Determinar o conjunto solução do sistema \left\{\begin{array}{l}x+y+z=6 \\\\ 4 x+2 y-z=5 \\\ x+3 y+2 z=13\end{array}\right..
Soluçāo:
\begin{aligned} & A=\left[\begin{array}{ccc} 1 & 1 & 1 \\\\ 4 & 2 & -1 \\\\ 1 & 3 & 2 \end{array}\right] \Rightarrow \operatorname{det} A=8 ; \quad A_{x}=\left[\begin{array}{ccc} 6 & 1 & 1 \\\\ 5 & 2 & -1 \\\\ 13 & 3 & 2 \end{array}\right] \Rightarrow \operatorname{det} A_{x}=8 ; \\\ & A_{y}=\left[\begin{array}{ccc} 1 & 6 & 1 \\\\ 4 & 5 & -1 \\\\ 1 & 13 & 2 \end{array}\right] \Rightarrow \operatorname{det} A_{y}=16 ; \quad A_{z}=\left[\begin{array}{ccc} 1 & 1 & 6 \\\\ 4 & 2 & 5 \\\\ 1 & 3 & 13 \end{array}\right] \Rightarrow \operatorname{det} A_{y}=24 ; \end{aligned}
i) \mathrm{x}=\frac{\operatorname{det} \mathrm{A}_{\mathrm{x}}}{\operatorname{det} \mathrm{A}}=\frac{8}{8}=1;
ii) \mathrm{y}=\frac{\operatorname{det} \mathrm{A}_{y}}{\operatorname{det} \mathrm{~A}}=\frac{16}{8}=2;
iii) \mathrm{z}=\frac{\operatorname{det} \mathrm{A}_{2}}{\operatorname{det} \mathrm{~A}}=\frac{24}{8}=3
Logo, o conjunto soluçāo do sistema é S=\{(1,2,3)\}

Passo 1

O enunciado pede o conjunto-solução do sistema linear
\begin{cases} x + y + z = 6,\\\\ 4x + 2y - z = 5,\\\ x + 3y + 2z = 13. \end{cases}
O conteúdo aqui é resolução de sistema 3\times3 e usaremos a Regra de Cramer: montamos a matriz dos coeficientes A e as matrizes A_x,A_y,A_z substituindo a coluna correspondente pelos termos independentes; depois calculamos determinantes e aplicamos
x = \frac{\det A_x}{\det A},\quad y = \frac{\det A_y}{\det A},\quad z = \frac{\det A_z}{\det A}.

Passo 2

Montamos
A = \begin{pmatrix} 1 & 1 & 1\\\\ 4 & 2 & -1\\\\ 1 & 3 & 2 \end{pmatrix}
e calculamos \det A por expansão na primeira linha:
\det A =1\cdot\det\begin{pmatrix}2 & -1\\\ 3 & 2\end{pmatrix} -1\cdot\det\begin{pmatrix}4 & -1\\\ 1 & 2\end{pmatrix} +1\cdot\det\begin{pmatrix}4 & 2\\\ 1 & 3\end{pmatrix}.
Cada menor é
\det\begin{pmatrix}2 & -1\\\ 3 & 2\end{pmatrix}=2\cdot2-(-1)\cdot3=4+3=7,
\det\begin{pmatrix}4 & -1\\\ 1 & 2\end{pmatrix}=4\cdot2-(-1)\cdot1=8+1=9,
\det\begin{pmatrix}4 & 2\\\ 1 & 3\end{pmatrix}=4\cdot3-2\cdot1=12-2=10.
Logo
\det A = 1\cdot7 - 1\cdot9 + 1\cdot10 = 7 - 9 + 10 = 8.

Passo 3

Construímos
A_x = \begin{pmatrix} 6 & 1 & 1\\\\ 5 & 2 & -1\\\\ 13 & 3 & 2 \end{pmatrix}
e calculamos \det A_x por expansão na primeira linha:
\det A_x =6\cdot\det\begin{pmatrix}2 & -1\\\ 3 & 2\end{pmatrix} -1\cdot\det\begin{pmatrix}5 & -1\\\ 13 & 2\end{pmatrix} +1\cdot\det\begin{pmatrix}5 & 2\\\ 13 & 3\end{pmatrix}.
Já sabemos \det\begin{pmatrix}2 & -1\\\ 3 & 2\end{pmatrix}=7.
\det\begin{pmatrix}5 & -1\\\ 13 & 2\end{pmatrix}=5\cdot2-(-1)\cdot13=10+13=23,
\det\begin{pmatrix}5 & 2\\\ 13 & 3\end{pmatrix}=5\cdot3-2\cdot13=15-26=-11.
Então
\det A_x =6\cdot7 -1\cdot23 +1\cdot(-11)=42-23-11=8.

Passo 4

Construímos
A_y = \begin{pmatrix} 1 & 6 & 1\\\\ 4 & 5 & -1\\\\ 1 & 13 & 2 \end{pmatrix}
e calculamos \det A_y por expansão na primeira linha:
\det A_y =1\cdot\det\begin{pmatrix}5 & -1\\\ 13 & 2\end{pmatrix} -6\cdot\det\begin{pmatrix}4 & -1\\\ 1 & 2\end{pmatrix} +1\cdot\det\begin{pmatrix}4 & 5\\\ 1 & 13\end{pmatrix}.
Já temos \det\begin{pmatrix}5 & -1\\\ 13 & 2\end{pmatrix}=23 e \det\begin{pmatrix}4 & -1\\\ 1 & 2\end{pmatrix}=9.
\det\begin{pmatrix}4 & 5\\\ 1 & 13\end{pmatrix}=4\cdot13-5\cdot1=52-5=47.
Logo
\det A_y =1\cdot23 -6\cdot9 +1\cdot47 =23-54+47=16.

Passo 5

Construímos
A_z = \begin{pmatrix} 1 & 1 & 6\\\\ 4 & 2 & 5\\\\ 1 & 3 & 13 \end{pmatrix}
e calculamos \det A_z por expansão na primeira linha:
\det A_z =1\cdot\det\begin{pmatrix}2 & 5\\\ 3 & 13\end{pmatrix} -1\cdot\det\begin{pmatrix}4 & 5\\\ 1 & 13\end{pmatrix} +6\cdot\det\begin{pmatrix}4 & 2\\\ 1 & 3\end{pmatrix}.
\det\begin{pmatrix}2 & 5\\\ 3 & 13\end{pmatrix}=2\cdot13-5\cdot3=26-15=11,
\det\begin{pmatrix}4 & 5\\\ 1 & 13\end{pmatrix}=47\quad(\text{já calculado}),á
\det\begin{pmatrix}4 & 2\\\ 1 & 3\end{pmatrix}=4\cdot3-2\cdot1=12-2=10.
Portanto
\det A_z =1\cdot11 -1\cdot47 +6\cdot10 =11-47+60=24.

Passo 6

Aplicando a Regra de Cramer:
x = \frac{\det A_x}{\det A} = \frac{8}{8} = 1,\quad y = \frac{\det A_y}{\det A} = \frac{16}{8} = 2,\quad z = \frac{\det A_z}{\det A} = \frac{24}{8} = 3.

Resposta

O conjunto-solução é
S = \{(1,2,3)\}.