Non-Homogeneous System [Problem] "Find a general solution of the system and use that solution to find a general solution of the associated homogeneous system and a particular solution of  the given system."
$\begin{bmatrix}3 & 4 & 1 & 2 \\ 6 & 8 & 2 & 5\\9 & 12 & 3 & 10 \end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{bmatrix} = \begin{bmatrix}3 \\ 7 \\ 13\end{bmatrix}$
So we solve it like its a homogenous system. I end up getting to $\begin{bmatrix} 3 & 4& 1 & 2\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\end{bmatrix}$.So would $x_2$ and $x_3$ be free variables? I can set them to $x_2 = \alpha$ and $x_3 = t$, $x_4 = 0$ then we set up:
$\begin{cases}3x_1 + 4x_2 + x_3 + 2x_4 = 0\\0+0+0+x_4 = 0 \\0+0+0+0 = 0\end{cases}$
So, $\underline{v} = \begin{pmatrix}3 \\ 7 \\ 13\end{pmatrix} + \begin{pmatrix}\frac{1}{3}(-4\alpha -t - 2_4) \end{pmatrix}$
I don't think this is right. What would $x_4$ be? I'm confused
 A: For the augmented RREF matrix, I get:
$$\begin{bmatrix}1 & \dfrac{4}{3} & \dfrac{1}{3} & 0 & \dfrac{1}{3} \\ 0 & 0 & 0 & 1 & 1\\0 & 0 & 0 & 0 & 0\end{bmatrix}$$
This means we can write:
$$x_4 = 1$$
$$x_1 =  \dfrac{1}{3}-\dfrac{4}{3}x_2 - \dfrac{1}{3}x_3$$
So, we have two free variables, $x_2$ and $x_3$.
Can you now write this in a general form?
A: Try row reducing the augmented matrix to reduced row echelon form:
$$
\left[\begin{array}{cccc|c}
3 & 4 & 1 & 2 & 3\\
6 & 8 & 2 & 5 & 7\\
9 & 12 & 3 & 10 & 13
\end{array}\right]
\sim
\left[\begin{array}{cccc|c}
3 & 4 & 1 & 2 & 3\\
0 & 0 & 0 & 1 & 1\\
0 & 0 & 0 & 4 & 1
\end{array}\right]
\sim
\left[\begin{array}{cccc|c}
3 & 4 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 1\\
0 & 0 & 0 & 0 & -3
\end{array}\right]
$$
From the last row, we find that $0=-3$, a contradiction. So the system has no solution. Perhaps you made a typo?
A: The method we were taught was to set up an augmented matrix like so:
$$
\left[\begin{array}{cccc|c}
3 & 4 & 1 & 2 & 3\\
6 & 8 & 2 & 5 & 7\\
9 & 12 & 3 & 10 & 13\\
\end{array}\right]
$$
And then solve the matrix to the left of the vertical bar like a homogeneous system.
$$
\left[\begin{array}{cccc|c}
3 & 4 & 1 & 2 & 3\\
0 & 0 & 0 & 1 & 1\\
0 & 0 & 0 & 0 & 0\\
\end{array}\right] \rightarrow \left[\begin{array}{cccc|c}
3 & 4 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 1\\
0 & 0 & 0 & 0 & 0\\
\end{array}\right]
$$
So $3x_1+4x_2+x_3 = 1$, and $x_4 = 1$
