Finding out vectors that screw up linearly independence when given a set I want to

Find the vector space spanned by
$A =$ {$(1,1,0,1),(1,2,-1,1),(3,4,-1,3),(-1,-3,-2,-1)$}

By definition it's all the linear combinations I can make with those 4 vectors, However I was told that I could only do that in the case they are linearly independent. if not I must choose only the linearly independent ones and make the linear combination with them.
So I went to check if they are linearly independent, They are not. However How do I know which one is causing the trouble?. I tried making linear combinations between them but It didn't take me anywhere. So If anyone could give me any hint it would be much appreciated. Thanks for reading.
 A: Rename your vectors $a,b,c,d$. Most commonly, to find out if they are LI or not you calculate the determinant of the matrix $M=\begin{pmatrix}a\\b\\c\\d\end{pmatrix}$, that is, the rows of $M$ are your vectors (maybe you consider columns instead of rows, but it doesn't really matter). If the determinant of $M$ is $0$ then your vectors are Linearly dependent. This is the same as $M$ being invertible, and there is another way to check if $M$ is invertible: Row reduction.
When we do row reduction, we change the rows of $M$ by linear combinations of the row and the other ones. If we get zero, this means that the row we start with is a linear combination of the other ones.
Let's see in your example: \begin{align*}M&=\begin{pmatrix}1&1&0&1\\1&2&-1&1\\3&4&-1&3\\-1&-3&-2&-1\end{pmatrix}\overset{\substack{R_2-R_1\\R_3-3R_1\\R_4+R_1}}{\longrightarrow}\begin{pmatrix}1&1&0&1\\0&1&-1&0\\0&1&-1&0\\0&-2&-2&0\end{pmatrix}
\end{align*}
And now changing the third row by $R_3-R_2$ we get zero. This means the third row, which corresponds to the third vector, is the "problematic" one. In fact, we kept track of the operations: The second row of the last matrix is $b-a$, and the third row is $c-3a$
$$b-a=c-3a,\qquad\text{ that is }\qquad c=2a+b$$
You can then verify that $a,b,d$ are LI.
(Remark: The choice to say that $c$ is the "problematic" one was rather arbitrary. We only choose it because we put $c$ below $a$ and $b$ in the matrix $M$. Actually, the equation $2a+b-c=0$ implies that we can discard either $a$,$b$ or $c$, and in fact the sets $\{a,b,d\}$, $\{a,c,d\}$ and $\{b,c,d\}$ are all LI.)
