# Find basis and dimension of system of linear equations

I have the homogeneous system of linear equations

$$2x_1 + 2x_2 + 4x_3 - 2x_4 = 0,$$ $$x_1 + 2x_2 + x_3 + 2x_4 = 0,$$ $$-x_1 + x_2 + 4x_3- 2x_4 = 0.$$

I row reduced to $$\begin{bmatrix}1 & 0 & 0 & -.625\\0 & 1 & 0 & 1.875\\0 & 0 & 1 & -1.125\end{bmatrix}$$

And came up with the general solution: $$\begin{bmatrix}x_1 \\x_2\\x_3\\x_4\end{bmatrix} = \begin{bmatrix}-.625 \\ 1.875 \\ -1.125 \\ t\end{bmatrix}$$ t = any real number

So I believe that the basis will be:
$$\begin{bmatrix}2\\1\\-1\end{bmatrix} + \begin{bmatrix}2\\2\\1\end{bmatrix} + \begin{bmatrix}4\\1\\4\end{bmatrix} + {\begin{bmatrix}-2\\2\\-2\end{bmatrix}}$$

Dimension = # of vectors in the basis so It would be 4.

My question is did I handle the free variable $$x_4$$ correctly in my answer? Or should I exclude the last vector, $${\begin{bmatrix}-2\\2\\-2\end{bmatrix}}$$, because they are all free variables, which would make the Dimension = 3

If the RREF is $$\begin{bmatrix}1 & 0 & 0 & -.625\\0 & 1 & 0 & 1.875\\0 & 0 & 1 & -1.125\end{bmatrix}$$
And came up with the general solution: $$\begin{bmatrix}x_1 \\x_2\\x_3\\x_4\end{bmatrix} = \begin{bmatrix}0.625t \\ -1.875t \\ 1.125t \\ t\end{bmatrix}=t\begin{bmatrix}0.625 \\ -1.875 \\ 1.125 \\ 1\end{bmatrix}$$