4x-y+ 3z =1
3x+2y+4z=8
2x-2y+4z=0
решение:
4 -1 3
Δ = 3 2 4 = 32 - 18 - 8 - ( 12 -32 -12) = 6 +32 = 38
2 -2 4
1 -1 3
Δ х = 8 2 4 = 8 -48 +0 -( 0 -8 -32 ) = -40 +40 = 0
0 -2 4
4 1 3
Δ у= 3 8 4 = 128 +8 +0 - ( 48 +0 +12) = 136 -60 = 78
2 0 4
4 -1 1
Δ z = 3 2 8 = 0 -6 -16 - ( 4 -64 -0) = -22 + 60 = 38
2 -2 0
х = Δ х/Δ = 0/38 = 0
у = Δ у/Δ = 76/38 = 2
z= Δz / Δ = 38/38 = 1
Ответ: (0; 2; 1)