Compute the LU factorization

**wopashui** · January 6th 2011, 06:17 PM

Compute the LU factorization of the matrix
A=
[2 1 1]
[4 1 0]
[-2 2 1]

and solve A[ $x_1$ $x_2$ $x_3$ $]^T$ = ][8 11 3 $]^T$
using back substitution.

Let A=
[1 1 1]
[1 1 2]
[1 2 5]
Find a permutation matrix $P$ $\in$ $R^{3,3}$ so that PA has an LU factoriza-
tion. Compute L and U.

this is should be an easy question, but it's been so long since i ve done matrix, so any help will be appreaciated..

**dwsmith** · January 6th 2011, 07:44 PM

$\displaystyle LU=\begin{bmatrix} l_{11} & 0 & 0\\ l_{21} & l_{22} & 0\\ l_{31} & l_{32} & l_{33} \end{bmatrix} \begin{bmatrix} u_{11} & u_{12} & u_{13}\\ 0 & u_{22} & u_{23}\\ 0 & 0 & u_{33} \end{bmatrix}=\begin{bmatrix}2&1&1\\4&1&0\\-2&2&1\end{bmatrix}=A$

$\displaystyle A=\sum_{i=1}^{3}L_{ji}U_{ik}$

**pickslides** · January 6th 2011, 08:08 PM

$\displaystyle \begin{bmatrix}2&1&1\\4&1&0\\-2&2&1\end{bmatrix}= \begin{bmatrix} l_{11} & 0 & 0\\ l_{21} & l_{22} & 0\\ l_{31} & l_{32} & l_{33} \end{bmatrix} \begin{bmatrix} u_{11} & u_{12} & u_{13}\\ 0 & u_{22} & u_{23}\\ 0 & 0 & u_{33} \end{bmatrix}$

If this gives too many variables to solve for, make $\displaystyle l_{nn}=1$ i.e.

$\displaystyle \begin{bmatrix}2&1&1\\4&1&0\\-2&2&1\end{bmatrix}= \begin{bmatrix} 1 & 0 & 0\\ l_{21} & 1 & 0\\ l_{31} & l_{32} & 1 \end{bmatrix} \begin{bmatrix} u_{11} & u_{12} & u_{13}\\ 0 & u_{22} & u_{23}\\ 0 & 0 & u_{33} \end{bmatrix}$

**Random Variable** · January 6th 2011, 08:15 PM

$\displaystyle \begin{bmatrix} 2 & 1 & 1\\ 4 & 1 & 0\\ -2 & 2 & 1 \end{bmatrix}$ $\displaystyle \rightarrow \displaystyle \begin{bmatrix} 2 & 1 & 1\\ 0 & -1 & -2\\ 0 & 3 & 2 \end{bmatrix}$ $\rightarrow \displaystyle \begin{bmatrix} 2 & 1 & 1\\ 0 & -1 & -2\\ 0 & 0 & -4 \end{bmatrix} = U$

$L = \displaystyle \begin{bmatrix} 1 & 0 & 0\\ 2 & 1 & 0\\ -1 & -3 & 1 \end{bmatrix}$ (where 2,-1,and -3 are the multipliers used to eliminate those three entires of the original matrix)

first solve $\displaystyle Lc = \begin{bmatrix} 1 & 0 & 0\\ 2 & 1 & 0\\ -1 & -3 & 1 \end{bmatrix} \begin{bmatrix} c_{1}\\ c_{2}\\ c_{3} \end{bmatrix}= \begin{bmatrix} 8\\ 11\\ 3 \end{bmatrix}$

then solve $\displaystyle Ux = \begin{bmatrix} 2 & 1 & 1\\ 0 & -1 & -2\\ 0 & 0 & -4 \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}= \begin{bmatrix} c_{1}\\ c_{2}\\ c_{3} \end{bmatrix}$

**wopashui** · January 8th 2011, 07:59 PM

Originally Posted by Random Variable

$\displaystyle \begin{bmatrix} 2 & 1 & 1\\ 4 & 1 & 0\\ -2 & 2 & 1 \end{bmatrix}$ $\displaystyle \rightarrow \displaystyle \begin{bmatrix} 2 & 1 & 1\\ 0 & -1 & -2\\ 0 & 3 & 2 \end{bmatrix}$ $\rightarrow \displaystyle \begin{bmatrix} 2 & 1 & 1\\ 0 & -1 & -2\\ 0 & 0 & -4 \end{bmatrix} = U$

$L = \displaystyle \begin{bmatrix} 1 & 0 & 0\\ 2 & 1 & 0\\ -1 & -3 & 1 \end{bmatrix}$ (where 2,-1,and -3 are the multipliers used to eliminate those three entires of the original matrix)

first solve $\displaystyle Lc = \begin{bmatrix} 1 & 0 & 0\\ 2 & 1 & 0\\ -1 & -3 & 1 \end{bmatrix} \begin{bmatrix} c_{1}\\ c_{2}\\ c_{3} \end{bmatrix}= \begin{bmatrix} 8\\ 11\\ 3 \end{bmatrix}$

then solve $\displaystyle Ux = \begin{bmatrix} 2 & 1 & 1\\ 0 & -1 & -2\\ 0 & 0 & -4 \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}= \begin{bmatrix} c_{1}\\ c_{2}\\ c_{3} \end{bmatrix}$

thx, I got the idea, so how do I find the permutation matrix, is there something to do with the identity matrix?

**Random Variable** · January 9th 2011, 03:10 AM

Partial pivoting is about exchanging rows so that you are always pivoting on the entry in the column with largest absolute value. It's to try to prevent what's referred to as swamping.

Since all the entries in the first column are the same, you don't need to exchange any rows.

$\displaystyle \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 2\\ 1 & 2 & 5 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & 1\\ 0(1) & 0 & 1\\ 0(1) & 1 & 4 \end{bmatrix}$

It's a good idea to put the multipliers in parenthesis so that you can keep track of them when you exchange rows.

In this problem it's not about swamping, but rather the fact that you can't pivot on a zero. So exchange rows 2 and 3. The permuatation matrix for such a exchange is $\displaystyle \begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}$

$\displaystyle \rightarrow \begin{bmatrix} 1 & 1 & 1\\ 0(1) & 1 & 4\\ 0(1) & 0 & 1 \end{bmatrix}$

then $\displaystyle PA = \begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 2\\ 1 & 2 & 5 \end{bmatrix}$ $\displaystyle = LU = \begin{bmatrix} 1 & 0 & 0\\ 1 & 1 & 0\\ 1 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 & 1\\ 0 & 1 & 4\\ 0 & 0 & 1 \end{bmatrix}$

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