# Chapter 1 - Linear Equations in Linear Algebra - 1.7 Exercises: 16

These vectors are linearly independent.

#### Work Step by Step

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.

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