Solving a System of Equations Linear Algebra For one of my homework problems I have the following system of equations.
$$
\left\{ 
\begin{array}{c}
x_1+2x_2-2x_3-3x_5+2x_6=-4 \\ 
-x_4+4x_5+3x_6=-9 \\ 
x_1+2x_2-5x_5+4x_6=-4
\end{array}
\right. 
$$
The answer is to be formatted as follows: 
. I got the following when I created the augmented matrix and put it in row echelon form.
$$
        \begin{matrix}
        1 & 2 & 0 & 0 & -5 & 4 & -4 \\
        0 & 0 & 1 & 0 & -1 & 1 & 0  \\
        0 & 0 & 0 & 1 & -4 & -3 & 9 \\
        \end{matrix}
$$
I am not sure how to continue from this point. My main confusion is how exactly the variables s, t, and u are suppose to be considered. Thank you very much for your help.
 A: Guide:
Look at the non-pivot column, that is column that doesn't have leading $1$. That corressponds to $x_2, x_5, x_6 $.
Let $x_2=s$, $x_5 = t$, $x_6 =u$.
Now the last row says that $$x_4 -4t-3u=9$$
Now solve for $x_4$ in terms of $s, t, u$. 
Perform similar trick for $x_1$ and $x_3$.
A: Because the number of leading ones or the rank of the matrix is less than the number of unknowns, this tells us that there are infinite solutions. This means we have to set some x values as arbitrary values: s, t, u
After finding the RREF, one way to find what these values should be is to solve for the variables with leading one's as their coefficients.
So in this case, the leading ones are in columns 1,3,4 which indicate that we can try to solve for $x_1,x_3,x_4$

Lets take a look at the first row. If we convert that back into a equation we get
$$x_1 + 2x_2 -5x_5 + 4x_6 = -4$$
Solving for $x_1$ we reduce it to
$$x_1= -4 - 2x_2 +5x_5 - 4x_6 $$
We can do the same things for rows 2 and 3 and solve for $x_3$ and $x_4$
Which result in the following reduced equations
$$x_3 = x_5 -x_6 $$
And
$$x_4 =9 + 4x_5 + 3x_6$$

Now to finally use s,t, and u. For the variables that didn't have the leading one as a coefficient, we can randomly choose them to have arbitrary values: $x_2 = s,x_5=t,x_6=u$ where $s,t,u$ are arbitrary
Now combining this with the equations from above, we can change this to matrix form:
$$
        \begin{bmatrix}
        x_1\\
        x_2\\
        x_3\\
        x_4\\
        x_5\\
        x_6\\
        \end{bmatrix}
=
        \begin{bmatrix}
        -4 - 2x_2 +5x_5 - 4x_6\\
        s\\
        x_5 -x_6\\
        9 + 4x_5 + 3x_6\\
        t\\
        u\\
        \end{bmatrix}
$$
We can further 'reduce' by substituting $x_2 = s,x_5=t,x_6=s$ in to get
$$
        \begin{bmatrix}
        x_1\\
        x_2\\
        x_3\\
        x_4\\
        x_5\\
        x_6\\
        \end{bmatrix}
=
\begin{bmatrix}
        -4 - 2s +5t - 4u\\
        s\\
        t -u\\
        9 + 4t + 3u\\
        t\\
        u\\
        \end{bmatrix}
$$
This further can be reduced into a linear combination of three vectors with arbitrary scalars s,t,u and a constant vector.
$$
        \begin{bmatrix}
        x_1\\
        x_2\\
        x_3\\
        x_4\\
        x_5\\
        x_6\\
        \end{bmatrix}
=
        \begin{bmatrix}
        -4\\
        0\\
        0\\
        9\\
        0\\
        0\\
        \end{bmatrix}
+ s
        \begin{bmatrix}
        -2\\
        1\\
        0\\
        0\\
        0\\
        0\\
        \end{bmatrix}
+ t
\begin{bmatrix}
        5\\
        0\\
        1\\
        4\\
        1\\
        0\\
        \end{bmatrix}
+u
\begin{bmatrix}
        -4\\
        0\\
        -1\\
        3\\
        0\\
        1\\
        \end{bmatrix}
$$
This method can be done in much easier ways without having to write as much once it is understood well. E.g the constant vector is determined by column 7 in the augmented matrix.
