Método de Jacobi

De início, vamos demonstrar o Método de Jacobi através de um exemplo, partindo do sistema linear a seguir:

\[ \left[\begin{matrix}8 &-2 &5 \\1 &-5 &3 \\-1 &3 &-10 \end{matrix} \right. \left| \begin{matrix} 43 \\ 3 \\ -28 \end{matrix} \right] \]

E inicialmente vamos criar a matriz de resultado \(x\) com os valores iguais a 0 (para lembrar, os índices estão dispostos abaixo do valores):

\[x =\underset{\color{red}\begin{matrix} x_1\textrm{ } & x_2\textrm{ } & x_3\textrm{ } \end{matrix} }{\left[\begin{matrix} 0\textrm{,} & 0\textrm{,} & 0 \end{matrix} \right]}\]

Iteração 1:

Matriz Operação
\[\left[\begin{matrix} \color{blue} 8 &\color{green} -2 \color{black} &\color{orange} 5 \color{black} \\1 &-5 &3 \\-1 &3 &-10 \end{matrix} \right. \left| \begin{matrix} \color{red} 43 \\ 3 \\ -28 \end{matrix} \right]\] \[\begin{matrix}x_1=\frac{1}{\color{blue}8\color{black}}(\color{red}43\color{black} - ((\color{green} -2 \times 0 \color{black}) +(\color{orange} 5 \times 0 \color{black}))) \end{matrix} = \colorbox{grey}{5.375} \] \[x_{novo} = \left[\begin{matrix} \colorbox{grey}{5.375}\color{black} &\textrm{,} &\textrm{,} \end{matrix} \right]\]
\[\left[\begin{matrix}8 &-2 &5 \\\color{green} 1 \color{black} & \color{blue} -5 &\color{orange} 3 \color{black} \\-1 &3 &-10 \end{matrix} \right. \left| \begin{matrix} 43 \\ \color{red} 3 \\ -28 \end{matrix} \right]\] \[\begin{matrix}x_2=\frac{1}{\color{blue}-5\color{black}}(\color{red}3\color{black} - ((\color{green} 1 \times 0 \color{black}) +(\color{orange} 3 \times 0 \color{black}))) \end{matrix} = \colorbox{grey}{-0.6} \] \[x_{novo} = \left[\begin{matrix}5.375, & \colorbox{grey}{-0.6}\color{black} &\textrm{,} \end{matrix} \right]\]
\[\left[\begin{matrix}8 &-2 &5 \\1 &-5 &3 \\\color{green} -1 \color{black} &\color{orange} 3 \color{black} & \color{blue} -10 \end{matrix} \right. \left| \begin{matrix} 43 \\ 3 \\ \color{red} -28 \end{matrix} \right]\] \[\begin{matrix}x_3=\frac{1}{\color{blue}-10\color{black}}(\color{red}-28\color{black} - ((\color{green} -1 \times 0 \color{black}) +(\color{orange} 3 \times 0 \color{black}))) \end{matrix} = \colorbox{grey}{2.8} \] \[x_{novo} = \left[\begin{matrix}5.375, &-0.6, & \colorbox{grey}{2.8}\color{black} \end{matrix} \right]\]

Ao final dessa primeira iteração temos os seguintes valores de \(x\):

\[x_{novo} = \left[\begin{matrix}5.375 &-0.6 &2.8 \end{matrix} \right]\]

Cálculo do erro na iteração 1:

