Determine whether the list is linearly independent. $S=\{(1+i,1-i,1),(i,0,1),(-2,-1+i,0)\}$

I am having trouble solving the coefficient matrix, where column 1 is vector 1, column 2 is vector 2, and column 3 is vector 3 and set that equal to 0. 

If you can, provide a step by step.
You can check for linear independence (over either of $\Bbb R$ or $\Bbb C$) by performing row operations on the matrix whose rows are these vectors.

If you are testing for independence over $\Bbb R$, you can only multiply rows with real scalars, and if you are testing over $\Bbb C$ you can then use complex scalars. If you get to row-echelon form and there are no zero rows, then they are independent: otherwise, they're dependent. 

Strategically you should test for $\Bbb C$ linear dependence first because if you find they are $\Bbb C$ independent, then they have to be $\Bbb R$ independent as well. (See why?) But anyhow, you've said in a comment you wanted to check for $\Bbb C$ dependence anyway, so we'll definitely focus on that.

Here is a start checking the matrix over $\Bbb C$:

$$\begin{bmatrix}1+i&amp;1-i&amp;1\\
i&amp;0&amp;1\\
-2&amp;-1+i&amp;0\end{bmatrix}\rightarrow
\begin{bmatrix}
1&amp;1-i&amp;0\\
i&amp;0&amp;1\\
-2&amp;-1+i&amp;0\end{bmatrix}\rightarrow
\begin{bmatrix}
1&amp;1-i&amp;0\\
i&amp;0&amp;1\\
0&amp;1-i&amp;0\end{bmatrix}\rightarrow
\begin{bmatrix}
0&amp;1-i&amp;i\\
i&amp;0&amp;1\\
0&amp;1-i&amp;0\end{bmatrix}$$

The first operation was to replace $R_1$ with $R_1+R_2$, then the second operation was to replace $R_3$ with $2R_1+R_3$, and in the third $R_1$ got replaced with $R_1+iR_2$. In just a few more steps, you should have your answer.

I will hand it off to you at this point! Let me know in comments if you get stuck.
Start by notating
$$a=(1+i,1-i,1)\\b=(i,0,1)\\c=(-2,-1+i,0)$$
Now we need to solve the equation $z_1 a+z_2 b+z_3 c = 0$ with non-trivial solution. put $z_n=a_n+ib_n$ and get:
$$(a_1+ib_1)*(1+i,1-i,1)+(a_2+ib_2)*(i,0,1)+(a_3+ib_3)*(-2,-1+i,0)=(0,0,0)\\
(a_1-b_1+i(a_1+b_1),a_1+b_1-i(a_1+b_1),a_1+ib_1)+(-b_2+ia_2,0,a_2+ib_2)+(-2a_3-2ib_3,-a_3-b_3+i(a_3-b_3),0)=(0,0,0)$$
Which gives six equation (each element in the vector must be equal in both imagenary and real parts):
$$
a_1-b_1-b_2-2a_3=0\\
a_1+b_1+a_2-2b_3=0\\
a_1+b_1-a_3-b_3=0\\
-a_1-b_1+a_3-b_3=0\\
a_1+a_2=0\\
b1+b2=0
$$
The vector are independent iff the determinant is not 0:
\begin{vmatrix}
  1 &amp; -1 &amp; 0 &amp; -1 &amp; -2 &amp; 0  \\
  1 &amp;  1 &amp; 1 &amp;  0 &amp;  0 &amp; -2 \\
  1 &amp;  1 &amp; 0 &amp;  0 &amp; -1 &amp; -1 \\
 -1 &amp; -1 &amp; 0 &amp;  0 &amp;  1 &amp; -1 \\
  1 &amp;  0 &amp; 1 &amp;  0 &amp;  0 &amp; 0  \\
  0 &amp;  1 &amp; 0 &amp;  1 &amp;  0 &amp; 0  \\
\end{vmatrix}

If the det is 0, there is non-trivial solution and the vector is depended.

The determinant can be calculated. Enjoy!