## Linear Algebra and Its Applications (5th Edition)

By inspection, we see that the second vector is $\frac{3}{2}$ times the first vector. Hence: $\frac{3}{2}\begin{bmatrix}4\\-2\\6\end{bmatrix}-\begin{bmatrix}6\\-3\\9\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$. This is a nontrivial solution to the homogeneous vector equation, so the vectors are linearly dependent.