Question

Does anyone know how to solve 3 by 3 linear systems?

In the system below, use equation (1) with equation (2) to eliminate x. Then use equation (1) with equation (3) to eliminate x.

x-y-2z=4 (1)

-x+3y-z=8 (2)

-2x-y-4z=-1 (3)

Answer 1

x = 3, y = 3, z = - 2

Explanation:

Given the 3 equations

x - y - 2z = 4 → (1)

- x + 3y - z = 8 → (2)

- 2x - y - 4z = - 1 → (3)

Add (1) and (2) term by term to eliminate x

2y - 3z = 12 → (4)

Multiply (1) by 2 and add to (3) to eliminate x

2x - 2y - 4z = 8 → (5)

Add (3) and (5) term by term to eliminate x

- 3y - 8z = 7 → (6)

Multiply (4) by 3 and (6) by 2

6y - 9z = 36 → (7)

- 6y - 16z = 14 → (8)

Add (7) and (8) term by term to eliminate y

- 25z = 50 ( divide both sides by - 25 )

z = - 2

Substitute z = - 2 into (7)

6y + 18 = 36 ( subtract 18 from both sides )

6y = 18 ( divide both sides by 6 )

y = 3

Substitute y = 3 and z = - 2 into (1)

x - 3 + 4 = 4

x + 1 = 4 ( subtract 1 from both sides )

x = 3

Solution is x = 3, y = 3, z = - 2