# Math Help - Gauss-Jordan-Elimination-Row

1. ## Gauss-Jordan-Elimination-Row

Ok, I have actually posted this question before but I've made some typo and it's not the same as the question in my Assignment.

Using the working out and advice given before I've made it somewhere but not quite there.

I need to solve

w + 2x + 2y + 6z = 6
x + y + z = 6
w -x + y + 9z = -2

(Attached)

This is what I came up with so far, but it looks messy and I have doubt that it's wrong.

How can I make it neater? Or how can I make it in terms of some value t?

2. ## Re: Gauss-Jordan-Elimination-Row

Hello, Newbie999!

You're off to a good start . . . you didn't go far enough.

$\text{Solve: }\:\begin{array}{ccc}w + 2x + 2y + 6z &=& 6 \\ \qquad x \;+\; y \;+\; z &=& 6 \\ w -x + y + 9z &=& \text{-}2 \end{array}$

$\text{We have: }\:\left|\begin{array}{cccc|c}1&2&2&6&6 \\ 0&1&1&1&6 \\ 1&\text{-}1&1&9&\text{-}2 \end{array}\right|$

$\begin{array}{c}\\ \\ R_3-R_1 \end{array}\left|\begin{array}{cccc|c} 1&2&2&6&6 \\ 0&1&1&1&6 \\ 0&\text{-}3&\text{-}1&3&\text{-}8\end{array}\right|$

$\begin{array}{c}\\ \\ R_3+3R_2 \end{array} \left|\begin{array}{cccc|c}1&2&2&6&6 \\ 0&1&1&1&6 \\ 0&0&2&6&10 \end{array}\right|$

. . . . $\begin{array}{c}\\ \\ \frac{1}{2}R_3 \end{array} \left|\begin{array}{cccc|c} 1&2&2&6&6 \\ 0&1&1&1&6 \\ 0&0&1&3&5 \end{array}\right|$

$\begin{array}{c}R_1-2R_3 \\ R_2-R_3 \\ \\ \end{array} \left|\begin{array}{cccc|c} 1&2&0&0&\text{-}4 \\ 0&1&0&\text{-}2 & 1 \\ 0&0&1&3&5 \end{array}\right|$

We have: . $\begin{Bmatrix} w+2x \:=\:\text{-}4 & \Rightarrow & w \:=\:\text{-}4 - 2x & [1] \\ x - 2z \:=\:1 &\Rightarrow& x \:=\:1 + 2z & [2] \\ y + 3z \:=\:5 &\Rightarrow& y \:=\:5-3z & [3]\end{Bmatrix}$

Substitute [2] into [1]: . $w \:=\:\text{-}4-2(1+2z)$

We have: . $\begin{Bmatrix}w &=& \text{-}6 - 4z \\ x &=& 1 + 2z \\ y &=& 5 - 3z \\ z &=& z \end{Bmatrix}$

On the right, replace $z$ with a parameter $t.$

Therefore: . $\begin{Bmatrix}w &=& \text{-}6 - 4t \\ x &=& 1 + 2t \\ y &=& 5 - 3t \\ z &=& t \end{Bmatrix}$