Determining number of solutions for system of equations I'm having difficulty determining the number of solutions of a system of equations.
Suppose I want to solve the following system of equations
$$\begin{bmatrix}
5x_1+3x_2+2x_3\\
2x_1-8x_2+6x_3\\
7x_1+4x_2+15x_3
\end{bmatrix}
=\begin{bmatrix} 0\\0\\0 \end{bmatrix}
$$
Solving by elimination
$$
\begin{split}
2R_1&:\space 10x_1 &+ 6x_2 &+4x_3   &=0\\
5R_2&:\space 10x_1 &-40x_2 &+30x_3 &=0\\
2R_1-5R_2&:\space  &+46x_2 &-26x_3  &=0
\end{split}
$$
A solution to this would be $x_2=\frac{13}{23}$, $x_3=1$
From $R_1$, $x_1=-\frac{17}{23}$
Since these 3 values satisfy $R_1$ and $R_2$, but don't satisfy $R_3$, does this automatically mean that the system only has the trivial solution $R_1=R_2=R_3=0$?
 A: No, it means you picked  the "wrong" solution of the reduced equation. Maybe there is a different value of $x_2,x_3$ which would force a different value for $x_1$...
UPDATE
To understand if the system will admit non-trivial solutions, eliminate the variables one by one from the equations in a systematic way. For example, you combined $R_1$ and $R_2$ to get $$46x_2 -26x_3 = 0$$ and in a similar way, compute $2R_3 - 7R_2$ to eliminate $x_1$ as well, getting the equation $$64x_2 -12x_3 = 0$$
Now note your new system (once you divide top equation by 2 and bottom one by 4 to simplify) is
$$
\begin{split}
23x_2 &- 13x_3 &= 0\\
16x_2 &- 3x_3 &= 0
\end{split}
$$
It is now easy to eliminate $x_2$. Note from the last equation, $x_2 = 3/16 x_3$ and plugging that into the top equation:
$$
0 = 23x_2 - 13x_3 = 23 \times \frac{3}{16} x_3 - 13 x_3
  = x_3\left(\frac{23 \cdot 3}{16} - 13\right)
$$
which means $x_3=0$ is the only possible value. Hence,
$$
x_2 = \frac{3}{16} x_3 = 0
$$
and you can plug into any of the original equations to prove $x_1=0$ as well, and the system only admits the trivial solution.
A: Ok,dividing the second equation by $2$ we get
$$x_1-4x_2+3x_3=0$$
Multiplying this equation by $-5$ and adding to the first we get
$$23x_2-13x_3=0$$ and analogously
$$16x_2-3x_3=0$$ and from $$x_3=\frac{16}{3}x_2$$ and the equation above we get
$$x_1=x_2=x_3=0$$
