I'm trying to solve a system using Gaussian Elimination to get to RE form (the one where the leading numbers are not reduced to one's) and then back substituting. However, I got the wrong answer can someone point out where I went wrong?
My work:
$$ \left[ \begin{array}{ccc|c} 1&-3&-2&0\\ -1&2&1&0\\ 2&4&6&0\\ \end{array} \right]$$
R2 = R2 + R1
R3 = R3 - 2R1
$$ \left[ \begin{array}{ccc|c} 1&-3&-2&0\\ 0&-1&-1&0\\ 0&10&10&0\\ \end{array} \right]$$
R3 = R3 + 10R2
$$ \left[ \begin{array}{ccc|c} 1&-3&-2&0\\ 0&-1&-1&0\\ 0&0&0&0\\ \end{array} \right]$$
$-y - t = 0 $
-> $ y = -t$
$x - 3y - 2t=0$
-> $x - 3(-t) -2t = 0 $
-> $x = -t $
My answer:
$$ \left[ \begin{array}{c|c} -1\\ -1\\ \end{array} \right]$$
The correct answer:
$$ \left[ \begin{array}{c|c} 1\\ 1\\ -1 \end{array} \right]t$$
You forgot that $t$ represents the third component of the vector. This means your answer should be $[x,y, z] = [-t, -t, t]$, which is equivalent to the answer given (since $t$ is an arbitrary real number).