Why am I getting inconsistent results measuring the area of this skew triangle? Consider the triangle $(0,0),(4,0),(4,3)$. This has an area of 6 by the .5bh rule. Consider the following transformation 
\begin{equation}
\begin{bmatrix} 
-2 & -1 \\
3 & 5 
\end{bmatrix} 
\begin{bmatrix} 
0 & 4 & 4\\
0 & 0 & 3
\end{bmatrix} 
=
\begin{bmatrix} 
0 & -8 & -11\\
0 & 12 & 27
\end{bmatrix} 
\end{equation} 
Thus, we have a new triangle $(0,0) (-8,12) (-11,27)$. 
Taking the determinant of the linear transformation yields $-7$ and, if we take the absolute value of this and multiply by $6$, the area of the preimage, we get $42$.
Yet .5bh yields a different result. If you draw the triangle, you get a skew triangle. You can divide this into two triangles and 
$.5(3)(15) + .5(3)(12) = 40.5$ 
My Question: 
Why am I getting inconsistent results from these two methods?
 A: The area of the preimage triangle is $6$ (calculated from $\frac 12bh$ formula).  The area of the image triangle is $42$ (calculated from $\frac 12bh$ formula).  The determinant of your transformation matrix is $-7$.  So your matrix transformed an area of $6$ to an area of $|6\cdot -7| = 42$.
So what's the problem here?
NOTE: If you're having problems calculating the area of the triangle from the points, just use this calculator.

Edit: Here's how I found it using the area formula:
The length of the side from $(0,0)$ to $(-11,27)$ is clearly just $\sqrt{11^2+27^2}$.  That is the base.  The height is then the distance from the line containing those two points to the point $(-8,12)$.  Using the distance formula we can see that that is $$d(27x+11y=0,(-8,12))= \frac{|(27)(-8)+(11)(12)|}{\sqrt{27^2+11^2}}$$
Then $$\require{cancel}\frac 12bh = \frac 12\cancel{\sqrt{11^2+27^2}}\frac{|(27)(-8)+(11)(12)|}{\cancel{\sqrt{27^2+11^2}}} = \color{red}{42}$$
A: If you cut the triangle vertically at the line $x=-8,$ you need to find the length of the vertical "base" which goes from the point $(-8,12)$ to the point on the line joining $(0,0)$ to $(-11,27),$ which has equation $y=-27x/11.$ This means the change in $y$ coordinates, which is the length of the base when cut this way, is $(-27/11)*(-8)-12=84/11.$ Then the two triangle areas are $(1/2)(84/11)*3=126/11$ for the left triangle and $(1/2)(84/11)*8=336/11,$ for a combined area of $42.$
The triangle could also be naturally cut using a horizontal slice, but again the length of the "base" is fractional, I think it was $34/9.4
A: Coffeemath's answer explains your error. Here's an easy way to find the area of the skew triangle: the magnitude of the cross product of two vectors is the area of the parallellogram they span, so the area of your transformed triangle is $\frac12||\left<-8,12,0\right>\times \left<-11,27,0\right>||=\frac12{\tiny\left|Det\pmatrix{-8 & -11 \cr 12 & 27}\right|}=42=6\cdot 7$, as expected.
