PX153 - K9 - LU decomposition method

\begin{aligned} A & = [\begin{array}{c} 1 & 0 & 0 \\ L_{21} & 1 & 0 \\ L_{31} & L_{32} & 1 \end{array}] [\begin{array}{c} U_{11} & U_{12} & U_{13} \\ 0 & U_{22} & U_{23} \\ 0 & 0 & U_{33} \end{array}] \\ = [\begin{array}{c} U_{11} & U_{12} & U_{13} \\ L_{21} U_{11} & L_{21} U_{12} + U_{22} & L_{21} U_{13} + U_{23} \\ L_{31} U_{11} & L_{31} U_{12} + L_{32} U_{22} & L_{31} U_{13} + L_{32} U_{23} + U_{33} \end{array}] \end{aligned}

det A = det L \cdot det U = U_{11} \times U_{22} \times U_{33}

usually the fastest method to find the determinant, $| A |$ , and the solution vector, $\vec{x}$ , for simultaneous equations
$O (n \times n)$ operations
eg: $A = [\begin{matrix} 2 & 4 & 3 \\ 1 & - 2 & - 2 \\ - 3 & 3 & 2 \end{matrix}], \vec{x} = [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}], \vec{b} = [\begin{matrix} 4 \\ 0 \\ - 7 \end{matrix}]$
- $1^{s t}$ row $: U_{11} = 2$ , $U_{12} = 4$ , $U_{13} = 3$
- $2^{n d}$ row $: L_{21} U_{11} = 1 ⟹ L_{21} = \frac{1}{2}$
- $3^{r d}$ row $: L_{31} U_{11} = - 3 ⟹ L_{31} = - \frac{3}{2}$
- remaining:
  $L_{21} U_{12} + U_{22} = - 2 ⟹ U_{22} = - 4$
  $L_{21} U_{13} + U_{23} = - 2 ⟹ U_{23} = - \frac{7}{2}$
  $L_{31} U_{12} + L_{32} U_{22} = 3 ⟹ L_{32} = - \frac{3}{4}$
  $L_{31} U_{13} + L_{32} U_{23} + U_{33} = 2 ⟹ U_{33} = - \frac{11}{8}$

A = [\begin{matrix} 1 & 0 & 0 \\ \frac{1}{2} & 1 & 0 \\ - \frac{3}{2} & - \frac{9}{4} & 1 \end{matrix}] [\begin{matrix} 2 & 4 & 3 \\ 0 & - 4 & - \frac{7}{3} \\ 0 & 0 & - \frac{11}{8} \end{matrix}]

det A = (1) \times (2 \times (- 4) \times (- \frac{11}{8})) = - 11

\begin{matrix} A \vec{x} = \vec{b} \\ L U \vec{x} = \vec{b} \\ l e t, U \vec{x} = \vec{y} ⟹ L \vec{y} = b \\ L \vec{y} = b \\ [\begin{matrix} 1 & 0 & 0 \\ \frac{1}{2} & 1 & 0 \\ - \frac{3}{2} & - \frac{9}{4} & 1 \end{matrix}] [\begin{matrix} y_{1} \\ y_{2} \\ y_{3} \end{matrix}] = [\begin{matrix} 4 \\ 0 \\ - 7 \end{matrix}] \\ y_{1} = 4 \\ y_{2} = - 2 \\ y_{3} = - \frac{11}{2} \\ U \vec{x} = \vec{y} \\ [\begin{matrix} 2 & 4 & 3 \\ 0 & - 4 & - \frac{7}{3} \\ 0 & 0 & - \frac{11}{8} \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 4 \\ - 2 \\ - \frac{11}{2} \end{matrix}] \\ x_{3} = 4 \\ x_{2} = - 3 \\ x_{1} = 2 \\ ∴ \vec{x} = [\begin{matrix} 2 \\ - 3 \\ 4 \end{matrix}] \end{matrix}