Question

3x-4y=1 x+3y=9 cramer

Answer 1

$\text{Primeiro vamos reescrever o sistema em sua forma matricial:}\\\\\star\text{Relembrar produto entre matrizes.}\\\\\\\left\{\begin{matrix} 3x-4y=1\\\\x+3y=9 \end{matrix}\right.\\\\\\\\\underbrace{\begin{pmatrix}3~~~-4\\1~~~~~~~3\end{pmatrix}}_{matriz~incompleta}\cdot\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}1\\9\end{pmatrix}\\\\\\\star\text{Pelo teorema de Crammer, se o determinante dessa matriz for}\\\text{diferente de zero, o sistema possuir\'a solu\c{c}\~ao \'unica.}$

$D=\begin{vmatrix} 3~~~-4\\ 1~~~~~~~3 \end{vmatrix}=(3\cdot3)-(-4\cdot1)=9-(-4)=9+4=13 \neq 0\\\\\\\text{O sistema possui solu\c{c}\~ao \'unica.}\\\\\star\text{As solu\c{c}\~oes s\~ao obtidas pela seguinte raz\~ao:}\\\\x=\dfrac{D_1}{D}~~\text{onde}~~D_1~~\text{\'e o determinante da matriz obtida da matriz}\\\text{incompleta substituindo a \underline{primeira} coluna pela coluna dos termos}\\\text{independentes do sistema.}$

$y=\dfrac{D_2}{D}~~\text{onde}~~D_2~~\text{\'e o determinante da matriz obtida da matriz}\\\text{incompleta substituindo a \underline{segunda} coluna pela coluna dos termos}\\\text{independentes do sistema.}$

$D_1=\begin{vmatrix}1~~~-4 \\9~~~~~~3 \end{vmatrix}=(1\cdot3)-(9\cdot(-4))=3-(-36)=3+36=39\\\\\\x=\dfrac{D_1}{D}~\Leftrightarrow~\dfrac{39}{13}=\fbox{ 3 }\\\\\\D_2=\begin{vmatrix}3~~~~~1 \\1~~~~~9\end{vmatrix}=(3\cdot9)-(1\cdot1)=27-1=26\\\\\\y=\dfrac{D_2}{D}~\Leftrightarrow~\dfrac{26}{13}=\fbox{ 2 }$

$\text{Assim, a solu\c{c}\~ao para sistema \'e}\\\\\\S=\left\{\begin{matrix}x=3\\\\y=2\end{matrix}\right.$