<s>[INST] Determine if vectors are linearly independent Determine if the following set of vectors is linearly independent:
$$\left[\begin{array}{r}2\\2\\0\end{array}\right],\left[\begin{array}{r}1\\-1\\1\end{array}\right],\left[\begin{array}{r}4\\2\\-2\end{array}\right]$$
I've done the following system of equations, and I think I did it right... It's been such a long time since I did this sort of thing...
Assume the following:
\begin{equation*}
a\left[\begin{array}{r}2\\2\\0\end{array}\right]+b\left[\begin{array}{r}1\\-1\\1\end{array}\right]+c\left[\begin{array}{r}4\\2\\-2\end{array}\right]=\left[\begin{array}{r}0\\0\\0\end{array}\right]
\end{equation*}
Determine if $a=b=c=0$:
\begin{align}
2a+b+4c&=0&&(1)\\
2a-b+2c&=0&&(2)\\
b-2c&=0&&(3)
\end{align}
Subtract $(2)$ from $(1)$:
\begin{align}
b+c&=0&&(4)\\
b-2c&=0&&(5)
\end{align}
Substitute $(5)$ into $(4)$, we get $c=0$.
So now what do I do with this fact? I'm tempted to say that only $c=0$, and $a$ and $b$ can be something else, but I don't trust that my intuition is right.
 [/INST] You just stopped too early:
Since you have 3 varibles with 3 equations, you can simply obtain $a,b,c$ by substituting $c = 0$ back into the two equations: 


*

*From equation $(3)$, $c = 0 \implies b = 0$. 

*With $b = 0, c = 0$ substituted into equation $(1)$ or $(2)$, $b = c = 0 \implies a = 0$. 
So in the end, since 
$$\begin{equation*}
a\left[\begin{array}{r}2\\2\\0\end{array}\right]+b\left[\begin{array}{r}1\\-1\\1\end{array}\right]+c\left[\begin{array}{r}4\\2\\-2\end{array}\right]=\left[\begin{array}{r}0\\0\\0\end{array}\right] \end{equation*}\implies a = b = c = 0, $$ 
the vectors are linearly independent, based on the definition(shown below).

The list of vectors is said to be linearly independent if the only $c_1,...,c_n$ solving the equation $0=c_1v_1+...+c_nv_n$ are $c_1=c_2=...=c_n=0.$

You could have, similarly, constructed a $3\times 3$ matrix $M$ with the three given vectors as its columns, and computed the determinant of $M$.  Why would this help?  Because we know that if $\det M \neq 0$, the given vectors are linearly independent. (However, this method applies only when the number of vectors is equal to the dimension of the Euclidean space.)
$$M = \begin{bmatrix} 2 & 1 & 4 \\ 2 & -1 & 2 \\ 0 & 1 & -2 \end{bmatrix}$$
$$\det M = 12 \neq 0 \implies\;\text{linear independence of the columns}.$$
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