Question

alguém me ajuda pfv???​

Answer 1

a)

$\mathrm{Dado\ o\ sistema}\ \text{linear:}\ \begin{cases}3x-2y=-12 \\5x+6y=8\end{cases}$

$\mathrm{Podemos\ resolv\hat{e}}\text{-lo}\ \mathrm{atrav\acute{e}s\ da}\ \textbf{Regra\ de\ Cramer.}$

$\mathrm{Primeiramente,\ calculam}\text{-se}\ \mathrm{os\ determinantes}\ D,\ D_x\ \text{e}\ D_y.$

$D=\left|\begin{array}{cc}3&-2\\5&6\end{array}\right|=18-(-10)\Longrightarrow D=28$

$D_x=\left|\begin{array}{cc}-12&-2\\8&6\end{array}\right|=-72-(-16)\Longrightarrow D_x=-56$

$D_y=\left|\begin{array}{cc}3&-12\\5&8\end{array}\right|=24-(-60)\Longrightarrow D_y=84$

$\mathrm{Agora,\ \acute{e}\ poss\acute{\i}vel\ obter\ os\ valores\ das\ inc\acute{o}gnitas}\ x\ \text{e}\ y.$

$x=\dfrac{D_x}{D}\Longrightarrow x=-\dfrac{56}{28}\ \therefore\ \boxed{x=-2}$

$y=\dfrac{D_y}{D}\Longrightarrow y=\dfrac{84}{28}\ \therefore\ \boxed{y=3}$

$\mathrm{A\ solu\c{c}\tilde{a}o\ do\ sistema\ linear\ \acute{e}}\ (x,y)=(-2,3).$

b)

$\text{Dado o sistema linear:}\ \begin{cases}x+4y+7z=2 \\2x+3y+6z=2 \\5x+y-z=8\end{cases}$

$\mathrm{Podemos\ resolv\hat{e}}\text{-lo}\ \mathrm{atrav\acute{e}s\ da}\ \textbf{Regra\ de\ Cramer.}$

$\mathrm{Primeiramente,\ calculam}\text{-se}\ \mathrm{os\ determinantes}\ D,\ D_x,\ D_y\ \text{e}\ D_z.$

$D=\left|\begin{array}{ccc}1&4&7\\2&3&6\\5&1&-1\end{array}\right|\Longrightarrow D=28$

$D_x=\left|\begin{array}{ccc}2&4&7\\2&3&6\\8&1&-1\end{array}\right|\Longrightarrow D_x=28$

$D_y=\left|\begin{array}{ccc}1&2&7\\2&2&6\\5&8&-1\end{array}\right|\Longrightarrow D_y=56$

$D_z=\left|\begin{array}{ccc}1&4&2\\2&3&2\\5&1&8\end{array}\right|\Longrightarrow D_z=-28$

$\mathrm{Agora,\ \acute{e}\ poss\acute{\i}vel\ obter\ os\ valores\ das\ inc\acute{o}gnitas}\ x,\ y\ \text{e}\ z.$

$x=\dfrac{D_x}{D}\Longrightarrow x=\dfrac{28}{28}\ \therefore\ \boxed{x=1}$

$y=\dfrac{D_y}{D}\Longrightarrow y=\dfrac{56}{28}\ \therefore\ \boxed{y=2}$

$z=\dfrac{D_z}{D}\Longrightarrow z=-\dfrac{28}{28}\ \therefore\ \boxed{z=-1}$

$\mathrm{A\ solu\c{c}\tilde{a}o\ do\ sistema\ linear\ \acute{e}}\ (x,y,z)=(1,2,-1).$