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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. –  ncmathsadist Jan 23 '12 at 2:16
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Something is amiss. You have a $4\times 3$ matrix multiplied by a $4 \times 1$ matrix, which is undefined. –  Austin Mohr Jan 23 '12 at 2:17
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Try writing $A$ in its reduced-row-echelon form. –  JavaMan Jan 23 '12 at 2:19
    
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. –  intervade Jan 23 '12 at 3:41
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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 $Ax=0$? –  N. S. Jan 23 '12 at 3:42

1 Answer 1

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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Ok, I understand the concept of the free variables in this situation. However, I do not understand what Ax = 0 has to do with the matrix only having one solution. I realize this is the Ker(A) or Null(a). With that said, we were not introduced to kernels until the next chapter. –  intervade Jan 23 '12 at 4:07
    
@Dalton Interesting. Perhaps there is another way, then? To your question: if $c_1$ and $c_2$ are such that $Ac_1 = Ac_2 = b$, then $A(c_1 - c_2) = b - b = 0$, so $c_1 - c_2 \in \operatorname{null}A$. If the kernel is trivial, then this means that $c_1 = c_2$, so the solution is unique. –  Dylan Moreland Jan 23 '12 at 4:11
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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. –  Dylan Moreland Jan 23 '12 at 4:14

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