Solve Using a Matrix by Elimination 2x+y-z=5 , x+4y+2z=16 , 15x+6y-2z=12

Problem

{[2*x+y−z=5],[x+4*y+2*z=16],[15*x+6*y−2*z=12])

Solution

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

[[2,1,−1,5],[1,4,2,16],[15,6,−2,12]]

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

[[1,4,2,16],[2,1,−1,5],[15,6,−2,12]]

3. Eliminate the first column entries below the pivot by performing (R_2)−2*(R_1)→(R_2) and (R_3)−15*(R_1)→(R_3)

[[1,4,2,16],[0,−7,−5,−27],[0,−54,−32,−228]]

4. Simplify the third row by dividing by −2 to make the numbers more manageable (−1/2*(R_3)→(R_3).

[[1,4,2,16],[0,−7,−5,−27],[0,27,16,114]]

5. Eliminate the second column entry below the second pivot by performing 7*(R_3)+27*(R_2)→(R_3)

[[1,4,2,16],[0,−7,−5,−27],[0,0,−23,69]]

6. Solve for z using the third row.

−23*z=69

z=−3

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

−7*y−5*(−3)=−27

−7*y+15=−27

−7*y=−42

y=6

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

x+4*(6)+2*(−3)=16

x+24−6=16

x+18=16

x=−2

Final Answer

(x,y,z)=(−2,6,−3)

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