# 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.

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.