Métodos de Gauss-Jordan

O método de Gauss-Jordan consiste no escalonamento da matriz através das operações elementares até que se obtenha a matriz diagonal equivalente

Veja um exemplo, resolve o sistema linear através do método de Gauss-Jordan

\[ \left[\begin{matrix}1 &1 &-1 &-1 \\1 &2 &-3 &-4 \\-2 &-4 &7 &10 \\-2 &-6 &12 &19 \end{matrix} \right. \left| \begin{matrix} 4 \\ 7 \\ -15 \\ -24 \end{matrix} \right] \]
Operação Comando HP PPL \(i\) \(j\)
\( \left[\begin{matrix} \color{blue} 1 &1 &-1 &-1 \\ \color{red} 1 &2 &-3 &-4 \\-2 &-4 &7 &10 \\-2 &-6 &12 &19 \end{matrix} \right. \left| \begin{matrix} 4 \\ 7 \\ -15 \\ -24 \end{matrix} \right] \begin{matrix} \textrm{ } \\ \Rightarrow L_2= L_2 - \frac{\color{red}M_{21}}{\color{blue}M_{11}} L_1 \left| L_2= L_2 - \frac{\color{red}1}{\color{blue}1} L_1 \right. \\ \textrm{ } \\ \textrm{ } \end{matrix} \) \( SCALEADD(M1, -1, 1, 2); \) \(1\) \(2\)
\( \left[\begin{matrix} \color{blue} 1 &1 &-1 &-1 \\0 &1 &-2 &-3 \\ \color{red} -2 &-4 &7 &10 \\-2 &-6 &12 &19 \end{matrix} \right. \left| \begin{matrix} 4 \\ 3 \\ -15 \\ -24 \end{matrix} \right] \begin{matrix} \textrm{ } \\ \textrm{ } \\ \Rightarrow L_3= L_3 - \frac{\color{red}M_{31}}{\color{blue}M_{11}} L_1 \left| L_3= L_3 - \frac{\color{red}-2}{\color{blue}1} L_1 \right. \\ \textrm{ } \end{matrix} \) \( SCALEADD(M1, 2, 1, 3); \) \(1\) \(3\)
\( \left[\begin{matrix} \color{blue} 1 &1 &-1 &-1 \\0 &1 &-2 &-3 \\0 &-2 &5 &8 \\ \color{red} -2 &-6 &12 &19 \end{matrix} \right. \left| \begin{matrix} 4 \\ 3 \\ -7 \\ -24 \end{matrix} \right] \begin{matrix} \textrm{ } \\ \textrm{ } \\ \textrm{ } \\ \Rightarrow L_4= L_4 - \frac{\color{red}M_{41}}{\color{blue}M_{11}} L_1 \left| L_4= L_4 - \frac{\color{red}-2}{\color{blue}1} L_1 \right. \end{matrix} \) \( SCALEADD(M1, 2, 1, 4); \) \(1\) \(4\)
\( \left[\begin{matrix}1 & \color{red} 1 &-1 &-1 \\0 & \color{blue} 1 &-2 &-3 \\0 &-2 &5 &8 \\0 &-4 &10 &17 \end{matrix} \right. \left| \begin{matrix} 4 \\ 3 \\ -7 \\ -16 \end{matrix} \right] \begin{matrix} \Rightarrow L_1= L_1 - \frac{\color{red}M_{12}}{\color{blue}M_{22}} L_2 \left| L_1= L_1 - \frac{\color{red}1}{\color{blue}1} L_2 \right. \\ \textrm{ } \\ \textrm{ } \\ \textrm{ } \end{matrix} \) \( SCALEADD(M1, -1, 2, 1); \) \(2\) \(1\)
\( \left[\begin{matrix}1 &0 &1 &2 \\0 & \color{blue} 1 &-2 &-3 \\0 & \color{red} -2 &5 &8 \\0 &-4 &10 &17 \end{matrix} \right. \left| \begin{matrix} 1 \\ 3 \\ -7 \\ -16 \end{matrix} \right] \begin{matrix} \textrm{ } \\ \textrm{ } \\ \Rightarrow L_3= L_3 - \frac{\color{red}M_{32}}{\color{blue}M_{22}} L_2 \left| L_3= L_3 - \frac{\color{red}-2}{\color{blue}1} L_2 \right. \\ \textrm{ } \end{matrix} \) \( SCALEADD(M1, 2, 2, 3); \) \(2\) \(3\)
