Mengenai Saya
- airinda ektria
- airinda orang yang imut,hehe tapi punya kekurangan yakni sifat ngmbek....
Minggu, 15 Maret 2009
1. Sistem Persamaan Linier 3 variabel
Misal dik :
Ex + Fy + Gz = H ……(1)
Px + Qy + Rz = S ……(2)
Tx + Uy + Vz = W ……(3)
Penyelesaian :
a. eliminasi persamaan (1) dan (2)
Ex + Fy + Gz = H |*R| (ER)x + (FR)y + (GR)z = HR
Px + Qy + Rz = S |*G| (GP)x + (GQ)y + (GR)z = GS
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(ER–GP)x + (FR–GQ)y = (HR–GS) ……pers.(4)
b. eliminasi persamaan (1) dan (3)
Ex + Fy + Gz = H |*V| (EV)x + (FV)y + (GV)z = HV
Tx + Uy + Vz = W |*G| (GT)x + (GU)y + (GV)z = GW
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(EV–GT)x + (FV–GU)y = (HV–GW) ...pers. (5)
c. eliminasi persamaan (3) dan (4)
(ER–GP)x + (FR–GQ)y = (HR–GS) |*(FV–GU)| {(ER–GP)(FV–GU)}x + {(FR–GQ)(FV–GU)}y = {(HR–GS)(FV–GU)}
(EV–GT)x + (FV–GU)y = (HV–GW) |*(FR–GQ) {(EV-GT)(FR–GQ)}x + {(FV–GU)(FR–GQ)}y = {(HV–GW)(FR–GQ)}
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{(ER-GP)(FV-GU) - (EV-GT)(FR-GQ)}x = {(HR-GS)(FV-GU)} - {(HV–GW)(FR–GQ)}
Maka :
x = {(HR-GS)(FV-GU)} - {(HV–GW)(FR–GQ)}/{(ER-GP)(FV-GU) - (EV-GT)(FR-GQ)}
d. dari persamaan (4)
(ER–GP)x + (FR–GQ)y = (HR–GS)
y = {(HR-GS) - (ER-GP)x}/(FR-GQ)
e. dari persamaan (1)
Ex + Fy + Gz = H
z = {H - (Ex+Fy)}/G
Langganan:
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