Given governing equations, determine the condition for a solution and if they exist, solve the system. Given the governing equations:
1a + 2b = 2c
2a + 4b = d
2a + 5b = e
3a + 9b = f
What conditions are required for a solution x = [a,b]^T to exist?
If these conditions exist, solve the system for a and b.
I set up Ax=b, did some Gaussian Elimination and came up with:
1a + 2b = 2c
0a + 1b = e - 4c
0a + 0b = d - 4c
0a + 0b = f - 3e + 6c
I'm not sure how to answer the questions....  I understand that the last two equations should be my constraint equations.  ??  Any help would be greatly appreciated!  :)
 A: First, Gaussian eliminate a bit further.  You can get rid of the $2b$ in the first equation by subtracting twice the second equation.
Now you're left with $a$ as a function of $e$ and $c$, $b$ as a function of $e$ and $c$, $d$ as a function of $c$, and $f$ as a function of $e$ and $c$.  Can you turn this into a two dimensional subspace of $\mathbb{R}^6$ in which solutions exist?

(REPLY TO COMMENT)
So $d$ and $f$ are arbitrary here - they can be anything we want, and so long as $a,b,c,$ and $e$ obey the relations you've obtained there, the tuple $$\left(\begin{array}{c}a\\b\\c\\d\\e\\f\end{array}\right)=\left(\begin{array}{c}(7/2)d+(2/3)f\\(-1/2)d+(1/3)f\\(1/4)d\\d\\(-1/2)d+(1/3)f\\f\end{array}\right)$$ will be a solution to the governing equation.  Note that we can separate this: $$\left(\begin{array}{c}(7/2)d+(2/3)f\\(-1/2)d+(1/3)f\\(1/4)d\\d\\(-1/2)d+(1/3)f\\f\end{array}\right)=\left(\begin{array}{c}7/2\\-1/2\\1/4\\1\\-1/2\\0\end{array}\right)d+\left(\begin{array}{c}2/3\\1/3\\0\\0\\1/3\\1\end{array}\right)f.$$
Now, let me define something for you:

Definition. Given vectors $v_1,\ldots, v_r$, the span of $v_1,\ldots, v_r$ is the set $\{\sum_{i=1}^rav_i|a_i\in\mathbb{R}\}$.

In other words, the span of vectors $v_1,\ldots, v_r$ is all vectors which are linear combinations of $v_1,\ldots, v_r$.
You'll notice that this definition relates to the above problem: my claim is that the solutions to the governing equations is the span of the vectors $$v=\left(\frac{7}{2},-\frac{1}{2},\frac{1}{4},1,-\frac{1}{2},0\right)^\intercal\text{ and }w=\left(\frac{2}{3},\frac{1}{3},0,0,\frac{1}{3},1\right)^\intercal.$$  (The relationship between "span" and the terminology "$2$-dimensional subspace" is a bit more involved, but you can read about it on wikipedia.  Loosely, the span of vectors is always a subspace, and it is two dimensional because there are two vectors here which are not multiples of one another.)  So there are infinitely many solutions to your governing equations, and the solutions are exactly those vectors contained in the span of $v$ and $w$.
