Resolva os sistemas, usando a regra de Cramer.
\begin{cases}x+y & =1 \\\\ 2 x+z & =1 \\\ x+2 y & =-1\end{cases}

Passo 1

O enunciado pede para resolver o sistema
\begin{cases} x + y + 0\cdot z = 1,\\\\ 2x + 0\cdot y + z = 1,\\\ x + 2y + 0\cdot z = -1 \end{cases}
usando a regra de Cramer. Vamos montar a matriz de coeficientes, calcular o determinante principal D e os determinantes D_x, D_y, D_z, e então obter x, y e z.

Passo 2

Escrevemos a matriz dos coeficientes e o vetor dos termos independentes:
A = \begin{pmatrix} 1 & 1 & 0\\\\ 2 & 0 & 1\\\\ 1 & 2 & 0 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix}1\\\ 1\\\ -1\end{pmatrix}.
Pela regra de Cramer,
x = \frac{D_x}{D},\quad y = \frac{D_y}{D},\quad z = \frac{D_z}{D},
onde D = \det(A) e D_x (resp. D_y, D_z) é o determinante obtido trocando a coluna de x (resp. y, z) por \mathbf{b}.

Passo 3

Calculamos o determinante principal D:
D = \det\begin{pmatrix} 1 & 1 & 0\\\\ 2 & 0 & 1\\\\ 1 & 2 & 0 \end{pmatrix} = 1\cdot\det\begin{pmatrix}0 & 1\\\ 2 & 0\end{pmatrix} -1\cdot\det\begin{pmatrix}2 & 1\\\ 1 & 0\end{pmatrix} +0\cdot\bigl(\cdots\bigr).
Como
\det\begin{pmatrix}0 & 1\\\ 2 & 0\end{pmatrix} = 0\cdot0 - 1\cdot2 = -2, \quad \det\begin{pmatrix}2 & 1\\\ 1 & 0\end{pmatrix} = 2\cdot0 - 1\cdot1 = -1,
vem
D = 1\cdot(-2) - 1\cdot(-1) = -2 + 1 = -1.

Passo 4

Determinante D_x, trocando a coluna de x por \mathbf{b}:
D_x = \det\begin{pmatrix} 1 & 1 & 0\\\\ 1 & 0 & 1\\\\ -1 & 2 & 0 \end{pmatrix} = 1\cdot\det\begin{pmatrix}0 & 1\\\ 2 & 0\end{pmatrix} -1\cdot\det\begin{pmatrix}1 & 1\\\ -1 & 0\end{pmatrix} +0\cdot\bigl(\cdots\bigr).
Aqui
\det\begin{pmatrix}0 & 1\\\ 2 & 0\end{pmatrix} = -2, \quad \det\begin{pmatrix}1 & 1\\\ -1 & 0\end{pmatrix} = 1\cdot0 - 1\cdot(-1) = 1,
logo
D_x = 1\cdot(-2) - 1\cdot1 = -2 - 1 = -3.

Passo 5

Determinante D_y, trocando a coluna de y por \mathbf{b}:
D_y = \det\begin{pmatrix} 1 & 1 & 0\\\\ 2 & 1 & 1\\\\ 1 & -1 & 0 \end{pmatrix} = 1\cdot\det\begin{pmatrix}1 & 1\\\ -1 & 0\end{pmatrix} -1\cdot\det\begin{pmatrix}2 & 1\\\ 1 & 0\end{pmatrix} +0\cdot\bigl(\cdots\bigr).
Calculando
\det\begin{pmatrix}1 & 1\\\ -1 & 0\end{pmatrix} = 1\cdot0 - 1\cdot(-1) = 1, \quad \det\begin{pmatrix}2 & 1\\\ 1 & 0\end{pmatrix} = -1,
obtemos
D_y = 1\cdot1 - 1\cdot(-1) = 1 + 1 = 2.

Passo 6

Determinante D_z, trocando a coluna de z por \mathbf{b}:
D_z = \det\begin{pmatrix} 1 & 1 & 1\\\\ 2 & 0 & 1\\\\ 1 & 2 & -1 \end{pmatrix} = 1\cdot\det\begin{pmatrix}0 & 1\\\ 2 & -1\end{pmatrix} -1\cdot\det\begin{pmatrix}2 & 1\\\ 1 & -1\end{pmatrix} +1\cdot\det\begin{pmatrix}2 & 0\\\ 1 & 2\end{pmatrix}.
Aqui
\det\begin{pmatrix}0 & 1\\\ 2 & -1\end{pmatrix} = 0\cdot(-1) - 1\cdot2 = -2,
\det\begin{pmatrix}2 & 1\\\ 1 & -1\end{pmatrix} = 2\cdot(-1) - 1\cdot1 = -3,
\det\begin{pmatrix}2 & 0\\\ 1 & 2\end{pmatrix} = 2\cdot2 - 0\cdot1 = 4,
logo
D_z = 1\cdot(-2) - 1\cdot(-3) + 1\cdot4 = -2 + 3 + 4 = 5.

Passo 7

Aplicando Cramer:
x = \frac{D_x}{D} = \frac{-3}{-1} = 3,\quad y = \frac{D_y}{D} = \frac{2}{-1} = -2,\quad z = \frac{D_z}{D} = \frac{5}{-1} = -5.

Resposta

(x,y,z) = (3, -2, -5).

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