Solve Using a Matrix by Elimination 5x+4y-5z=4 , 3x-5y+3z=1 , 6x-2y+6z=10

Problem

{[5*x+4*y−5*z=4],[3*x−5*y+3*z=1],[6*x−2*y+6*z=10])

Solution

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

[[5,4,−5,|,4],[3,−5,3,|,1],[6,−2,6,|,10]]

2. Simplify the third row by dividing all elements in (R_3) by 2 to make the numbers smaller.

(R_3)→1/2*(R_3)

[[5,4,−5,|,4],[3,−5,3,|,1],[3,−1,3,|,5]]

3. Create a zero in the first column of the third row by subtracting the second row from the third row.

(R_3)→(R_3)−(R_2)

[[5,4,−5,|,4],[3,−5,3,|,1],[0,4,0,|,4]]

4. Solve for y using the simplified third row.

4*y=4

y=1

5. Substitute y into the second row to create an equation with x and z

3*x−5*(1)+3*z=1

3*x+3*z=6

x+z=2

6. Substitute y into the first row to create another equation with x and z

5*x+4*(1)−5*z=4

5*x−5*z=0

x−z=0

7. Solve the system of two variables by adding the two simplified equations.

(x+z)+(x−z)=2+0

2*x=2

x=1

8. Solve for z using the equation x−z=0

1−z=0

z=1

Final Answer

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

Want more problems? Check here!