Finding bases for a vector space 
How would one go about solving this? We know a basis is both linearly independent and spans $V$. My first thought was to represent {${u, v, w}$} as the matrices \begin{bmatrix}1\\0\\0\end{bmatrix} \begin{bmatrix}0\\1\\0\end{bmatrix}\begin{bmatrix}0\\0\\1\end{bmatrix} and inputting the other elements accordingly, i.e. {${-u+v-w}$} would be \begin{bmatrix}-1\\1\\-1\end{bmatrix}
Then solve these top find if they're linearly independent. Am I on the right track?
 A: Eliminate as many of the options as you can by inspection before resorting to grinding through a calculation. The first and fourth options don’t have three elements, so can’t be bases for $V$. The third choice can also be eliminated because $-\mathbf u+\mathbf v-\mathbf w=-(\mathbf u-\mathbf v+\mathbf w)$, so it’s not linearly independent. That leaves only the second choice.  
If you’re being extra-careful you might verify that the remaining set is linearly independent by constructing a matrix as you described and either computing its determinant or row-reducing, but if you’re guaranteed that the answer must be one of the choices, answer (ii) and move on to the next question.
A: The first and the fourth sets are not bases, since they don't have exactly $3$ elements.
For the other $2$ sets, yes, your method works. I would determine whether they are linearly independent or not by computing the determinant of the matrix whose columns are those vectors that you mentioned.
A: As already noted case (i) and (iv) can be immediately excluded. For (ii) and (iii) we can perform RREF on the coefficents matrices 
$$M_{(ii)}=\begin{bmatrix}-\frac12&2&-1\\2&4&1\\-1&1&\frac32 \end{bmatrix}\quad M_{(iii)}=\begin{bmatrix}-1&1&-1\\1&-1&1\\10&-2&2 \end{bmatrix}$$
and check whether the rank is $3$ (note that the sum of the first $2$ rows for $M_{(iii)}$ leads to the zero vector).
