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ESCUELA DE INGENIERÍA DE PETROLEOS
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.
ESCUELA DE INGENIERÍA DE PETROLEOS ,[object Object],[object Object],[object Object],[object Object],[object Object]
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.
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.
ESCUELA DE INGENIERÍA DE PETROLEOS ,[object Object],[object Object],[object Object],[object Object]
ESCUELA DE INGENIERÍA DE PETROLEOS ,[object Object],[object Object]
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
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
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
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
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
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
ESCUELA DE INGENIERÍA DE PETROLEOS ,[object Object],[object Object],[object Object],[object Object],[object Object]

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LU Decomposition Fula

  • 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.
  • 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.
  • 7.
  • 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.