On Friday (13 January), Brenda asked a good question: why is the determinant of a matrix, ad−bc, the measure of how much the transformation represented by that matrix expands areas?
Here’s a proof.
Since matrices represent linear transformations, if we find out how much the transformation expands a standard unit area, then we know how much it expands all areas.
Here is a standard unit area, a 1×1 square, with corners at (0,0), (1,0), (0,1), and (1,1).
The matrix transforms the unit square into a parallelogram with corners at (0,0), (a,c), (a+b,c+d), and (b,d).
We need to find the area of that parallelogram.
That area depends partly on how big a, b, c, and d are, but also partly on the angles in the parallelogram – on how wide the angle is between the line from (0,0) to (a,c) and the line from (0,0) to (b,d). If (0,0), (a,c), and (b,d) are exactly in line, then the area of the image is zero however big a, b, c, and d are.
The slope of the line from (0,0) to (a,c) is c/a
The slope of the line from (0,0) to (b,d) is d/b
The difference in the slopes, which is a measure of the angle between them, is d/b−c/a
But d/b−c/a is (ad−bc)/ab
There it is – ad−bc! The minus comes from the fact that the determinant has to include some measure of how much the transformation represented by the matrix leaves the axes skew to each other, and how much it narrows down the area by bringing them closer to each other.
To see this exactly, look at the diagram of the parallelogram.
If we add the bottom red-shaded area to the parallelogram, and subtract the top red-shaded area from it;
and add the left-hand red-shaded area, and subtract the right-hand green-shaded area
then we transform the parallelogram into the rectangle with sides a and d and area ad.
Except that adding and subtracting is not quite right.
To get to ad, we have subtracted the area marked 2, though it was never in the parallelogram
We have subtracted the area marked 1, though it was never in the parallelogram
We have subtracted the area 3 twice, although it was in the parallelogram only once
In other words, we have “over-subtracted” by the total of the areas 1, 2, and 3, which is the same as the area of the top-right rectangle, bc
Area of the parallelogram = expansion factor for areas = ad−bc ▇