Let A = [1 1 1 -1 -2 3 -2 -4 6] and b = [1 6 12]. Use an LU-decomposition to solve Ax = b.

Question

Let [latex]A = \left [ \begin{matrix} 1 & 1 & 1 \ -1 & -2 & 3 \ -2 & -4 & 6 \end{matrix} ight ] [/latex] and [latex]\vec{b} = \left [ \begin{matrix} 1 \ 6 \ 12 \end{matrix} ight ] [/latex]. Use an LU-decomposition to solve [latex]A\vec{x} = \vec{b}[/latex].

Accepted Answer

We first find an LU-decomposition for A. Row reducing gives

[latex]\left [ \begin{matrix} 1 & 1 & 1 \ -1 & -2 & 3 \ -2 & -4 & 6 \end{matrix} ight ] \begin{matrix} \ R_{2} + R_{1} \R_{3} + 2 R_{1} \end{matrix} \sim \left [ \begin{matrix} 1 & 1 & 1 \ 0 & -1 & 4 \ 0 & -2 & 8 \end{matrix} ight ] \begin{matrix} \ \R_{3} - 2R_{2} \end{matrix} \sim \left [ \begin{matrix} 1 & 1 & 1 \ 0 & -1 & 4 \ 0 & 0 & 0 \end{matrix} ight ] = U[/latex]

From our elementary row operations, we find that [latex]L = \left [ \begin{matrix} 1 & 0 & 0 \ -1 & 1 & 0 \ -2 & 2 & 1 \end{matrix} ight ] [/latex].

We let [latex]\vec{y} = U\vec{x}[/latex] and solve [latex]L\vec{y} = \vec{b}[/latex]. This gives

[latex]y_{1} = 1[/latex]

[latex]-y_{1} + y_{2} = 6[/latex]

[latex]-2y_{1} + 2y_{2} + y_{3} = 12[/latex]

Hence, [latex]y_{1} = 1, y_{2} = 6 + 1 = 7[/latex], and [latex]y_{3} = 12 + 2 - 14 = 0[/latex]. Next we solve [latex]U\vec{x} = \left [ \begin{matrix} 1 \ 7 \ 0 \end{matrix} ight ] [/latex].

[latex]x_{1} + x_{2} + x_{3} = 1[/latex]

[latex]-x_{2} + 4x_{3} = 7[/latex]

[latex]0x_{3} = 0[/latex]

This gives [latex]x_{3} = t \in \mathbb{R}, x_{2} = -7 + 4t,[/latex] and [latex]x_{1} = 1 + (7 - 4t) - t = 8 - 5t[/latex]. Thus,

[latex]\vec{x} = \left [ \begin{matrix} 8 - 5t \ -7 + 4t \ t \end{matrix} ight ] = \left [ \begin{matrix} 8 \ -7 \ 0 \end{matrix} ight ] + t \left [ \begin{matrix} -5 \ 4 \ 1 \end{matrix} ight ] ,\ \ \ \ \ t \in \mathbb{R}[/latex]

Question 3.7.5: Let A = [1 1 1 -1 -2 3 -2 -4 6] and b = [1 6 12]. Use an LU-...

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