Question

Resolva o sistema usando regra de Cramer:​

Answer 1

Resposta:

x= -1, y=1 e z= -1

Explicação passo-a-passo:

Regra de Cramer

$\displaystyle \left[\begin{array}{ccc}3&2&-1\\1&3&1\\2&2&-2\end{array}\right] .\left[\begin{array}{ccc}x\\y\\x\end{array}\right] =\left[\begin{array}{ccc}0\\1\\2\end{array}\right]$

$\displaystyle \displaystyle \Delta=\left[\begin{array}{ccc}3&2&-1\\1&3&1\\2&2&-2\end{array}\right] \\\\\ \Delta=3.3.(-2)+2.1.2+1.2.(-1)-2.3.(-1)-1.2.(-2)-2.1.3=-18 + 4-2 + 6 + 4-6=-12$

Cálculo de x:

$\displaystyle \Delta~x= \left[\begin{array}{ccc}0&2&-1\\1&3&1\\2&2&-2\end{array}\right] =(0).(3).(-2)+(2).(1).(2)+(1).(2).(-1)-(2).(3).(-1)-(1).(2).(-2)-(2).(1).(0)=2+(10)=12\\\\x=\frac{ \Delta~x}{ \Delta} =\frac{12}{-12} =-1$

Cálculo de y:

$\displaystyle\Delta~y= \left[\begin{array}{ccc}3&0&-1\\1&1&1\\2&2&-2\end{array}\right] =(3).(1).(-2)+(0).(1).(2)+(1).(2).(-1)-(2).(1).(-1)-(1).(0).(-2)-(2).(1).(3)=-8+(-4)=-12\\\\y=\frac{\Delta~y}{\Delta} =\frac{-12}{-12} =1$

Cálculo de z:

$\displaystyle \Delta~z=\left[\begin{array}{ccc}3&2&0\\1&3&1\\2&2&2\end{array}\right] =(3).(3).(2)+(2).(1).(2)+(1).(2).(0)-(2).(3).(0)-(1).(2).(2)-(2).(1).(3)=22+(-10)=12\\\\z=\frac{\Delta~z}{\Delta} =\frac{12}{-12} =-1$