## Elementary Linear Algebra 7th Edition

$x_{1}$-$x_{2}$=0, $3x_{1}$-$2x_{2}$=-1 $3\times$$x_{1}-x_{2}=0 -1\times$$3x_{1}$-$2x_{2}$=-1 $3x_{1}$-$3x_{2}$=0 $-3x_{1}$+$2x_{2}$=1 0 $-x_{2}$=1 $x_{2}$=-1 $x_{1}$-$x_{2}$=0 $x_{1}$=$x_{2}$ $x_{1}$=-1
Considering equations 1 and 2: First step: Use the reduction system and multiply the equation 1 by 3 and the equation 2 by -1. Second step: Add the now multiplied equations and that way you eliminate $3x_{1}$ and $-3x_{1}$. With the remaining result you can get $x_{1}$ by adding $-3x_{2}$ and $2x_{2}$ and 0 and 1. That way you get that $x_{2}$ is -1. Third step: Plug in the value of $x_{2}$ in the first equation and you will get that $x_{1}$ is -1 as well.