Show that vectors x=(1, 2,3) and y=(3, 2,1) are linearly independent. Let L be a span of x,y. Show that z=(-5, 2,0) is NOT contained in L. 
"Show that vectors $x=(1, 2,3)$ and $y=(3, 2,1)$ are linearly independent. Let L be a span of x,y. Show that $z=(-5, 2,0)$ is NOT contained in L."

Hey i'm a bit confused i dunno, we weren't left with a chapter to read (which i use a lot when this happens). I know we're supposed to give these a good whack before we post them here and i'd do so right after i get a hint. 
I dunno this week we were talking about bijective and surjective functions for algebra. Someone mentioned basis and how it's in the work somewhere but did not get thought. I'm guessing by looking around that i can set the vectors into a 3 by 2 matrix with each column equalling to zero.
$\begin{bmatrix} 1&3 \\ 2&2 \\ 3&1  \end{bmatrix}$ = $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$
Somehow solving this, but i feel like we can't solve a 3 by 2 matrix. Proving that it does in fact solve to a vector {0,0,0} to show its independence. 
What i have no idea on is how z=(-5,2,0) is not contained in L. 
 A: Surjectivity and bijectivity are not concepts that will help with this problem. They come in later when you're working with transformations and matrix multiplication.
So for this problem what we want to do is figure out if x is some multiple of y, in other words are these 2 vectors some multiple of each other. You had the right idea by placing it into a matrix however, you want to place the 2 vectors in a 2x3 matrix instead. This allows you to do is reduce the matrix. The key for determining whether or not they are linearly independent is whether or not you have and zero rows. A zero row indicated that one of those vectors is colinear to some other vector in your set. This works in for more vectors as well. Assume I give you 2 more vectors, $a=(2,4,6)$ and $b=(2,4,7)$. Using the same method of putting the 4 vectors in a matrix can you determine whether the 4 are linearly independent? 
The second part of the question requires you to remember that given 2 vectors (i.e. lines) one can define a plane. A plane in ${R}^2$ is defined by the equation $ax + by + cz + d = 0 $ where normal is defined to be $n=(a,b,c)$. In order to find the normal we use the cross product of x and y. Find d is easy because you already have a vector you know is on the plane so you plug it into that equation. To show z is not on that plane L you plug the vector z into your equation and you will see that some non zero number = 0 from which you can conclude that z is not on L.
Sorry for the wordy response (its my first one) and no answers, I forgot how to use LaTeX for matrices. Hope it helped :)
A: for the first question you can see that the two vectors are linearly dependent if :
$\begin{bmatrix} 1 \\ 2 \\ 3  \end{bmatrix}$ = $\alpha\begin{bmatrix}3 \\2 \\ 1 \end{bmatrix}$
So $\alpha=\frac{1}{3}$ from the first line and $\alpha=1\neq \frac{1}{3}$ so $\begin{bmatrix} 1 \\ 2 \\ 3  \end{bmatrix}$ &  $\begin{bmatrix}3 \\2 \\ 1 \end{bmatrix}$ are linearly independent.
2) If $\begin{bmatrix} -5 \\ 2 \\ 0  \end{bmatrix}\in L$ so it exist $a,b\in \mathbb{R}$
$$
a \begin{bmatrix} 1 \\ 2 \\ 3  \end{bmatrix}+b \begin{bmatrix}3 \\2 \\ 1 \end{bmatrix}=\begin{bmatrix} -5 \\ 2 \\ 0  \end{bmatrix}
$$
so 
$$
\left\{\begin{array}{l}
a+3b=-5\\
a+b=1\\
3a+b=0
\end{array} \right.
$$
and this system don't have a solution, because the system formed by the two first equations $
\left\{\begin{array}{l}
a+3b=-5\\
a+b=1
\end{array} \right.
$ have as solution $(a,b)=(4 ,-3 )$ and  the system formed by the two last equations $
\left\{\begin{array}{l}
a+b=1 \\
3a+b=0
\end{array} \right.
$ have as solution $(a,b)=(-\frac{1}{2} ,\frac{3}{2} )$
