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Okay so consider a matrix $$ A = \left( \begin{array}{ccc} 0 & 1 & 1 \\ 1 & -2 & -1 \\ 1 & -1 & 1 \end{array} \right)$$ so obviously to get the upper and lower traiangular matrix for LU factorization I first pivot and then eliminate so I get $$ A = \left( \begin{array}{ccc} 1 & -2 & -1 \\ 0 & 1 & 1 \\ 1 & -1 & 1 \end{array} \right)$$ then making everything 0 under the diagonal I get the following

$$ U = \left( \begin{array}{ccc} 1 & -2 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right)$$ and $$ L = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0\\ 1 & 1 & 1 \end{array} \right)$$

Then since we pivoted we also get a permutation matrix $P$ $$ P = \left( \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right)$$

But if I try to go back using $A = P^TLU = (P^{-1}L)U$ I do not get my original matrix back. Note: the question is only asking to find the permutation matrix. I got the right permutation matrix and it matches the answer at the back of the book but I have a feeling I am just getting lucky.

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"I do not get my original matrix back." - really? Could you show your multiplication and the result? –  J. M. Dec 7 '11 at 4:37
    
My mistake. How do iclose this without marking an answer –  Tyler Hilton Dec 7 '11 at 4:50
    
There should be a delete link below your question and above the comments. –  J. M. Dec 7 '11 at 4:51
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You considered the matrix $A$
$$ A = \left( \begin{array}{ccc} 0 & 1 & 1 \\ 1 & -2 & -1 \\ 1 & -1 & 1 \end{array} \right)$$ and you multiplied by the matrix $P$ to get $$ PA = \left( \begin{array}{ccc} 1 & -2 & -1 \\ 0 & 1 & 1 \\ 1 & -1 & 1 \end{array} \right)$$

Your computations for $A$, $P$, $L$ and $U$ are right. Since $P = P^T = P^{-1}$ these are all the same. Computing the product $LU$ and $PA$ gives the same answer. I don't know what went wrong with your things.

Hope that helps! Sorry I couldn't say much more, but I think you just have to look at your computations.

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