Definition D1
The solution set of E₁ : a₁x + b₁y = c₁ and E₂ : a₂x + b₂y = c₂ is the set of all (x, y) making both equations true.
Proof.
(1)Each equation represents a line; that is, (a₁, b₁) ≠ (0, 0) and (a₂, b₂) ≠ (0, 0).convention
(2)The solution set is the intersection of the solution sets of the two equations.convention
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Definition D2
D = a₁b₂ − a₂b₁ is called the determinant of this system.
Proof.
D is determined by the four coefficients alone and is independent of the constants c₁, c₂.convention
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Lemma L1
Every solution satisfies Dx = c₁b₂ − c₂b₁ and Dy = a₁c₂ − a₂c₁.
Proof.
(1)Let (x, y) be a solution.assumption, D1
(2)Multiply both sides of E₁ by b₂ and both sides of E₂ by b₁.(1), E
(3)a₁b₂x + b₁b₂y = c₁b₂ and a₂b₁x + b₁b₂y = c₂b₁(2), A1, A2
(4)Subtracting the two equations gives (a₁b₂ − a₂b₁)x = c₁b₂ − c₂b₁(3), A11, A1, A3, A4, A5
(5)That is, Dx = c₁b₂ − c₂b₁(4), D2
(6)Multiplying E₂ by a₁ and E₁ by a₂ and subtracting gives, in the same way, Dy = a₁c₂ − a₂c₁(1), E, A1, A2, A11, A3, A4, A5, D2
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Theorem T1
If D ≠ 0, there is exactly one solution.
Proof.
(1)Suppose D ≠ 0.assumption
(2)If (x, y) and (x′, y′) are solutions, then Dx = Dx′ and Dy = Dy′.L1
(3)Since D ≠ 0, x = x′ and y = y′. There is at most one solution.(1), (2), A7
(4)Set x₀ = (c₁b₂ − c₂b₁)(1/D), y₀ = (a₁c₂ − a₂c₁)(1/D).(1), A8
(5)a₁x₀ + b₁y₀ = (a₁(c₁b₂ − c₂b₁) + b₁(a₁c₂ − a₂c₁))(1/D)(4), A1, A2
(6)In the numerator, −a₁c₂b₁ and b₁a₁c₂ cancel, leaving c₁(a₁b₂ − a₂b₁) = c₁D.(5), A1, A2, A3, A4, A5
(7)a₁x₀ + b₁y₀ = c₁D(1/D) = c₁ × 1 = c₁(5), (6), A2, A8, A6
(8)In the same way, a₂x₀ + b₂y₀ = c₂D(1/D) = c₂(4), A1, A2, A3, A4, A5, A8, A6
(9)(x₀, y₀) is a solution, and by (3) it is the only one.(7), (8), (3), D1
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Theorem T2
If D = 0, there is no solution or there are infinitely many.
Proof.
(1)Suppose D = 0.assumption
(2)If there is no solution, the claim already holds. Suppose there is a solution (x₀, y₀).assumption, D1
(3)For each real s, take the point (x₀ − b₁s, y₀ + a₁s).convention
(4)a₁(x₀ − b₁s) + b₁(y₀ + a₁s) = (a₁x₀ + b₁y₀) + (−a₁b₁s + b₁a₁s)(3), A1, A2, A3, A4
(5)Since −a₁b₁s + b₁a₁s = 0, the value is c₁. The first equation holds.(2), (4), A2, A5
(6)a₂(x₀ − b₁s) + b₂(y₀ + a₁s) = (a₂x₀ + b₂y₀) + (a₁b₂ − a₂b₁)s(3), A1, A2, A3, A4
(7)(a₁b₂ − a₂b₁)s = Ds = 0 × s = 0, so the value is c₂. The second equation holds as well.(1), (2), (6), D2, A9, A5
(8)Since (a₁, b₁) ≠ (0, 0), if s ≠ s′, then b₁s ≠ b₁s′ or a₁s ≠ a₁s′.D1, A7
(9)Hence distinct s give distinct solutions.(8), A3, A4, A5
(10)There are infinitely many reals, so there are infinitely many solutions.(9), A12
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Theorem T3
When D = 0, both cases actually occur.
Proof.
(1)For x + y = 0 and x + y = 1, we have D = 1×1 − 1×1 = 0.D2, A6, A5
(2)A common solution would give 0 = 1, a contradiction. There is no solution.(1), D1, A10
(3)For x + y = 0 and 2x + 2y = 0, we have D = 1×2 − 2×1 = 0.D2, A2, A6, A5
(4)(s, −s) satisfies both equations for every real s.A1, A5
(5)Distinct s give distinct solutions, so there are infinitely many.(4), A12
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Corollary C
If b₁ ≠ 0 and b₂ ≠ 0, then D = 0 means the two lines have the same slope.
Proof.
(1)Since b₁ ≠ 0, E₁ can be written y = (−a₁)(1/b₁)x + c₁(1/b₁), with m₁ = (−a₁)(1/b₁).A8, A1, A2, A6
(2)In the same way, m₂ = (−a₂)(1/b₂).A8, A1, A2, A6
(3)Multiplying both sides of m₁ = m₂ by b₁b₂ gives −a₁b₂ = −a₂b₁(1), (2), E, A2, A8, A6
(4)Hence a₁b₂ = a₂b₁, that is, D = 0.(3), A5, D2
(5)Conversely, if D = 0, then a₁b₂ = a₂b₁, and multiplying both sides by (1/b₁)(1/b₂) gives m₁ = m₂.D2, A5, E, A8, A2, A6
(6)The algebraic condition that the determinant is 0 is the geometric condition that the slopes are equal.(4), (5)
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