This is a homework problem that I would love some direction on!
I'm given $A = \begin{bmatrix} 3 & 7 & -4\\ 5 & -2 & 6\\ 2 & 1 & -1\\ 4 & 1 & 2 \end{bmatrix}$
The question: Let $\vec{b}$ be a vector in $R^4$ such that the system $A\vec{x} = \vec{b}$ has a solution. Explain why it has only one solution.
Now, I've started off attempting to actually solve the system using the vector $b$:
$\begin{bmatrix} 3 & 7 & -4\\ 5 & -2 & 6\\ 2 & 1 & -1\\ 4 & 1 & 2 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} b_{1} \\ b_{2} \\ b_{3} \\ b_{4} \end{bmatrix}$
This proved to be a huge mess so I'm going to guess that this was the wrong way to go about it. Then I thought about relating it to pivots/pivot positions but I don't fully understand all of that yet. Can anyone offer me some suggestions?
EDIT:
$A =\begin{bmatrix} 3 & 7 & -4\\ 5 & -2 & 6\\ 2 & 1 & -1\\ 4 & 1 & 2 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} $
So based on the reduced form above, can I assume this matrix only has one solution because there are no free variables?