Show that the linear combination is unique? I have the following question:
Let S be a subset of the vector space $\Bbb R^3$ defined by
$$
S =  \left\{  ~
\begin{bmatrix}2\\-1\\ 0\end{bmatrix},
\begin{bmatrix}1\\3\\ -2\end{bmatrix}, 
\begin{bmatrix}1\\1\\ 4\end{bmatrix} 
~\right\}
$$  


*

*show that 
$v= \begin{bmatrix}-4\\4\\ -6\end{bmatrix}$ is in $\text{span}(S)$ by constructing it as a linear combination of the vectors in $S$.

*Show that this linear combination is unique.


This is what I got after solving part 1: 
$$
c_1\begin{bmatrix}2\\-1\\ 0\end{bmatrix}
 + c_2\begin{bmatrix}1\\3\\ -2\end{bmatrix} 
+ c_3\begin{bmatrix}1\\1\\ 4\end{bmatrix} = \begin{bmatrix}-4\\4\\ -6\end{bmatrix}$$
after solving this I got  $c_1= -2$, $c_2= 1$, $c_3= -1$.
How do I show that the linear combination is unique? 
 A: Hint: If the linear combination is not unique (so there exist two sets of coefficients $c_1,c_2,c_3$ and $d_1,d_2,d_3$), show that this implies the vectors in $S$ are linearly dependent. Then show that this is not the case.
A: You said you solved the equation for $c_1, c_2, c_3$.
If, by this, you mean that you found the complete set of all solutions for these variables (as people often mean by "solve"), then you've already proven the linear combination is unique, since there was only one solution.
A: Show that the null space of the matrix 
$$\begin{bmatrix}
2&1&1\\
-1&3&1\\
0&-2&4
\end{bmatrix}
$$
is just the zero vector so that the columns of the matrix are linearly independent. Alternatively depending on how you found your scalars (say using row-reduction and you found a unique solution to the system) then there is nothing to show.
A: you have to solve the system
$$2c_1+c_2+c_3=-4$$
$$-c_1+3c_2+c_3=4$$
$$-2c_2+4c_3=-6$$
by multiplication of the second line by 2 und adding to the first line we get
$$2c_1+c_2+c_3=4$$
$$7c_2+3c_3=4$$
$$-2c_2+4c_3=-6$$
multiplying the second equation by 2 and the third by 7 we get 
$$34c_3=-34$$
$$c_3=-1$$
and we get $$c_2=1$$ and $$c_1=-2$$
