Question

Use Gauss-Jordan elimination to solve the following linear system.

x – 6y – 3z = 4
–2x – 3z = –8
–2x + 2y – 3z = –14

Answer 1

(-2,-3,4)

Explanation:

Answer 2

`\begin{cases}x-6y-3z=4&(1)\\-2x-3z=-8&(2)\\-2x+2y-3z=-14&(3)\end{cases}`

First eliminate
`y` by adding
`(1)` to three times
`(3)`. This gives


`(x-6y-3z)+3(-2x+2y-3z)=4+3(-14)\iff -5x-12z=-38`

This reduces the system of one of two equations and two unknowns:


`\begin{cases}-5x-12z=-38&(1)^*\\-2x-3z=-8&(2)^*\end{cases}`

You can eliminate
`z` by subtracting
`(1)^*` and four times
`(2)^*` to get


`(-5x-12z)-4(-2x-3z)=-38-4(-8)\iff3x=-6\implies x=-2`

Back-substitute to find
`z` and
`y`. You should end up with
`(x,y,z)=(-2,-3,4)`.