To check if the vectors $\mathbf{v_1} = (1, 2, 3)$, $\mathbf{v_2} = (4, 1, 2)$, and $\mathbf{v_3} = (3, -1, -1)$ are linearly independent or dependent, we can set up the following linear combination equation:

$c_1 \mathbf{v_1} + c_2 \mathbf{v_2} + c_3 \mathbf{v_3} = \mathbf{0}$

This expands to:

$c_1 (1, 2, 3) + c_2 (4, 1, 2) + c_3 (3, -1, -1) = (0, 0, 0)$

Which gives the system of equations:

1. \ & c_1 + 4c_2 + 3c_3 = 0 \\ 2. \ & 2c_1 + c_2 - c_3 = 0 \\ 3. \ & 3c_1 + 2c_2 - c_3 = 0 \end{aligned}$$ We will solve this system to determine whether there are non-trivial solutions (i.e., solutions where not all the $$c_i$$'s are zero). If such solutions exist, the vectors are linearly dependent; otherwise, they are linearly independent. I'll solve this now.The solution to the system of equations is: $$c_1 = c_3, \quad c_2 = -c_3$$ This means there is a non-trivial solution where $$c_3$$ can be any scalar, which implies that not all coefficients are zero. Therefore, the vectors $$\mathbf{v_1}$$, $$\mathbf{v_2}$$, and $$\mathbf{v_3}$$ are **linearly dependent**. Would you like further details or have any questions? Here are 5 related questions: 1. How do you interpret the concept of linear dependence in vector spaces? 2. What is the geometric meaning of linearly dependent vectors? 3. Can linearly dependent vectors form a basis for a vector space? 4. How does the rank of a matrix relate to linear independence of vectors? 5. What is the significance of finding non-trivial solutions in such systems? **Tip:** Linearly dependent vectors lie in the same plane or line in space, meaning one vector can be expressed as a combination of the others.