Solve Using a Matrix by Elimination x+y+z=6 , 2x-y+3z=8 , x+2y=4

Problem

{[x+y+z=6],[2*x−y+3*z=8],[x+2*y=4])

Solution

1. Write the augmented matrix representing the system of linear equations.

[[1,1,1,|,6],[2,−1,3,|,8],[1,2,0,|,4]]

2. Eliminate the x-terms from the second and third rows using the first row.

(R_2)−2*(R_1)⇒(R_2)

[[1,1,1,|,6],[0,−3,1,|,−4],[1,2,0,|,4]]

(R_3)−(R_1)⇒(R_3)

[[1,1,1,|,6],[0,−3,1,|,−4],[0,1,−1,|,−2]]

3. Swap rows to make the calculations easier for the next step.

(R_2)↔(R_3)

[[1,1,1,|,6],[0,1,−1,|,−2],[0,−3,1,|,−4]]

4. Eliminate the y-term from the third row using the second row.

(R_3)+3*(R_2)⇒(R_3)

[[1,1,1,|,6],[0,1,−1,|,−2],[0,0,−2,|,−10]]

5. Solve for z by dividing the third row by −2

−1/2*(R_3)⇒(R_3)

[[1,1,1,|,6],[0,1,−1,|,−2],[0,0,1,|,5]]

z=5

6. Solve for y using back-substitution into the second row equation y−z=−2

y−5=−2

y=3

7. Solve for x using back-substitution into the first row equation x+y+z=6

x+3+5=6

x+8=6

x=−2

Final Answer

(x,y,z)=(−2,3,5)

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