# Thread: Gauss Jordan Elimination and Augmented Matrices

1. ## Gauss Jordan Elimination and Augmented Matrices

Not a specially question but I am really bad at them so I would want to know if there is any quick ways or methods to do them except technology. Or like any ways to check the answers as I seem to get various answers

Just for example, how to do

1 2 1
2 -3 4

Those different colors are the ones equal to!

Like for that example how do you know that you cannot go any further(no solutions))

2. Hello, someone21!

I'm not sure what your question is,
. . but I can solve the system for you . . .

$\displaystyle \left|\begin{array}{cc|c}1 & 2 & 1 \\ 2 & \text{-}3 & 4\end{array}\right|$

We have: .$\displaystyle \left|\begin{array}{cc|c}1 & 2 & 1 \\ 2 & \text{-}3 & 4 \end{array}\right|$

. $\displaystyle \begin{array}{c} \\ R_2-2R_1\end{array} \left|\begin{array}{cc|c}1 & 2 & 1 \\ 0 & \text{-}7 & 2 \end{array}\right|$

. . . . $\displaystyle \begin{array}{c} \\ \text{-}\frac{1}{7}R_2 \end{array} \left|\begin{array}{cc|c}1 & 2 & 1 \\ 0 & 1 & \text{-}\frac{2}{7} \end{array}\right|$

. $\displaystyle \begin{array}{c} R_1-2R_2 \\ \\ \end{array}\left|\begin{array}{cc|c}1 & 0 & \frac{11}{7} \\ \\[-4mm] 0 & 1 & \text{-}\frac{2}{7}\end{array}\right|$

Solution: .$\displaystyle \boxed{x \:=\:\frac{11}{7}\quad y \:=\:\text{-}\frac{2}{7}}$

3. Originally Posted by Soroban
Hello, someone21!

I'm not sure what your question is,
. . but I can solve the system for you . . .

We have: .$\displaystyle \left|\begin{array}{cc|c}1 & 2 & 1 \\ 2 & \text{-}3 & 4 \end{array}\right|$

. $\displaystyle \begin{array}{c} \\ R_2-2R_1\end{array} \left|\begin{array}{cc|c}1 & 2 & 1 \\ 0 & \text{-}7 & 2 \end{array}\right|$

. . . . $\displaystyle \begin{array}{c} \\ \text{-}\frac{1}{7}R_2 \end{array} \left|\begin{array}{cc|c}1 & 2 & 1 \\ 0 & 1 & \text{-}\frac{2}{7} \end{array}\right|$

. $\displaystyle \begin{array}{c} R_1-2R_2 \\ \\ \end{array}\left|\begin{array}{cc|c}1 & 0 & \frac{11}{7} \\ \\[-4mm] 0 & 1 & \text{-}\frac{2}{7}\end{array}\right|$

Solution: .$\displaystyle \boxed{x \:=\:\frac{11}{7}\quad y \:=\:\text{-}\frac{2}{7}}$

Thanks but the thing is there are so many ways to do them so how we do it the fastest

i mean like in that example do we try to make the top left one =1 then make the exact lower row 0 or make the number next to it 0