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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?

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    What do you know about rank or invertability? You can ditch one of the equations and see that a system based on the remaining three always has a unique solution.2012-01-23
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    Something is amiss. You have a $4\times 3$ matrix multiplied by a $4 \times 1$ matrix, which is undefined.2012-01-23
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    Try writing $A$ in its reduced-row-echelon form.2012-01-23
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    AustinMohr I think I fixed that issue, sorry. ncmathsadist, I know about rank, but have not learned about invertability yet. JavaMan, I've posted what I think is the correct idea based on your suggestion.2012-01-23
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    {\bf Hint}: If there are two different solutions, then their difference would be a non-trivial solution for the corresponding homogeneous system. Can you solve $Ax=0$?2012-01-23
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    I think we should only have $(x_1, x_2, x_3)$ and $(b_1, b_2, b_3)$.2012-01-23

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