We're going to take a look at how determinants can actually be used to compute the area of
a parallelogram.
So here is the parallelogram that we're going to work.
It's just in 2D.
It has vertices at 0, 0, 7, 2, 5, 9, and 12, 11.
So these are the vertices of the parallelogram.
And we are going to sketch this parallelogram, and we're also going to compute its area.
So sketching, it's not too bad.
This is just two-dimensional.
I know that I'm going to have a point or a vertex at 0, 0, so here's the point 0, 0.
I also have a vertex at the point 7, 2, which is this point right here.
There's a vertex at, 5, 9, which is that point right there, and there's a vertex at 12, 11,
which is this point right here.
So these are the vertices of my parallelogram.
I can now ahead go and draw lines connecting the vertices to get the sides of the parallelogram.
So here is this two-dimensional parallelogram, and what I'd like to do is I'd like to compute
the surface area of this parallelogram.
Turns out I can do that using determinants pretty easily.
The area of this parallelogram is really determined by just two vectors-- the vector 5, 9 and
the vector 7, 2.
If I actually compute the determinant of the 2-by-2 matrix, whose columns are these vectors,
I will actually get the surface area.
So what's the determinant of this?
It's the determinant of 5 times 10 minus 63, this is the absolute value of the determinant,
mind you.
The absolute value of 10 minus 63 is absolute value of a negative 53, which is 53 square
units.
So this is the area of this parallelogram.
It's not incredibly intuitive as to why that's working and we're not going to go into the
theory why that's working.
We can do some simple computations that are to show that this does indeed work out.
If we weren't going to use determinants, how would you compute the area of this parallelogram?
Well, one way to do it might be as follows-- first is to redraw our parallelogram.
One way to do it might be to think of the area of this kind of orange square, whose
area is pretty easy to compute, it's a rectangular area.
Its area is just 12 times 11, which is 132, but then subtract out the pieces that don't
correspond to the parallelogram.
So for instance, this triangle right here isn't part of that parallelogram.
So I want to take that away.
And this triangle right here, which I'll call T2, that's not part of the parallelogram,
so I would need to take that away.
Triangle 3, triangle 4, and then also these two strips-- strip 1 and strip 2, those aren't
part of the parallelogram.
If I was to compute the total area then take away these four triangles and two strips,
what I'm left with should be the area of the parallelogram.
So the parallelogram area is actually equal to the total area minus the sum of the triangles
minus the sum of the strips.
So if we just do some basic geometry area computations, we should get this answer we
got just a minute ago when we used determinants.
So let's see if that actually works out.
So let's compute the area of triangle 1.
Triangles are easy to compute their area, it's 1/2 base times height.
So 1/2 9 times 5 is 45/2.
Triangle 2, 1/2 base times height is 1/2 times 7 times 2, which is a 7.
Triangle 3, 1/2 base times height is 1/2 times 9 times 5, which is 45/2.
And triangle 4, is 1/2 base times height, which is 1/2 times 7 times 2, which is 7.
If I add all that up, the amount of area that the triangles contribute is 59.
Now let's go ahead do the strips.
Those are even easier to compute, it's just length times height.
And both of them are the same.
They're both 2 by 5, so they each have an area of 10 square units.
Total contribution will be 20.
So now we know what the triangles add up to in terms of area, we know what these strips
add up to in terms of area.
Let's go back to our equation and finish doing a little computation for the parallelogram
area.
It should equal the total area of 132 minus the total area of the triangles, which is
59 minus the area of the strips, which is 20, which is equal to 53.
So doing this kind of more conventional way, we get the exact same answer that we had just
a minute ago when doing the determinant computation to get the area of a parallelogram.
You can make some-- so this was just a two-dimensional parallelogram.
The math kind of goes to higher dimensions as well.
It just gets a lot harder to visualize and draw.
You can compute the three-dimensional volume of a parallelepiped, using determinant computations
and you can continue going to higher dimensions as well.