Trouble determining L of a matrix Here is the matrix:  
$$\mathbf Α = \begin{bmatrix} 
3  & -6  & 6 & -3\\
6  & -10 & 8 & -3\\
-1 & -4  & 10& -3
\end{bmatrix}.
$$
I am supposed to find an LU factorization** of the matrix $\mathbf A$. 
i.e $\mathbf A \mathbf x = \mathbf b$ is equal to $\mathbf L(\mathbf U \mathbf x)=\mathbf b$ if I'm correct.
Here is my U (U is the upper triangle, I row reduced without scaling and note that pivots aren't supposed to be $1$):
$$\mathbf{U} = \begin{bmatrix}
3  & -6  & 6  & -3\\
0  &  2  & -4 &  3\\
0  &  0  &  0 & 5
\end{bmatrix}.
$$
Now I must find L, I understand $\mathbf L$ must form the lower triangle, but strangely pivots must now be $1$? 
I've been having trouble determining L?
 A: Matrix $\mathbf L$ is a $3\times 3$ matrix in the form $\mathbf L = \begin{bmatrix} 1 & 0 & 0\\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{bmatrix}$, where $l_{ij}$ are the multipliers. How are $l_{ij}$ defined?
In order to find the $\mathbf{U}$ matrix, we do row operations. For example, in order to find the second row of the $\mathbf{U}$ matrix we do something like: 
$$ \mathbf{U}_2 := R_2 - l_{21} R_1,\tag 1$$
 where $\mathbf{U}_k$ is the $k-$ row of matrix $\mathbf U$ and $R_k$ is the $k-$ row of matrix $A$. In a similar way, we have that: 
$$\mathbf{U}_3: = (R_3 - l_{31}R_1) - l_{32} \mathbf{U}_2.\tag 2$$
Also, it is obvious that $$\mathbf{U}_1 = R_1.\tag 3$$
Now, taking advantage of $(1),(2),(3)$ we can easily verify that $$\mathbf L \mathbf U = \begin{bmatrix} 1 & 0 & 0\\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{bmatrix}\cdot \begin{bmatrix}\mathbf U_1 \\ \mathbf{U}_2 \\ \mathbf{U}_3\end{bmatrix} = \begin{bmatrix} R_1 \\ R_2 \\ R_3 \end{bmatrix} = A.$$
Below is the desirable matrix $\mathbf L$.

 Matrix $\mathbf L = \begin{bmatrix} 1 & 0 & 0 \\2 & 1 & 0 \\ -1/3 & -3 & 1\end{bmatrix}$.

