The document discusses LU decomposition, which writes a matrix as the product of a lower triangular matrix and an upper triangular matrix. It may also include a permutation matrix. LU decomposition is used to solve systems of linear equations or calculate the determinant. The document provides details on performing LU decomposition step-by-step, including finding the L and U matrices, and then using those matrices to solve for x, the unknown variables.

1. ESCUELA DE INGENIERÍA DE PETROLEOS

2. ESCUELA DE INGENIERÍA DE PETROLEOS The LU decomposition is a matrix decomposition which writes a matrix as the product of a lower triangular matrix and an upper triangular matrix. The product sometimes includes a permutation matrix as well. This decomposition is used in numerical analysis to solve systems of linear equations or calculate the determinant.

3. ESCUELA DE INGENIERÍA DE PETROLEOS Get the lower triangular matrix L and upper triangular matrix U. Solve Ly = b (to find y). The result of the previous step is saved in a new array of name y. Make Ux = y (to find x). The result of the previous step is stored in a new array called x, which provides the values for the unknowns of the equation.

4. ESCUELA DE INGENIERÍA DE PETROLEOS Let A be a square matrix. An LU decomposition is a decomposition of the form where D is a diagonal matrix and L and U are unit triangular matrices, meaning that all the entries on the diagonals of L and U are one. An LUP decomposition (also called a LU decomposition with partial pivoting) is a decomposition of the form where L and U are again lower and upper triangular matrices and P is a permutation matrix, i.e., a matrix of zeros and ones that has exactly one entry 1 in each row and column.

5. ESCUELA DE INGENIERÍA DE PETROLEOS An LU decomposition with full pivoting (Trefethen and Bau) takes the form Above we required that A be a square matrix, but these decompositions can all be generalized to rectangular matrices as well. In that case, L and P are square matrices which each have the same number of rows as A , while U is exactly the same shape as A . Upper triangular should be interpreted as having only zero entries below the main diagonal, which starts at the upper left corner.

6. ESCUELA DE INGENIERÍA DE PETROLEOS Make zero all values below the pivot without turning this into a. To achieve the above is required to obtain a factor which is necessary to make clear the values below the pivot. This factor is equal to the number you want to make the number zero pivot. This factor multiplied by -1 is then multiplied by the pivot and this result is added value that is in the position to change (the value in the position to become zero).

7. ESCUELA DE INGENIERÍA DE PETROLEOS Build an array of the same order that the original matrix with ones on the main diagonal and zeros for the items that match j> i. As the elements below the main diagonal are located on multiple Gauss decomposition used to achieve the "zero" in the corresponding position.

8. ESCUELA DE INGENIERÍA DE PETROLEOS 1. Find the values of x1, x2 and x3 for the following system of equations: 4X1 -2X2 -X3 = 9 5X1 +X2 -X3 = 7 X1 +2X2 -X3 = 12 4 -2 -1 9 A = 5 1 -1 b = 7 1 2 -1 12

9. ESCUELA DE INGENIERÍA DE PETROLEOS SOLUTION 1. Se halla “U”: 4 -2 -1 5 1 -1 R2 R2 – (5/4)*R1 1 2 -1 R3 R3 – (1/4)*R1 4 -2 -1 0 7/2 ¼ 0 5/2 -3/4 R3 R3 – (5/2)/(7/2)*R2

10. ESCUELA DE INGENIERÍA DE PETROLEOS 1. Se halla “U”: 4 -2 -1 U = 0 7/2 ¼ 0 0 -13/14 2. Se halla “L”: 1 0 0 1 0 0 L = ? 1 0 L = 5/4 1 0 ? ? 1 ¼ 5/7 1

11. ESCUELA DE INGENIERÍA DE PETROLEOS 3. Is verified L*U = A 1 0 0 4 -2 -1 5/4 1 0 x 0 7/2 ¼ = ¼ 5/7 1 0 0 -13/14 4+0+0 -2+0+0 -1+0+0 4 -2 -1 5+0+0 -5/2+7/2+0 -5/4+1/4+0 = 5 1 -1 1+0+0 -1/2 +5/2+0 -1/4+5/28-13/14 1 2 -1

12. ESCUELA DE INGENIERÍA DE PETROLEOS 4. Now is clear “Y” of L*Y = b 1 0 0 Y1 9 5/4 1 0 * Y2 = 7 ¼ 5/7 1 Y3 12 Y1 = 9 Y1 = 9 5/4Y1 + Y2 = 7 Y2 = -17/4 1/4Y1 + 5/7Y2 +Y3 = 12 Y3 = 179/14

13. ESCUELA DE INGENIERÍA DE PETROLEOS 5. Se despeja “X” de U*X = Y 4 -2 -1 X1 9 0 14/4 ¼ * X2 = -17/4 0 0 -13/14 X3 179/14 4X1 -2X2 -X3 = 9 X1 = -17/13 14/4X2 +1/4X3 = -17/4 X2 = -3/13 RTA -13/14X3 = 179/14 X3 = -179/13

14. ESCUELA DE INGENIERÍA DE PETROLEOS http://www.unizar.es/aragon_tres/unidad6/Matrices/u6matte20.pdf http://en.wikipedia.org/wiki/LU_decomposition http://www.ditutor.com/matrices/matriz_simetrica.html http://www.cramster.com/reference/wiki.aspx?wiki_name=Band_matrix Eduardo Carrillo, Class en Presentation Ppt Methods Numeric's. Universidad Industrial de Santander 2010.