X1+x2-x3=3 2x-x2+x3=0 x1-2x2=1 как решить

х ₁ + x ₂ - x ₃ = 3

2x ₁ - x ₂ + x ₃ = 0

x ₁ - 2x ₂ = 1

Решим с помощью правила Крамера:

1 1 - 1

Δ = 2 - 1 1 = 0 + 4 + 1 - (1 - 2+0) = 5 + 1 = 6

1 - 2 0

3 1 - 1

Δх = 0 - 1 1 = 0 + 0 + 1 - (1 - 6 + 0) = 1 + 5 = 6

1 - 2 0

1 3 - 1

Δу = 2 0 1 = 0 - 2 + 3 - (0 + 1 + 0) = 1 - 1 = 0

1 1 0

1 1 3

Δz = 2 - 1 0 = - 1 - 12 + 0 - (-3 + 0+2) = - 13 + 1 = - 12

1 - 2 1

x = Δx/Δ = 1

y = Δy/Δ = 0

z = Δz/Δ = - 2

Ответ: (1; 0; - 2)