Linear equation System What are the solutions of the following system:
$  14x_1 + 35x_2 -  7x_3 -  63x_4 = 0 $
$ -10x_1 - 25x_2 +  5x_3 +  45x_4 = 0 $
$ 26x_1 + 65x_2 - 13x_3 - 117x_4 = 0 $


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*4 unknowns (n), 3 equations


$ Ax=0: $
$ \begin{pmatrix} 14 & 35 & -7 & -63 & 0 \\ -10 & -25 & 5 & 45 & 0 \\ 26 & 65 & -13 & -117& 0  \end{pmatrix} $
Is there really more than one solution because of:


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*$\operatorname{rank}(A) = \operatorname{rank}(A') = 1 < n $


What irritates me:
http://www.wolframalpha.com/input/?i=LinearSolve%5B%7B%7B14%2C35%2C-7%2C-63%7D%2C%7B-10%2C-25%2C5%2C45%7D%2C%7B26%2C65%2C-13%2C-117%7D%7D%2C%7B0%2C0%2C0%7D%5D
Row reduced form:
$ \begin{pmatrix} 1 & 2.5 & -0.5 & -9.2 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 &    \end{pmatrix} $
How to find a solution set ?
 A: Yes, there are infinitely many real solutions. Since there are more unknowns than equations, this system is called underdetermined. Underdetermined systems can have either no solutions or infinitely many solutions. Trivially the zero vector solves the equation:$$Ax=0$$
This is sufficient to give that the system must have infinitely many solutions. To find these solutions, it suffices to find the row reduced echelon form of the augmented matrix for the system. As you have already noted, the augmented matrix is:
\begin{pmatrix} 14 & 35 & -7 & -63 & 0 \\ -10 & -25 & 5 & 45 & 0 \\ 26 & 65 & -13 & -117& 0  \end{pmatrix}
Row reducing this we obtain:
\begin{pmatrix} 1 & \frac{5}{2} & \frac{-1}{2} & \frac{-9}{2} & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0& 0  \end{pmatrix}
This corresponds to the equation $$x_1+\frac{5}{2}x_2-\frac{1}{2}x_3-\frac{9}{2}x_4=0$$ 
You can then express this solution set with
 $$x = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}=\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \frac{2}{9}x_1 +\frac{5}{9}x_2 -\frac{9}{4}x_3 \end{pmatrix}$$
As you have already noted, the rank(A) = 1, giving you $n-1=3$ free parameters. That is, you can supply any real values for $x_1,x_2,x_3$ and $x_4$ will be given as above. The choice to let $x_1,x_2,x_3$ be free parameters here is completely arbitrary. One could also freely choose $x_1,x_2,x_4$ and have $x_3$ be determined by solving for $x_3$ in the above equation.
The wikipedia article on underdetermined systems has some more details and explanation than what I've provided: http://en.wikipedia.org/wiki/Underdetermined_system
Row reduced echelon forms can be computed using Wolfram Alpha by entering "row reduce" following by the matrix. If you're interested, Gauss-Jordan elimination is a pretty good method for calculating reduced-row echelon forms by hand.
