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Why is there a second pivot?

  • Thread starter aaronfue
  • Start date
  • #1
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Homework Statement



Use LU factorization with partial pivoting for the following set of equations:

3x1 - 2x2 + x3 = -10
2x1 + 6x2 - 4x3 = 44
-8x1 - 2x2 + 5x3 = -26


The Attempt at a Solution


I made an attempt to solve this problem, but my answer was wrong compared to the book. There was an additional partial pivot after setting elements 21 & 31 equal to zero. I just would like to know why?

The following is what I got for my L and U matrices:

U=
[ -8 -2 5 ]
[ 0 5.5 -3.25]
[ 0 0 1.25]

L=
[1 0 0]
[.25 1 0]
[.775 0.5 1]
 

Answers and Replies

  • #2
SteamKing
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Do your L and U matrices, when multiplied together, give the original matrix of coefficients? The first row of your U is identical to the third row of the coefficient matrix. Coincidence?
 
  • #3
Ray Vickson
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Homework Statement



Use LU factorization with partial pivoting for the following set of equations:

3x1 - 2x2 + x3 = -10
2x1 + 6x2 - 4x3 = 44
-8x1 - 2x2 + 5x3 = -26


The Attempt at a Solution


I made an attempt to solve this problem, but my answer was wrong compared to the book. There was an additional partial pivot after setting elements 21 & 31 equal to zero. I just would like to know why?

The following is what I got for my L and U matrices:

U=
[ -8 -2 5 ]
[ 0 5.5 -3.25]
[ 0 0 1.25]

L=
[1 0 0]
[.25 1 0]
[.775 0.5 1]
Please show your work details, step-by-step. When I do it (with [-8,-2,5] in row 1 and [3,-2,1] in row 3) I get a different U from yours and do not need any more "partial" pivots; straight pivoting works perfectly well. Or, maybe, I have not understood your question---but I still get a different U.
 

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