Are there linear mappings for the following vectors, if so, what are they? (a)Is there a linear mapping $f : \mathbb{R}^{3} \rightarrow \mathbb{R}^3$ with
$ f(1,1,1)=(3,2,7), f(0,2,1)=(2,1,-1), f(2,0,1)=(1,0,0)$?
(b)Is there a linear mapping $f :\mathbb{R}^{3} \rightarrow \mathbb{R}^3$ with 
$ f(0,2,-1)=(3,-1,3), f(1,0,2)=(0,-2,1), f(-2,2,-5)=(3,3,1)$?
Substantiate your statements and if it exists, provide such a linear mapping.
My attempt so far:
(a) is it first important to determine whether $(1,1,1), (0,2,-1), (2,0,1)$ (called vectors, right??) are a basis for $\mathbb{R}^{3}$? in that case my understanding is a basis is linear independent vectors which span a vectorspace. Since $2(1,1,1)-(0,2,1)=(2,0,1)$ i would like to conclude that these vectors are linearly dependent $\Rightarrow$ do not form a basis for $\mathbb{R}^{3} \Rightarrow$ there is no linear map for these vectors. In case my reasoning is correct, I just wanted to double check is that i could have reached the linear dependence conclusion by turning the vectors into a matrix and seeing if that yielded a row of zeros or a free variable, right? (i happened to just look at it and recognize that this time).
(b)actually after writing my (a) attempt i realize that (b) would also be linearly dependent, correct? i still haven't learned to create a matrix with latex yet, but in case someone is skeptical that it is linearly dependent i could show you my steps. however, first i'd prefer to know if i am even on the right track ;)
as a note i'm always happy to hear if i am using terminology improperly. thanks in advance for all your help!
 A: Hint: Suppose we have a linear map $f:\mathbb{R^3}\to \mathbb{R^3}$. Say, $f_1(x,y,z)=a_1x+b_1y+c_1z$, $f_2(x,y,z)=a_2x+b_2y+c_2z$, $f_3(x,y,z)=a_3x+b_3y+c_3z$ where $f_1,f_2,f_3$ are the $x,y,z$ components of $f$. Substituting the values $(1,1,1),(0,2,1),(2,0,1)$ in this map, we get the corresponding values of the output if such a function exists. Now this gives us $9$ equations in $9$ unknowns, $a_i,b_i,c_i$ with $i=1,2,3$. If these equations form a determined system, we have a unique solution and hence a map. If the system is overdetermined, i.e. one of the equations is a linear combination of the others, then we have several such maps. If the system is inconsistent, i.e. the equations cannot be satisfied simulatenously, then we have no solution and no such map. So it just boils down to solving these equations simultaneously.
Edit: Adrian makes a useful remark, see his comment below.
A: If you define $f(x,y,z)=A\cdot \left(
                        \begin{array}{}
                         x \\
                         y \\
                         z \\
                        \end{array}
                      \right)$, where the matrix $A$ is given for  the 9 ecuations as:
$$ A\cdot \left(
                        \begin{array}{}
                         1 \\
                         1 \\
                         1 \\
                        \end{array}
                      \right) = \left(
                        \begin{array}{}
                         3 \\
                         2 \\
                         7 \\
                        \end{array}
                      \right)$$
You can find all coefficients of the matrix $A$.
So each relation as the previous we find 3 ecuation, in total 9 ecuations with 9 unknowns, then you problem is solves .
But remember this depends of that ecuations system can be solved !!!!
A: For (a), your reasoning is incomplete. For example consider the problem of finding $f$ when
$f(1,0,0) = (0,0,4)$, $f(2,0,0) = (0,0,8)$ and $f(3,0,0) = (0,0,12)$
There is clearly a linear function $f$ that satisfies this (infinitely many, in fact), even though the vectors $(1,0,0), (2,0,0), (3,0,0)$ are linearly dependent.
If you show that the "output" vectors are linearly independent, however, your justification would be good.
Timothy Wagner's answer is slightly incomplete: you have to solve the 3 equations in 3 unknowns problem 3 times: once for each output dimension.
A: Given the linear dependence you observed in the input vectors in (a), you could check if the same dependence applies to the output vectors.  As a linear map requires that f(ax+b)=af(x)+f(b), if the output vectors do not satisfy the same dependence there is no map.  This would save you trying to solve the equations.
On the other hand, if you prove the input vectors independent, there is guaranteed to be a linear map.
A: A brute-force approach: try to write using different notation: e.g., say you are searching for a matrix $A$ that satisfies $AX=Y$, where the columns of $X$ are given by the different vectors that are arguments to your $f$, and the columns of $Y$ are the different vectors that you obtain on applying $f$. Then, see if you can "solve" to get $A=XY^{-1}$.
