1/x-1/y+1/z=4,
2/x+1/y-3/z=0,
1/x+1/y+1/z=2. Let Δ=
| 1 -1 1 |
| 2 1 -3 | = 1(1+3)+1(2+3)+1(2-1)=4+5+1=10
| 1 1 1 |
Δx=
| 4 -1 1 |
| 0 1 -3 | = 4(1+3)+1(0+6)+1(0-2)=16+6-2=20
| 2 1 1 |
1/x=Δx/Δ=20/10=2, so x=½.
Δy=
| 1 4 1 |
| 2 0 -3 | = 1(0+6)-4(2+3)+1(4-0)=6-20+4=-10
| 1 2 1 |
1/y=Δy/Δ=-10/10=-1, so y=-1.
Δz=
| 1 -1 4 |
| 2 1 0 | = 1(2-0)+1(4-0)+4(2-1)=2+4+4=10
| 1 1 2 |
1/z=Δz/Δ=10/10=1, so z=1.
SOLUTION. (x,y,z)=(½,-1,1).