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? 
 A: \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:
$$
\begin{bmatrix}
    1 & -3 & h \\
    0 & i & -5i \\
    0 & 0 & 7+2h 
    \end{bmatrix}\xrightarrow{R_2\leftarrow -iR_2} 
\begin{bmatrix}
1 & -3 & h \\
0 & 1 & -5 \\
0 & 0 & 7+2h 
    \end{bmatrix}\xrightarrow{R_1\leftarrow R_1+3R_2} 
\begin{bmatrix}
1 & 0 & h-15 \\
0 & 1 & -5 \\
0 & 0 & 7+2h 
    \end{bmatrix}.$$
A: If  the vector $y$ lies in the plane generated by the vectors $v$ and $w$ then $y$ can be expressed as a linear combination of $v$ and $w$ i.e. $y=c_1v+c_2w$.
Then we must have:
$$
    \begin{vmatrix}
    1 & -3 & h \\
    0 & i & -5i \\
    -2 & 8 & -3 \\
    \end{vmatrix}=0
$$ which gives the value of $h$.
A: You want to write $y = av + bw$ for some scalars $a$ and $b$, and find the value of $h$ that makes this possible. Note that since the other two terms of the vector don't involve $h$, you can actually work out what the linear combination $av + bw = y$ should be without using $h$.
Since the middle entry of $v$ is $0$, you'll definitely need to take $-5$ lots of $w$, so $b = -5$. Then, looking at the bottom entry, you'll get $-2a + (-5)(8) = -3$, and so $a=  -37/2$. Now check the top entry: $-37/2 + (-5)(-3) = h$, so $h = -7/2$.
