Considere o sistema
\begin{array}{llc} 4 x+y+z+w & = & 6 \\\\ 3 x+7 y-z+w & = & 1 \\\\ 7 x+3 y-5 z+8 w & = & -3 \\\ x+y+z+2 w & = & 3 \end{array}
Resolva o sistema usando o Método das Matrizes Elementares.

Passo 1

Temos o sistema
\begin{cases} 4x + y + z + w = 6,\\\\ 3x + 7y - z + w = 1,\\\\ 7x + 3y -5z + 8w = -3,\\\ x + y + z + 2w = 3.
O objetivo é escrever Ax=b, montar a matriz aumentada [A\mid b] e aplicar operações de linha (cada uma correspondente a multiplicar por uma matriz elementar) para triangularizar a matriz e, depois, resolver por substituição retroativa.

Passo 2

Escrevemos a matriz aumentada inicial:
\left[\begin{array}{cccc|c} 4 & 1 & 1 & 1 & 6\\\\ 3 & 7 & -1 & 1 & 1\\\\ 7 & 3 & -5 & 8 & -3\\\\ 1 & 1 & 1 & 2 & 3 \end{array}\right]

Passo 3

Eliminação de x nas linhas 2, 3 e 4 usando o pivô a_{11}=4:
R2 \gets R2 -\tfrac{3}{4}R1,
R3 \gets R3 -\tfrac{7}{4}R1,
R4 \gets R4 -\tfrac{1}{4}R1.
Fica
\left[\begin{array}{cccc|c} 4 & 1 & 1 & 1 & 6\\\\ 0 & \tfrac{25}{4} & -\tfrac{7}{4} & \tfrac{1}{4} & -\tfrac{7}{2}\\\\ 0 & \tfrac{5}{4} & -\tfrac{27}{4} & \tfrac{25}{4} & -\tfrac{27}{2}\\\\ 0 & \tfrac{3}{4} & \tfrac{3}{4} & \tfrac{7}{4} & \tfrac{3}{2} \end{array}\right]

Passo 4

Eliminação de y nas linhas 3 e 4 usando o pivô a_{22}=\tfrac{25}{4}:
R3 \gets R3 -\tfrac{\tfrac{5}{4}}{\tfrac{25}{4}}R2 = R3 -\tfrac15R2,
R4 \gets R4 -\tfrac{\tfrac{3}{4}}{\tfrac{25}{4}}R2 = R4 -\tfrac{3}{25}R2.
Obtemos
\left[\begin{array}{cccc|c} 4 & 1 & 1 & 1 & 6\\\\ 0 & \tfrac{25}{4} & -\tfrac{7}{4} & \tfrac{1}{4} & -\tfrac{7}{2}\\\\ 0 & 0 & -\tfrac{32}{5} & \tfrac{31}{5} & -\tfrac{64}{5}\\\\ 0 & 0 & \tfrac{24}{25} & \tfrac{43}{25} & \tfrac{48}{25} \end{array}\right]

Passo 5

Eliminação de z na linha 4 com pivô a_{33}=-\tfrac{32}{5}:
R4 \gets R4 -\tfrac{\tfrac{24}{25}}{-\tfrac{32}{5}}R3 = R4 +\tfrac{3}{20}R3.
Agora a forma triangular superior é
\left[\begin{array}{cccc|c} 4 & 1 & 1 & 1 & 6\\\\ 0 & \tfrac{25}{4} & -\tfrac{7}{4} & \tfrac{1}{4} & -\tfrac{7}{2}\\\\ 0 & 0 & -\tfrac{32}{5} & \tfrac{31}{5} & -\tfrac{64}{5}\\\\ 0 & 0 & 0 & \tfrac{53}{20} & 0 \end{array}\right]

Passo 6

Resolvemos agora o sistema triangular por substituição retroativa:
Da última linha,
\tfrac{53}{20}\,w = 0 \quad\Longrightarrow\quad w = 0.
Na terceira linha,
-\tfrac{32}{5}\,z + \tfrac{31}{5}\,w = -\tfrac{64}{5} \quad\Longrightarrow\quad -\tfrac{32}{5}\,z = -\tfrac{64}{5} \quad\Longrightarrow\quad z = 2.
Na segunda linha,
\tfrac{25}{4}\,y - \tfrac{7}{4}\,z + \tfrac{1}{4}\,w = -\tfrac{7}{2} \quad\Longrightarrow\quad \tfrac{25}{4}\,y - \tfrac{7}{4}\cdot2 = -\tfrac{7}{2} \quad\Longrightarrow\quad \tfrac{25}{4}\,y = 0 \quad\Longrightarrow\quad y = 0.
Na primeira linha,
4x + y + z + w = 6 \quad\Longrightarrow\quad 4x + 0 + 2 + 0 = 6 \quad\Longrightarrow\quad x = 1.

Resposta

x = 1,\quad y = 0,\quad z = 2,\quad w = 0.

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