Question

Questão 2 - Resolva o sistema abaixo usando a regra de Cramer e marque a solução:


a) S = { 1, 1, 1 }

b) S = { 1, -1/2, -½ }

c) S = { 1, - ⅓ , - ⅓ }

d) S = { 1 , 3 , 4 }

Answer 1

$\Large\boxed{\begin{array}{l}\begin{cases}\sf x+3y-z=0\\\sf 2x+y+z=1\\\sf 3x-y+z=3\end{cases}\\\sf A=\begin{bmatrix}\sf1&\sf3&\sf-1\\\sf2&\sf1&\sf1\\\sf3&\sf-1&\sf1\end{bmatrix}\\\sf det\,A=1\cdot(1+1)-3\cdot(2-3)-1\cdot(-2-3)\\\sf det\,A=2+3+5\\\sf det\,A=10\end{array}}$

$\Large\boxed{\begin{array}{l}\sf A_x=\begin{bmatrix}\sf0&\sf3&\sf-1\\\sf1&\sf1&\sf1\\\sf3&\sf-1&\sf1\end{bmatrix}\\\sf det\,A_x=-3\cdot(1-3)-1\cdot(-1-3)\\\sf det\,A_x=6+4\\\sf det\,A_x=10\end{array}}$

$\Large\boxed{\begin{array}{l}\sf A_y=\begin{bmatrix}\sf1&\sf0&\sf-1\\\sf2&\sf1&\sf1\\\sf3&\sf3&\sf1\end{bmatrix}\\\sf det\,A_y=1\cdot(1-3)-1\cdot(6-3)\\\sf det\,A_y=-2-3\\\sf det\,A_y=-5\end{array}}$$\Large\boxed{\begin{array}{l}\sf A_z=\begin{bmatrix}\sf1&\sf3&\sf0\\\sf2&\sf1&\sf1\\\sf3&\sf-1&\sf3\end{bmatrix}\\\sf det\,A_z=1\cdot(3+1)-3\cdot(6-3)\\\sf det\,A_z=4-9\\\sf det\,A_z=-5\end{array}}$

$\Large\boxed{\begin{array}{l}\sf x=\dfrac{det\,A_x}{det\,A}\\\\\sf x=\dfrac{10}{10}=1\\\\\sf y=\dfrac{det\,A_y}{det\,A}\\\\\sf y=\dfrac{-5\div5}{10\div5}=-\dfrac{1}{2}\\\\\sf z=\dfrac{det\,A_z}{det\,A}\\\\\sf z=\dfrac{-5\div5}{10\div5}\\\\\sf z=-\dfrac{1}{2}\\\\\sf S=\bigg\{1,-\dfrac{1}{2},-\dfrac{1}{2}\bigg\}\\\huge\boxed{\boxed{\boxed{\boxed{\sf\red{\maltese}~\blue{alternativa~b}}}}}\end{array}}$