Question

Resolver os sistemas lineares utilizando a regra de Cramer..

Answer 1

Trata-se de uma questão envolvendo equações lineares, assunto do segundo ano do E.M. O conteúdo abordado é a aplicação da Regra de Cramer em sistemas 3x3.

Lembre-se: um determinante é dado pela subtração entre a somatória dos produtos de sua diagonal principal pela somatória dos produtos de sua diagonal secundária.

Raízes da equação A) (5, -2, -2).

Raízes da equação B) (0, 0, 0).

Answer 2

• A)

$\sf \begin{cases}\sf 2x+y+2z=4\\\\\ \sf x+2y+z=-1\\\\\sf 3x+5y+2z=1\end{cases}$

$\sf {Matriz\:de\:coeficientes}\\\\\\\sf M=\begin{pmatrix}\sf 2&\sf 1&\sf 2\\ \sf 1&\sf 2&\sf 1\\ \sf 3&\sf 5&\sf 2\end{pmatrix}\\\\\\\sf \begin{pmatrix}\sf 4\\ \sf -1\\ \sf 1\end{pmatrix}\\\\\\\sf {Substitua\:}x-valores\:da\:coluna\:com\:valores\:da\:coluna\:de\:respostas}\\\\\\\sf M_x=\begin{pmatrix}\sf 4&\sf 1&\sf 2\\ \sf -1&\sf 2&\sf 1\\ \sf 1&\sf 5&\sf 2\end{pmatrix}\\\\\\\sf {Substitua\:}y-valores\:da\:coluna\:com\:valores\:da\:coluna\:de\:respostas}$

$\sf M_y=\begin{pmatrix}\sf 2&\sf 4&\sf 2\\ \sf 1&\sf -1&\sf 1\\ \sf 3&\sf 1&\sf 2\end{pmatrix}$

$\sf {Substitua\:}z-valores\:da\:coluna\:com\:valores\:da\:coluna\:de\:respostas}$

$\sf M_z=\begin{pmatrix}\sf 2&\sf 1&\sf 4\\ \sf 1&\sf 2&\sf -1\\ \sf 3&\sf 5&\sf 1\end{pmatrix}$

$\sf D=-3\\\\\\\sf D_x=-15\\\\\\\sf D_y=6\\\\\\\sf D_z=6$

$\sf \bullet Agora \ Resolva\:usando\:a\:regra\:de\:Cramer$

$\sf \displaystyle x=\frac{D_x}{D}=\frac{-15}{-3}\\\\\\\boxed{\sf S=\{ 5 \}}$

$\sf y=\dfrac{D_y}{D}=\frac{6}{-3}\\\\\\\boxed{\sf S=\{ -2\}}$

$\sf z=\dfrac{D_z}{D}=\dfrac{6}{-3}\\\\\\\boxed{\sf S=\{ -2 \} }$

$\sf \boxed{\sf S=\{ 5,-2,-2\}}$

• B)

$\sf \sf \begin{cases}\sf x+3y-z=0\\\\\ \sf 2y+2z=0\\\\\sf x+y+z=0\end{cases}$

$\sf M=\begin{pmatrix}\sf 1&\sf 3&\sf -1\\ \sf 0&\sf 2&\sf 2\\ \sf 1&\sf 1&\sf 1\end{pmatrix}$

$\sf {Coluna\:de\:respostas:} \begin{pmatrix}\sf 0\\ \sf 0\\ \sf 0\end{pmatrix}$

$\sf M_x=\begin{pmatrix}\sf 0&\sf 3&\sf -1\\ \sf 0&\sf 2&\sf 2\\ \sf 0&\sf 1&\sf 1\end{pmatrix}\\\\\\\sf M_y=\begin{pmatrix}\sf 1&\sf 0&\sf -1\\ \sf 0&\sf 0&\sf 2\\ \sf 1&\sf 0&\sf 1\end{pmatrix}\\\\\\\sf M_z=\begin{pmatrix}\sf 1&\sf 3&\sf 0\\ \sf 0&\sf 2&\sf 0\\ \sf 1&\sf 1&\sf 0\end{pmatrix}$

$\sf D=8\\\\\\\sf D_x=0\\\\\\\sf D_y=0\\\\\\\sf D_z=0$

$\sf Regra\:de\:Cramer:\\\\\\\sf \displaystyle x=\frac{D_x}{D},\:y=\frac{D_y}{D},\:z=\frac{D_z}{D}\\\\\\\sf D\:{denota\:o\:determinante}\\\\\\\sf x=\frac{D_x}{D}=\frac{0}{8}$

$\boxed{\sf S=\{0\}}$

$\sf y=\dfrac{D_y}{D}=\dfrac{0}{8}\\\\\\\sf \boxed{\sf S=\{ 0\}}$

$\sf z=\dfrac{D_z}{D}=\dfrac{0}{8}\\\\\\\boxed{\sf S=\{0\}}$

$\boxed{\sf S=\{0,0,0\}}$