Discussion
Consider solving the following system of simultaneous equations using LU decomposition.
\[ \begin{aligned} x_{1} + x_{2} - 2x_{3} &= 4 \\ x_{1} + 3x_{2} - x_{3} &= 7 \\ 2x_{1} + x_{2} - 5x_{3} &= 7 \end{aligned} \]
where \(L\) and \(U\) are denoted as
\[ L = \begin{pmatrix} L_{11} & 0 & 0 \\ L_{21} & L_{22} & 0 \\ L_{31} & L_{32} & L_{33} \end{pmatrix}, \]
\[ U = \begin{pmatrix} U_{11} & U_{12} & U_{13} \\ 0 & U_{22} & U_{23} \\ 0 & 0 & U_{33} \end{pmatrix} \]
Which one of the following is the correct combination of values for \(L_{32}\), \(U_{33}\), and \(x_{1}\)?
Answer: D
The values are \(L_{32} = -\frac{1}{2}\), \(U_{33} = -\frac{1}{2}\) and \(x_{1} = 0\).
Use Doolittle's form, with 1s on the diagonal of \(L\). The first row of \(U\) is the first equation, (1, 1, −2), and the multipliers are \(L_{21} = 1\), \(L_{31} = 2\).
\[ \begin{aligned} R_{2} - R_{1} &: (0,\ 2,\ 1) \\ R_{3} - 2R_{1} &: (0,\ -1,\ -1) \\ L_{32} &= \tfrac{-1}{2} = -\tfrac{1}{2} \\ U_{33} &= -1 - \left(-\tfrac{1}{2}\right)(1) = -\tfrac{1}{2} \end{aligned} \]
Forward substitution with \(Ly = b\) gives \(y = (4,\ 3,\ \tfrac{1}{2})\), and back substitution with \(Ux = y\) gives:
\[ \begin{aligned} x_{3} &= \tfrac{1/2}{-1/2} = -1 \\ 2x_{2} + x_{3} &= 3 \Rightarrow x_{2} = 2 \\ x_{1} + x_{2} - 2x_{3} &= 4 \Rightarrow x_{1} = 0 \end{aligned} \]
Check in the third equation: \(0 + 2 + 5 = 7\). The options with \(L_{32} = 2\) take the multiplier of row 1 instead of row 2, and \(U_{33} = 2\) or \(x_{1} = -1\) come from sign errors in the elimination.
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