 Posted Thursday, February 19, 2009 9:55 AM
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Hello. I believe that the way your calculator parses input is giving a flawed result for LU decomposition. Tell me if I'm wrong, but I believe that LU decompositions are only possible on matrices that use only Type III elementary operations. Here is an example, I typed in:
A =
2 1 1
6 4 5
4 1 3
I believe your calculator parses that input and optimizes it to:
6 4 5
4 1 3
2 1 1
I believe this will cause a flawed result because you performed a Type I elementary operation by changing the rows. The U that your calculator returned was:
6.000 4.000 5.000 <---- Type I operation
0.000 -1.667 -0.333
0.000 0.000 -0.600
I did it by hand by only using Type III elementary operations and I got U to be:
2 1 1
0 1 2
0 0 3
I may be wrong, but if I'm correct, I hope this helps. Nice work altogether though!
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Posted Thursday, February 19, 2009 11:24 AM
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| Thanks for posting here. For numerical stability reasons, our implementation of LU decomposition uses pivoting. Therefore the original matrix equals to A= P x L x U where P is a permatation Matrix. You may try verifying the results (increase the number of decimals) using matrix multiplication and you will get the original matrix.
Trifon Triantafillidis | Lead Developer | 
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Posted Wednesday, June 15, 2011 4:22 AM
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Indeed very useful for me, Thanks for sharing.
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