Resolver os sistemas pela regra de Cramer:
\left\{\begin{array}{l}-x+y-z=5 \\\ x+2 y+4 z=4 \\\\ 3 x+y-2 z=-3\end{array}\right.

Passo 1

O enunciado pede para resolver o sistema
\begin{cases} -\,x + y - z = 5,\\\ x + 2y + 4z = 4,\\\\ 3x + y -2z = -3 \end{cases}
usando a regra de Cramer. Vamos montar a matriz dos coeficientes, calcular seu determinante principal D e os determinantes D_x, D_y e D_z obtidos trocando-se as colunas pelas constantes do lado direito. Em seguida, aplicaremos
x = \frac{D_x}{D},\quad y = \frac{D_y}{D},\quad z = \frac{D_z}{D}.

Passo 2

A matriz dos coeficientes é
A = \begin{pmatrix} -1 & 1 & -1\\\\ 1 & 2 & 4\\\\ 3 & 1 & -2 \end{pmatrix}.
Seu determinante D = \det(A) vale
D = (-1)\det\begin{pmatrix}2 & 4\\\ 1 & -2\end{pmatrix} + 1\det\begin{pmatrix}1 & 4\\\ 3 & -2\end{pmatrix} +(-1)\det\begin{pmatrix}1 & 2\\\ 3 & 1\end{pmatrix}.
Calculando cada menor:
\det\begin{pmatrix}2 & 4\\\ 1 & -2\end{pmatrix}=2\cdot(-2)-4\cdot1=-8,
\det\begin{pmatrix}1 & 4\\\ 3 & -2\end{pmatrix}=1\cdot(-2)-4\cdot3=-14,
\det\begin{pmatrix}1 & 2\\\ 3 & 1\end{pmatrix}=1\cdot1-2\cdot3=-5.
Logo,
D = (-1)(-8) + 1\bigl(-14\bigr) +(-1)(-5) =8 -14 +5 =27.

Passo 3

Determinante D_x: trocamos a primeira coluna por \begin{pmatrix}5\\\ 4\\\ -3\end{pmatrix},
D_x = \det\begin{pmatrix} 5 & 1 & -1\\\\ 4 & 2 & 4\\\\ -3 & 1 & -2 \end{pmatrix} =5\det\begin{pmatrix}2 & 4\\\ 1 & -2\end{pmatrix} -1\det\begin{pmatrix}4 & 4\\\ -3 & -2\end{pmatrix} +(-1)\det\begin{pmatrix}4 & 2\\\ -3 & 1\end{pmatrix}.
Calculando:
\det\begin{pmatrix}2 & 4\\\ 1 & -2\end{pmatrix}=-8,\quad \det\begin{pmatrix}4 & 4\\\ -3 & -2\end{pmatrix}=4,\quad \det\begin{pmatrix}4 & 2\\\ -3 & 1\end{pmatrix}=10.
Assim,
D_x = 5(-8) -1(4) +(-1)(10) = -40 -4 -10 = -54.

Passo 4

Determinante D_y: trocamos a segunda coluna por \begin{pmatrix}5\\\ 4\\\ -3\end{pmatrix},
D_y = \det\begin{pmatrix} -1 & 5 & -1\\\\ 1 & 4 & 4\\\\ 3 & -3 & -2 \end{pmatrix} = -1\det\begin{pmatrix}4 & 4\\\ -3 & -2\end{pmatrix} -5\det\begin{pmatrix}1 & 4\\\ 3 & -2\end{pmatrix} +(-1)\det\begin{pmatrix}1 & 4\\\ 3 & -3\end{pmatrix}.
Calculando:
\det\begin{pmatrix}4 & 4\\\ -3 & -2\end{pmatrix}=4,\quad \det\begin{pmatrix}1 & 4\\\ 3 & -2\end{pmatrix}=-14,\quad \det\begin{pmatrix}1 & 4\\\ 3 & -3\end{pmatrix}=-15.
Logo,
D_y = -1(4) -5(-14) +(-1)(-15) = -4 +70 +15 =81.

Passo 5

Determinante D_z: trocamos a terceira coluna por \begin{pmatrix}5\\\ 4\\\ -3\end{pmatrix},
D_z = \det\begin{pmatrix} -1 & 1 & 5\\\\ 1 & 2 & 4\\\\ 3 & 1 & -3 \end{pmatrix} = -1\det\begin{pmatrix}2 & 4\\\ 1 & -3\end{pmatrix} -1\det\begin{pmatrix}1 & 4\\\ 3 & -3\end{pmatrix} +5\det\begin{pmatrix}1 & 2\\\ 3 & 1\end{pmatrix}.
Calculando:
\det\begin{pmatrix}2 & 4\\\ 1 & -3\end{pmatrix}=-10,\quad \det\begin{pmatrix}1 & 4\\\ 3 & -3\end{pmatrix}=-15,\quad \det\begin{pmatrix}1 & 2\\\ 3 & 1\end{pmatrix}=-5.
Portanto,
D_z = -1(-10) -1(-15) +5(-5) =10 +15 -25 =0.

Passo 6

Aplicando a regra de Cramer:
x = \frac{D_x}{D} = \frac{-54}{27} = -2,\quad y = \frac{D_y}{D} = \frac{81}{27} = 3,\quad z = \frac{D_z}{D} = \frac{0}{27} = 0.

Resposta

x = -2,\;y = 3,\;z = 0.

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