Find an LU-decomposition of A = [2 -1 4 4 -1 6 -1 -1 3].

Question

Find an LU-decomposition of [latex]A = \left [ \begin{matrix} 2 & -1 & 4 \ 4 & -1 & 6 \ -1 & -1 & 3 \end{matrix} ight ] [/latex].

Accepted Answer

Row reducing and keeping track of our row operations gives

[latex]\left [ \begin{matrix} 2 & -1 & 4 \ 4 & -1 & 6 \ -1 & -1 & 3 \end{matrix} ight ] \begin{matrix} \ R_{2} - 2R_{1} \R_{3} + \frac{1}{2} R_{1} \end{matrix} \sim \left [ \begin{matrix} 2 & -1 & 4 \ 0 & 1 & -2 \ 0 & -3/2 & 5 \end{matrix} ight ] \begin{matrix} \ \R_{3} + \frac{3}{2} R_{2} \end{matrix} \sim \left [ \begin{matrix} 2 & -1 & 4 \ 0 & 1 & -2 \ 0 & 0 & 2 \end{matrix} ight ] = U[/latex]

We have [latex]E_{3} E_{2} E_{1} A = U[/latex], so [latex]A = E_{1} ^{-1} E_{2}^{-1} E_{3}^{-1} U[/latex], where

[latex]E_{1} = \left [ \begin{matrix} 1 & 0 & 0 \ -2 & 1 & 0 \ 0 & 0 & 1 \end{matrix} ight ] , E_{2} = \left [ \begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 1/2 & 0 & 1 \end{matrix} ight ] , E_{3} = \left [ \begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 3/2 & 1 \end{matrix} ight ] [/latex]

Hence, we let

[latex]L = E_{1} ^{-1} E_{2}^{-1} E_{3}^{-1} = \left [ \begin{matrix} 1 & 0 & 0 \ 2 & 1 & 0 \ 0 & 0 & 1 \end{matrix} ight ] \left [ \begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ -1/2 & 0 & 1 \end{matrix} ight ] \left [ \begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & -3/2 & 1 \end{matrix} ight ] [/latex]

[latex]= \left [ \begin{matrix} 1 & 0 & 0 \ 2 & 1 & 0 \ 0 & 0 & 1 \end{matrix} ight ] \left [ \begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ -1/2 & -3/2 & 1 \end{matrix} ight ] = \left [ \begin{matrix} 1 & 0 & 0 \ 2 & 1 & 0 \ -1/2 & -3/2 & 1 \end{matrix} ight ][/latex]

And we get A = LU.

Question 3.7.1: Find an LU-decomposition of A = [2 -1 4 4 -1 6 -1 -1 3].

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