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 $a,b,c,d,e,f$