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I am trying to find the LU decomposition for a given matrix $A$:

\begin{bmatrix} 1 & -1 & 3 & 4 & -2\\ -1 & 1 & -4 & -5 & 3\\ 2 & -2 & 5 & 9 & -5 \end{bmatrix}

$r_1 + r_2$ and $-2r_1 + r_3$ yields

\begin{bmatrix} 1 & -1 & 3 & 4 & -2\\ 0 & 0 & -1 & -1 & 1\\ 0 & 0 & -1 & 1 & -1 \end{bmatrix}

$-r_2 + r_3$ yields

\begin{bmatrix} 1 & -1 & 3 & 4 & -2\\ 0 & 0 & -1 & -1 & 1\\ 0 & 0 & 0 & 2 & -2 \end{bmatrix}

$2r_2 + r_3$ yields

\begin{bmatrix} 1 & -1 & 3 & 4 & -2\\ 0 & 0 & -1 & -1 & 1\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}

This is our $U.$

Hence our lower triangular matrix, $L$ is \begin{bmatrix} 1 & 0 & 0\\ -1 & 1 & 0\\ 2 & 1 & 1 \end{bmatrix}

However, from our last row operation, our $L$ is now \begin{bmatrix} 1 & 0 & 0\\ -1 & 1 & 0\\ 2 & -2 & 1 \end{bmatrix}

However, when I do $L \cdot U$ I get \begin{bmatrix} 1 & -1 & 3 & 4 & -2\\ -1 & 1 & -4 & -5 & 3\\ 2 & -2 & 8 & 10 & -6 \end{bmatrix}

Note that the last three entries in the third row are all off by $1$, so obviously I messed with my $L$ a bit in the last row, to no avail though. Any help?

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your $2r_2+r_3$ is neither correct nor necessary. –  Maesumi Feb 19 '13 at 1:56
    
Mind explaining why it's not necessary? Why don't I need use elimination to make the $a_{3,4}$ a zero? –  Joe Feb 19 '13 at 2:02
    
Umm, check out what happens in that step - the third column, in particular! –  kcrisman Feb 19 '13 at 2:12
    
Ah, I see it now. Thanks. –  Joe Feb 19 '13 at 2:19
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