linear algebra - Showing that these vectors are linearly dependent - Have I done enough?

Question

In $\mathbb R^3$, let $$ v_1 =
\begin{pmatrix}
1 \\
1 \\
2 \\
\end{pmatrix}
$$
$$ v_2 =
\begin{pmatrix}

0 \\
-1 \\
-2\\
\end{pmatrix}
$$

$$ v_3 =
\begin{pmatrix}
0 \\
1 \\

3 \\
\end{pmatrix}
$$

$$ v_4 =
\begin{pmatrix}
1 \\
1 \\
1 \\
\end{pmatrix}

$$

Show that they are linearly dependent in $\mathbb R^3$

Attempt

Let $a_1, a_2, a_3, a_4 \in \mathbb R$ such that $ a_1 v_1 + a_2 v_2 + a_3 v_3 + a_4 v_4 = \mathbf 0 $

When I expand on this I get the following system of equations:

$$a_1 + a_4 = 0$$
$$a_1 - a_2 + a_3 + a_4 = 0$$
$$2a_1 - 2a_2 + 3a_3 + a_4 = 0$$

Ultimately, I get the following answers:

$$ a_1 = -a_2 = -a_3 = -a_4$$

i.e.

$$ a_1
\begin{pmatrix}
1 \\
-1 \\
-1 \\
-1 \\
\end{pmatrix}
=
\begin{pmatrix}
a_1 \\

a_2\\
a_3\\
a_4 \\
\end{pmatrix}
$$

Is this enough? How do I conclude from here that this is linearly dependent? I mean obviously $a_1 \ne 0$ (well in at least one instance) - but how do I show that further? Help would be awesome.

Answer

You can either argue that since $\;\dim_{\Bbb R}\Bbb R^3=3\;$ then the maximal number of lin. ind. vectors is three and thus your four vectors must be lin. dependent, or else do what you did and observe tha the condition $\,a_1=-a_2=-a_3=-a_4\;$ does not force all the coefficients to be zero and thus there are non-zero coeff. s.t. $\;a_1v_1+\ldots+a_4v_4=0\;$ and by definition this means the vectors are l.i.