Método de Gauss-Seidel
O que muda nesse método é a forma de usar a variável \(x_i\) calculada, aplicando imediatamente para o cálculo do próximo valor.
Veja o exemplo, a partir da matriz:
\[ \left[\begin{matrix}3 &0 &2 \\-1 &6 &-4 \\-2 &2 &-11 \end{matrix} \right. \left| \begin{matrix} 23 \\ 15 \\ -42 \end{matrix} \right] \]Iniciamos a matriz de valores de \(x\):
\[x =\underset{\color{red}\begin{matrix} x_1\textrm{ } & x_2\textrm{ } & x_3\textrm{ } \end{matrix} }{\left[\begin{matrix} 0\textrm{,} & 0\textrm{,} & 0 \end{matrix} \right]}\]Iteração 1:
| Matriz | Operação |
|---|---|
| \[\left[\begin{matrix} \color{blue} 3 &\color{green} 0 \color{black} &\color{orange} 2 \color{black} \\-1 &6 &-4 \\-2 &2 &-11 \end{matrix} \right. \left| \begin{matrix} \color{red} 23 \\ 15 \\ -42 \end{matrix} \right]\] | \[\begin{matrix}x_1=\frac{1}{\color{blue}3\color{black}}(\color{red}23\color{black} - ((\color{green} 0 \times 0 \color{black}) +(\color{orange} 2 \times 0\color{black}))) \end{matrix} = \colorbox{grey}{7.66667} \] \[x = \left[\begin{matrix} 7.66667 &\color{green} 0 \color{black} &\color{orange} 0 \color{black} \end{matrix} \right]\] |
| \[\left[\begin{matrix}3 &0 &2 \\\color{green} -1 \color{black} & \color{blue} 6 &\color{orange} -4 \color{black} \\-2 &2 &-11 \end{matrix} \right. \left| \begin{matrix} 23 \\ \color{red} 15 \\ -42 \end{matrix} \right]\] | \[\begin{matrix}x_2=\frac{1}{\color{blue}6\color{black}}(\color{red}15\color{black} - ((\color{green} -1 \times 7.66667 \color{black}) +(\color{orange} -4 \times 0 \color{black}))) \end{matrix} = \colorbox{grey}{3.77778} \] \[x = \left[\begin{matrix}\color{green} 7.66667 \color{black} & 3.77778 &\color{orange} 0 \color{black} \end{matrix} \right]\] |
| \[\left[\begin{matrix}3 &0 &2 \\-1 &6 &-4 \\\color{green} -2 \color{black} &\color{orange} 2 \color{black} & \color{blue} -11 \end{matrix} \right. \left| \begin{matrix} 23 \\ 15 \\ \color{red} -42 \end{matrix} \right]\] | \[\begin{matrix}x_3=\frac{1}{\color{blue}-11\color{black}}(\color{red}-42\color{black} - ((\color{green} -2 \times 7.66667 \color{black}) +(\color{orange} 2 \times 3.77778 \color{black}))) \end{matrix} = \colorbox{grey}{3.11111} \] \[x = \left[\begin{matrix}\color{green} 7.66667 \color{black} &\color{orange} 3.77778 \color{black} & 3.11111 \end{matrix} \right]\] |
Perceba que os valores calculados já foram utilizados nos próximos cálculos, com isso, o vetor \(x\) ao final da primeira iteração:
\[x = \left[\begin{matrix}7.66667 & 3.77778 & 3.11111 \end{matrix} \right]\]Cálculo do erro na iteração 1:
O cálculo de erro continua da mesma forma:
| Matriz | Operação |
|---|---|
