Solve the system using Gaussian elimination with back-substitution or Gauss-Jordan elimination Here is the system:
$$
\left\{
\begin{aligned}
x_1-3x_3&=-2 \\
3x_1+x_2-2x_3&=5 \\
2x_1+2x_2+x_3&=4
\end{aligned}
\right.
$$
This is my very first problem actually using a matrix so here is my attempt
First I setup the augmented matrix:
$$ \left[
      \begin{array}{ccc|c}
        1&0&-3&-2\\
        3&1&-2&5 \\
        2&2&1&4
      \end{array}
    \right]$$
Then I did $(-3)R_1+R_2->R_2$ which produced:
$$ \left[
      \begin{array}{ccc|c}
        1&0&-3&-2\\
        0&1&7&11 \\
        2&2&1&4
      \end{array}
    \right]$$
And then I did $(-2)R_1+R_3->R_3$:
$$ \left[
      \begin{array}{ccc|c}
        1&0&-3&-2\\
        0&1&7&11 \\
        0&2&7&8
      \end{array}
    \right]$$
And then $(1/2)R_3->R_3$:
$$ \left[
      \begin{array}{ccc|c}
        1&0&-3&-2\\
        0&1&7&11 \\
        0&1&\frac{7}{2}&4
      \end{array}
    \right]$$
And then $(-1)R_2+R_3->R_3$
$$ \left[
      \begin{array}{ccc|c}
        1&0&-3&-2\\
        0&1&7&11 \\
        0&0&\frac{-7}{2}&7
      \end{array}
    \right]$$
And finally $\left(\frac{-1}{7}\right)R_3->R_3$
$$ \left[
      \begin{array}{ccc|c}
        1&0&-3&-2\\
        0&1&7&11 \\
        0&0&1&-1
      \end{array}
    \right]$$
But this is not correct according to the solution.  Can someone tell me where I went wrong?  Clearly I did something really wrong here.

From your step: And then $(-1)R_2+R_3->R_3$
If you look at where you have $−11+4$. 