A próxima etapa é calcular o erro, dado pelo módulo da diferença do valor de \(b\) pela soma dos valores dos elementos: \(a_{ij}\times x_{j}\).
De forma genérica:

\[\xi_{i} =\left| b_i-\sum\limits_{j=1}^{n}{a_{ij}\times x_{j}} \right| \]
Matriz Operação
\[\left[\begin{matrix}\color{green} 8 \color{black} &\color{orange} -2 \color{black} &\color{purple} 5 \color{black} \\1 &-5 &3 \\-1 &3 &-10 \end{matrix} \right. \left| \begin{matrix} \color{red} 43 \\ 3 \\ -28 \end{matrix} \right]\] \[\xi_{1}=\left|\color{red}43 - ((\color{green} 8 \times 5.375 \color{black}) +(\color{orange} -2 \times -0.6 \color{black}) +(\color{purple} 5 \times 2.8 \color{black})) \right| \] \[\xi_{1}=\left|-15.2\right| = 15.2\]
\[\left[\begin{matrix}8 &-2 &5 \\\color{green} 1 \color{black} &\color{orange} -5 \color{black} &\color{purple} 3 \color{black} \\-1 &3 &-10 \end{matrix} \right. \left| \begin{matrix} 43 \\ \color{red} 3 \\ -28 \end{matrix} \right]\] \[\xi_{2}=\left|\color{red}3 - ((\color{green} 1 \times 5.375 \color{black}) +(\color{orange} -5 \times -0.6 \color{black}) +(\color{purple} 3 \times 2.8 \color{black})) \right| \] \[\xi_{2}=\left|-13.775\right| = 13.775\]
\[\left[\begin{matrix}8 &-2 &5 \\1 &-5 &3 \\\color{green} -1 \color{black} &\color{orange} 3 \color{black} &\color{purple} -10 \color{black} \end{matrix} \right. \left| \begin{matrix} 43 \\ 3 \\ \color{red} -28 \end{matrix} \right]\] \[\xi_{3}=\left|\color{red}-28 - ((\color{green} -1 \times 5.375 \color{black}) +(\color{orange} 3 \times -0.6 \color{black}) +(\color{purple} -10 \times 2.8 \color{black})) \right| \] \[\xi_{3}=\left|7.175\right| = 7.175\]
\[\xi = \left[\begin{matrix}15.2 &13.775 &7.175 \end{matrix} \right]\] Maior Erro: \(15.2\)

Iteração 2:

Com os valores de \(x\) calculados anteriomente, executamos mais um passo detalhado:

\[x = \left[\begin{matrix}5.375 &-0.6 &2.8 \end{matrix} \right]\]
Matriz Operação
\[\left[\begin{matrix} \color{blue} 8 &\color{green} -2 \color{black} &\color{orange} 5 \color{black} \\1 &-5 &3 \\-1 &3 &-10 \end{matrix} \right. \left| \begin{matrix} \color{red} 43 \\ 3 \\ -28 \end{matrix} \right]\] \[\begin{matrix}x_1=\frac{1}{\color{blue}8\color{black}}(\color{red}43\color{black} - ((\color{green} -2 \times -0.6 \color{black}) +(\color{orange} 5 \times 2.8 \color{black}))) \end{matrix} = \colorbox{grey}{3.475} \] \[x_{novo} = \left[\begin{matrix} \colorbox{grey}{3.475}\color{black} &\textrm{,} &\textrm{,} \end{matrix} \right]\]
\[\left[\begin{matrix}8 &-2 &5 \\\color{green} 1 \color{black} & \color{blue} -5 &\color{orange} 3 \color{black} \\-1 &3 &-10 \end{matrix} \right. \left| \begin{matrix} 43 \\ \color{red} 3 \\ -28 \end{matrix} \right]\] \[\begin{matrix}x_2=\frac{1}{\color{blue}-5\color{black}}(\color{red}3\color{black} - ((\color{green} 1 \times 5.375 \color{black}) +(\color{orange} 3 \times 2.8 \color{black}))) \end{matrix} = \colorbox{grey}{2.155} \] \[x_{novo} = \left[\begin{matrix}3.475, & \colorbox{grey}{2.155}\color{black} &\textrm{,} \end{matrix} \right]\]
\[\left[\begin{matrix}8 &-2 &5 \\1 &-5 &3 \\\color{green} -1 \color{black} &\color{orange} 3 \color{black} & \color{blue} -10 \end{matrix} \right. \left| \begin{matrix} 43 \\ 3 \\ \color{red} -28 \end{matrix} \right]\] \[\begin{matrix}x_3=\frac{1}{\color{blue}-10\color{black}}(\color{red}-28\color{black} - ((\color{green} -1 \times 5.375 \color{black}) +(\color{orange} 3 \times -0.6 \color{black}))) \end{matrix} = \colorbox{grey}{2.0825} \] \[x_{novo} = \left[\begin{matrix}3.475, &2.155, & \colorbox{grey}{2.0825}\color{black} \end{matrix} \right]\]

