How do I complete solving this homogeneous system of linear equations using Gaussian elimination? $${ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }=0\\ { x }_{ 1 }+{ x }_{ 2 }+2{ x }_{ 3 }-{ x }_{ 4 }=0\\ 2{ x }_{ 1 }+2{ x }_{ 2 }+{ x }_{ 3 }+{ x }_{ 4 }=0\\ -{ x }_{ 1 }-{ x }_{ 2 }+{ x }_{ 3 }-2{ x }_{ 4 }=0\\ $$
Steps I took:
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 1 & 0 &  0 \\
    1 & 1 & 2 & -1 &  0 \\
    2 & 2 & 1 & 1  & 0 \\
   -1 & -1 & 1 & -2 & 0
  \end{array}\right]$$
$${ R }_{ 2 }={ 2R }_{ 2 }-{ R }_{ 3 }\\ \qquad \rightarrow $$
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 1 & 0 &  0 \\
    0 & 0 & 3 & -3 &  0 \\
    2 & 2 & 1 & 1  & 0 \\
   -1 & -1 & 1 & -2 & 0
  \end{array}\right]$$
$${ R }_{ 3 }={ R }_{ 3 }+{ 2R }_{ 4 }\\ \qquad \rightarrow $$
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 1 & 0 &  0 \\
    0 & 0 & 3 & -3 &  0 \\
    0 & 0 & 3 & -3  & 0 \\
   -1 & -1 & 1 & -2 & 0
  \end{array}\right]$$
$${ R }_{ 4 }={ R }_{ 1 }+{ R }_{ 4 }\\ \qquad \rightarrow $$
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 1 & 0 &  0 \\
    0 & 0 & 3 & -3 &  0 \\
    0 & 0 & 3 & -3  & 0 \\
    0 & 0 & 2 & -2 & 0
  \end{array}\right]$$
$${ R }_{ 3 }={ R }_{ 3 }-{ R }_{ 2 }\\ \qquad \rightarrow $$
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 1 & 0 &  0 \\
    0 & 0 & 3 & -3 &  0 \\
    0 & 0 & 0 &  0  & 0 \\
    0 & 0 & 2 & -2 & 0
  \end{array}\right]$$
$${ R }_{ 2 }=2{ R }_{ 2 }-3{ R }_{ 4 }\\ \qquad \rightarrow $$
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 1 & 0 &  0 \\
    0 & 0 & 3 & -3 &  0 \\
    0 & 0 & 0 &  0  & 0 \\
    0 & 0 & 0 &  0 & 0
  \end{array}\right]$$
At this point I am stuck. I am having a lot of trouble figuring out what I need to do once I complete all the necessary row operations on my matrix. I would like to be guided through the process as of this point. 
Please explain it to me in layman terms. I am hoping to use what I learn from being guided through the final steps of this problem in order to understand how to solve ones without any hand holding. 
I'm also really frustrated with the fact that I am doing these operations like a computer without actually understanding what is going on with the "independent" and "dependent" variables.
 A: So I'm going to solve this using Gauss-Jordan elimination as opposed to just Gaussian elimination. It's basically the same exact thing, except you get $0$'s above all of the leading $1$'s as well.
First off, since this is a homogeneous system of equations, you don't need to use an augmented matrix. The $0$'s on the right hand side of the line will stay the same no matter what row operation you use.
$
        \left(\begin{matrix}
        1 & 1 & 1 & 0   \\
        1 & 1 & 2 & -1  \\
        2 & 2 & 1 & 1   \\
        -1 & -1 & 1 & -2\\
        \end{matrix}\right)
$
$
        \begin{matrix}
        \\
        R_2-R_1\\
        R_3-2R_1\\
        R_4+R_1\\
        \end{matrix}
$
$
        \left(\begin{matrix}
        1 & 1 & 1 & 0   \\
        0 & 0 & 1 & -1\\
        0 & 0 & -1 & 1\\
        0 & 0 & 2 & -2\\
        \end{matrix}\right)
$
$
        \begin{matrix}
        R_1-R_2\\
        \\
        R_3+R_2\\
        R_4-2R_2\\
        \end{matrix}
$
$
        \left(\begin{matrix}
        1 & 1 & 0 & 1\\
        0 & 0 & 1 & -1\\
        0 & 0 & 0 & 0\\
        0 & 0 & 0 & 0\\
        \end{matrix}\right)
$
I did multiple row operations at once for brevity, but this is the same answer that you came up with. Taking your answer and dividing $R_2$ by $3$ and then subtract $R_2$ from $R_1$ and you get this Gauss-Jordan solution.
