Convert tricky system of equations into coefficient matrix I have the following system of equations (taken from this wiki page https://en.wikipedia.org/wiki/YUV#SDTV_with_BT.601), and I am trying to understand how they convert this:
$W_R = 0.299$
$W_G = 1 - W_R - W_B$
$W_B = 0.114$
$U_\max = 0.436$
$V_\max = 0.615$
$ $
$Y' = W_RR + W_GG + W_BB$
$U = U_\max \frac{B-Y'}{1-W_B}$
$V = V_\max \frac{R-Y'}{1-W_R}$
Into the matrix:
$$
\left[
\begin{matrix}
Y'\\
U \\
V \\
\end{matrix}
\right] =
\left[
\begin{matrix}
0.299& 0.587 & 0.114 \\
-0.14713 & -0.28886 & 0.436 \\
0.615 & -0.51499 & -0.10001 \\
\end{matrix}
\right]
\left[
\begin{matrix}
R\\
G \\
B \\
\end{matrix}
\right]
$$
I can see how the $Y'$ row is constructed by simply moving out the coefficients, but no amount of simplification that I am able to understand  has helped me in re-arranging the $U$ and $V$ equations to similar effect.
I tried breaking the $Y'$ constant in the $U$ equation back into its component pieces, and then simplifying (through wolfram alpha) gave me:
$U_\max \frac{B - (W_RR + W_GG + W_BB)}{1-W_B}$
Which I re-arranged to the following:
$(-RU_\max W_R) + (-GU_\max W_G) + (-BU_\max W_B) + (BU_\max) + (-U_\max W_B)$
Which allowed be to remove the $R$ and $G$ components, but I don't know how to then remove $B$. I'm sure I'm not doing things correctly.
Can anyone offer any assistance?
Thanks
 A: You have:
$$U = U_\max \frac{B-Y'}{1-W_B} = 0.436 \dfrac{B - Y'}{1 - 0.114} = 0.492099(B -(0.114 B+0.587 G+0.299 R))$$
This simplifies to (the second row):
$$0.492099(0.886 B-0.587 G-0.299 R) =  0.436 B-0.288862 G-0.147138 R$$
Rearranging the last expression to be in the matrix order:
$$-0.147138 R -0.288862 G + 0.436 B$$
Seeing this, can you do the third row?
A: It's not as complicated as you might think. The first row should be easy: $$Y' = 0.299R+0.587G+0.114B$$
For the second row, we know the expression for $U$:
$$\begin{align} U &= \frac{U_{\text{max}}}{1-W_B}B\,-\frac{U_{\text{max}}}{1-W_B}Y' \\
&= -\frac{U_{\text{max}}}{1-W_B}(0.299R)-\frac{U_{\text{max}}}{1-W_B}(0.587G)-\frac{U_{\text{max}}}{1-W_B}(0.114-1)B
\end{align}$$
where we have substituted $Y'$ and reorganized into the components $R,G,B$. Do note that $\frac{U_{\text{max}}}{1-W_B}=0.492$ is just a number.
For the third row, the expression for $V$:
$$\begin{align} V &= \frac{V_{\text{max}}}{1-W_R}R\,-\frac{V_{\text{max}}}{1-W_R}Y' \\ 
&= \frac{V_{\text{max}}}{1-W_R}(1-0.299)R - \frac{V_{\text{max}}}{1-W_R}(0.587G) -\frac{V_{\text{max}}}{1-W_R}(0.114B)
\end{align}$$
where once again, the expression for $Y'$ was substituted, and $\frac{V_{\text{max}}}{1-W_R}=0.877$ can be used to evaluate each component. 
Finally, we can compose the three equations into a simple matrix equation, which is what you have:
$$
\left[
\begin{matrix}
Y'\\
U \\
V \\
\end{matrix}
\right] =
\left[
\begin{matrix}
0.299& 0.587 & 0.114 \\
-0.14713 & -0.28886 & 0.436 \\
0.615 & -0.51499 & -0.10001 \\
\end{matrix}
\right]
\left[
\begin{matrix}
R\\
G \\
B \\
\end{matrix}
\right]$$
Let me know if there is any step that needs clarification.
A: With a little help from the above posters (and a quick maths lesson from @quick7silver ), I realised where I was going wrong. My initial problem wasn't quite as I had described, as I also wanted my result matrix to incorporate the original variables (as I was transcribing the formula into a computer program), which meant I couldn't perform the initial variable substitution which greatly simplified the problem for the other answerers.
My final working for $U$ was as follows:
$U = U_\max \frac{B-(W_RR+W_GG+W_BB)}{1-W_B}$
$\phantom{U} = U_\max \frac{B+(-W_RR-W_GG-W_BB)}{1-W_B}$
$\phantom{U} = U_\max \frac{-W_RR - W_GG + (1-W_B)B}{1-W_B}$
$\phantom{U} = \frac{U_\max}{1-W_B} -W_RR-W_GG+(1-W_B)B$
$\phantom{U} = \frac{U_\max.-W_RR}{1-W_B} - \frac{U_\max.W_GG}{1-W_B} + \frac{U_\max.(1-W_B)B}{1-W_B}$
$\phantom{U} = -U_\max\frac{W_RR}{1-W_B} - U_\max\frac{W_GG}{1-W_B} + U_\max B$
Which when incorporated into the final matrix:
$$
\left[
\begin{matrix}
W_R & W_G & W_B \\
\frac{-U_\max W_R}{1-W_B} & \frac{-U_\max W_G}{1-W_B} & U_\max \\
V_\max & \frac{-V_\max W_G}{1-W_R} & \frac{-V_\max W_B}{1-W_R} \\
\end{matrix}
\right]
$$
Thanks a lot for all your help!
