4- Resolva \left\{\begin{array}{c}x-3 y+z=4 \\\\ 2 x-y=-2 \\\\ 4 x-3 z=0\end{array}\right. usando a regra de Cramer.

Passo 1

O enunciado pede para resolver o sistema
\begin{cases} x - 3y + z = 4,\\\\ 2x - y + 0\cdot z = -2,\\\\ 4x + 0\cdot y - 3z = 0 \end{cases}
usando a regra de Cramer, que consiste em calcular o determinante da matriz dos coeficientes e os determinantes das matrizes obtidas ao substituir cada coluna de incógnita pelo vetor dos termos independentes.

Passo 2

Vamos chamar de \Delta o determinante da matriz dos coeficientes
A = \begin{pmatrix} 1 & -3 & 1 \\[6pt] 2 & -1 & 0 \\[6pt] 4 & 0 & -3 \end{pmatrix}
Calculando
\Delta = \det(A) = 1\cdot\det\begin{pmatrix}-1 & 0 \\[3pt]0 & -3\end{pmatrix} -(-3)\cdot\det\begin{pmatrix}2 & 0 \\[3pt]4 & -3\end{pmatrix} +1\cdot\det\begin{pmatrix}2 & -1 \\[3pt]4 & 0\end{pmatrix}
\det\begin{pmatrix}-1 & 0 \\[3pt]0 & -3\end{pmatrix}=3,\quad \det\begin{pmatrix}2 & 0 \\[3pt]4 & -3\end{pmatrix}=-6,\quad \det\begin{pmatrix}2 & -1 \\[3pt]4 & 0\end{pmatrix}=4
logo
\Delta = 1\cdot3 -(-3)\cdot(-6) + 1\cdot4 = 3 - 18 + 4 = -11.]

Passo 3

Agora calculamos \Delta_x,\Delta_y,\Delta_z substituindo as colunas de x,y,z por (4,-2,0).
\Delta_x = \det \begin{pmatrix} 4 & -3 & 1 \\[3pt] -2 & -1 & 0 \\[3pt] 0 & 0 & -3 \end{pmatrix} =4\cdot3 +3\cdot6 +1\cdot0 = 12 + 18 + 0 = 30,
\Delta_y = \det \begin{pmatrix} 1 & 4 & 1 \\[3pt] 2 & -2 & 0 \\[3pt] 4 & 0 & -3 \end{pmatrix} =1\cdot6 -4\cdot(-6) +1\cdot8 = 6 + 24 + 8 = 38,
\Delta_z = \det \begin{pmatrix} 1 & -3 & 4 \\[3pt] 2 & -1 & -2 \\[3pt] 4 & 0 & 0 \end{pmatrix} =0 +3\cdot8 +4\cdot4 = 0 +24 +16 = 40.]

Passo 4

Pela regra de Cramer temos
x = \frac{\Delta_x}{\Delta} = \frac{30}{-11} = -\frac{30}{11},\quad y = \frac{\Delta_y}{\Delta} = \frac{38}{-11} = -\frac{38}{11},\quad z = \frac{\Delta_z}{\Delta} = \frac{40}{-11} = -\frac{40}{11}.

Resposta

\bigl(x,y,z\bigr) = \Bigl(-\tfrac{30}{11},\,-\tfrac{38}{11},\,-\tfrac{40}{11}\Bigr).

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