Resultado da Iteração 2:

\[x_{novo} = \left[\begin{matrix}3.475 &2.155 &2.0825 \end{matrix} \right]\]

Cálculo do erro na iteração 2:

Matriz Operação
\[\left[\begin{matrix}\color{green} 8 \color{black} &\color{orange} -2 \color{black} &\color{purple} 5 \color{black} \\1 &-5 &3 \\-1 &3 &-10 \end{matrix} \right. \left| \begin{matrix} \color{red} 43 \\ 3 \\ -28 \end{matrix} \right]\] \[\xi_{1}=\left|\color{red}43 - ((\color{green} 8 \times 3.475 \color{black}) +(\color{orange} -2 \times 2.155 \color{black}) +(\color{purple} 5 \times 2.0825 \color{black})) \right| \] \[\xi_{1}=\left|9.0975\right| = 9.0975\]
\[\left[\begin{matrix}8 &-2 &5 \\\color{green} 1 \color{black} &\color{orange} -5 \color{black} &\color{purple} 3 \color{black} \\-1 &3 &-10 \end{matrix} \right. \left| \begin{matrix} 43 \\ \color{red} 3 \\ -28 \end{matrix} \right]\] \[\xi_{2}=\left|\color{red}3 - ((\color{green} 1 \times 3.475 \color{black}) +(\color{orange} -5 \times 2.155 \color{black}) +(\color{purple} 3 \times 2.0825 \color{black})) \right| \] \[\xi_{2}=\left|4.0525\right| = 4.0525\]
\[\left[\begin{matrix}8 &-2 &5 \\1 &-5 &3 \\\color{green} -1 \color{black} &\color{orange} 3 \color{black} &\color{purple} -10 \color{black} \end{matrix} \right. \left| \begin{matrix} 43 \\ 3 \\ \color{red} -28 \end{matrix} \right]\] \[\xi_{3}=\left|\color{red}-28 - ((\color{green} -1 \times 3.475 \color{black}) +(\color{orange} 3 \times 2.155 \color{black}) +(\color{purple} -10 \times 2.0825 \color{black})) \right| \] \[\xi_{3}=\left|-10.165\right| = 10.165\]
\[\xi = \left[\begin{matrix}9.0975 &4.0525 &10.165 \end{matrix} \right]\]

Maior Erro: \(10.165\)

Iteração 3:

Com os valores de \(x\) calculados anteriomente, executamos mais um passo detalhado:

