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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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    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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If $c \in \mathbf R^3$ is a vector such that $Ac = b$, then the solutions of $Ax = b$ are precisely $ c + \operatorname{null} A = \{c + d : d \in \mathbf R^3 \text{ and } Ad = 0\}. $ If you can use or justify this, then all you need to do is show that the homogeneous system $Ax = 0$ has only the trivial solution $x = (0, 0, 0)^T$. This is true if and only if after performing elementary row operations to $A$ to get a matrix in row echelon form there are exactly three (the maximum possible number) pivots. If you had fewer pivots, then there would be free variables.

Do you know how to put a matrix in row echelon form? We can certainly go over that.

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    @Dalton I agree with your reduced form, by the way (It has to be that matrix, or else the problem is false!). We can talk about why that shows that the kernel is trivial, but it seems like you have a good handle on that.2012-01-23