Direct Numerical Analysis

LU Decomposition

A system of equations

$$a_1x_1+b_1x_2+c_1x_3=A$$
$$a_2x_1+b_2x_2+c_2x_3=B$$
$$a_3x_1+b_3x_2+c_3x_3=C$$

can be written in matrix form as

$$[A] \{ x \}=\{C\}$$

where
$$
[A]=
\begin{bmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3 \\
\end{bmatrix}
$$

$$
\{ x \}= \left\{
\begin{matrix}
x_1 \\
x_2 \\
x_3 \\
\end{matrix} \right\}
$$

$$
\{ C \}= \left\{
\begin{matrix}
A \\
B \\
C \\
\end{matrix} \right\}
$$

Matrix \( [A] \) can be decomposed into a product of two matrices \( [L][U] \) where

$$
[L]=
\begin{bmatrix}
L_{11} & 0 & 0 \\
L_{21} & L_{22} & 0 \\
L_{31} & L_{32} & L_{33} \\
\end{bmatrix}
$$

$$
[U]=
\begin{bmatrix}
1 & U_{12} & U_{13} \\
0 & 1 & U_{23} \\
0 & 0 & 1 \\
\end{bmatrix}
$$

And $$[U]\{ x \} = \{ D \}$$
$$[L]\{ D \} = \{ C \}$$

\( \{ D \} \) can then be found by forward substitution. Giving

$$[U] \{ x \} = \{ D \}$$

Where \( \{ x \} \) can be found by backwards substitution.

Cramer's Rule

For a system

$$[A]\{ x \} = \{ B \}$$

Where \( [A] \) is reversible (\( det[A] \neq 0 \)). \( \{ x \} \) can be found by

$$x_i = \frac{det[A_i]}{det[A]}$$

Where \( [A_i] \) is the matrix \( [A] \) with the \(i\)th column replaced by the column matrix \( \{ B \} \).

eg. For

$$\begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} \begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = \begin{Bmatrix} B_1 \\ B_2 \end{Bmatrix}$$

$$x_1 = \frac{\begin{vmatrix} \ B_1 & A_{12} \\ \ B_2 & A_{22} \end{vmatrix}}{\begin{vmatrix}A_{11} & A_{12} \\ A_{21} & A_{22} \end{vmatrix}}$$

Inverse Method

For

$$[A] \{ x \} = \{ C \}$$
$$[A]^{-1} [A] \{ x \} = \{ C \} [A]^{-1}$$
$$[I] \{ x \} = \{ C \} [A]^{-1}$$
$$\{ x \} = \{ C \} [A]^{-1}$$