Solve Using a Matrix by Elimination 2x+y-z=2 , x+3y+2z=1 , x+y+z=2

Problem

{[2*x+y−z=2],[x+3*y+2*z=1],[x+y+z=2])

Solution

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

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

2. Swap rows to place a 1 in the first pivot position ((R_1)↔(R_3).

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

3. Eliminate the first column entries below the first pivot using row operations (R_2)−(R_1)→(R_2) and (R_3)−2*(R_1)→(R_3)

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

4. Swap rows to make the second pivot easier to work with ((R_2)↔(R_3).

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

5. Eliminate the second column entry below the second pivot using (R_3)+2*(R_2)→(R_3)

[[1,1,1,|,2],[0,−1,−3,|,−2],[0,0,−5,|,−5]]

6. Solve for z using the third row.

−5*z=−5

z=1

7. Solve for y using back-substitution into the second row.

−y−3*(1)=−2

−y=1

y=−1

8. Solve for x using back-substitution into the first row.

x+(−1)+1=2

x=2

Final Answer

(x,y,z)=(2,−1,1)

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