How to show this set is a basis? I want to show that a basis for $\mathbb{R}^4$ is: $ \begin{pmatrix} 1 \\ 2 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 2 \\ 2 \\ 1 \\ -2 \end{pmatrix}, \begin{pmatrix} 1 \\ 4 \\ 0 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}$. 
I know that I need to show linear independence and that these span $\mathbb{R}^4$. However, this was a past exam question and I am not really sure that row reduction/solving systems (for only 4 marks) is what they want. Is there an easier/alternative way to show that they span the space and are LI? 
Thank you :)

While I suggest to show this using the determinant, here is one easy and quick way to do it without determinants:
As we have 4 vectors, we only need to show that they are linearly independent. Let $a,b,c,d\in \mathbb R$ with
$$a\cdot\begin{pmatrix} 1 \\ 2 \\ 0 \\ 0 \end{pmatrix}+b\cdot\begin{pmatrix} 2 \\ 2 \\ 1 \\ -2 \end{pmatrix} + c\cdot \begin{pmatrix} 1 \\ 4 \\ 0 \\ 0 \end{pmatrix}+d\cdot\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}.$$
It follows immediately that $b=0$ by looking at the third row of each vector, so this leaves us with
$$a\cdot\begin{pmatrix} 1 \\ 2 \\ 0 \\ 0 \end{pmatrix} + c\cdot \begin{pmatrix} 1 \\ 4 \\ 0 \\ 0 \end{pmatrix}+d\cdot\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}.$$
The same argument yields $d=0$ by looking at the fourth row and from
$$a\cdot\begin{pmatrix} 1 \\ 2 \\ 0 \\ 0 \end{pmatrix} + c\cdot \begin{pmatrix} 1 \\ 4 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$
we can easily conclude $a=c=0$ (otherwise we would get a contradiction with $a=-c$ and $a=-2c$). 
