Question

x+2y-z=2
2x-y+z=3
x+y+z=6
calcule o valor da variáveis indicadas no sistema

Answer 1

X+2y-z=2 (i)
2x-y+z=3  (ii)
x+y+z=6   (iii)

(i)-(iii)

y-2z=-4  (iv)

(ii) -2(iii)

-3y-z=-9 ==>3y+z=9   ==>vezes 2 ==> 6y+2z=18  (v)

(iv)+(v) ==> 7y=14 ==>y=2 

Usando (iv)  ==>2-2z=-4 ==>z=6/2=3


Usando (i)  x+2y-z=2 ==>x+4-3=2 ==>x=1

Resposta x=1,y=2,z=3

Answer 2

{x + 2y - z = 2
{2x - y + z = 3
{x + y + z = 6

D

$\left|\begin{array}{ccc}1&2& - 1\\2&- 1&1\\1&1&1\end{array}\right| = \left|\begin{array}{ccc}- 1 - (2.2)&1 - [(2.(- 1)]\\1 - (2.1)&1 - [(1.(- 1)]\end{array}\right| = \left|\begin{array}{ccc} - 5&3\\- 1&2\end{array}\right| = - 7$

Dx

$\left|\begin{array}{ccc}2&2& - 1\\3& - 1&1\\6&1&1\end{array}\right| \\ \\ \\ A_3_1 = 1\: . \: \left|\begin{array}{ccc}2& - 1\\- 1&1\end{array}\right| \\ \\ A_3_1 = 1 \\ \\ A_3_2 = (- 1)\: . \: \left|\begin{array}{ccc}2&- 1\\3&1\end{array}\right| \\ \\ A_3_2 = (- 1).5 = - 5 \\ \\ A_3_3 = 1\: .\: \left|\begin{array}{ccc}2&2\\3& - 1\end{array}\right| \\ \\ A_3_3 = - 8 \\ \\ D_x = 6.1 + 1.(- 5) + 1.(- 8) \\ D_x = - 7$

Dy

$\left|\begin{array}{ccc}1&2& - 1\\2&3&1\\1&6&1\end{array}\right| = \left|\begin{array}{ccc}- 1&3\\4&2\end{array}\right| = - 2 - 12 = - 14$

Dz

$\left|\begin{array}{ccc}1&2&2\\2& - 1&3\\1&1&6\end{array}\right| = \left|\begin{array}{ccc}- 5&- 1\\- 1&4\end{array}\right| = - 20 - 1 = - 21$

x = Dx/D = - 7/ - 7 = 1

y = Dy/D = - 14/- 7 = 2

z = Dz/D = - 21/- 7 = 3

S = {(1, 2, 3)}