Question

02-Resolver o sistema abaixo pela Regra de Cramer.
x+2y-z = 2
2x - y +z = 3
x + y+ z = 6

Answer 1

coeficiente :
$\left[\begin{array}{ccc}1&2&-1\\2&-1&1\\1&1&1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}2\\3\\6\end{array}\right]$

x = $\frac{ \left[\begin{array}{ccc}2&2&-1\\3&-1&1\\6&1&1\end{array}\right] \left[\begin{array}{ccc}2&2\\3&-1\\6&1\end{array}\right] }{ \left[\begin{array}{ccc}1&2&-1\\2&-1&1\\1&1&1\end{array}\right] \left[\begin{array}{ccc}1&2\\2&-1\\1&1\end{array}\right] }$ 

   = $\frac{(-2+12-3)-(6+2+6)}{(-1+2-2)-(4+1+1)} = \frac{7-14}{-1-6}$

   = $\frac{-7}{-7} = 1$

y = $\frac{ \left[\begin{array}{ccc}1&2&-1\\2&3&1\\1&6&1\end{array}\right] \left[\begin{array}{ccc}1&2\\2&3\\1&6\end{array}\right] }{ \left[\begin{array}{ccc}1&2&-1\\2&-1&1\\1&1&1\end{array}\right] \left[\begin{array}{ccc}1&2\\2&-1\\1&1\end{array}\right] }$
 
   = $\frac{(3+2-12)-(4+6-3)}{-7} = \frac{-7-7}{-7}$
 
   = $\frac{-14}{-7} = 2$
   
z = $\frac{ \left[\begin{array}{ccc}1&2&2\\2&-1&3\\1&1&6\end{array}\right] \left[\begin{array}{ccc}1&2\\2&-1\\1&1\end{array}\right] }{ \left[\begin{array}{ccc}1&2&-1\\2&-1&1\\1&1&1\end{array}\right] \left[\begin{array}{ccc}1&2\\2&-1\\1&1\end{array}\right] }$ 

   = $\frac{(-6+6+4)-(24+3-2)}{-7} = \frac{4-25}{-7}$
   
   = $\frac{-21}{-7} = 3$

x = 1, y =2 , z = 3