The system in question is the following:
$x_1 – x_5 = 0$
$x_2 – x_5 – x_6=0$
$x_3 -x_5 – x_6=0$
$x_4-x_5-x_6=0$
Letting $x_5$ and $x_6$ be some real numbers $k$ and $t$, respectively, I can find a general solution:
$S = \left \{k\begin{bmatrix}1\\1\\1\\1\\1\\0\end{bmatrix} + t\begin{bmatrix}0\\1\\1\\1\\0\\1\end{bmatrix}| k, t \in \mathbb{R} \right \}.$
But how do I find all the solutions such that each $x_i$ is equal only to $1, 0$ or $-1$? Furthermore, how might I show that those solutions are linear combinations of the general solution?
Thanks!
Best Answer
First:
The idea of a general solution is to be a function which, if you substitute in all (in this case real) values, yields all solutions. Every substitution of values into the general solution (i.e. each substitution of $k,t$) is a linear combination.
Therefore, all possible solutions are linear combinations of the general solution, and by that especially all solutions containing only values in $\{-1,0,1\}$.
Second:
To be completely formal, the restriction on the solutions you want can be seen as a series of systems of equations:
Starting from the initial series of equations, your general solution is technically nothing different than another system of equation (though one which uses the minimal amount of necessary variables). So, one method would be to do for every vector $c\in \{-1,0,1\}^6$: $$ k\begin{bmatrix}1\\1\\1\\1\\1\\0\end{bmatrix} + t\begin{bmatrix}0\\1\\1\\1\\0\\1\end{bmatrix} = c $$
Given that this system of equations is oversaturated, there is only a finite amount of solutions, so for each $c$ you could calculate them all.
However, that's a lot of work to do. An easier method would be to look only at the first and second row in the system of equations and solve it for $\{-1,0,1\}$. As we've got only 2 variables, this already yields all possible solutions, we now only have to filter those, who don't work out.