Resolva os seguintes sistemas lineares utilizando a Regra de Cramer:
\left\{\begin{array}{l}x+2 y+3 z=1 \\\\ 2 x-y+z=0 \\\\ 3 x+y-z=2\end{array}\right.

Passo 1

O enunciado pede para resolver o sistema
\begin{cases} x + 2y + 3z = 1,\\\\ 2x - y + z = 0,\\\\ 3x + y - z = 2 \end{cases}
usando a Regra de Cramer. O conteúdo envolve determinantes de matrizes 3\times3 e a fórmula
x = \frac{D_x}{D},\quad y = \frac{D_y}{D},\quad z = \frac{D_z}{D},
onde D é o determinante da matriz de coeficientes e D_x,D_y,D_z são determinantes obtidos substituindo a coluna correspondente pelos termos independentes.

Passo 2

Primeiro, montamos a matriz de coeficientes
A = \begin{pmatrix} 1 & 2 & 3\\\\ 2 & -1 & 1\\\\ 3 & 1 & -1 \end{pmatrix}
e calculamos seu determinante D = \det(A) pela Regra de Sarrus:
D = 1\cdot\begin{vmatrix}-1 & 1 \\\\ 1 & -1\end{vmatrix} -2\cdot\begin{vmatrix}2 & 1 \\\\ 3 & -1\end{vmatrix} +3\cdot\begin{vmatrix}2 & -1 \\\\ 3 & 1\end{vmatrix}.
Agora cada menor:
\begin{vmatrix}-1 & 1 \\\\ 1 & -1\end{vmatrix} =(-1)(-1)-(1)(1)=0, \quad \begin{vmatrix}2 & 1 \\\\ 3 & -1\end{vmatrix} =2\cdot(-1)-1\cdot3=-5,
\begin{vmatrix}2 & -1 \\\\ 3 & 1\end{vmatrix} =2\cdot1-(-1)\cdot3=5.
Logo
D = 1\cdot0 -2\cdot(-5) +3\cdot5 = 0 +10 +15 = 25.

Passo 3

Agora D_x substitui a primeira coluna por (1,0,2):
D_x = \det\begin{pmatrix} 1 & 2 & 3\\\\ 0 & -1 & 1\\\\ 2 & 1 & -1 \end{pmatrix} =1\cdot\begin{vmatrix}-1 & 1\\\ 1 & -1\end{vmatrix} -2\cdot\begin{vmatrix}0 & 1\\\ 2 & -1\end{vmatrix} +3\cdot\begin{vmatrix}0 & -1\\\ 2 & 1\end{vmatrix}.
Calculamos:
\begin{vmatrix}-1 & 1\\\ 1 & -1\end{vmatrix}=0, \quad \begin{vmatrix}0 & 1\\\ 2 & -1\end{vmatrix}=0\cdot(-1)-1\cdot2=-2,
\begin{vmatrix}0 & -1\\\ 2 & 1\end{vmatrix}=0\cdot1-(-1)\cdot2=2.
Então
D_x = 1\cdot0 -2\cdot(-2) +3\cdot2 = 0 +4 +6 = 10.

Passo 4

Para D_y substituímos a segunda coluna por (1,0,2):
D_y = \det\begin{pmatrix} 1 & 1 & 3\\\\ 2 & 0 & 1\\\\ 3 & 2 & -1 \end{pmatrix} =1\cdot\begin{vmatrix}0 & 1\\\ 2 & -1\end{vmatrix} -1\cdot\begin{vmatrix}2 & 1\\\ 3 & -1\end{vmatrix} +3\cdot\begin{vmatrix}2 & 0\\\ 3 & 2\end{vmatrix}.
Os menores são
\begin{vmatrix}0 & 1\\\ 2 & -1\end{vmatrix}=-2, \quad \begin{vmatrix}2 & 1\\\ 3 & -1\end{vmatrix}=-5, \quad \begin{vmatrix}2 & 0\\\ 3 & 2\end{vmatrix}=4.
Logo
D_y = 1\cdot(-2) -1\cdot(-5) +3\cdot4 = -2 +5 +12 = 15.

Passo 5

Para D_z substituímos a terceira coluna por (1,0,2):
D_z = \det\begin{pmatrix} 1 & 2 & 1\\\\ 2 & -1 & 0\\\\ 3 & 1 & 2 \end{pmatrix} =1\cdot\begin{vmatrix}-1 & 0\\\ 1 & 2\end{vmatrix} -2\cdot\begin{vmatrix}2 & 0\\\ 3 & 2\end{vmatrix} +1\cdot\begin{vmatrix}2 & -1\\\ 3 & 1\end{vmatrix}.
Os menores são
\begin{vmatrix}-1 & 0\\\ 1 & 2\end{vmatrix}=-2, \quad \begin{vmatrix}2 & 0\\\ 3 & 2\end{vmatrix}=4, \quad \begin{vmatrix}2 & -1\\\ 3 & 1\end{vmatrix}=5.
Assim
D_z = 1\cdot(-2) -2\cdot4 +1\cdot5 = -2 -8 +5 = -5.

Passo 6

Aplicando a Regra de Cramer:
x = \frac{D_x}{D} = \frac{10}{25} = \frac{2}{5},\quad y = \frac{D_y}{D} = \frac{15}{25} = \frac{3}{5},\quad z = \frac{D_z}{D} = \frac{-5}{25} = -\frac{1}{5}.

Resposta

(x,y,z) = \Bigl(\tfrac{2}{5},\,\tfrac{3}{5},\,-\tfrac{1}{5}\Bigr).

Exercícios relacionados