Question

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

Answer 1

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

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 $(x,y,z)=(-2,-3,4)$.

Answer 2

Answer:

(-2,-3,4)

Step-by-step explanation: