Question

Resolver sistema linear

x+y+z=3
2x-3y+z=-1
-x+y-2z=-3

Answer 1

Oi Jv.

$x+y+z=3\quad (I)\\ 2x-3y+z=-1\quad (II)\\ -x+y-2z=-3\quad (III)\\ \\ (I)+(III)\\ x+y+z=3\\ -x+y-2z=-3\\ 2y-z=0\\ \\ (-2*I)+(II)\\ -2x-2y-2z=-6\\ 2x-3y+z=-1\\ -5y-z=-7\\ \\ 2y-z=0\quad \\ -5y-z=-7\quad (-1)\\ \\ 2y-z=0\\ 5y+z=7\\ 7y=7\\ y=\frac { 7 }{ 7 } \Leftrightarrow 1\\ \\ 2(1)-z=0\\ 2-z=0\\ 2=z\\ \\ x+1+2=3\\ x=3-3\\ x=0$