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

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

{\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 \dots

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}