Let B = [2 1 -1 -4 3 3 6 8 -3] and b = [3 -13 4]. Use an LU-decomposition of B to solve Bx = b.

Question

Let [latex]B = \left [ \begin{matrix} 2 & 1 & -1 \ -4 & 3 & 3 \ 6 & 8 & -3 \end{matrix} ight ] [/latex] and [latex]\vec{b} = \left [ \begin{matrix} 3 \ -13 \ 4 \end{matrix} ight ] [/latex]. Use an LU-decomposition of B to solve [latex]B\vec{x} = \vec{b}[/latex].

Accepted Answer

In Example 3.7.2 we found an LU-decomposition of B. We write [latex]B\vec{x} = \vec{b}[/latex] as [latex]LU\vec{x} = \vec{b}[/latex] and take [latex]\vec{y} = U\vec{x}[/latex]. Writing out the system [latex]L\vec{y} = \vec{b}[/latex], we get

[latex]y_{1} = 3[/latex]

[latex]-2y_{1} + y_{2} = -13[/latex]

[latex]3y_{1} + y_{2} + y_{3} = 4[/latex]

Using forward-substitution, we find that [latex]y_{1} = 3,[/latex] so [latex]y_{2} = -13 + 2(3) = -7[/latex] and [latex]y_{3} = 4 - 3(3) - (-7) = 2[/latex]. Hence, [latex]\vec{y} = \left [ \begin{matrix} 3 \ -7 \ 2 \end{matrix} ight ] [/latex].

Thus, our system [latex]U\vec{x} = \vec{y}[/latex] is

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

[latex]5x_{2} + x_{3} = -7[/latex]

[latex]-x_{3} = 2[/latex]

Using back-substitution, we get [latex]x_{3} = -2, 5x_{2} = -7 - (-2) ⇒ x_{2} = -1[/latex] and [latex]2x_{1} = 3 - (-1) + (-2) ⇒ x_{1} = 1[/latex]. Thus, the solution is [latex]\vec{x} = \left [ \begin{matrix} 1 \ -1 \ -2 \end{matrix} ight ] [/latex].

Question 3.7.4: Let B = [2 1 -1 -4 3 3 6 8 -3] and b = [3 -13 4]. Use an LU-...

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