I was asked to find LU decomposition of the following matrix without using a permutation:
$$A=\begin{bmatrix}1 & 2 & 3 \\ 1 & 2 & 4\\ 1 & 2 & 5 \end{bmatrix}$$
I found one of them but what is the second, any help?
$$L=\begin{bmatrix}1 & 0 & 0 \\1 & 1 & 0\\ 1 & 0 & 1 \end{bmatrix}, U=\begin{bmatrix}1 & 2 & 3 \\0 & 0 & 1\\ 0 & 0 & 2 \end{bmatrix}$$
You have found that:
$$A=\begin{bmatrix}1 & 2 & 3 \\ 1 & 2 & 4\\ 1 & 2 & 5 \end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\1 & 1 & 0\\ 1 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 2 & 3 \\0 & 0 & 1\\ 0 & 0 & 2 \end{bmatrix}=LU$$
Let $E$ corresponds to the elementary matrix of adding the third column to the second column.
Then we have $A=(LE)(E^{-1}U)$.
Notice that $E^{-1}U$ is still upper triangular, and $LE$ is still lower triangular. Furthermore, the diagonal of $LE$ consists of $1$ (if you impose such constraint). If you do not impose such constraint, consider $LU=(kL)(\frac1kU), k \ne 0$.
Alternatively,
$$\begin{bmatrix}1 & 0 & 0 \\1 & 1 & 0\\ 1 & x & 1 \end{bmatrix}\begin{bmatrix}1 & 2 & 3 \\0 & 0 & 1\\ 0 & 0 & y \end{bmatrix}$$
solve for $x$ and $y$.