1. Aplique Cramer (se possível):
\left\{\begin{array}{l} 4 x-3 y+z=10 \\ 2 x+y-2 z=0 \\ x+2 y+3 z=6 \end{array}\right.

Passo 1

O enunciado pede para resolver o sistema
\begin{cases} 4x - 3y + z = 10,\\\\ 2x + y - 2z = 0,\\\ x + 2y + 3z = 6 \end{cases}
usando a Regra de Cramer. Para isso, vamos calcular o determinante da matriz de coeficientes D e os determinantes obtidos ao substituir cada coluna pelos termos independentes D_x, D_y, D_z, e então usar
x = \frac{D_x}{D},\quad y = \frac{D_y}{D},\quad z = \frac{D_z}{D}.

Passo 2

Calculando D, determinante da matriz de coeficientes:
D = \begin{vmatrix} 4 & -3 & 1\\\\ 2 & 1 & -2\\\\ 1 & 2 & 3 \end{vmatrix}.
Usando a regra de Sarrus ou expansão pela primeira linha:
D = 4\cdot\begin{vmatrix}1 & -2\\\ 2 & 3\end{vmatrix} -(-3)\cdot\begin{vmatrix}2 & -2\\\ 1 & 3\end{vmatrix} +1\cdot\begin{vmatrix}2 & 1\\\ 1 & 2\end{vmatrix}.
Agora cada subdeterminante:
\begin{vmatrix}1 & -2\\\ 2 & 3\end{vmatrix} = 1\cdot3 -(-2)\cdot2 = 7, \quad \begin{vmatrix}2 & -2\\\ 1 & 3\end{vmatrix} = 2\cdot3 -(-2)\cdot1 = 8, \quad \begin{vmatrix}2 & 1\\\ 1 & 2\end{vmatrix} = 2\cdot2 -1\cdot1 = 3.
Logo,
D = 4\cdot7 +3\cdot8 +1\cdot3 = 28 + 24 + 3 = 55.

Passo 3

Calculando D_x, substituindo a primeira coluna pelos termos independentes (10,0,6):
D_x = \begin{vmatrix} 10 & -3 & 1\\\\ 0 & 1 & -2\\\\ 6 & 2 & 3 \end{vmatrix} = 10\cdot\begin{vmatrix}1 & -2\\\ 2 & 3\end{vmatrix} -(-3)\cdot\begin{vmatrix}0 & -2\\\ 6 & 3\end{vmatrix} +1\cdot\begin{vmatrix}0 & 1\\\ 6 & 2\end{vmatrix}.
Calculando esses subdeterminantes:
\begin{vmatrix}1 & -2\\\ 2 & 3\end{vmatrix}=7, \quad \begin{vmatrix}0 & -2\\\ 6 & 3\end{vmatrix}=0\cdot3 -(-2)\cdot6 = 12, \quad \begin{vmatrix}0 & 1\\\ 6 & 2\end{vmatrix}=0\cdot2 -1\cdot6 = -6.
Assim,
D_x = 10\cdot7 +3\cdot12 +1\cdot(-6) = 70 + 36 - 6 = 100.
E portanto
x = \frac{D_x}{D} = \frac{100}{55} = \frac{20}{11}.

Passo 4

Calculando D_y, substituindo a segunda coluna pelos termos independentes:
D_y = \begin{vmatrix} 4 & 10 & 1\\\\ 2 & 0 & -2\\\\ 1 & 6 & 3 \end{vmatrix} = 4\cdot\begin{vmatrix}0 & -2\\\ 6 & 3\end{vmatrix} -10\cdot\begin{vmatrix}2 & -2\\\ 1 & 3\end{vmatrix} +1\cdot\begin{vmatrix}2 & 0\\\ 1 & 6\end{vmatrix}.
Os subdeterminantes são
\begin{vmatrix}0 & -2\\\ 6 & 3\end{vmatrix}=12, \quad \begin{vmatrix}2 & -2\\\ 1 & 3\end{vmatrix}=8, \quad \begin{vmatrix}2 & 0\\\ 1 & 6\end{vmatrix}=12.
Logo
D_y = 4\cdot12 -10\cdot8 +1\cdot12 = 48 - 80 + 12 = -20,
de onde
y = \frac{D_y}{D} = \frac{-20}{55} = -\frac{4}{11}.

Passo 5

Calculando D_z, substituindo a terceira coluna pelos termos independentes:
D_z = \begin{vmatrix} 4 & -3 & 10\\\\ 2 & 1 & 0\\\\ 1 & 2 & 6 \end{vmatrix} = 4\cdot\begin{vmatrix}1 & 0\\\ 2 & 6\end{vmatrix} -(-3)\cdot\begin{vmatrix}2 & 0\\\ 1 & 6\end{vmatrix} +10\cdot\begin{vmatrix}2 & 1\\\ 1 & 2\end{vmatrix}.
Os subdeterminantes são
\begin{vmatrix}1 & 0\\\ 2 & 6\end{vmatrix}=6, \quad \begin{vmatrix}2 & 0\\\ 1 & 6\end{vmatrix}=12, \quad \begin{vmatrix}2 & 1\\\ 1 & 2\end{vmatrix}=3.
Então
D_z = 4\cdot6 +3\cdot12 +10\cdot3 = 24 + 36 + 30 = 90,
e
z = \frac{D_z}{D} = \frac{90}{55} = \frac{18}{11}.

Resposta

x = \tfrac{20}{11},\quad y = -\tfrac{4}{11},\quad z = \tfrac{18}{11}.

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