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

$Solving by Gaussian elimination$ $using the following system of$ $linear equation {\begin{cases} x - 3 y - 2 z = 6 \\ 2 x - 4 y - 3 z = 8 \\ - 3 x + 6 y + 8 z = - 5 \end{cases}$
	Answered by bobhans last updated on 06/Oct/20
	$Solving by Gaussian elimination using the following system of linear equation { ((x−3y−2z = 6)),((2x−4y−3z = 8 )),((−3x+6y+8z = −5)) :} { ((L_1 : x−3y−2z=6)),((L_2 : 2x−4y−3z=8)),((L_3 : −3x+6y+8z=−5)) :} These step yield (−2)L_1 : −2x+6y+4z=−12 L_2 : 2x−4y−3z=8 ____________________ + L_2 ^∗ : 2y+z = −4 3L_1 : 3x−9y−6z=18 L_3 :−3x +6y+8z=−5 ___________________ + L_3 ^∗ : −3y+2z = 13 Thus ,the original system is replaced by the following system ′ L_1 : x−3y−2z=6 L_2 : 2y+z =−4 L_3 : −3y+2z=13 Next step yield 3L_2 : 6y+3z = −12 2L_3 : −6y+4z=26 _______________ + L_3 ^(∗∗) : 7z = 14 Thus our system is replaced by the following system : L_1 : x−3y−2z=6 L_2 : 2y+z=−4 L_3 : 7z=14 The system is now triangular form so Part A is completed. Part B. The values for unknowns are obtained in reverse order z,y,x by back−substitution Specifically, { ((7z=14→z=2)),((2y+z=−4→2y=−6,y=−3)),((x−3y−2z=6,x+9−4=6,x=1)) :} Thus the solution of the triangular system and hence the original system is as follows x = 1; y=−3 ; z=2$
	$Solving by Gaussian elimination using$ $the following system of linear equation$ ${\begin{cases} x - 3 y - 2 z = 6 \\ 2 x - 4 y - 3 z = 8 \\ - 3 x + 6 y + 8 z = - 5 \end{cases}$ ${\begin{cases} L_{1} : x - 3 y - 2 z = 6 \\ L_{2} : 2 x - 4 y - 3 z = 8 \\ L_{3} : - 3 x + 6 y + 8 z = - 5 \end{cases}$ $These step yield$ $(- 2) L_{1} : - 2 x + 6 y + 4 z = - 12$ $L_{2} : 2 x - 4 y - 3 z = 8$ $____________________+$ $L_{2}^{} : 2 y + z = - 4$ ${3 L}_{1} : 3 x - 9 y - 6 z = 18$ $L_{3} : - 3 x + 6 y + 8 z = - 5$ $___________________+$ $L_{3}^{} : - 3 y + 2 z = 13$ $Thus, the original system is replaced by the$ $following system^{'}$ $L_{1} : x - 3 y - 2 z = 6$ $L_{2} : 2 y + z = - 4$ $L_{3} : - 3 y + 2 z = 13$ $Next step yield$ ${3 L}_{2} : 6 y + 3 z = - 12$ ${2 L}_{3} : - 6 y + 4 z = 26$ $_______________+$ $L_{3}^{* *} : 7 z = 14$ $Thus our system is replaced by the following$ $system : L_{1} : x - 3 y - 2 z = 6$ $L_{2} : 2 y + z = - 4$ $L_{3} : 7 z = 14$ $The system is now triangular form$ $so Part A is completed .$ $Part B . The values for unknowns are obtained$ $in reverse order z, y, x by back - substitution$ $Specifically, {\begin{cases} 7 z = 14 \to z = 2 \\ 2 y + z = - 4 \to 2 y = - 6, y = - 3 \\ x - 3 y - 2 z = 6, x + 9 - 4 = 6, x = 1 \end{cases}$ $Thus the solution of the triangular system$ $and hence the original system is as follows$ $x = 1; y = - 3; z = 2$
	Commented by bemath last updated on 06/Oct/20

	$gave kudos . . .$
	Answered by 1549442205PVT last updated on 06/Oct/20

	$\| \begin{matrix} 1 & - 3 & - 2 & 6 \\ 2 & - 4 & - 3 & 8 \\ - 3 & 6 & 8 & - 5 \end{matrix} \|$ $(multiplying first row by 2 then substract from second row$ $next : multiplying first row by 3 then adding to third row$ $\sim \| \begin{matrix} 1 & - 3 & - 2 & 6 \\ 0 & 2 & 1 & - 4 \\ 0 & - 3 & 2 & 13 \end{matrix} \|$ $Multiplying second row by \frac{3}{2} then adding to third row$ $\sim \| \begin{matrix} 1 & - 3 & - 2 & 6 \\ 0 & 2 & 1 & - 4 \\ 0 & 0 & \frac{7}{2} & 7 \end{matrix} \|$ $We get the triangle system :$ ${\begin{cases} x - 3 y - 2 z = 6 \\ 2 y + z = - 4 \\ \frac{7}{2} z = 7 \end{cases} \Leftrightarrow {\begin{cases} z = 2 \\ y = - 3 \\ x = 1 \end{cases}$
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