Elimination Matricies can be used to derive a LU factorization: First, this gives an upper triangular matrix
\begin{equation} U = M_{n-1, n-1} \dots M_{22} M_{11} A \end{equation}
We can also create the inverses of each of these:
\begin{equation} A = L_{11} L_{22} \dots L_{n-1,n-1} \cdot M_{n-1,n-1} \dots M_{22} \cdot M_{11} \cdot A \end{equation}
The first half L_{j} composes a lower triangular matrix; the second half M_{j} which composes a upper triangular matrix. Then, this helps solve:
\begin{equation} Ac = b \end{equation}
Because we can factor first to:
\begin{equation} \left(LU\right) c = b \end{equation}
Then, this makes it really easy to solve, because we can.
\begin{equation} \hat{c} = Uc \end{equation}
then, we can solve L\hat{c} = b using forward substitution; then we can solve Uc = \hat{c} using back substitution. Notice that for every new b, we don’t need to perform Gaussian elimination.