# Thread: echelon method problem...

1. ## echelon method problem...

Can someone please help me solve this problem using the Gauss-Jordan Method:

3x-5y-2z=-9
-4x+3y+z=11
8x-5y+4z=6

2. Hello, halloweengrl24!

$\displaystyle \text{Solve using the Gauss-Jordan Method:}$

. . . $\displaystyle \begin{array}{cccc}3x-5y-2z&=& -9 & [1] \\ -4x+3y+z &=& 11 & [2] \\8x-5y+4z &=& 6 & [3] \end{array}$

I worked ahead a few steps and ran into some difficulty.
Looking back, I found these preliminary steps to be helpful.

Multiply [2] by -1: .$\displaystyle 4x - 3y - z \:=\:\text{-}11$

And reorder the equations: .$\displaystyle \begin{array}{ccc}4x-3y - z &=& \text{-}11 \\ 8x - 5y + 4z &=& 6 \\ 3x - 5y - 2z &=& \text{-}9 \end{array}$

$\displaystyle \text{We have: } \;\left|\begin{array}{ccc|c}4 & \text{-}3 & \text{-}1 & \text{-}11 \\ 8 & \text{-}5 & 9 & 6 \\ 3 & \text{-}5 & \text{-}2 & \text{-}9 \end{array}\right|$

$\displaystyle \begin{array}{c}R_1-R_3 \\ \\ \\ \end{array} \left|\begin{array}{ccc|c} 1 & 2 & 1 & \text{-}2 \\ 8 & \text{-}5 & 4 & 6 \\ 3 & \text{-}5 & \text{-}2 & \text{-}9 \end{array}\right|$

$\displaystyle \begin{array}{c} \\ R_2 - 8R_1 \\ R_3 - 3R_1\end{array} \left|\begin{array}{ccc|c} 1 & 2 & 1 & \text{-}2 \\ 0 & \text{-}21 & \text{-}4 & 22 \\ 0 & \text{-}11 & \text{-}5 & \text{-}3 \end{array} \right|$

$\displaystyle \begin{array}{c} \\ R_2-2R_3 \\ (\text{-}1)R_3 \end{array} \left|\begin{array}{ccc|c} 1 & 2 & 1 & \text{-}2 \\ 0 & 1 & 6 & 28 \\ 0 & 11 & 5 & 3 \end{array}\right|$

$\displaystyle \begin{array}{c}R_1 - 2R_2 \\ \\ R_3 - 11R_2 \end{array} \left| \begin{array}{ccc|c} 1 & 0 & \text{-}11 & \text{-}58 \\ 0 & 1 & 6 & 28 \\ 0 & 0 & \text{-}61 & \text{-}305 \end{array}\right|$

. . $\displaystyle \begin{array}{c} \\ \\ (\text{-}\frac{1}{61})R_3 \end{array} \left| \begin{array}{ccc|c} 1 & 0 & \text{-}11 & \text{-}58 \\ 0 & 1 & 6 & 28 \\ 0 & 0 & 1 & 5 \end{array} \right|$

$\displaystyle \begin{array}{c} R_1 + 11R_3 \\ R_2 - 6R_3 \\ \\ \end{array} \left| \begin{array}{ccc|c} 1 & 0 & 0 & \text{-}3 \\ 0 & 1 & 0 & \text{-}2 \\ 0 & 0 & 1 & 5 \end{array}\right|$

$\displaystyle \text{Therefore: }\;\begin{Bmatrix}x &=& \text{-}3 \\ y &=& \text{-}2 \\ z &=& 5 \end{Bmatrix}$