(IME-77 CG) Dada equação
matricial \left[\begin{array}{cccc}1 & 2 & 0 & 1 \\\\ 3 & 5 & 2 & 1 \\\\ 0 & -3 & 1 & -5 \\\\ -1 & 1 & -1 & 1\end{array}\right]\left[\begin{array}{c}x_{1} \\\ x_{2} \\\ x_{3} \\\ x_{4}\end{array}\right]=\left[\begin{array}{c}-5 \\\\ -10 \\\\ 6 \\\\ -4\end{array}\right].
Determine a matriz X=\left[\begin{array}{c}x_{1} \\\ x_{2} \\\ x_{3} \\\ x_{4}\end{array}\right]

Passo 1

O enunciado nos dá o sistema linear
A\,X = b,
onde
A = \begin{pmatrix} 1 & 2 & 0 & 1 \\\\ 3 & 5 & 2 & 1 \\\\ 0 & -3 & 1 & -5 \\\\ -1 & 1 & -1 & 1 \end{pmatrix}, \quad X = \begin{pmatrix}x_{1}\\\ x_{2}\\\ x_{3}\\\ x_{4}\end{pmatrix}, \quad b = \begin{pmatrix}-5\\\\ -10\\\\ 6\\\\ -4\end{pmatrix}.
Vamos resolver por eliminação de Gauss, obtendo uma forma triangular superior para depois usar substituição regressiva.

Passo 2

Formamos a matriz aumentada e eliminamos a primeira coluna em R_2 e R_4 usando R_1:
\left[\begin{array}{cccc|c} 1 & 2 & 0 & 1 & -5 \\\\ 3 & 5 & 2 & 1 & -10 \\\\ 0 & -3 & 1 & -5 & 6 \\\\ -1 & 1 & -1 & 1 & -4 \end{array}\right] \;\xrightarrow{\,R_2\to R_2-3R_1,\;R_4\to R_4+R_1\,}\; \left[\begin{array}{cccc|c} 1 & 2 & 0 & 1 & -5 \\\\ 0 & -1 & 2 & -2 & 5 \\\\ 0 & -3 & 1 & -5 & 6 \\\\ 0 & 3 & -1 & 2 & -9 \end{array}\right].

Passo 3

Eliminamos a segunda coluna em R_3 e R_4 usando a nova R_2:
\left[\begin{array}{cccc|c} 1 & 2 & 0 & 1 & -5 \\\\ 0 & -1 & 2 & -2 & 5 \\\\ 0 & -3 & 1 & -5 & 6 \\\\ 0 & 3 & -1 & 2 & -9 \end{array}\right] \;\xrightarrow{\,R_3\to R_3-3R_2,\;R_4\to R_4+3R_2\,}\; \left[\begin{array}{cccc|c} 1 & 2 & 0 & 1 & -5 \\\\ 0 & -1 & 2 & -2 & 5 \\\\ 0 & 0 & -5 & 1 & -9 \\\\ 0 & 0 & 5 & -4 & 6 \end{array}\right].

Passo 4

Eliminamos a terceira coluna em R_4 usando R_3:
\left[\begin{array}{cccc|c} 1 & 2 & 0 & 1 & -5 \\\\ 0 & -1 & 2 & -2 & 5 \\\\ 0 & 0 & -5 & 1 & -9 \\\\ 0 & 0 & 5 & -4 & 6 \end{array}\right] \;\xrightarrow{\,R_4\to R_4+R_3\,}\; \left[\begin{array}{cccc|c} 1 & 2 & 0 & 1 & -5 \\\\ 0 & -1 & 2 & -2 & 5 \\\\ 0 & 0 & -5 & 1 & -9 \\\\ 0 & 0 & 0 & -3 & -3 \end{array}\right].
Agora temos um sistema triangular superior:
\begin{cases} -3\,x_{4} = -3,\\\\ -5\,x_{3} + 1\,x_{4} = -9,\\\\ -\,x_{2} + 2\,x_{3} -2\,x_{4} = 5,\\\ x_{1} + 2\,x_{2} + x_{4} = -5. \end{cases}

Passo 5

Aplicamos substituição regressiva:
-3\,x_{4} = -3 \quad\Rightarrow\quad x_{4} = 1,
-5\,x_{3} + (1)\cdot1 = -9 \quad\Rightarrow\quad -5\,x_{3} = -10 \quad\Rightarrow\quad x_{3} = 2,
-\,x_{2} + 2\cdot2 -2\cdot1 = 5 \quad\Rightarrow\quad -\,x_{2} +4 -2 = 5 \quad\Rightarrow\quad -\,x_{2} = 3 \quad\Rightarrow\quad x_{2} = -3,
x_{1} + 2\cdot(-3) + 1\cdot1 = -5 \quad\Rightarrow\quad x_{1} -6 +1 = -5 \quad\Rightarrow\quad x_{1} = 0.

Resposta

X = \begin{pmatrix} 0\\\\ -3\\\\ 2\\\\ 1 \end{pmatrix}.

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