\( \left[\begin{matrix}1 &0 &1 &2 \\0 & \color{blue} 1 &-2 &-3 \\0 &0 &1 &2 \\0 & \color{red} -4 &10 &17 \end{matrix} \right. \left| \begin{matrix} 1 \\ 3 \\ -1 \\ -16 \end{matrix} \right] \begin{matrix} \textrm{ } \\ \textrm{ } \\ \textrm{ } \\ \Rightarrow L_4= L_4 - \frac{\color{red}M_{42}}{\color{blue}M_{22}} L_2 \left| L_4= L_4 - \frac{\color{red}-4}{\color{blue}1} L_2 \right. \end{matrix} \) \( SCALEADD(M1, 4, 2, 4); \) \(2\) \(4\)
\( \left[\begin{matrix}1 &0 & \color{red} 1 &2 \\0 &1 &-2 &-3 \\0 &0 & \color{blue} 1 &2 \\0 &0 &2 &5 \end{matrix} \right. \left| \begin{matrix} 1 \\ 3 \\ -1 \\ -4 \end{matrix} \right] \begin{matrix} \Rightarrow L_1= L_1 - \frac{\color{red}M_{13}}{\color{blue}M_{33}} L_3 \left| L_1= L_1 - \frac{\color{red}1}{\color{blue}1} L_3 \right. \\ \textrm{ } \\ \textrm{ } \\ \textrm{ } \end{matrix} \) \( SCALEADD(M1, -1, 3, 1); \) \(3\) \(1\)
\( \left[\begin{matrix}1 &0 &0 &0 \\0 &1 & \color{red} -2 &-3 \\0 &0 & \color{blue} 1 &2 \\0 &0 &2 &5 \end{matrix} \right. \left| \begin{matrix} 2 \\ 3 \\ -1 \\ -4 \end{matrix} \right] \begin{matrix} \textrm{ } \\ \Rightarrow L_2= L_2 - \frac{\color{red}M_{23}}{\color{blue}M_{33}} L_3 \left| L_2= L_2 - \frac{\color{red}-2}{\color{blue}1} L_3 \right. \\ \textrm{ } \\ \textrm{ } \end{matrix} \) \( SCALEADD(M1, 2, 3, 2); \) \(3\) \(2\)
\( \left[\begin{matrix}1 &0 &0 &0 \\0 &1 &0 &1 \\0 &0 & \color{blue} 1 &2 \\0 &0 & \color{red} 2 &5 \end{matrix} \right. \left| \begin{matrix} 2 \\ 1 \\ -1 \\ -4 \end{matrix} \right] \begin{matrix} \textrm{ } \\ \textrm{ } \\ \textrm{ } \\ \Rightarrow L_4= L_4 - \frac{\color{red}M_{43}}{\color{blue}M_{33}} L_3 \left| L_4= L_4 - \frac{\color{red}2}{\color{blue}1} L_3 \right. \end{matrix} \) \( SCALEADD(M1, -2, 3, 4); \) \(3\) \(4\)
\( \left[\begin{matrix}1 &0 &0 &\color{red}0 \\0 &1 &0 & 1 \\0 &0 &1 &2 \\0 &0 &0 & \color{blue} 1 \end{matrix} \right. \left| \begin{matrix} 2 \\ 1 \\ -1 \\ -2 \end{matrix} \right] \begin{matrix} \Rightarrow L_1= L_1 - \frac{\color{red}M_{14}}{\color{blue}M_{44}} L_4 \left| L_1= L_1 - \frac{\color{red}0}{\color{blue}1} L_4 \right. \\ \textrm{ } \\ \textrm{ } \\ \textrm{ } \end{matrix} \) \( SCALEADD(M1, 0, 4, 1); \) \(4\) \(1\)
\( \left[\begin{matrix}1 &0 &0 &0 \\0 &1 &0 & \color{red} 1 \\0 &0 &1 &2 \\0 &0 &0 & \color{blue} 1 \end{matrix} \right. \left| \begin{matrix} 2 \\ 1 \\ -1 \\ -2 \end{matrix} \right] \begin{matrix} \textrm{ } \\ \Rightarrow L_2= L_2 - \frac{\color{red}M_{24}}{\color{blue}M_{44}} L_4 \left| L_2= L_2 - \frac{\color{red}1}{\color{blue}1} L_4 \right. \\ \textrm{ } \\ \textrm{ } \end{matrix} \) \( SCALEADD(M1, -1, 4, 2); \) \(4\) \(2\)
\( \left[\begin{matrix}1 &0 &0 &0 \\0 &1 &0 &0 \\0 &0 &1 & \color{red} 2 \\0 &0 &0 & \color{blue} 1 \end{matrix} \right. \left| \begin{matrix} 2 \\ 3 \\ -1 \\ -2 \end{matrix} \right] \begin{matrix} \textrm{ } \\ \textrm{ } \\ \Rightarrow L_3= L_3 - \frac{\color{red}M_{34}}{\color{blue}M_{44}} L_4 \left| L_3= L_3 - \frac{\color{red}2}{\color{blue}1} L_4 \right. \\ \textrm{ } \end{matrix} \) \( SCALEADD(M1, -2, 4, 3); \) \(4\) \(3\)

