Solve Using a Matrix by Elimination x+y-z=2 , x+2y+3z=7 , x+3y-z=4

Problem

{[x+y−z=2],[x+2*y+3*z=7],[x+3*y−z=4])

Solution

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

[[1,1,−1,|,2],[1,2,3,|,7],[1,3,−1,|,4]]

2. Eliminate the x-terms from the second and third rows by performing row operations (R_2)−(R_1)→(R_2) and (R_3)−(R_1)→(R_3)

[[1,1,−1,|,2],[0,1,4,|,5],[0,2,0,|,2]]

3. Eliminate the y-term from the third row by performing the row operation (R_3)−2*(R_2)→(R_3)

[[1,1,−1,|,2],[0,1,4,|,5],[0,0,−8,|,−8]]

4. Solve for z using the third row equation −8*z=−8

z=1

5. Solve for y by substituting z=1 into the second row equation y+4*z=5

y+4*(1)=5

y=1

6. Solve for x by substituting y=1 and z=1 into the first row equation x+y−z=2

x+1−1=2

x=2

Final Answer

(x,y,z)=(2,1,1)

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