Need help with a linear equation with a free variable? Is $a_1, a_2, a_3$ a linear combination of $b$?
$a_1 = (1, -2, 0), a_2 = (0, 1, 2), a_3 = (5, -6, 8), b = (2, -1, 6)$
I used Gaussian elimination to get to.
$$
\left[
\begin{array}{@{}ccc|c@{}}
1&0&5 & 2 \\
0&1&4 & 3 \\
0&0&0 & 0 \\
\end{array}
\right]
$$
so now $x_3$, is a free variable and I can't tell if it is a linear has a linear combination? Can someone tell me if it does and why?

Assuming you did the reduction properly, you just solve the remaining system:
$$x_1 + 5x_3 = 2$$
$$x_2 + 4x_3 = 3$$
Solving this, you get $x_2 = 3 -4 x_3$ and $x_1 = 2 - 5x_3$.
So, yes, you are free to choose $x_3$.
Update
Given any system of equations there are exactly three possibilities for the solution:


*

*(1) There will not be a solution.

*(2) There will be exactly one solution.

*(3) There will be infinitely many solutions.
When you do RREF and you get a row of zeros, what does that mean? It means that you have infinitely many solutions (hence, free variables).
How can you tell the remaining two cases from the augmented system? Can you relate this back to other methods, with say, the determinant? It is very good to ask these questions, just as you are doing.
Update 2
$$
\left[
\begin{array}{@{}ccc|c@{}}
1 & -2 & 0 & 2 \\
0 & 1 & 2 & -2 \\
5 & -6 & 8 & 6 \\
\end{array}
\right]
$$
If you RREF, using the following:


*

*-5 R1 + R3 $\rightarrow$ R3

*R4/4 $\rightarrow$ R4

*-R2 + R3 $\rightarrow$ R3 and then Negate and Divide by 2 $\rightarrow$ R3

*2R2 + R1 $\rightarrow$ R1, we end up with:
$$
\left[
\begin{array}{@{}ccc|c@{}}
1 & 0 & 4 & 0 \\
0 & 1 & 2 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right]
$$
So, I am confused if you have written the problem incorrectly or if you calculated the RREF incorrectly. Which is it? 
In this solution, the last equation is showing $0 = 1$, which is nonsense, so there is no solution.
Of course, if we took:
$$\text{det} 
\left[
\begin{array}{@{}ccc|c@{}}
1 & -2 & 0  \\
0 & 1 & 2  \\
5 & -6 & 8  \\
\end{array}
\right] = 0
$$
we already knew there would be no solutions, assuming you wrote your vectors correctly.
Regards
