Solve Using a Matrix by Row Operations 4x+3y+z=7 , x-3y+2z=5 , 11x-2y+3z=10

Problem

{[4*x+3*y+z=7],[x−3*y+2*z=5],[11*x−2*y+3*z=10])

Solution

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

[[4,3,1,|,7],[1,−3,2,|,5],[11,−2,3,|,10]]

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

[[1,−3,2,|,5],[4,3,1,|,7],[11,−2,3,|,10]]

3. Eliminate the first column entries below the pivot by performing (R_2)−4*(R_1)→(R_2) and (R_3)−11*(R_1)→(R_3)

[[1,−3,2,|,5],[0,15,−7,|,−13],[0,31,−19,|,−45]]

4. Normalize the second row by dividing by 15 ((R_2)/15→(R_2).

[[1,−3,2,|,5],[0,1,−7/15,|,−13/15],[0,31,−19,|,−45]]

5. Eliminate the second column entry below the pivot by performing (R_3)−31*(R_2)→(R_3)

[[1,−3,2,|,5],[0,1,−7/15,|,−13/15],[0,0,−68/15,|,−272/15]]

6. Solve for z by normalizing the third row ((R_3)⋅(−15/68)→(R_3).

[[1,−3,2,|,5],[0,1,−7/15,|,−13/15],[0,0,1,|,4]]

7. Back-substitute to find y using (R_2)+7/15*(R_3)→(R_2)

y=−13/15+7/15*(4)=15/15=1

8. Back-substitute to find x using (R_1)+3*(R_2)−2*(R_3)→(R_1)

x=5+3*(1)−2*(4)=0

Final Answer

(x,y,z)=(0,1,4)

Want more problems? Check here!