Implementasi Eliminasi Gauss

07/06/2022

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IMPLEMENTASI ELIMINASI GAUSS

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• Y = Dependent variable

• X = Independent variable

• A = Constant (intercept)

• B = Regression coeficient

𝑌 = 𝐴 + 𝐵𝑋

𝐵 = 𝑁 σ_𝑖=1^𝑁 𝑋_𝑖𝑌_𝑖 − σ_𝑖=1^𝑁 𝑋_𝑖 . σ_𝑖=1^𝑁 𝑌_𝑖 𝑁 σ_𝑖=1^𝑁 𝑋_𝑖² − σ_𝑖=1^𝑁 𝑋_𝑖 ² 𝐴 = ത𝑌 − 𝐵 ത𝑋

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07/06/2022

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Gauss-Jordan Ellimination can be used to solve the above problem.

𝑌 = 𝑏₀ + 𝑏₁𝑋₁ + 𝑏₂𝑋₂

𝑁𝑏₀ + 𝑏₁ ෍

𝑖=1 𝑁

𝑋_1𝑖 + 𝑏₂෍

𝑖=1 𝑁

𝑋_2𝑖 = ෍

𝑖=1 𝑁

𝑌_𝑖

𝑏₀෍

𝑖=1 𝑁

𝑋_1𝑖 + 𝑏₁෍

𝑖=1 𝑁

𝑋_1𝑖 ² + 𝑏₂෍

𝑖=1 𝑁

𝑋_1𝑖. 𝑋_2𝑖 = ෍

𝑖=1 𝑁

𝑋_1𝑖. 𝑌_𝑖

𝑏₀෍

𝑖=1 𝑁

𝑋_2𝑖 + 𝑏₁෍

𝑖=1 𝑁

𝑋_1𝑖. 𝑋_2𝑖 + 𝑏₂෍

𝑖=1 𝑁

𝑋_2𝑖 ² = ෍

𝑖=1 𝑁

𝑋_2𝑖. 𝑌_𝑖

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Gauss-JordanEllimination can be used to solve the above problem.

𝑌 = 𝑏₀ + 𝑏₁𝑋₁ + 𝑏₂𝑋₂ + 𝑏₃𝑋₃

𝑁𝑏₀+ 𝑏₁෍

𝑖=1 𝑁

𝑋_1𝑖+ 𝑏₂෍

𝑖=1 𝑁

𝑋_2𝑖+ 𝑏₃෍

𝑖=1 𝑁

𝑋_3𝑖 = ෍

𝑖=1 𝑁

𝑌_𝑖

𝑏₀෍

𝑖=1 𝑁

𝑋_1𝑖+ 𝑏₁෍

𝑖=1 𝑁

𝑋_1𝑖 ²+ 𝑏₂෍

𝑖=1 𝑁

𝑋_1𝑖. 𝑋_2𝑖 + 𝑏₃෍

𝑖=1 𝑁

𝑋_1𝑖. 𝑋_3𝑖 = ෍

𝑖=1 𝑁

𝑋_1𝑖. 𝑌_𝑖

𝑏₀෍

𝑖=1 𝑁

𝑋_2𝑖+ 𝑏₁෍

𝑖=1 𝑁

𝑋_1𝑖. 𝑋_2𝑖 + 𝑏₂෍

𝑖=1 𝑁

𝑋_2𝑖 ²+ 𝑏₃෍

𝑖=1 𝑁

𝑋_2𝑖. 𝑋_3𝑖 = ෍

𝑖=1 𝑁

𝑋_2𝑖. 𝑌_𝑖

𝑏₀෍

𝑖=1 𝑁

𝑋_3𝑖+ 𝑏₁෍

𝑖=1 𝑁

𝑋_1𝑖. 𝑋_3𝑖 + 𝑏₂෍

𝑖=1 𝑁

𝑋_2𝑖. 𝑋_3𝑖 + 𝑏₃෍

𝑖=1 𝑁

𝑋_3𝑖 ²= ෍

𝑖=1 𝑁

𝑋_3𝑖. 𝑌_𝑖

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