Improve/extend my attempted intuitive explanation for why terms in determinant calculations have alternating signs The determinant of a shape defined by points $(a,b)$ and $(c,d)$ as labelled in the gif below is 
$\left|\begin{matrix}a&c\\b&d\end{matrix}\right| = ad-bc$
The following process is the result of my trying to intuitively understand why this is, knowing that, for the two dimensional case, the determinant is simply the area.

First calculate the height of the dark blue triangle, denoted $h$. ($\theta$ is the lower left angle of that triangle.)
$\begin{align*}
\theta = \arctan\left(\frac{b}{a}\right)
\end{align*}$
$\begin{align*}
h &= c\tan\theta\\
&=c\frac{b}{a}
\end{align*}$
In the final arrangement, the area of the shape is now clearly given by
$\begin{align*}
Area &= a(d-h)\\
&= a \left(d - \frac{cb}{a}\right)\\
&= ad - bc
\end{align*}$
Was this a reasonable way go about intuitively demonstrating the formula to oneself? Are there any more obvious/simpler ways I overlooked?

How would this (or an alternate intuitive demonstration) be extended to a three dimensional case, where the formula is the following and calculates volume?
$\begin{align*}
|A| &= \left| \begin{matrix} a&b&c\\d&e&f\\g&h&i\end{matrix}\right|\\
&=a(ei-hf) - b(di - gf) + c(dh - ge)
\end{align*}$
I'm particularly interested in why the second term (prefixed $b$) is negative.
Also, why would the $d$ and $h$ terms be negative, if you expanded along the second row: $|A| = -d(bi-hc) + e(ai-gc) - f(ah-gb)$?

In summary, I'd like to know: Why do the signs in the terms of determinant calculations alternate? Explain in a manner similar to my 'intuitive' attempt, if possible.


How would this (or an alternate intuitive demonstration) be extended to a three dimensional case, where the formula is the following and calculates volume?

Yes.  Just as the determinant of the $2\times 2$ matrix is the area of the parallelogram formed with the two column vectors as sides, so too is the determinant of a $3\times 3$ matrix the area of a parallelepiped with formed with the three column vectors as sides.  And analogously for higher dimensional square matrices.
Your animation is an excellent demonstration of how this is so, and a similar construct can be developed for the third dimensional equivalent.
More precisely, the absolute value of the determinants equate to these measures.  The signage is an indication of the direction of the angles between one column vector and the next (widdershins or clockwise chirallity).

I'm particularly interested in why the second term (prefixed b) is negative.

It's an artifact of the order you're taught to extract the numbers. That is, take each of the numbers in the first row in turn, place fingers over its row and column, then multiply that numbers by the determinant of the $2\times 2$ matrix you see, and alternatively add and subtract the resulting terms.
$$\begin{align}
|A| &= \begin{vmatrix} a&b&c\\d&e&f\\g&h&i\end{vmatrix}\\
&=a(ei-hf) - b(di - gf) + c(dh - ge)
\end{align}$$
Instead, imagine the first columns repeat, and multiply each of the three number in the first column by determinant of the two by two matrix just to its lower right, and just add them together.  
Or, alternatively, add up the three top-left-to-bottom-right diagonal products and subtract the next three top-right-to-bottom-left diagonal products.
Well, if that's as transparent as mud, can you see what I mean from the following?
$$\begin{align}
|A| & = \left.\begin{vmatrix} a&b&c\\d&e&f\\g&h&i\end{vmatrix}\color{gray}{\begin{matrix} a&b&c\\d&e&f\\g&h&i\end{matrix}}\right|
\\
&=a(ei-fh) + b(fg - di) + c(dh - eg)
\\
&= aei+bfg+cdh-(afh+bdi+ceg)
\end{align}$$
