Usando o método de Gauss (operações por linha) resolva o seguinte sistema de equações lineares.
\begin{aligned} 2 x_{1}-x_{2} & +3 x_{3}+4 x_{4}=-1 \\\ x_{1} & +2 x_{3}+x_{4}=0 \\\\ 3 x_{1} & +x_{3}+2 x_{4}=1 \\\\ 2 x_{1}+x_{2} & +4 x_{3}+x_{4}=0 \end{aligned}

Passo 1

O enunciado pede para resolver pelo método de eliminação de Gauss o sistema
\begin{cases} 2x_{1}-x_{2}+3x_{3}+4x_{4}=-1,\\\ x_{1}+2x_{3}+x_{4}=0,\\\\ 3x_{1}+x_{3}+2x_{4}=1,\\\\ 2x_{1}+x_{2}+4x_{3}+x_{4}=0. \end{cases}
Vamos montar a matriz aumentada e aplicar pivoteamentos e eliminações linha a linha até obter a forma escalonada reduzida.

Passo 2

Montamos a matriz aumentada inicial
\left[\begin{array}{cccc|c} 2 & -1 & 3 & 4 & -1\\\\ 1 & 0 & 2 & 1 & 0\\\\ 3 & 0 & 1 & 2 & 1\\\\ 2 & 1 & 4 & 1 & 0 \end{array}\right]
Para ter pivô 1 na primeira coluna, trocamos R_{1}\leftrightarrow R_{2}:
\left[\begin{array}{cccc|c} 1 & 0 & 2 & 1 & 0\\\\ 2 & -1 & 3 & 4 & -1\\\\ 3 & 0 & 1 & 2 & 1\\\\ 2 & 1 & 4 & 1 & 0 \end{array}\right]
Agora eliminamos x_{1} em R_{2},R_{3},R_{4}:
R_{2}\gets R_{2}-2R_{1},\;R_{3}\gets R_{3}-3R_{1},\;R_{4}\gets R_{4}-2R_{1}, resultando em
\left[\begin{array}{cccc|c} 1 & 0 & 2 & 1 & 0\\\\ 0 & -1& -1 & 2 & -1\\\\ 0 & 0 & -5 & -1 & 1\\\\ 0 & 1 & 0 & -1 & 0 \end{array}\right]

Passo 3

Pivô na segunda coluna: dividimos R_{2} por -1 para obter 1:
\left[\begin{array}{cccc|c} 1 & 0 & 2 & 1 & 0\\\\ 0 & 1 & 1 & -2 & 1\\\\ 0 & 0 & -5 & -1 & 1\\\\ 0 & 1 & 0 & -1 & 0 \end{array}\right]
Eliminamos x_{2} em R_{4}:
R_{4}\gets R_{4}-R_{2}, que dá
\left[\begin{array}{cccc|c} 1 & 0 & 2 & 1 & 0\\\\ 0 & 1 & 1 & -2 & 1\\\\ 0 & 0 & -5 & -1 & 1\\\\ 0 & 0 & -1 & 1 & -1 \end{array}\right]

Passo 4

Pivô na terceira coluna: dividimos R_{3} por -5:
\left[\begin{array}{cccc|c} 1 & 0 & 2 & 1 & 0\\\\ 0 & 1 & 1 & -2 & 1\\\\ 0 & 0 & 1 & \tfrac{1}{5} & -\tfrac{1}{5}\\\\ 0 & 0 & -1 & 1 & -1 \end{array}\right]
Eliminamos x_{3} em R_{1},R_{2},R_{4}:
\begin{aligned} R_{1}&\gets R_{1}-2R_{3},\\\ R_{2}&\gets R_{2}-R_{3},\\\ R_{4}&\gets R_{4}+R_{3}, \end{aligned}
ficando
\left[\begin{array}{cccc|c} 1 & 0 & 0 & \tfrac{3}{5} & \tfrac{2}{5}\\\\ 0 & 1 & 0 & -\tfrac{11}{5} & \tfrac{6}{5}\\\\ 0 & 0 & 1 & \tfrac{1}{5} & -\tfrac{1}{5}\\\\ 0 & 0 & 0 & \tfrac{6}{5} & -\tfrac{6}{5} \end{array}\right]

Passo 5

Pivô na quarta coluna: dividimos R_{4} por \tfrac{6}{5}:
\left[\begin{array}{cccc|c} 1 & 0 & 0 & \tfrac{3}{5} & \tfrac{2}{5}\\\\ 0 & 1 & 0 & -\tfrac{11}{5} & \tfrac{6}{5}\\\\ 0 & 0 & 1 & \tfrac{1}{5} & -\tfrac{1}{5}\\\\ 0 & 0 & 0 & 1 & -1 \end{array}\right]
Eliminamos x_{4} em R_{1},R_{2},R_{3}:
\begin{aligned} R_{1}&\gets R_{1}-\tfrac{3}{5}R_{4},\\\ R_{2}&\gets R_{2}+\tfrac{11}{5}R_{4},\\\ R_{3}&\gets R_{3}-\tfrac{1}{5}R_{4}, \end{aligned}
obtemos a forma reduzida
\left[\begin{array}{cccc|c} 1 & 0 & 0 & 0 & 1\\\\ 0 & 1 & 0 & 0 & -1\\\\ 0 & 0 & 1 & 0 & 0\\\\ 0 & 0 & 0 & 1 & -1 \end{array}\right]

Resposta

x_{1}=1, x_{2}=-1, x_{3}=0, x_{4}=-1.

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