LU Decomposition of $A$ I am currently working with a practice problem in which we are asked to find the $LU$ factorization of some matrix $A$, given below:
$$
  A= \begin{pmatrix}
    1 & 2 & 3 \\
    2 & 4 & 7 \\
    3 & 6 & 10 \\
    3 & 6 & 10 \\
   \end{pmatrix}
$$
In the first step, I added $-2$ of the first row to the second, $-3$ of the first row to the third, and $-3$ of the first row to the forth to give the matrix:
$$
    \begin{pmatrix}
    1 & 2 & 3 \\
    0 & 0 & 1 \\
    0 & 0 & 1 \\
    0 & 0 & 1 \\
    \end{pmatrix}
$$
In the second step, I added $-1$ of the second row to the third row, and $-1$ of the second row to the fourth row, to give the matrix $U$:
$$
   U = \begin{pmatrix}
    1 & 2 & 3 \\
    0 & 0 & 1 \\
    0 & 0 & 0 \\
    0 & 0 & 0 \\
    \end{pmatrix}
$$
Given these steps, I determined the matrix $L$ to be:
$$
   L= \begin{pmatrix}
    1 & 0 & 0 & 0 \\
    -2 & 1 & 0 & 0 \\
    -3 & -1 & 1 & 0 \\
    -3 & -1 & 0 & 1 \\
    \end{pmatrix}
$$
However, although the answer $U$ was found to be correct, the answer $L$ was given the following form:
$$
   L= \begin{pmatrix}
    1 & 0 & 0 & 0 \\
    2 & 1 & 0 & 0 \\
    3 & 1 & 1 & 0 \\
    3 & 1 & 0 & 1 \\
    \end{pmatrix}
$$
However, I don't understand why all of the terms that I found to be negative are positive in the given answer. Can anyone help me figure it out? Thanks!
(Note: this is not homework, but independent study.)

Using notation from here, you obtained that
$$L_{24}(-1)L_{23}(-1)L_{14}(-3)L_{13}(-3)L_{12}(-2)A = U$$
so
$$A = L_{12}(-2)^{-1}L_{13}(-3)^{-1}L_{14}(-3)^{-1}L_{23}(-1)^{-1}L_{24}(-1)^{-1}U = L_{12}(2)L_{13}(3)L_{14}(3)L_{23}(1)L_{24}(1)U$$
where $L = L_{12}(2)L_{13}(3)L_{14}(3)L_{23}(1)L_{24}(1)$ is lower triangular with unit diagonal.
Applying these transformations in order to the identity matrix $I$ yields the correct $L$.
