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Question Number 220366 by klipto last updated on 11/May/25
solve the system of equation  using gaussian elimination method  x+2y+3z=10  2x−3y+z=1  3x+y−2z=9
$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{system}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{equation}} \\ $$$$\boldsymbol{\mathrm{using}}\:\boldsymbol{\mathrm{gaussian}}\:\boldsymbol{\mathrm{elimination}}\:\boldsymbol{\mathrm{method}} \\ $$$$\boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{y}}+\mathrm{3}\boldsymbol{\mathrm{z}}=\mathrm{10} \\ $$$$\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{3}\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{z}}=\mathrm{1} \\ $$$$\mathrm{3}\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}−\mathrm{2}\boldsymbol{\mathrm{z}}=\mathrm{9} \\ $$
Answered by Frix last updated on 11/May/25
(2)−2×(1)     −7y  −5z=−19  (3)−3×(1)     −5y−11z=−21    7y+5z=19  5y+11z=21  7×(B)−5×(A)     52z=52  ⇒ z=1  ⇒ 7y+5=19  ⇒ y=2  ⇒ x+4+3=10  ⇒ x=3
$$\left(\mathrm{2}\right)−\mathrm{2}×\left(\mathrm{1}\right)\:\:\:\:\:−\mathrm{7}{y}\:\:−\mathrm{5}{z}=−\mathrm{19} \\ $$$$\left(\mathrm{3}\right)−\mathrm{3}×\left(\mathrm{1}\right)\:\:\:\:\:−\mathrm{5}{y}−\mathrm{11}{z}=−\mathrm{21} \\ $$$$ \\ $$$$\mathrm{7}{y}+\mathrm{5}{z}=\mathrm{19} \\ $$$$\mathrm{5}{y}+\mathrm{11}{z}=\mathrm{21} \\ $$$$\mathrm{7}×\left({B}\right)−\mathrm{5}×\left({A}\right)\:\:\:\:\:\mathrm{52}{z}=\mathrm{52} \\ $$$$\Rightarrow\:{z}=\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{7}{y}+\mathrm{5}=\mathrm{19} \\ $$$$\Rightarrow\:{y}=\mathrm{2} \\ $$$$\Rightarrow\:{x}+\mathrm{4}+\mathrm{3}=\mathrm{10} \\ $$$$\Rightarrow\:{x}=\mathrm{3} \\ $$
Answered by ea last updated on 12/May/25

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