Area calculation for transformed rectangle seems too small. I came upon this problem in Khan Academy precalculus, in the unit on matrices.  In the video, he shows this image of a rectangle and asks you to determine its area after being transformed by a matrix.  The determinant of the transformation, multiplied times the original area, should equal the transformed area.

I followed the instructions and got his answer.
$Area = 7 * 5 = 35$
$det=(5*8)-(9*4) = 4$
$Area' = 35 * 4 = 140$
But I didn't believe the answer, because the matrix seems to transform the unit vectors by a large factor.  So I reframed the problem to imagine the rectangle beginning with its corner on the origin.

I defined a new vector for this rectangle, $A$.  If I then apply the same transformation matrix ...
$$ \begin{pmatrix}5 & 9\\\ 4 & 8\end{pmatrix}\begin{pmatrix}7\\\ 5\end{pmatrix}$$
$$7\begin{pmatrix}5 \\\ 4 \end{pmatrix}+5\begin{pmatrix}9 \\\ 8 \end{pmatrix}=\begin{pmatrix}35 \\\ 28 \end{pmatrix}+\begin{pmatrix}45 \\\ 40 \end{pmatrix}=\begin{pmatrix}80 \\\ 68 \end{pmatrix}$$
I get a rectangle of the dimensions $80*68=5440$
I'm certain that Khan is correct and I am wrong, but I don't understand why.  How can the area of the transformed image be so small when I think it should be very large?
 A: Lots of misconceptions here. First, you should be multiplying the matrix by the vector on the right:
$$\left[\begin{matrix}5 & 9 \\ 4 & 8\end{matrix}\right]\left[\begin{matrix}a\\b\end{matrix}\right].$$
Next, you are assuming the rectangle maps to a rectangle, so that you can look just at the diagonal of the rectangle. This linear mapping does not map the original rectangle to a rectangle. It maps it to a parallelogram whose edges are
$$\left[\begin{matrix}5 & 9 \\ 4 & 8\end{matrix}\right]\left[\begin{matrix}7\\0\end{matrix}\right] = \left[\begin{matrix} 35 \\ 28\end{matrix}\right] \quad\text{and}\quad
\left[\begin{matrix}5 & 9 \\ 4 & 8\end{matrix}\right]\left[\begin{matrix}0\\5\end{matrix}\right] = \left[\begin{matrix} 45 \\ 40\end{matrix}\right].$$
The (signed) area of this parallelogram is given by the determinant
$$\left|\begin{matrix} 35 & 45 \\ 28 & 40 \end{matrix}\right| = 35\cdot 40-28\cdot 45 = 140.$$
A: Ah, I think I found it.  I thought of deleting the question but I'll post an answer now anyways in case anyone else is as thick-skulled as me.  My calculation was correct for finding that the $(7,5)$ vertex of the rectangle would be at $(80,68)$.  But I made my area claim based on the assumption that the transformed image would be a rectangle.  It will in fact be a parallelogram, and in this case a very skinny one!  If I apply the same transformation to all four vertices, I get a shape with vertices at $(80,68)$, $(35,28)$, $(0,0)$, and $(45,40)$.  It looks about like this, and the area is indeed $140$.

