- #1

mathmari

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MHB

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We have the matrix $$A=\begin{pmatrix}1 & -2 & 1 \\ 3 & -1 & 2 \\ -2 & -2 & 1\end{pmatrix}$$ I want to apply the LR decomposition with column pivoting. First we permutate the first two rows and we get $$A=\begin{pmatrix}3 & -1 & 2 \\ 1 & -2 & 1 \\ -2 & -2 & 1\end{pmatrix}$$ Then we apply the Gauss algorithm and we get $$A=\begin{pmatrix}3 & -1 & 2 \\ 0 & -5/3 & 1/3 \\ 0 & -8/3 & 7/3\end{pmatrix}$$ Then the largest value of the submatrix in absolute value is $8/3$ so we permutate the second and third row and we get $$A=\begin{pmatrix}3 & -1 & 2 \\ 0 & -8/3 & 7/3 \\ 0 & -5/3 & 1/3 \end{pmatrix}$$ Then we apply again the Gauss algorithm and we get $$A=\begin{pmatrix}3 & -1 & 2 \\ 0 & -8/3 & 7/3 \\ 0 & 0 & -9/8 \end{pmatrix}$$ From the above we get the matrices $$R=\begin{pmatrix}3 & -1 & 2 \\ 0 & -8/3 & 7/3 \\ 0 & 0 & -9/8 \end{pmatrix} \ \text{ and } \ L=\begin{pmatrix}1 & 0 & 0 \\ 1/3 & 1 & 0 \\ -2/3 & 5/8 & 1 \end{pmatrix}$$ To check if the results are correct we have to check if $LR=PA$, where $P=P_1P_0$ where $P_0$ and $P_1$ are the permutation matrices, or not? (Wondering)

How are the permutation matrices $P_0$ and $P_1$ defined in this case? $P_0$ is the one that permutated the first two rows and $P_1$ is the one that permutated the second and third row, right? But how can we calculate the matric $P$ ? (Wondering)