So I'm doing an a bit of a pre-assessment for something, and I feel like I am missing something on this question:
Now I know how to solve a normal 3 variable system, but with this they are all set to 0, how exactly would I solve it? Or would I just solve it / explain it by saying that there are an infinite number of solutions since any variable == 0 would mean the other two could be anything. Maybe I'm just brain farting, but some help would be greatly appreciated.
I looked for similar questions, where all parts of the system were set to 0, but I didn't see any.
Generally speaking, for this system of equations, whether there is a unique solution, no solution or there is an infinite number of solutions depend on whether one of the $3$ equations can be deduced from the other two or conflicting the other two.
Case $1$: If one of the equations can be deduced from the other two, then there is an infinite number of solutions.
Case $2$: If one of the equations is conflicting the other two, then there is no solution. For example, $x_1 + x_2 + x_3 = 0$ and $x_1 + x_2 + x_3 = 1$.
Case $3$: If any of the equations cannot be deduced from the other two and there are no conflicts, then there is a unique solution.
For the three equations in your problem, there should be a unique solution only.
One way to find whether one of the equations can be deduced from the others or have conflict with the others, is to write down the coefficients of the three equations as a matrix like this:
$ M= \left[ {\begin{array}{cc} 1 & -2 & 3 & | & 0 \\ -2 & -3 & -4 & | & 0 \\ 2 & -4 & 4 & | & 0\\ \end{array} } \right] $
and then try to simplify the $3 \times 3$ matrix to see if the identity matrix $I$ can be obtained.
$ = \left[ {\begin{array}{cc} 1 & -2 & 3 & | & 0 \\ 0 & 1 & 0 & | & 0 \\ 2 & -4 & 4 & | & 0\\ \end{array} } \right] $ (Row $2 \rightarrow$ Row $2 +$ Row $3$)
$ = \left[ {\begin{array}{cc} 1 & -2 & 3 & | & 0 \\ 0 & 1 & 0 & | & 0 \\ 0 & 0 & -2 & | & 0\\ \end{array} } \right] $ (Row $3 \rightarrow$ Row $3 - 2 \times $ Row $1$)
$ = \left[ {\begin{array}{cc} 1 & -2 & 3 & | & 0 \\ 0 & 1 & 0 & | & 0 \\ 0 & 0 & 1 & | & 0\\ \end{array} } \right] $ (Row $3 \rightarrow -0.5 \times $ Row $3$)
$ = \left[ {\begin{array}{cc} 1 & 0 & 0 & | & 0 \\ 0 & 1 & 0 & | & 0 \\ 0 & 0 & 1 & | & 0\\ \end{array} } \right] $ (Row $1 \rightarrow $ Row $1 + 2\times $ Row $2 - 3\times $Row $3$)
Hence there is only one unique solution, i.e. $x_1 = x_2 = x_3 = 0$.
In case you cannot get the identity matrix $I$, e.g.
$ \left[ {\begin{array}{cc} 1 & 0 & 0 & | & k_1 \\ 0 & 1 & 1 & | & k_2 \\ 0 & 0 & 0 & | & 0\\ \end{array} } \right] $ ($k_1$ and $k_2$ can be any number)
There is an infinite number of solutions.
Finally, if you cannot get an identity matrix $I$, but instead something like this:
$ \left[ {\begin{array}{cc} 1 & 0 & 0 & | & k_1 \\ 0 & 1 & 1 & | & k_2 \\ 0 & 0 & 0 & | & k_3\\ \end{array} } \right] $ ($k_3 \neq 0$, and $k_1$ and $k_2$ can be any number)
The last row implies $0 x_1 + 0 x_2 + 0 x_3 \neq 0$, which is impossible. So there is no solution (one of the original equations conflicts with the other two).