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Realizando os cálculos, encontramos que o conjunto solução do sistema é:

[tex]\boxed{\large\displaystyle\text{$\mathsf{S=(2, 2, 1)}$}}[/tex]

A Regra de Cramer é uma técnica utilizada para resolver sistemas de equações.

• Neste método, os valores das variáveis ​​do sistema devem ser calculados usando os determinantes das matrizes.

A Regra de Cramer estabelece as seguinte relações:

[tex]\boxed{\large\displaystyle\text{$\mathsf{x=\dfrac{Dx}{D}, y=\dfrac{Dy}{D}, z=\dfrac{Dz}{D}}$}}[/tex]

Em que:

• D é o determinante da matriz formada pelos coeficientes do sistema.
• Dx, Dy e Dz são os determinantes da matriz formadas pelos coeficientes do sistema, porém com a coluna dos coeficientes de x, y e z, respectivamente, substituídas pelos termos independentes.

Temos o seguinte sistema de equações:

[tex]\large\displaystyle\text{$\mathsf{\left\{\begin{matrix} x+y-3z=1\\ 2x+y+0=6 \\ x+0+z=3 \end{matrix}\right.}$}[/tex]

A matriz formada pelos coeficientes do sistema será a seguinte:

[tex]\large\displaystyle\text{$\mathsf{\begin{vmatrix} 1 & 1 & -3\\ 2 & 1 & 0\\ 1& 0 & 1 \end{vmatrix}}$}[/tex]

Com isso, podemos calcular o determinante "D":

[tex]\large\displaystyle\text{$\mathsf{\begin{vmatrix} 1 & 1 & -3\\ 2 & 1 & 0\\ 1& 0 & 1 \end{vmatrix}}$}\\\\\\\\\large\displaystyle\text{$\mathsf{\begin{vmatrix} 1 & 1 & -3\\ 2 & 1 & 0\\ 1& 0 & 1 \end{vmatrix}\left.\begin{matrix} 1 & 1\\ 2 & 1\\ 1 & 0 \end{matrix}\right|}$}[/tex]

[tex]\large\displaystyle\text{$\mathsf{D=\underbrace{1~.~1~.~1}_{1}+\underbrace{1~.~0~.~1}_{0}+\underbrace{(-3)~.~2~.~0}_{0}-\underbrace{(-3)~.~1~.~1}_{-3}-\underbrace{1~.~0~.~0}_{0}-\underbrace{1~.~2~.~1}_{2}}$}\\\\\\\large\displaystyle\text{$\mathsf{D=1+0+0-(-3)-0-2}$}\\\large\displaystyle\text{$\mathsf{D=1+3-2}$}\\\boxed{\large\displaystyle\text{$\mathsf{D=2}$}}[/tex]

Agora devemos calcular o determinante Dx. Lembre-se de substituir a coluna com os coeficientes de x (a primeira) pelos termos independentes.

[tex]\large\displaystyle\text{$\mathsf{\begin{vmatrix} \red{1} & 1 & -3\\ \red{2} & 1 & 0\\ \red{1}& 0 & 1 \end{vmatrix} \rightarrow \begin{vmatrix} \red{1} & 1 & -3\\ \red{6} & 1 & 0\\ \red{3}& 0 & 1 \end{vmatrix} }$}\\\\\\\large\displaystyle\text{$\mathsf{\begin{vmatrix} 1 & 1 & -3\\ 6 & 1 & 0\\ 3& 0 & 1 \end{vmatrix} \left.\begin{matrix} 1 & 1\\ 6& 1\\ 3& 0 \end{matrix}\right|}$}\\\\\\[/tex]

[tex]\large\displaystyle\text{$\mathsf{Dx=\underbrace{1~.~1~.~1}_{1}+\underbrace{1~.~0~.~3}_{0}+\underbrace{(-3)~.~6~.~0}_{0}-\underbrace{(-3)~.~1~.~3}_{-9}-\underbrace{1~.~0~.~0}_{0}-\underbrace{1~.~6~.~1}_{6}}$}\\\\\\\large\displaystyle\text{$\mathsf{Dx=1+0+0-(-9)-0-6}$}\\\large\displaystyle\text{$\mathsf{Dx=1+9-6}$}\\\boxed{\large\displaystyle\text{$\mathsf{Dx=4}$}}[/tex]

Calcularemos agora o determinante Dy. A coluna dos coeficientes de y (a segunda) deve ser substituída pelos termos independentes.

