Exploring the Need for a Second Pivot in LU Factorization with Partial Pivoting

[aaronfue]

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]

 

[SteamKing]

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?

 

[Ray Vickson]

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.