| \[\left[\begin{matrix}\color{green} 3 \color{black} &\color{orange} 0 \color{black} &\color{purple} 2 \color{black} \\-1 &6 &-4 \\-2 &2 &-11 \end{matrix} \right. \left| \begin{matrix} \color{red} 23 \\ 15 \\ -42 \end{matrix} \right]\] | \[\xi_{1}=\left|\color{red}23 - ((\color{green} 3 \times 7.66667 \color{black}) +(\color{orange} 0 \times 3.77778 \color{black}) +(\color{purple} 2 \times3.11111 \color{black})) \right| \] \[\xi_{1}=\left|-6.22222\right| = 6.22222\] |
| \[\left[\begin{matrix}3 &0 &2 \\\color{green} -1 \color{black} &\color{orange} 6 \color{black} &\color{purple} -4 \color{black} \\-2 &2 &-11 \end{matrix} \right. \left| \begin{matrix} 23 \\ \color{red} 15 \\ -42 \end{matrix} \right]\] | \[\xi_{2}=\left|\color{red}15 - ((\color{green} -1 \times 7.66667 \color{black}) +(\color{orange} 6 \times 3.77778 \color{black}) +(\color{purple} -4 \times 3.11111 \color{black})) \right| \] \[\xi_{2}=\left|12.4444\right| = 12.4444\] |
| \[\left[\begin{matrix}3 &0 &2 \\-1 &6 &-4 \\\color{green} -2 \color{black} &\color{orange} 2 \color{black} &\color{purple} -11 \color{black} \end{matrix} \right. \left| \begin{matrix} 23 \\ 15 \\ \color{red} -42 \end{matrix} \right]\] | \[\xi_{3}=\left|\color{red}-42 - ((\color{green} -2 \times 7.66667 \color{black}) +(\color{orange} 2 \times 3.77778 \color{black}) +(\color{purple} -11 \times 3.11111 \color{black})) \right| \] \[\xi_{3}=\left|0\right| = 0\] |
Continuando de forma resumida
Os próximos passos de forma resumida:
| Iteração | \(x\) | \(\xi\) | Maior erro |
|---|---|---|---|
| 2 | \(x = \left[\begin{matrix}5.59259 &5.50617 &3.80247 \end{matrix} \right]\) | \(\xi = \left[\begin{matrix}1.38272 &2.76543 &0 \end{matrix} \right]\) | \(2.76543 > 0.1\) |
| 3 | \(x = \left[\begin{matrix}5.13169 &5.89026 &3.9561 \end{matrix} \right]\) | \(\xi = \left[\begin{matrix}0.30727 &0.61454 &0 \end{matrix} \right]\) | \(0.61454 > 0.1\) |
| 4 | \(x = \left[\begin{matrix}5.02926 &5.97561 &3.99025 \end{matrix} \right]\) | \(\xi = \left[\begin{matrix}0.0682823 &0.136565 &0 \end{matrix} \right]\) | \(0.136565 > 0.1\) |
| 5 | \(x = \left[\begin{matrix}5.0065 &5.99458 &3.99783 \end{matrix} \right]\) | \(\xi = \left[\begin{matrix}0.0151738 &0.0303477 &0 \end{matrix} \right]\) | \(0.0303477 \ngtr 0.1\) |
Portanto a resposta final é:
\[x = \left[\begin{matrix}5.0065 &5.99458 &3.99783 \end{matrix} \right]\]ou, aproximadamente:
\[x \approx \left[\begin{matrix}5 &6 &4\end{matrix} \right]\]Atividade 1
Resolva os sistemas lineares a seguir de acordo com o método de Gauss-Seidel com \(\xi <0.1 \) (erro < 0.1) :
- \[ \left[\begin{matrix}3 &1 &-1 \\1 &4 &-1 \\-3 &-5 &12 \end{matrix} \right. \left| \begin{matrix} 14 \\ 15 \\ -15 \end{matrix} \right] \]
- \[ \left[\begin{matrix}4 &-1 &-2 \\2 &-6 &-3 \\-1 &1 &5 \end{matrix} \right. \left| \begin{matrix} -10 \\ -28 \\ 31 \end{matrix} \right] \]
- \[ \left[\begin{matrix}3 &0 &-2 \\1 &3 &-1 \\4 &2 &-11 \end{matrix} \right. \left| \begin{matrix} -8 \\ -2 \\ -67 \end{matrix} \right] \]