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Question
Solve the following systems of linear equations by Cramer’s rule:
3x + 3y – z = 11, 2x – y + 2z = 9, 4x + 3y + 2z = 25
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Solution
Δ = `|(3, 3, -1),(2, -1, 2),(4, 3, 2)|`
= 3(– 2 – 6) – 3(4 – 8) –1(6 + 4)
= 3(– 8) – 3(– 4) – 1(10)
= – 24 + 12 – 10
= – 22 ≠ 0
Δx = `|(11, 3, -1),(9, -1, 2),(25, 3, 2)|`
= 11 (– 2 – 6) – 3(18 – 50) – 1(27 + 25)
= 11(– 8) – 3(32) – 1(52)
= – 88 + 96 – 52
= – 44
Δy = `|(3, 11, -1),(2, 9, 2),(4, 25, 2)|`
= 3(18 – 50) – 11(4 – 8) – 1(50 – 36)
= 3(32) – 11(4) – 1(14)
= – 96 + 44 – 14
= – 66
Δx = `|(3, 3, 11),(2, -1, 9),(4, 3, 25)|`
= 3(– 25 – 27) – 3(50 – 36) + 11(6 + 4)
= 3(– 52) – 3(14) + 11(10)
= – 156 – 42 + 110
= – 88
By Cramer's rule x = `Delta_x/Delta = (-44)/(-22)` = 2
y = `Delta_y/Delta = (-66)/(-22)` = 3
z = `Delta_z/Delta = (-88)/(-22)` = 4
∴ x = 2, y = 3, z = 4.
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