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Solving-by-Gaussian-elimination-using-the-following-system-of-linear-equation-x-3y-2z-6-2x-4y-3z-8-3x-6y-8z-5-

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Question Number 116695 by bemath last updated on 06/Oct/20
Solving by Gaussian elimination using the following system of linear equation { ((x−3y−2z=6)),((2x−4y−3z=8)),((−3x+6y+8z=−5)) :}
$$\mathrm{Solving}\:\mathrm{by}\:\mathrm{Gaussian}\:\mathrm{elimination} \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{following}\:\mathrm{system}\:\mathrm{of} \\ $$$$\mathrm{linear}\:\mathrm{equation}\:\begin{cases}{\mathrm{x}−\mathrm{3y}−\mathrm{2z}=\mathrm{6}}\\{\mathrm{2x}−\mathrm{4y}−\mathrm{3z}=\mathrm{8}}\\{−\mathrm{3x}+\mathrm{6y}+\mathrm{8z}=−\mathrm{5}}\end{cases} \\ $$
Answered by bobhans last updated on 06/Oct/20
Solving by Gaussian elimination using the following system of linear equation { ((x−3y−2z = 6)),((2x−4y−3z = 8 )),((−3x+6y+8z = −5)) :} { ((L_1 : x−3y−2z=6)),((L_2 : 2x−4y−3z=8)),((L_3 : −3x+6y+8z=−5)) :} These step yield (−2)L_1 : −2x+6y+4z=−12 L_2 : 2x−4y−3z=8 ____________________ + L_2 ^∗ : 2y+z = −4 3L_1 : 3x−9y−6z=18 L_3 :−3x +6y+8z=−5 ___________________ + L_3 ^∗ : −3y+2z = 13 Thus ,the original system is replaced by the following system ′ L_1 : x−3y−2z=6 L_2 : 2y+z =−4 L_3 : −3y+2z=13 Next step yield 3L_2 : 6y+3z = −12 2L_3 : −6y+4z=26 _______________ + L_3 ^(∗∗) : 7z = 14 Thus our system is replaced by the following system : L_1 : x−3y−2z=6 L_2 : 2y+z=−4 L_3 : 7z=14 The system is now triangular form so Part A is completed. Part B. The values for unknowns are obtained in reverse order z,y,x by back−substitution Specifically, { ((7z=14→z=2)),((2y+z=−4→2y=−6,y=−3)),((x−3y−2z=6,x+9−4=6,x=1)) :} Thus the solution of the triangular system and hence the original system is as follows x = 1; y=−3 ; z=2
$$\:\mathrm{Solving}\:\mathrm{by}\:\mathrm{Gaussian}\:\mathrm{elimination}\:\mathrm{using} \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{system}\:\mathrm{of}\:\mathrm{linear}\:\mathrm{equation} \\ $$$$\:\begin{cases}{\mathrm{x}−\mathrm{3y}−\mathrm{2z}\:=\:\mathrm{6}}\\{\mathrm{2x}−\mathrm{4y}−\mathrm{3z}\:=\:\mathrm{8}\:}\\{−\mathrm{3x}+\mathrm{6y}+\mathrm{8z}\:=\:−\mathrm{5}}\end{cases} \\ $$$$\:\begin{cases}{\mathrm{L}_{\mathrm{1}} \::\:\mathrm{x}−\mathrm{3y}−\mathrm{2z}=\mathrm{6}}\\{\mathrm{L}_{\mathrm{2}} \::\:\mathrm{2x}−\mathrm{4y}−\mathrm{3z}=\mathrm{8}}\\{\mathrm{L}_{\mathrm{3}} \::\:−\mathrm{3x}+\mathrm{6y}+\mathrm{8z}=−\mathrm{5}}\end{cases} \\ $$$$\mathrm{These}\:\mathrm{step}\:\mathrm{yield}\: \\ $$$$\:\left(−\mathrm{2}\right)\mathrm{L}_{\mathrm{1}} \::\:\:−\mathrm{2x}+\mathrm{6y}+\mathrm{4z}=−\mathrm{12} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{L}_{\mathrm{2}} \::\:\:\:\:\:\mathrm{2x}−\mathrm{4y}−\mathrm{3z}=\mathrm{8} \\ $$$$\:\:\:\:\:\:\:\:\:\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\:+\: \\ $$$$\:\:\:\:\:\:\:\mathrm{L}_{\mathrm{2}} ^{\ast} \::\:\:\:\:\:\:\:\:\mathrm{2y}+\mathrm{z}\:=\:−\mathrm{4} \\ $$$$\:\:\mathrm{3L}_{\mathrm{1}} \::\:\:\:\:\mathrm{3x}−\mathrm{9y}−\mathrm{6z}=\mathrm{18} \\ $$$$\:\:\:\:\:\:\mathrm{L}_{\mathrm{3}} \::−\mathrm{3x}\:+\mathrm{6y}+\mathrm{8z}=−\mathrm{5} \\ $$$$\:\:\:\:\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\:+ \\ $$$$\:\:\:\:\:\:\mathrm{L}_{\mathrm{3}} ^{\ast} \::\:−\mathrm{3y}+\mathrm{2z}\:=\:\mathrm{13} \\ $$$$\mathrm{Thus}\:,\mathrm{the}\:\mathrm{original}\:\mathrm{system}\:\mathrm{is}\:\mathrm{replaced}\:\mathrm{by}\:\mathrm{the}\: \\ $$$$\mathrm{following}\:\mathrm{system}\:' \\ $$$$\mathrm{L}_{\mathrm{1}} \::\:\mathrm{x}−\mathrm{3y}−\mathrm{2z}=\mathrm{6} \\ $$$$\mathrm{L}_{\mathrm{2}} \::\:\:\:\:\:\:\:\:\:\mathrm{2y}+\mathrm{z}\:=−\mathrm{4} \\ $$$$\mathrm{L}_{\mathrm{3}} \::\:\:\:−\mathrm{3y}+\mathrm{2z}=\mathrm{13} \\ $$$$\mathrm{Next}\:\mathrm{step}\:\mathrm{yield}\: \\ $$$$\:\mathrm{3L}_{\mathrm{2}} \::\:\:\:\:\mathrm{6y}+\mathrm{3z}\:=\:−\mathrm{12} \\ $$$$\:\mathrm{2L}_{\mathrm{3}} \::\:−\mathrm{6y}+\mathrm{4z}=\mathrm{26} \\ $$$$\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\:+\: \\ $$$$\mathrm{L}_{\mathrm{3}} ^{\ast\ast} \::\:\mathrm{7z}\:=\:\mathrm{14}\: \\ $$$$\mathrm{Thus}\:\mathrm{our}\:\mathrm{system}\:\mathrm{is}\:\mathrm{replaced}\:\mathrm{by}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{system}\::\:\mathrm{L}_{\mathrm{1}} \::\:\mathrm{x}−\mathrm{3y}−\mathrm{2z}=\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{L}_{\mathrm{2}} \::\:\:\:\:\:\:\:\:\:\mathrm{2y}+\mathrm{z}=−\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{L}_{\mathrm{3}} \::\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{7z}=\mathrm{14}\: \\ $$$$\mathrm{The}\:\mathrm{system}\:\mathrm{is}\:\mathrm{now}\:\mathrm{triangular}\:\mathrm{form}\: \\ $$$$\mathrm{so}\:\mathrm{Part}\:\mathrm{A}\:\mathrm{is}\:\mathrm{completed}. \\ $$$$\mathrm{Part}\:\mathrm{B}.\:\mathrm{The}\:\mathrm{values}\:\mathrm{for}\:\mathrm{unknowns}\:\mathrm{are}\:\mathrm{obtained} \\ $$$$\mathrm{in}\:\mathrm{reverse}\:\mathrm{order}\:\mathrm{z},\mathrm{y},\mathrm{x}\:\mathrm{by}\:\mathrm{back}−\mathrm{substitution} \\ $$$$\mathrm{Specifically},\:\begin{cases}{\mathrm{7z}=\mathrm{14}\rightarrow\mathrm{z}=\mathrm{2}}\\{\mathrm{2y}+\mathrm{z}=−\mathrm{4}\rightarrow\mathrm{2y}=−\mathrm{6},\mathrm{y}=−\mathrm{3}}\\{\mathrm{x}−\mathrm{3y}−\mathrm{2z}=\mathrm{6},\mathrm{x}+\mathrm{9}−\mathrm{4}=\mathrm{6},\mathrm{x}=\mathrm{1}}\end{cases} \\ $$$$\mathrm{Thus}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangular}\:\mathrm{system} \\ $$$$\mathrm{and}\:\mathrm{hence}\:\mathrm{the}\:\mathrm{original}\:\mathrm{system}\:\mathrm{is}\:\mathrm{as}\:\mathrm{follows} \\ $$$$\mathrm{x}\:=\:\mathrm{1};\:\mathrm{y}=−\mathrm{3}\:;\:\mathrm{z}=\mathrm{2} \\ $$
Commented by bemath last updated on 06/Oct/20
gave kudos...
