Proving a set of linear functionals is a basis for a dual space I've seen some similar problems on the stackexchange and I want to be sure I am at least approaching this in a way that is sensible. 
The problem as stated: 
Let $V= \Bbb R^3$ and define $f_1, f_2, f_3 \in V^*$ as follows: 
$f_1(x,y,z)= x-2y ,\; f_2(x,y,z)= x+y+z,\;f_3(x,y,z)= y-3z$
Prove that $f_1, f_2, f_3$ is a basis for $V^*$ and and then find a basis for V for which it is the dual basis. 
Here's my problem: the question feels a bit circular. But this is what I attempted: 
To show that the linear functionals $f$ are a basis, we want that $f_i(x_j)=\delta_{ij}$, or that $f_i(x_j)=1$ if $i=j$ and that it is zero otherwise. 
That means that  we want to set this up so that 
$$1= f_1(x,y,z)= x-2y$$ 
$$0= f_2(x,y,z)= x+y +z$$ 
$$0= f_3(x,y,z)=   y-3z$$
That gives us three equations and three unknowns. Solving them we get $2x-2z=1$ for $x-z=\frac{1}{2}$ and $z=x-\frac{1}{2}$ and subbing into the equation for $f_3$ I get $0=y-3x-\frac{3}{2}$ which gets us $1=x-6x+3$ or $x=\frac{2}{5}$. That gives us $y=\frac{-3}{10}$ and $z=\frac{-1}{10}$. 
OK, this is where I am stuck on the next step. I just got what should be a vertical matrix I think, with the values $(\frac{2}{5}, \frac{-3}{10}, \frac{-1}{10})$ but I am not sure where to go from here. I am not entirely sure I set this up correctly. 
thanks
EDIT: I do know that I have to show that $f_1, f_2, f_3 $ are linearly independent. That I think I can manage, but I am unsure how to fit it into the rest of the problem or if I am even approaching this right. 
 A: Ok, so it looks like you are on the right track.  But you have only worked out one basis vector in $\mathbb{R}^3$.  First, notice that we will have three vectors $v_1, v_2, v_3$ where $v_i = (x, y, z)$ for $i = 1, 2, 3$ (i.e. these are elements of $\mathbb{R}^3$).  So as you point out we want that
$$f_1(v_1) = 1, f_1(v_2) = 0, f_1(v_3) = 0$$
$$f_2(v_1) = 0, f_2(v_2) = 1, f_2(v_3) = 0$$
$$f_3(v_1) = 0, f_3(v_2) = 0, f_3(v_3) = 1$$
So you have some more equations to work with to find the other basis vectors (it looks like you have solved the first one). 
Edit: And also, I think you can do the part to check that $f_1, f_2, f_3$ are linearly independent, but I will just write out what you want to start with:
Suppose you have $c_1, c_2, c_3 \in \mathbb{R}$ such that
$$c_1f_1 + c_2f_2 + c_3f_3 = 0$$ i.e. $$c_1(x - 2y) + c_2(x + y + z) + c_3(y -3z) = 0$$
Then you want to show all $c_i = 0$ and that proves they are linearly independent.  (It is just a bit of manipulation to get that).
A: What about a direct approach? Suppose $\,a,b,c\in\Bbb R\,$ are such that
$$af_1+bf_2+cf_3=0\in V^*\implies\;\forall\,v:=(x,y,z)\in\Bbb R^3\;,\;\;af_1v+bf_2v+cf_3v=0\iff$$
$$a(x-2y)+b(x+y+z)+c(y-3z)=0\iff$$
$$\iff (a+b)x-(2a-b-c)y+(b-3c)z=0$$
As the above is true for all $\;x,y,z\in\Bbb R\,$ , we must have
$$\begin{align*}\text{I}&\;\;\;\;a+b=0\\\text{II}&\;\;\;\;2a-b-c=0\\\text{II}&\;\;\;\;b-3c=0\end{align*}\;\;\implies a=-b\;,\;\;c=3a=\frac13b\implies a=b=c=0$$
and we're done.
