Reduced row echelon with imaginary numbers Working on the following problem:
Let $v = \begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix}
w = \begin{bmatrix} -3 \\ i \\ 8 \\ \end{bmatrix}
y = \begin{bmatrix} h \\ -5i \\ -3 \\ \end{bmatrix}$
For what values of $h$ is the vector $y$ in the plane generated by the vectors $v$ and $w$?
My work so far: I know that for $y$ to be in the plane generated by $v$ and $w$, $y$ has to be a linear combination of $v$ and $w$. This means that I need to find the solution of the augmented matrix:
$$
    \begin{bmatrix}
    1 & -3 & h \\
    0 & i & -5i \\
    -2 & 8 & -3 \\
    \end{bmatrix}
$$
I assume that I need to find the reduced echelon form of this matrix. However, I don't know how to deal with the imaginary numbers when doing the row reduction. Could anyone give me some pointers? 

\begin{align}
\begin{bmatrix}
    1 & -3 & h \\
    0 & i & -5i \\
    -2 & 8 & -3 
    \end{bmatrix}\xrightarrow{R_3\leftarrow R_3+2R_1} 
\begin{bmatrix}
    1 & -3 & h \\
    0 & i & -5i \\
    0 & 2 & -3+2h 
    \end{bmatrix}\xrightarrow{R_3\leftarrow R_3+2iR_2} 
\begin{bmatrix}
    1 & -3 & h \\
    0 & i & -5i \\
    0 & 0 & 7+2h 
    \end{bmatrix}
\end{align}
so the condition is  $\;h=-\dfrac72$.
If you really need the reduced row echelon form, just proceed:
$$
\begin{bmatrix}
    1 & -3 & h \\
    0 & i & -5i \\
    0 & 0 & 7+2h 
    \end{bmatrix}\xrightarrow{R_2\leftarrow -iR_2} 
\begin{bmatrix}
1 & -3 & h \\
0 & 1 & -5 \\
0 & 0 & 7+2h 
    \end{bmatrix}\xrightarrow{R_1\leftarrow R_1+3R_2} 
\begin{bmatrix}
1 & 0 & h-15 \\
0 & 1 & -5 \\
0 & 0 & 7+2h 
    \end{bmatrix}.$$
