# Explain why a system of more than 2 equations has only one solution

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
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
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
{\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

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.
@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