Is my solution to the system of equations correct? If I'm told that $T(\vec x)=A\vec x=\vec b$ and $A=\left[\begin{matrix}1&-3&2\\ 3&-8&8\\ 0&1&2\\ 1&0&8\\\end{matrix}\right]$ and that $\vec b=\left[\begin{matrix}1\\6\\3\\10\end{matrix}\right]$. I need to find some vector $\vec x$ that whose image under $T$ is $\vec b$. So normally to do this we set up an augmented matrix like this and solve: $$A=\left[\begin{matrix}1&-3&2&1\\ 3&-8&8&6\\ 0&1&2&3\\ 1&0&8&10\\\end{matrix}\right]$$ I was able to row reduce this to the matrix: $$A=\left[\begin{matrix}1&0&8&10\\ 0&1&2&3\\ 0&0&0&0\\ 0&0&0&0\\\end{matrix}\right]$$ So would that mean that I found the vector $\vec x$ to be $\vec x=\left[\begin{matrix}10\\3\\0\end{matrix}\right]$ as one solution? This would mean that there must be infinite solutions since there is one free variable right?
 A: Best to present the solution in the form 
$$x_3 = \alpha \implies x_1 = 10 - 8 \alpha \implies x_2 = 3 - 2\alpha$$ 
$$ \implies \vec x =
\begin{pmatrix} \\ \\
10 - 8 \alpha \\ \\
3 - 2\alpha \\ \\
\alpha \\ \\
\end{pmatrix}
$$

added to answer additional questions posted: 
And yes, this indeed means that since there is a free variable we are denoting $\alpha$, there are infinitely many possible values for $\alpha$, hence an infinite number of solutions to the system.  
However, since the values of $x_1, x_2$ depend on $x_3 = \alpha$, we do have that for any fixed $\alpha$, the other variables are then determined. So there are some constraints on the solutions.
A: The row reduced matrix gives the following: $x_1+8 x_3 = 10$, $x_2+2x_3 = 3$, which can be written as $x_1 = 10-8 x_3$, $x_2 = 3-2 x_3$. So you can choose $x_3$ arbitrarily and then compute the $x_1, x_2$ that satisfies the equations.
The solutions are then given by $\begin{bmatrix}10\\3\\0\end{bmatrix}+ x_3 \begin{bmatrix}-8\\-2\\0\end{bmatrix} $, $x_3 \in \mathbb{R}$.