E a matriz resultante final:

\[ \left[\begin{matrix}1 &0 &0 &0 \\0 &1 &0 &0 \\0 &0 &1 &0 \\0 &0 &0 &1 \end{matrix} \right. \left| \begin{matrix} 2 \\ 3 \\ 3 \\ -2 \end{matrix} \right] \]

Equação genérica

Podemos perceber que de forma genérica podemos escrever a fórmula matemática e o comando de programação:

  • Forma matemática:\[L_j= L_j - \frac{M_{ji}}{M_{ii}} L_i\]
  • Programação:
    SCALEADD(M1,-(M1(j,i)/M1(i,i)),i,j)

Que é o mesmo que o Método de Gauss, porém muda a forma como são percorridas as linhas e colunas nos laços de repetição: 

EXPORT gauss()
BEGIN
	PRINT();
	LOCAL DIMENSAO := 4;

	FOR I FROM 1 TO DIMENSAO DO
		FOR J FROM 1 TO DIMENSAO DO
			IF (I <> J) THEN
				PRINT("i:" + I + ", j:" + J);
			END;
		END;
	END;
END;

Atividade 1

Resolva os sistemas lineares a seguir de acordo com o método de Gauss-Jordan, apresentando a lista de comandos SCALE e SCALEDD para isso:

  • \[ \left[\begin{matrix}3 &-4 &-10 \\-6 &10 &24 \\9 &-14 &-32 \end{matrix} \right. \left| \begin{matrix} -7 \\ 19 \\ -25 \end{matrix} \right] \]
  • \[ \left[\begin{matrix}3 &0 &0 \\-6 &2 &1 \\-18 &4 &3 \end{matrix} \right. \left| \begin{matrix} 3 \\ 3 \\ 5 \end{matrix} \right] \]
  • \[ \left[\begin{matrix}1 &3 &-2 &0 \\2 &9 &-4 &-1 \\2 &9 &-2 &-1 \\-8 &-36 &12 &5 \end{matrix} \right. \left| \begin{matrix} 4 \\ 7 \\ 12 \\ -34 \end{matrix} \right] \]
  • \[ \left[\begin{matrix}1 &-4 &1 &5 &0 \\-2 &10 &-3 &-13 &6 \\-6 &28 &-7 &-35 &6 \\2 &-10 &3 &14 &-6 \\6 &-28 &7 &34 &-3 \end{matrix} \right. \left| \begin{matrix} 5 \\ -6 \\ -25 \\ 8 \\ 26 \end{matrix} \right] \]