Reduced row echelon with imaginary numbers

Question

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?

Answer 1

\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: $$ \begin{bmatrix} 1 & -3 & h \\ 0 & i & -5i \\ 0 & 0 & 7+2h \end{bmatrix}\xrightarrow{R_2\leftarrow -iR_2} \begin{bmatrix} 1 & -3 & h \\ 0 & 1 & -5 \\ 0 & 0 & 7+2h \end{bmatrix}\xrightarrow{R_1\leftarrow R_1+3R_2} \begin{bmatrix} 1 & 0 & h-15 \\ 0 & 1 & -5 \\ 0 & 0 & 7+2h \end{bmatrix}.$$

Answer 2

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 $h$.

Answer 3

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 $h = -7/2$.