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- Problem Statement
- Find the LU Decomposition of the matrix below

- Relevant Equations
- M = LU

Here is the initial matrix M:

[tex] M = \begin{bmatrix} 3 & 1 & 6 \\ -6 & 0 & -16 \\ 0 & 8 & -17 \end{bmatrix} [/tex]

I have used the shortcut method outlined in this youtube video: LU Decomposition Shortcut Method.

Here are the row reductions that I went through in order to get my U matrix:

1. [itex] R_3 - 8 R_1 [/itex]

[tex] = \begin{bmatrix} 3 & 1 & 6 \\ -6 & 0 & -16 \\ -24 & 0 & -65 \end{bmatrix} [/tex]

2. [itex] R_3 - 4 R_2 [/itex]

[tex] = \begin{bmatrix} 3 & 1 & 6 \\ -6 & 0 & -16 \\ 0 & 0 & -1 \end{bmatrix} [/tex]

3. [itex] R_2 + 2 R_1 [/itex]

[tex] U = \begin{bmatrix} 3 & 1 & 6 \\ 0 & 2 & -4 \\ 0 & 0 & -1 \end{bmatrix} [/tex]

This yields the correct U matrix, however, I get a slightly different L matrix to the answer. My L matrix is:

[tex] L = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 4 & 8 & 1 \end{bmatrix} [/tex]

In the answer, the final row reads 0, 4, 1.

Why would this be the case?

Any help is greatly appreciated.

[tex] M = \begin{bmatrix} 3 & 1 & 6 \\ -6 & 0 & -16 \\ 0 & 8 & -17 \end{bmatrix} [/tex]

I have used the shortcut method outlined in this youtube video: LU Decomposition Shortcut Method.

Here are the row reductions that I went through in order to get my U matrix:

1. [itex] R_3 - 8 R_1 [/itex]

[tex] = \begin{bmatrix} 3 & 1 & 6 \\ -6 & 0 & -16 \\ -24 & 0 & -65 \end{bmatrix} [/tex]

2. [itex] R_3 - 4 R_2 [/itex]

[tex] = \begin{bmatrix} 3 & 1 & 6 \\ -6 & 0 & -16 \\ 0 & 0 & -1 \end{bmatrix} [/tex]

3. [itex] R_2 + 2 R_1 [/itex]

[tex] U = \begin{bmatrix} 3 & 1 & 6 \\ 0 & 2 & -4 \\ 0 & 0 & -1 \end{bmatrix} [/tex]

This yields the correct U matrix, however, I get a slightly different L matrix to the answer. My L matrix is:

[tex] L = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 4 & 8 & 1 \end{bmatrix} [/tex]

In the answer, the final row reads 0, 4, 1.

Why would this be the case?

Any help is greatly appreciated.