LU decomposition on 5 by 3 matrix. This is a problem given in a quiz. Even after reading up a related question, I cannot figure it out.

Determine an LU-factorization of $$
C = 
\begin{bmatrix}
3 & 1 & -4 \\
6 & -3 & 10 \\
-9 & 5 & -11 \\
-3 & 0 & -7 \\
6 & -4 & 2
\end{bmatrix}
$$

In the related question, I can acquire an upper triangular matrix with row operations and permute the rows. However, I can't turn this into an upper triangular matrix, only a lower one, but then it becomes UL-factorization.
Any insight is welcome.
 A: Because $C$ is not square, $U$ will not be in upper triangular form but rather in row echelon form.
In your case, proceed as you would for a square matrix, that is, reduce $C$ to row echelon form to get $U$, and keep track of the multipliers you used to get $L$:
Use the first row, $R_1$, to make the first entries of the other rows $0$:
$$\begin{bmatrix}
3 &  1 &  -4\\
0 & -5 &  18\\
0 &  8 & -23\\
0 &  1 & -11\\
0 & -6 &  10\end{bmatrix}
\begin{array}
\\ \\
R_2 - 2R_1\\
R_3 + 3R_1\\
R_4 + 1R_1\\
R_5 - 2R_1.\end{array}$$
Next, use the second row to make the second entries of subsequent rows $0$:
$$\begin{bmatrix}
3 &  1 &  -4  \\
0 & -5 &  18  \\
0 &  0 &  29/5\\
0 &  0 & -37/5\\
0 &  0 & -58/5\end{bmatrix}
\begin{array}
\\ \\ \\
R_3 + 8/5R_2\\
R_4 + 1/5R_2\\
R_5 - 6/5R_2.\end{array}$$
Finally, use the third row to make the third entries of subsequent rows $0$:
$$U = \begin{bmatrix}
3 &  1 &  -4  \\
0 & -5 &  18  \\
0 &  0 &  29/5\\
0 &  0 &   0  \\
0 &  0 &   0  \end{bmatrix}
\begin{array}
\\ \\ \\ \\
R_4 + 37/29R_3\\
R_5 + 2R_3.\end{array}$$
The lower entries of $L$ are the negatives of the multipliers you used to reduce $A$ to $U$, hence,
$$L = \begin{bmatrix}
 1 &  0   &  0     & 0 & 0\\
 2 &  1   &  0     & 0 & 0\\
-3 & -8/5 &  1     & 0 & 0\\
-1 & -1/5 & -37/29 & 1 & 0\\
 2 &  6/5 & -2     & 0 & 1\end{bmatrix}.$$
You can check that $LU = C.$
If you use a program that requires a square matrix for input, give $C$ dummy fourth and fifth columns.  
