To determine if the vectors $\mathbf{v}_1 = \langle -1, 0 \rangle$, $\mathbf{v}_2 = \langle 2, 3 \rangle$, and $\mathbf{v}_3 = \langle 4, -1 \rangle$ are linearly independent, we need to check if the following equation has only the trivial solution (i.e., all coefficients $c_1, c_2, c_3$ are zero):

$c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 = \mathbf{0}$

This translates to the system of linear equations: $c_1 \langle -1, 0 \rangle + c_2 \langle 2, 3 \rangle + c_3 \langle 4, -1 \rangle = \langle 0, 0 \rangle$

Breaking this down into two equations (for each component): $- c_1 + 2c_2 + 4c_3 = 0$ $0c_1 + 3c_2 - c_3 = 0$

We now solve this system. Let's solve it step by step:

1. From the second equation: $3c_2 - c_3 = 0 \quad \Rightarrow \quad c_3 = 3c_2$

2. Substitute $c_3 = 3c_2$ into the first equation: $-c_1 + 2c_2 + 4(3c_2) = 0$ $-c_1 + 2c_2 + 12c_2 = 0$ $-c_1 + 14c_2 = 0 \quad \Rightarrow \quad c_1 = 14c_2$

So, we have $c_1 = 14c_2$ and $c_3 = 3c_2$.

Since the only free variable is $c_2$, the solution is not the trivial solution unless $c_2 = 0$. If $c_2 = 0$, then $c_1 = 0$ and $c_3 = 0$ as well. However, if $c_2 \neq 0$, then the solution involves non-zero coefficients.

Thus, the vectors are linearly dependent, because there are non-trivial solutions to the system.

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Do you have any questions or want further details?

Here are some related questions to expand on the topic:

1. What does it mean for vectors to be linearly dependent or independent?
2. Can vectors in a two-dimensional space ever be linearly independent if there are more than two vectors?
3. How can we generalize the method for checking linear independence to vectors in higher dimensions?
4. What is the geometric interpretation of linearly dependent vectors in 2D space?
5. How does the concept of linear independence relate to matrix rank?

Tip: In 2D space, if you have more than two vectors, they will always be linearly dependent because the dimension of the space limits the number of independent directions.