We take this matrix to reduced row-echelon form by elementary operations on the rows of the matrix, for this, write the matrix and then an arrow. Above this arrow will indicate the operations we're making so that the reader can follow the development.

4

3

1

115

423

232

4

3

21

115

423

21231

4

3

1

115

423

13212

1f

23

29

21

272170

2112130

21231

4

3

21

115

423

2123131

2153

ffff

0 Development to obtain the reduced row echelon form.

13192

139

21

139600

131110

21231

3

139

21

7170

131110

212313217 ff

3

139

21

7170

131110

21231

23

29

21

272170

2112130

212313

22132f

f

2

139

21

100

131110

21231

13192

139

21

139600

131110

2123139613 f

2

1

21

100

010

0231

2

139

21

100

131110

2123113

23211311

ffff

2

1

1

100

010

001

2

1

21

100

010

02311223 ff

2

1

1

2

1

1

100

010

001

z

y

x

2

1

1

z

y

x

0 The final reduced row echelon matrix indicates that:The solution of the system is x = 1, y =- 1 and z = 2.

0 It is a kind of direct method for calculating matrices.

0 The system consists of taking a system of equations as an equation as a pivot in order to give a matrix identical to the system of equations. When you delete an unknown in an equation, Gauss-Jordan eliminates that unknown in the rest of the equations.

GAUSS-JORDANWITH PIVOTING

0 The number arr in the coefficient matrix A that is used to eliminate akr , where k = r + 1, r + 2, . . . , N, is called the r th pivotal element, and the r th row is called the pivot row.

0 The following example illustrates how to use the operations to obtain an equivalent upper-triangular system UX = Y from a linear system AX = B, where A is an N × N matrix.

EXAMPLE

6

20

28

13

233

224

342

42

4321

4321

431

4321

xxxx

xxxx

xxx

xxxx

0 Solve the following system of linear equations by Gauss-Jordan with pivoting method.

0 The augmented matrix is:

6

20

28

13

2313

1224

3402

4121

3

4

2

41

31

21

m

m

m

pivot

0 The first row is used to eliminate elements in the first column below the diagonal. We refer to the first row as the pivotal row and the element a11 = 1 is called the pivotal element. The values mk1 are the multiples of row 1 that are to be subtracted from row k for k = 2, 3, 4. The result after elimination is:

45

32

2

13

14670

15260

5240

4121

75.1

5.1

42

32

m

m

pivot

0 The second row is used to eliminate elements in the second column that lie below the diagonal. The second row is the pivotal row and the values mk2 are the multiples of row 2 that are to be subtracted from row k for k = 3, 4. The result after elimination is:

5.48

35

2

13

25.55.900

5.7500

5240

4121

9.143m

pivot

0 Finally, the multiple m43 = −1.9 of the third row is subtracted from the fourth row, and the result is the upper-triangular system.

18

35

2

13

95.900

5.7500

5240

4121

0 The back-substitution algorithm can be used to solve, and we get: x4 = 2, x3 = 4, x2 = −1, x1 = 3.

3

1

4

2

1

2

3

4

x

x

x

x

0 http://cbi.azc.uam.mx/archivos/varios/ProblemarioW.pdf

0 http://www.ecs.fullerton.edu/~mathews/n2003/gaussjordan/GaussianJordanProof.pdf

BIBLIOGRAPHY