Solve Using a Matrix by Elimination 7x-5y-z=18 , x-2y-5z=-33 , -4x+y+z=-17

Problem

{[7*x−5*y−z=18],[x−2*y−5*z=−33],[−4*x+y+z=−17])

Solution

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

[[7,−5,−1,|,18],[1,−2,−5,|,−33],[−4,1,1,|,−17]]

2. Swap rows to place a 1 in the first pivot position ((R_1)↔(R_2).

[[1,−2,−5,|,−33],[7,−5,−1,|,18],[−4,1,1,|,−17]]

3. Eliminate the first column entries below the pivot by performing (R_2)−7*(R_1)→(R_2) and (R_3)+4*(R_1)→(R_3)

[[1,−2,−5,|,−33],[0,9,34,|,249],[0,−7,−19,|,−149]]

4. Create a pivot in the second row by performing (R_2)+(R_3)→(R_2)

[[1,−2,−5,|,−33],[0,2,15,|,100],[0,−7,−19,|,−149]]

5. Eliminate the second column entry below the pivot by performing (R_3)+3.5*(R_2)→(R_3)

[[1,−2,−5,|,−33],[0,2,15,|,100],[0,0,33.5,|,201]]

6. Solve for z using the third row.

33.5*z=201

z=6

7. Back-substitute z=6 into the second row equation to solve for y

2*y+15*(6)=100

2*y+90=100

2*y=10

y=5

8. Back-substitute y=5 and z=6 into the first row equation to solve for x

x−2*(5)−5*(6)=−33

x−10−30=−33

x−40=−33

x=7

Final Answer

(x,y,z)=(7,5,6)

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