Question

Qual a solução do sistema? *

x – 3y + 2z = 14
2x+5y-z = -9
-3x – y +2z = 2

(-2, 2, 3)

(2, -2, 3)

(3, -2, 2)

(1, -3, -2).

(-2, -3, 1)

Answer 1

Resposta:

(2, -2, 3)

Explicação passo-a-passo:

Regra de Cramer:

$\left[\begin{array}{ccc}1&-3&2\\2&5&-1\\-3&-1&2\end{array}\right] .\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}14\\-9\\2\end{array}\right]$

$\displaystyle \Delta=\left[\begin{array}{ccc}1&-3&2\\2&5&-1\\-3&-1&2\end{array}\right] =(1).(5).(2)+(-3).(-1).(-3)+(2).(-1).(2)-(-3).(5).(2)-(2).(-3).(2)-(-1).(-1).(1)=-3+(41)=38$

$\displaystyle \Delta x= \left[\begin{array}{ccc}14&-3&2\\-9&5&-1\\2&-1&2\end{array}\right] =(14).(5).(2)+(-3).(-1).(2)+(-9).(-1).(2)-(2).(5).(2)-(-9).(-3).(2)-(-1).(-1).(14)=164+(-88)=76$

$\displaystyle \Delta y=\left[\begin{array}{ccc}1&14&2\\2&-9&-1\\-3&2&2\end{array}\right] =(1).(-9).(2)+(14).(-1).(-3)+(2).(2).(2)-(-3).(-9).(2)-(2).(14).(2)-(2).(-1).(1)=32+(-108)=-76$$\displaystyle \Delta z=\left[\begin{array}{ccc}1&-3&14\\2&5&-9\\-3&-1&2\end{array}\right] =(1).(5).(2)+(-3).(-9).(-3)+(2).(-1).(14)-(-3).(5).(14)-(2).(-3).(2)-(-1).(-9).(1)=-99+(213)=114$

x=Δx/Δ=76/38=2

y=Δy/Δ= -76/38= -2

z=Δz/Δ=114/38=3