\[x = \left[\begin{matrix}3.475 &2.155 &2.0825 \end{matrix} \right]\]
Matriz Operação
\[\left[\begin{matrix} \color{blue} 8 &\color{green} -2 \color{black} &\color{orange} 5 \color{black} \\1 &-5 &3 \\-1 &3 &-10 \end{matrix} \right. \left| \begin{matrix} \color{red} 43 \\ 3 \\ -28 \end{matrix} \right]\] \[\begin{matrix}x_1=\frac{1}{\color{blue}8\color{black}}(\color{red}43\color{black} - ((\color{green} -2 \times 2.155 \color{black}) +(\color{orange} 5 \times 2.0825 \color{black}))) \end{matrix} = \colorbox{grey}{4.61219} \] \[x_{novo} = \left[\begin{matrix} \colorbox{grey}{4.61219}\color{black} &\textrm{,} &\textrm{,} \end{matrix} \right]\]
\[\left[\begin{matrix}8 &-2 &5 \\\color{green} 1 \color{black} & \color{blue} -5 &\color{orange} 3 \color{black} \\-1 &3 &-10 \end{matrix} \right. \left| \begin{matrix} 43 \\ \color{red} 3 \\ -28 \end{matrix} \right]\] \[\begin{matrix}x_2=\frac{1}{\color{blue}-5\color{black}}(\color{red}3\color{black} - ((\color{green} 1 \times 3.475 \color{black}) +(\color{orange} 3 \times 2.0825 \color{black}))) \end{matrix} = \colorbox{grey}{1.3445} \] \[x_{novo} = \left[\begin{matrix}4.61219, & \colorbox{grey}{1.3445}\color{black} &\textrm{,} \end{matrix} \right]\]
\[\left[\begin{matrix}8 &-2 &5 \\1 &-5 &3 \\\color{green} -1 \color{black} &\color{orange} 3 \color{black} & \color{blue} -10 \end{matrix} \right. \left| \begin{matrix} 43 \\ 3 \\ \color{red} -28 \end{matrix} \right]\] \[\begin{matrix}x_3=\frac{1}{\color{blue}-10\color{black}}(\color{red}-28\color{black} - ((\color{green} -1 \times 3.475 \color{black}) +(\color{orange} 3 \times 2.155 \color{black}))) \end{matrix} = \colorbox{grey}{3.099} \] \[x_{novo} = \left[\begin{matrix}4.61219, &1.3445, & \colorbox{grey}{3.099}\color{black} \end{matrix} \right]\]
\[x_{novo} = \left[\begin{matrix}4.61219 &1.3445 &3.099 \end{matrix} \right]\]

Cálculo do erro na iteração 3:

Matriz Operação
\[\left[\begin{matrix}\color{green} 8 \color{black} &\color{orange} -2 \color{black} &\color{purple} 5 \color{black} \\1 &-5 &3 \\-1 &3 &-10 \end{matrix} \right. \left| \begin{matrix} \color{red} 43 \\ 3 \\ -28 \end{matrix} \right]\] \[\xi_{1}=\left|\color{red}43 - ((\color{green} 8 \times 4.61219 \color{black}) +(\color{orange} -2 \times 1.3445 \color{black}) +(\color{purple} 5 \times 3.099 \color{black})) \right| \] \[\xi_{1}=\left|-6.7035\right| = 6.7035\]
\[\left[\begin{matrix}8 &-2 &5 \\\color{green} 1 \color{black} &\color{orange} -5 \color{black} &\color{purple} 3 \color{black} \\-1 &3 &-10 \end{matrix} \right. \left| \begin{matrix} 43 \\ \color{red} 3 \\ -28 \end{matrix} \right]\] \[\xi_{2}=\left|\color{red}3 - ((\color{green} 1 \times 4.61219 \color{black}) +(\color{orange} -5 \times 1.3445 \color{black}) +(\color{purple} 3 \times 3.099 \color{black})) \right| \] \[\xi_{2}=\left|-4.18669\right| = 4.18669\]
\[\left[\begin{matrix}8 &-2 &5 \\1 &-5 &3 \\\color{green} -1 \color{black} &\color{orange} 3 \color{black} &\color{purple} -10 \color{black} \end{matrix} \right. \left| \begin{matrix} 43 \\ 3 \\ \color{red} -28 \end{matrix} \right]\] \[\xi_{3}=\left|\color{red}-28 - ((\color{green} -1 \times 4.61219 \color{black}) +(\color{orange} 3 \times 1.3445 \color{black}) +(\color{purple} -10 \times 3.099 \color{black})) \right| \] \[\xi_{3}=\left|3.56869\right| = 3.56869\]
\[\xi = \left[\begin{matrix}6.7035 &4.18669 &3.56869 \end{matrix} \right]\]

