Proof: Matrix determinant gives area of image of unit square under mapping | Matrices | Khan Academy

The goal of this video is to feel good about the connection that we've talked about between the absolute value of the determinant of a two by two matrix and the area of the parallelogram that's defined by the two column vectors of that matrix.

So, for example, I have this column vector right over here, ac. So that's this blue vector. So this distance right over here, it goes a in the x direction. So this distance right over here is a, and then it goes c in the y direction. So this distance right over here is equal to c. And so this distance up here is also equal to a, and this is also equal to c.

So we have this vector, and then we have the bd vector. The bd vector, in the x direction, it goes a distance of b right over there, or if we draw it over here, goes a distance of b. And in the vertical direction, it goes a distance of d. So this right over here is d, and this distance right over here is d.

We can see that the parallelogram created or defined by those two vectors, its area is right over there. Now, let's see if we can connect that to the determinant or the absolute value of the determinant of this matrix. We're just going to assume for the sake of simplicity that a, b, c, and d are positive values, although we can in the future do this same thing where some of them are not positive. But this will hopefully give you a clue of how we can prove it.

Now, how can we figure out the area of this parallelogram? Well, one technique would be to find the area of this larger rectangle right over here and then from that subtract out the parts that are not in the parallelogram. So, let's do that.

So what's the area of this larger rectangle? Let's see, the dimensions here are this length from here to here is a, and then from here to here is b. So this is a plus b on this side, and on this side up here, this part is d, and then this part is c right over here. So it's d plus c.

So, the area of the whole thing is going to be (a + b) times (d + c), which is equal to—we just do the distributive property a few times—it's going to be ad + ac + bd + bc. Now from that, we're going to want to subtract out all of these other parts that are not in the parallelogram.

So, let's do that. So you have this triangle right over here whose area would be ac/2, a times c/2, but you also have this one which has the same area. So if we subtract both of them out, we'd want to subtract out a total of ac. Each of those are ac/2. So to count both of them, let's subtract out an ac.

Then, of course, we could do these two triangles, and the area of each of these triangles is bd/2, b times d/2. But add them together, their combined area is bd. So let's subtract that out, minus bd.

And now what is the area of this right over here? Well, that is b times c, so minus b times c actually. And that's also the area of this right over here, so we have another b times c, so minus 2bc.

So let's see what's going on. If we subtract these out, that takes out that, that takes out that, and if you take bc minus 2bc, we're going to be left with just a negative bc. So all of this is going to be equal to ad. What we have there, bc minus 2bc, is just going to be a negative bc.

Well, this is going to be the determinant of our matrix: ad - bc. So this isn't a proof that for any a, b, c, or d, the absolute value of the determinant is equal to this area, but it shows you the case where you have a positive determinant and all of these values are positive.

So hopefully that feels somewhat satisfying. You could try, if you like, to prove the cases if you don't have a positive determinant or if some combination of these are negative.