LU factorization of a singular matrix I am trying to find the LU factorisation of the following matrix: $$A=\begin{pmatrix} 1 & 0 & 3 \\ 2 & 2 & 2 \\ 3 & 6 & -3 \end{pmatrix}.$$ Note that $A$ is singular. 
I proceed with Gaussian elimination:
$$A=\begin{pmatrix} 1 & 0 & 3 \\ 2 & 2 & 2 \\ 3 & 6 & -3 \end{pmatrix}\implies \begin{pmatrix} 1 & 0 & 3 \\ 0 & 2 & -4 \\ 0 & 6 & -12 \end{pmatrix}\implies \begin{pmatrix} 1 & 0 & 3 \\ 0 & 2 & -4 \\ 0 & 0 & 0 \end{pmatrix}.$$ Gaussian elimination has not provided an upper triangular matrix, $U$, so how can an LU factorisation be aquired? 

$$A=\begin{pmatrix} 1 & 0 & 3 \\ 2 & 2 & 2 \\ 3 & 6 & -3 \end{pmatrix}=\begin{pmatrix}
 a& 0 & 0\\ 
 b& c &0 \\ 
 d& e & f
\end{pmatrix}\times \begin{pmatrix}
 1&0  &3 \\ 
0 & 2 &-4 \\ 
0 & 0 & 0
\end{pmatrix}$$
 multiply it and equate to original matrix and find $a,b,c,d,e,f$
