# Testing for linear independence (Vectors -> SOE)

I know this may be an odd question, so my apologies if I am stating it poorly. I'll try to update it further if what I am asking does not make sense.

Lets say that i have the following question:

"Determine whether the vectors (1,-2,3), (5,6,-1), (3,2,1) are linearly independent."

The equation: $$c_1 (1,-2,3) + c_2 (5,6,-1) + c_3 (3,2,1)$$ gives a linear system

$$\begin{cases} c_1 + 5c_2 + 3c_3 & \quad = 0\\ -2c_1 + 6c_2 + 2c_3 & \quad = 0\\ 3c_1 - c_2 + c_3 & \quad = 0 \end{cases}$$

Why would i translate the three vectors each as columns multiplied by scalars? I am a little fuzzy on how we get from the first "Determine whether ..." to the equation and then linear system. What is the basis for doing this?

You add vectors by component and multiply the scalars across all the components so the equation $$c_1(1,-2,3)+c_2(5,6,-1)+c_3(3,2,1) =0$$ means $$\left(c_1+5c_2+3c_3,-2c_1+6c_2+2c_3,3c_1-c_2+c_3\right) = 0.$$ That $0$ on the RHS is shorthand for the zero vector, which in this context is $(0,0,0).$
Now the vector equation holds if and only if it holds for all its components. Setting each component of the vector $$\left(c_1+5c_2+3c_3,-2c_1+6c_2+2c_3,3c_1-c_2+c_3\right)$$ to zero gives the system of equations you wrote down.
I answered this thinking the question was to explain how to get from the equation to the system you wrote down, but reading a little closer it seems like you might actually be asking for motivation for examining this system if you're trying to answer the question about linear independence. We know that the system is linearly dependent (i.e. not linearly independent) if and only if the equation $$c_1(1,-2,3)+c_2(5,6,-1)+c_3(3,2,1) =0$$ has a nontrivial solution. As I showed in my original answer, this is equivalent to the system of equations you wrote down having a nontrivial solution.