I am at the last part of an extended proof I am working on and it's come down to a tricky system.
Given the following:
$af+bg=-2$
$ah+bi=0$
$df+eg=2$
$dh+ei=3$
I want to show that $-2dh+3af-2ah=-6$.
Or even that $-2dh+3af-2ah$ is fixed for any choice of $a,b,d,e,f,g,h,i$.
I've tested numbers and I know that this is the case but I haven't been able to show this formally. Much obliged!
Your system is equivalent to the matrix equation:
$$ (I)\qquad \begin{bmatrix}a&b\\ d&e\end{bmatrix} \cdot \begin{bmatrix}f&h\\ g&i\end{bmatrix} = \begin{bmatrix}-2&0\\ 2&3\end{bmatrix} $$ And you are interested in $-2dh+3af-2ah$ which can be written as $$ (II)\qquad \begin{bmatrix}a&b&d&e\end{bmatrix} \begin{bmatrix}3&0&-2&0\\0&0&0&0\\0&0&-2&0\\0&0&0&0\end{bmatrix} \begin{bmatrix}f\\g\\h\\i \end{bmatrix} $$
Now vectorizing $(I)$ yields
$$ (I')\qquad \begin{bmatrix}a&b&0&0\\d&e&0&0\\0&0&a&b\\0&0&d&e\end{bmatrix} \begin{bmatrix}f\\g\\h\\i \end{bmatrix} = \begin{bmatrix}-2\\2\\0\\3 \end{bmatrix} $$
The idea is now to choose $\begin{bmatrix}x&y&z&w\end{bmatrix}$ such that
$$ \begin{bmatrix}x&y&z&w\end{bmatrix} \begin{bmatrix}a&b&0&0\\d&e&0&0\\0&0&a&b\\0&0&d&e\end{bmatrix} = \begin{bmatrix}a&b&d&e\end{bmatrix} \begin{bmatrix}3&0&-2&0\\0&0&0&0\\0&0&-2&0\\0&0&0&0\end{bmatrix} $$
Or equivalently
$$ \begin{bmatrix}a&d&0&0\\b&e&0&0\\0&0&a&d\\0&0&b&e\end{bmatrix} \begin{bmatrix}x\\y\\z\\w\end{bmatrix} = \begin{bmatrix}3a\\0\\-2a-2d\\0\end{bmatrix} $$
Which is uniquely solveable if $\det\begin{bmatrix}a&b\\ d&e\end{bmatrix} \neq 0$, after which you only need to multiply $(I')$ by $\begin{bmatrix}x&y&z&w\end{bmatrix}$
EDIT: If my calculation is correct,
$$ \begin{bmatrix}x\\y\\z\\w\end{bmatrix} = \begin{bmatrix}3+3\frac{bd}{D}\\-3\frac{ab}{D}\\-2\frac{a+d}{a}(1+\frac{bd}{D})\\\frac{2(a+d)b}{D}\end{bmatrix} \qquad D = \det\begin{bmatrix}a&b\\ d&e\end{bmatrix} $$
And then
$$ \begin{bmatrix}x&y&z&w\end{bmatrix} \begin{bmatrix}-2\\2\\0\\3 \end{bmatrix} = -6-6\frac{bd}{D} - 6\frac{ab}{D} + 6\frac{(a+d)b}{D} = -6 $$
Q.E.D.