[tex]\large\displaystyle\text{$\mathsf{\begin{vmatrix} 1 & \red{1} & -3\\ 2 & \red{1} & 0\\ 1& \red{0} & 1 \end{vmatrix}\rightarrow \begin{vmatrix} 1 & \red{1} & -3\\ 2 & \red{6} & 0\\ 1& \red{3} & 1 \end{vmatrix}}$}\\\\\\\large\displaystyle\text{$\mathsf{\begin{vmatrix} 1 & 1 & -3\\ 2 & 6 & 0\\ 1& 3 & 1 \end{vmatrix}\left.\begin{matrix} 1 & 1\\ 2 & 6\\ 1 & 3 \end{matrix}\right|}$}[/tex]

[tex]\large\displaystyle\text{$\mathsf{Dy=\underbrace{1~.~6~.~1}_{6}+\underbrace{1~.~0~.~1}_{0}+\underbrace{(-3)~.~2~.~3}_{-18}-\underbrace{(-3)~.~6~.~1}_{-18}-\underbrace{1~.~0~.~3}_{0}-\underbrace{1~.~2~.~1}_{2}}$}\\\\\\\large\displaystyle\text{$\mathsf{Dy=6+0+(-18)-(-18)-0-2}$}\\\large\displaystyle\text{$\mathsf{Dy=6-18+18-2}$}\\\boxed{\large\displaystyle\text{$\mathsf{Dy=4}$}}[/tex]

Agora o determinante Dz. Novamente, substitua a colina dos coeficientes de z (a terceira) pelos termos independentes.

[tex]\large\displaystyle\text{$\mathsf{\begin{vmatrix} 1 & 1 & \red{-3}\\ 2 & 1 & \red{0}\\ 1& 0 & \red{1} \end{vmatrix} \rightarrow \begin{vmatrix} 1 & 1 & \red{1}\\ 2 & 1 & \red{6}\\ 1& 0 & \red{3} \end{vmatrix}}$}\\\\\\\large\displaystyle\text{$\mathsf{\begin{vmatrix} 1 & 1 & 1\\ 2 & 1 & 6\\ 1& 0 & 3 \end{vmatrix}\left.\begin{matrix} 1 & 1\\ 2 & 1\\ 1 & 0 \end{matrix}\right|}$}[/tex]

[tex]\large\displaystyle\text{$\mathsf{Dz=\underbrace{1~.~1~.~3}_{3}+\underbrace{1~.~6~.~1}_{6}+\underbrace{1~.~2~.~0}_{0}-\underbrace{1~.~1~.~1}_{1}-\underbrace{1~.~6~.~0}_{0}-\underbrace{1~.~2~.~3}_{6}}$}\\\\\\\large\displaystyle\text{$\mathsf{Dz=3+6+0-1-0-6}$}\\\boxed{\large\displaystyle\text{$\mathsf{Dz=2}$}}[/tex]

Tendo todos os determinaets, podemos estabelecer os valores das incognitas x, y e z:

[tex]\large\displaystyle\text{$\mathsf{x=\dfrac{Dx}{D}, y=\dfrac{Dy}{D}, z=\dfrac{Dz}{D}}$}\\\\\\\large\displaystyle\text{$\mathsf{x=\dfrac{4}{2}, y=\dfrac{4}{2}, z=\dfrac{2}{2}}$}\\\\\\\boxed{\large\displaystyle\text{$\mathsf{x=2, y=2, z=1}$}}[/tex]

O conjunto solução do sistema é, portanto:

[tex]\boxed{\boxed{\large\displaystyle\text{$\mathsf{S=(2, 2, 1)}$}}}[/tex]

⭐ Espero ter ajudado! ⭐

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