Each row is an equation that equals $0$ because this is a homogeneous system of equations.
The first column in the matrix is a representation of the $x_1$ variables in each equation, the second column is the $x_2$ variables, the third is $x_3$, and the fourth is $x_4$. The leading $1$'s are the pivot elements of the matrix, and are usually chosen as the dependent variables but they don't necessarily have to be. 
The next step in solving this problem is to express it back in the normal format:
$$
  \begin{align}
     x_1+x_2+x_4=0 \\
     x_3-x_4=0     \\
     0x_2=0        \\
     0x_4=0        \\
  \end{align}
$$
Notice that the last two equations show that anything you choose for $x_2$ or $x_4$ will give you a homogeneous solution to the system.
Next you want to solve for the dependent variables, and give parameters to the independent variables:
$$
  \begin{align}
     x_1 &=-x_2-x_4 \\
     x_3 &=x_4     \\
     x_2 &=s        \\
     x_4 &=t        \\
  \end{align}
$$
Finally, substituting the independent variables, $x_2$ and $x_4$, into the equations of the dependent variables will give you all of the solutions to the homogeneous system of equations:
$$
  \begin{align}
     x_1 &=-s-t \\
     x_3 &=t     \\
     x_2 &=s        \\
     x_4 &=t        \\
  \end{align} 
$$
With these equations, given any two arbitrary values for $x_2$ and $x_4$, you can determine values for $x_1$ and $x_3$ such that the homogeneous system of equations is solved.
Furthermore, notice that you could have solved the equations for $x_2$ and $x_4$ making them the dependent variables and giving $x_1$ and $x_3$ parameters:
$$
  \begin{align}
     x_2 &=-x_4-x_1 \\
     x_4 &=x_3     \\
     x_1 &=s        \\
     x_3 &=t        \\
  \end{align} 
$$
These equations make $x_1$ and $x_3$ the independent variables; and, with any arbitrary choice for those, you can find the dependent variables $x_2$ and $x_4$ that will solve the homogeneous system of equations. In other words, the row operations don't do anything to the dependent/independent variables, you choose which will be which.
A: You end up with 
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 0& 1&  0 \\
    0 & 0 & 1 & -1&  0 \\
  \end{array}\right].$$
All you have to do is paint in just straightening the tableau, i.e., create an upper triangle matrix whose diagonal elements are $\pm1$ in the following way. The first row of your reduced matrix has the first $1$  at position $1$ so that would be the first row, the second row's position of the leading $1$ is $3$ so that becomes the third row; just leave space for row $2$ and $4$:
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 0& 1 &  0 \\
\strut &&&&\\
    0 & 0 & 1& -1&  0 \\
\strut &&&&
  \end{array}\right].$$
Now fill the empty rows with zeroes except for the row number's position: you insert a $-1$: the second row consists of zeroes except for the second positon, you fill in a $-1$ here:
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 0& 1&  0 \\
0&-1&0&0&0\\
    0 & 0 & 1& -1&  0 \\
\strut &&&&
  \end{array}\right].$$
Row number four gets a $-1$ at position four:
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 0& 1&  0 \\
0&-1&0&0&0\\
    0 & 0 & 1& -1&  0 \\
0&0&0&-1&0
  \end{array}\right].$$
Now the columns with the numbers of the inserted rows, namely column two 
$\left[\begin{array}{c}
    1 \\-1\\ 0 \\0
  \end{array}\right]$
and column four
$\left[\begin{array}{r}
    1\\ 0\\ -1\\ -1
  \end{array}\right]$
span the kernel of the matrix.  Cute, isn't it?