Maior Erro: \(6.7035\)

Continuando mais resumidamente

E daqui para frente iremos demonstrar o processo resumido:

Iteração \(x\) \(\xi\) Maior erro
3 \(x_{novo} = \left[\begin{matrix}3.77425 &2.18184 &2.74213 \end{matrix} \right]\) \(\xi = \left[\begin{matrix}3.45902 &1.90854 &3.34995 \end{matrix} \right]\) \(3.45902 > 0.1\)
4 \(x_{novo} = \left[\begin{matrix}4.20663 &1.80013 &3.07713 \end{matrix} \right]\) \(\xi = \left[\begin{matrix}2.43839 &1.43736 &1.5775 \end{matrix} \right]\) \(2.43839 > 0.1\)
5 \(x_{novo} = \left[\begin{matrix}3.90183 &2.0876 &2.91938 \end{matrix} \right]\) \(\xi = \left[\begin{matrix}1.3637 &0.77805 &1.16722 \end{matrix} \right]\) \(1.3637 > 0.1\)
6 \(x_{novo} = \left[\begin{matrix}4.07229 &1.93199 &3.0361 \end{matrix} \right]\) \(\xi = \left[\begin{matrix}0.894828 &0.520627 &0.637292 \end{matrix} \right]\) \(0.894828 > 0.1\)
7 \(x_{novo} = \left[\begin{matrix}3.96044 &2.03612 &2.97237 \end{matrix} \right]\) \(\xi = \left[\begin{matrix}0.526897 &0.303041 &0.42423 \end{matrix} \right]\) \(0.526897 > 0.1\)
8 \(x_{novo} = \left[\begin{matrix}4.0263 &1.97551 &3.01479 \end{matrix} \right]\) \(\xi = \left[\begin{matrix}0.333331 &0.193131 &0.247687 \end{matrix} \right]\) \(0.333331 > 0.1\)
9 \(x_{novo} = \left[\begin{matrix}3.98463 &2.01413 &2.99002 \end{matrix} \right]\) \(\xi = \left[\begin{matrix}0.201096 &0.115972 &0.157545 \end{matrix} \right]\) \(0.201096 > 0.1\)
10 \(x_{novo} = \left[\begin{matrix}4.00977 &1.99094 &3.00578 \end{matrix} \right]\) \(\xi = \left[\begin{matrix}0.125162 &0.0724005 &0.0947205 \end{matrix} \right]\) \(0.125162 > 0.1\)
11 \(x_{novo} = \left[\begin{matrix}3.99412 &2.00542 &2.99631 \end{matrix} \right]\) \(\xi = \left[\begin{matrix}0.0763204 &0.0440613 &0.0590855 \end{matrix} \right]\) \(0.0763204 \ngtr 0.1\)

Portanto a resposta final é:

\[x = \left[\begin{matrix}3.99412 &2.00542 &2.99631 \end{matrix} \right]\]

ou, aproximadamente:

\[x \approx \left[\begin{matrix}4 &2 &3\end{matrix} \right]\]

Atividade 1

Resolva os sistemas lineares a seguir de acordo com o método de Jacobi com \(\xi <0.1 \) (erro < 0.1) :

  • \[ \left[\begin{matrix}7 &2 &-4 \\-1 &-5 &3 \\2 &2 &-9 \end{matrix} \right. \left| \begin{matrix} 41 \\ -8 \\ -9 \end{matrix} \right] \]
  • \[ \left[\begin{matrix}5 &-1 &3 \\-1 &7 &-4 \\1 &-2 &8 \end{matrix} \right. \left| \begin{matrix} 35 \\ 27 \\ 24 \end{matrix} \right] \]
  • \[ \left[\begin{matrix}3 &1 &-1 \\1 &4 &-1 \\-3 &-5 &12 \end{matrix} \right. \left| \begin{matrix} 14 \\ 15 \\ -15 \end{matrix} \right] \]