Gauss Jordan Elimination and Augmented Matrices

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October 25th 2008, 07:29 AM
someone21

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))

Please help, i need quick ways to do and check
October 25th 2008, 08:17 AM
Soroban

Hello, someone21!

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

Quote:

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

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

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

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

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

Solution: . $\boxed{x \:=\:\frac{11}{7}\quad y \:=\:\text{-}\frac{2}{7}}$
October 25th 2008, 08:42 AM
someone21

Quote:

Originally Posted by Soroban

Hello, someone21!

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

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

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

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

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

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

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