$$\mathrm{gave}\:\mathrm{kudos}… \\ $$
Answered by 1549442205PVT last updated on 06/Oct/20
determinant ((1,(−3),(−2),6),(2,(−4),(−3),8),((−3),6,8,(−5))) (multiplying first row by 2 then substract from second row next:multiplying first row by 3 then adding to third row ∼ determinant ((1,(−3),(−2),6),(0,2,1,(−4)),(0,(−3),2,(13))) Multiplying second row by (3/2)then adding to third row ∼ determinant ((1,(−3),(−2),6),(0,2,1,(−4)),(0,0,(7/2),7)) We get the triangle system: { ((x−3y−2z=6)),((2y+z=−4)),(((7/2)z=7)) :}⇔ { ((z=2)),((y=−3)),((x=1)) :}
$$\begin{vmatrix}{\mathrm{1}}&{−\mathrm{3}}&{−\mathrm{2}}&{\mathrm{6}}\\{\mathrm{2}}&{−\mathrm{4}}&{−\mathrm{3}}&{\mathrm{8}}\\{−\mathrm{3}}&{\mathrm{6}}&{\mathrm{8}}&{−\mathrm{5}}\end{vmatrix} \\ $$$$\left(\mathrm{multiplying}\:\mathrm{first}\:\mathrm{row}\:\mathrm{by}\:\mathrm{2}\:\mathrm{then}\:\mathrm{substract}\:\mathrm{from}\:\mathrm{second}\:\mathrm{row}\right. \\ $$$$\mathrm{next}:\mathrm{multiplying}\:\mathrm{first}\:\mathrm{row}\:\mathrm{by}\:\mathrm{3}\:\mathrm{then}\:\mathrm{adding}\:\mathrm{to}\:\mathrm{third}\:\mathrm{row} \\ $$$$\sim\begin{vmatrix}{\mathrm{1}}&{−\mathrm{3}}&{−\mathrm{2}}&{\mathrm{6}}\\{\mathrm{0}}&{\mathrm{2}}&{\mathrm{1}}&{−\mathrm{4}}\\{\mathrm{0}}&{−\mathrm{3}}&{\mathrm{2}}&{\mathrm{13}}\end{vmatrix} \\ $$$$\mathrm{Multiplying}\:\mathrm{second}\:\mathrm{row}\:\mathrm{by}\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{then}\:\mathrm{adding}\:\mathrm{to}\:\mathrm{third}\:\mathrm{row} \\ $$$$\sim\begin{vmatrix}{\mathrm{1}}&{−\mathrm{3}}&{−\mathrm{2}}&{\mathrm{6}}\\{\mathrm{0}}&{\mathrm{2}}&{\mathrm{1}}&{−\mathrm{4}}\\{\mathrm{0}}&{\mathrm{0}}&{\frac{\mathrm{7}}{\mathrm{2}}}&{\mathrm{7}}\end{vmatrix} \\ $$$$\mathrm{We}\:\mathrm{get}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{system}: \\ $$$$\begin{cases}{\mathrm{x}−\mathrm{3y}−\mathrm{2z}=\mathrm{6}}\\{\mathrm{2y}+\mathrm{z}=−\mathrm{4}}\\{\frac{\mathrm{7}}{\mathrm{2}}\mathrm{z}=\mathrm{7}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{z}=\mathrm{2}}\\{\mathrm{y}=−\mathrm{3}}\\{\mathrm{x}=\mathrm{1}}\end{cases} \\ $$

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