Write $x_1$ as the linear combination of the other vectors.
$x_1 = (1, 2, -1, 3)$
$x_2 = (0, 1, 2, 4)$
$x_3 = (4, -2, 0, -2)$
$x_4 = (6, 3, 0, 8)$
I've created a matrix and row reduced, but I don't think I'm getting the right answer. I've reduced the matrix to: \begin{bmatrix}1&0&0&-1/2\\0&1&0&-1/2\\0&0&1&1/2\\0&0&0&0\end{bmatrix}
And I think the last column is where the solution comes from, I'm not sure though.
A general approach to these kind of problems is the following:
Linear combination of $\vec{x_1}$ in your case is: $$ \vec{x_1} = a \vec{x_2} + b\vec{x_3} + c \vec{x_4} $$ for some a,b,c constants. From this you can obtain a linear system with a,b, and c as unknowns, namely: $$ \begin{bmatrix}x_1^{(1)}\\ x_2^{(1)} \\ x_3^{(1)} \\ x_4^{(1)}\end{bmatrix} = a \cdot \begin{bmatrix}x_1^{(2)}\\ x_2^{(2)} \\ x_3^{(2)} \\ x_4^{(2)}\end{bmatrix} + b \cdot \begin{bmatrix}x_1^{(3)}\\ x_2^{(3)} \\ x_3^{(3)} \\ x_4^{(3)}\end{bmatrix} + c \cdot \begin{bmatrix}x_1^{(4)}\\ x_2^{(4)} \\ x_3^{(4)} \\ x_4^{(4)}\end{bmatrix} $$ This will derive 4 linear equations, with 3 unknowns: $$ x_1^{(1)} = a \cdot x_1^{(2)} + b \cdot x_1^{(3)} + c \cdot x_1^{(4)}$$ $$ x_2^{(1)} = a \cdot x_2^{(2)} + b \cdot x_2^{(3)} + c \cdot x_2^{(4)}$$ $$ x_3^{(1)} = a \cdot x_3^{(2)} + b \cdot x_3^{(3)} + c \cdot x_3^{(4)}$$ $$ x_4^{(1)} = a \cdot x_4^{(2)} + b \cdot x_4^{(3)} + c \cdot x_4^{(4)}$$ where $x_i^{(j)}$ is the $i^{th}$ component of the $j^{th}$ vector.
Solve the system with either direct substitution or Gaussian Elimination or some other method and you get your answer. Keep in mind that some systems do not have a solution, i.e. the system $5b = 10$ and $b = 3$ is inconsistent. You can also see that by plotting the two different values of b, that is: $b = 3$, $b = 2$ and by observing that they do not intersect anywhere.
See more here: How to determine if a linear system is solvable
Good luck!
