For anyone who was wondering, the original problem, $\displaystyle x_1- 3x_2= 2$, $\displaystyle x_2+ x_3= -2$, $\displaystyle 2x_1- x_2+ 4x_3= 1$ is equivalent to the matrix equation
$\displaystyle \begin{pmatrix}1 & -3 & 0 \\ 0 & 2 & 1 \\ 2 & -1 & 4 \end{pmatrix}\begin{pmatrix}x_1\\ x_2\\ x_3\end{pmatrix}= \begin{pmatrix}2 \\ -2 \\ 1\end{pmatrix}$
Thinking of that as "Ax= b", multiply both sides by $\displaystyle A^{-1}$ to get $\displaystyle x= A^{-1}b$.
Set up your augmented matrix
$\displaystyle \displaystyle \begin{align*} \left[ \begin{matrix} 1 & -3 & 0 & \phantom{-}2 \\ 0 & \phantom{-}1 & 1 & -2 \\ 2 & -1 & 4 & \phantom{-}1 \end{matrix} \right] \end{align*}$
R3 - 2R1 -> R3
$\displaystyle \displaystyle \begin{align*} \left[ \begin{matrix} 1 & -3 & 0 & \phantom{-}2 \\ 0 & \phantom{-}1 & 1 & -2 \\ 0 & \phantom{-}5 & 4 & -3 \end{matrix} \right] \end{align*}$
R3 - 5R2 -> R3
$\displaystyle \displaystyle \begin{align*} \left[ \begin{matrix} 1 & -3 & \phantom{-}0 & \phantom{-}2 \\ 0 & \phantom{-}1 & \phantom{-}1 & -2 \\ 0 & \phantom{-}0 & -1 & \phantom{-}7 \end{matrix} \right] \end{align*}$
So it can be seen that $\displaystyle \displaystyle \begin{align*} -x_3 = 7 \implies x_3 = -7 \end{align*}$. Then
$\displaystyle \displaystyle \begin{align*} x_2 + x_3 &= -2 \\ x_2 - 7 &= -2 \\ x_2 &= 5 \\ \\ x_1 - 3x_2 &= 2 \\ x_1 - 3(5) &= 2 \\ x_1 - 15 &= 2 \\ x_1 &= 17 \end{align*}$