- A solução do sistema linear 3x3 calculado pela regra de Cramer é igual a:
[tex]\Large\text{$\begin{gathered}S=\left\{\left( -\frac{5}{6} \ ,\ -\frac{7}{6}\ ,\ \frac{1}{3}\right)\right\}.\end{gathered}$}[/tex]
Desejamos calcular pela regra de Cramer o seguinte sistema 3x3:
[tex]\Large\text{$\begin{gathered}\begin{cases} x+y+6z=0\\ x-y-z=5\\ 2x-6y-z=5\end{cases}\ \ \ \forall x,y,z \in \mathbb{Z}.\end{gathered}$}[/tex]
A regra de cramer é dada da seguinte forma:
[tex]\Large\text{$\begin{gathered} (x,y,z)= \left( \frac{\det (D_x)}{\det (D)} \ ,\ \frac{\det (D_y)}{\det (D)}\ ,\ \frac{\det (D_z)}{\det (D)}\right)\ \ \red{(I)}.\end{gathered}$}[/tex]
Onde temos que det(D) será a matriz dos coeficientes, ou seja:
[tex]\Large\text{$\begin{gathered} \det(D)=\left[\begin{array}{ccc}1&1&6\\1&-1&-1\\2&-6&-1\end{array}\right] \end{gathered}$}[/tex]
Pela regra de Sarrus, temos que:
[tex]\Large\text{$\begin{gathered} \det(D)=\left[\begin{array}{ccc}1&1&6\\1&-1&-1\\2&-6&-1\end{array}\right] \left[\begin{array}{ccc}1&1\\1&-1\\2&-6\end{array}\right] \end{gathered}$}[/tex]
[tex]\Large\text{$\begin{gathered} \det(D)=1-2-36-(-12+6-1) \end{gathered}$}[/tex]
[tex]\Large\text{$\begin{gathered} \therefore \boxed{\det(D)=-30}\ \ \ \red{(II)} .\end{gathered}$}[/tex]
Feito isso, vamos agora calcular o det(Dx). Para isso iremos fazer a mesma coisa do caso anterior, mas na fileira do x deveremos substituir pela fileira dos números do lado direito da igualdade. Com isso, temos que:
[tex]\Large\text{$\begin{gathered} \det(D_x)=\left[\begin{array}{ccc}0&1&6\\0&-1&-1\\5&-6&-1\end{array}\right] \left[\begin{array}{ccc}0&1\\0&-1\\5&-6\end{array}\right] \end{gathered}$}[/tex]
[tex]\Large\text{$\begin{gathered} \det(D_x)=-5-(-30) \end{gathered}$}[/tex]
[tex]\Large\text{$\begin{gathered} \therefore \boxed{\det(D_x)=25}\ \ \ \red{(III)} .\end{gathered}$}[/tex]
Agora vamos calcular o det(Dy() substituindo a fileira do y pela fileira dos números.
[tex]\Large\text{$\begin{gathered} \det(D_y)=\left[\begin{array}{ccc}1&0&6\\1&0&-1\\2&5&-1\end{array}\right] \left[\begin{array}{ccc}1&0\\1&0\\2&5\end{array}\right] \end{gathered}$}[/tex]
[tex]\Large\text{$\begin{gathered} \det(D_y)=30-(-5) \end{gathered}$}[/tex]
[tex]\Large\text{$\begin{gathered} \therefore \boxed{\det(D_y)=35} \ \ \ \red{(IV)}. \end{gathered}$}[/tex]
E por fim, vamos calcular o det(Dz) substituindo a fileira do z pela fileira dos números.
[tex]\Large\text{$\begin{gathered} \det(D_z)=\left[\begin{array}{ccc}1&1&0\\1&-1&0\\2&-6&5\end{array}\right] \left[\begin{array}{ccc}1&1\\1&-1\\2&-6\end{array}\right] \end{gathered}$}[/tex]
[tex]\Large\text{$\begin{gathered} \det(D_z)=-5-(5) \end{gathered}$}[/tex]
[tex]\Large\text{$\begin{gathered} \therefore \boxed{\det(D_z)=-10} \ \ \ \red{(V)}. \end{gathered}$}[/tex]
- Substituindo na equação I, temos que:
[tex]\Large\text{$\begin{gathered} (x,y,z)= \left( \frac{\det (D_x)}{\det (D)} \ ,\ \frac{\det (D_y)}{\det (D)}\ ,\ \frac{\det (D_z)}{\det (D)}\right)\ \ \end{gathered}$}[/tex]
[tex]\Large\text{$\begin{gathered} (x,y,z)= \left( -\frac{25}{30} \ ,\ -\frac{35}{30}\ ,\ \frac{10}{30}\right)\ \ \end{gathered}$}[/tex]
[tex]\Large\text{$\begin{gathered}\therefore \green{\underline{\boxed{(x,y,z)= \left( -\frac{5}{6} \ ,\ -\frac{7}{6}\ ,\ \frac{1}{3}\right)}}}\ \ (\checkmark).\end{gathered}$}[/tex]
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