 I want to say a little bit of why people care about determinants. We'll see some more applications in future lectures as well, but there's one short one we can mention right now. Determinants can be used to calculate area. If we have two vectors, U and V inside the plane, associated to any two vectors, we can form a parallelogram. This is what the parallelogram rule was about all before. With the parallelogram rule, if you put the two vectors, U and V together so that their tails come together, the diagonal of this parallelogram was the sum of the two vectors, and then the other diagonal is the difference of the vectors. This parallelogram, this geometric object is significantly related to the algebraic properties of these vectors, additions and subtraction. What I want to talk about now is actually the area of this parallelogram. It turns out that if we form the matrix A so that its first column is U and its second column is V, then the determinant of this matrix A will equal the area of the parallelogram. Now, unfortunately though, it does depend on who comes first, U or V, there's an orientation issue. If you choose them in the wrong order, you actually could get a negative number, which doesn't really make sense for area. But as you often see in calculus and such when you do integration, if you get a negative area that's easy to fix, just take the absolute value of that thing. The area of this parallelogram is equal to the absolute value of the determinant. This is also true with respect to three by three matrices. Associated to the three columns of the matrix, there is a parallel pipe head, which is like the three-dimensional analog of the parallelogram. The idea is you stack parallelograms. Green's not going to work there. You can stack parallelograms on top of each other and make some type of three-dimensional object. This is probably the grossest picture anyone's ever drawn in their lives. But you try to get this three-dimensional parallelogram. It's kind of like imagine if you have a stack of cards and you were to push it in a couple different directions. How does the card slant? You have a parallel pipe head. A rectangular prism is a special type of parallel pipe head when the angles are all right angles. Anyways, the volume of said parallel pipe head will equal the determinant of the associated matrix. The columns of the matrix are the vectors that span the parallel pipe head. But again, because of orientation reasons, you might have to take absolute values. And if we take higher-dimensional analogs of this, they hold as well. So if you take a four-by-four matrix, the determinant of that matrix up to sine change is equal to the hypervolume of the hyper-parallel pipe head. Sort of like the four-dimensional analog. You can do five dimensions, six dimensions, et cetera. The reason I want to mention this is that in calculus, we actually use this principle a lot. When we do change of variables in multivariable calculus, we have to calculate the Jacobian, which is the determinant of a change of... It was the determinant of the derivative matrix, the partial derivative matrix there. And so why does the determinant show up in these integrals? Well, because the determinant is calculating area, right? And, well, integrals calculate area. So it kind of makes sense that in order to calculate areas, you're gonna use areas. And the determinant is keeping track of the change of area as you go from one variable to another or change the volume or higher-dimensional analogs. And so determinants is how you calculate linear area. Integrals is how you calculate non-linear area. And integrals, as you're taking limits of linear approximations, it kind of suggests why the linear area formula, aka determinants, is showing up in those integrals. So I want to do an example of this thing. So let's just take a two-dimensional example, take a parallelogram. And so let's take the points negative two, negative two, zero, three, four, negative one, and six, four. How do we calculate the area of that parallelogram using determinants? So I'm gonna graph it right here. I'm gonna zoom out a little bit so we can see it a little, just a tiny bit better. And so let's do those points. So there is the point two, two, negative two, negative two right there. There is the point zero, three right here. There's the point four, negative one right there. And then there's the point six comma four, which is this one right there. In which case if we connect the dots, you can see that we have in fact formed a parallelogram. Okay, I mean it's not perfectly drawn but that's the right idea. Now in order to use the rule we just did a moment ago, you have to have a parallelogram that goes through the origin. So we actually want a point right here. And so what we can do is we can just perform an affine transformation for which we're just gonna translate everything by a factor. So if we have this point two, negative two, negative two, what we can do is we can actually just translate all the points so that negative two, negative two goes to the origin, right? And so basically what we're gonna do is we're just gonna add, we're just gonna do the translation map, translate by this factor two comma two. We're just gonna add two to all the coordinates. And so if we do that we'll take two, two here, two, two here, two, two here. Where do all my points go? Well, you're gonna get the point zero, zero. This point right here, which was, I'm gonna label these things now, you get zero, three. It's going to, if we add points to it, it's gonna become two comma five. I'm sorry, that's not that point. JK about that one. This was the point four, negative one. We get four, negative one. If we add two, two to that, we're gonna get six, one, which is this point right here, six, one. If we do three, zero, that's this one right here, three, zero. When you translate that one, that becomes two, five, which is this point right here, two comma five. And then lastly, if we take the point six, four, which is right there, when we translate that, that'll become the point eight comma six. And so this forms a parallelogram now. I'll try that again. I'll connect the dots one more time. We get a parallelogram. And this parallelogram will be congruent to the original parallelogram. And what we're interested in is these two vectors right here. So this is the vector U and this is the vector V. So to find the area of this thing, we have to calculate the determinant of the first column of the vector U since it goes from zero, zero to two, five, the column will just be two, five, right? And then the second one V, since it goes from zero, zero to six, one, the second column will look like six, one, like so. And so if we calculate the determinant, we get two minus 30, which gives us a negative 28, but as area should be positive, we really needed to take absolute values the whole way. And so we're gonna get 28 square units of area for this parallelogram right here. Now, admittedly in order to calculate the area, we don't have to draw the picture. I mostly wanted to draw the picture to kind of illustrate geometrically what we're doing. But this determinant is what we needed to do to calculate the area of this parallelogram. And that brings us to the end of our lecture today that concludes section 5.1 about introduction to determinants. We'll talk some more about determinants, of course, throughout this chapter. We'll talk about these more next time. But thanks for having you here. If you have any questions, feel free to post them in the comments below. I'll be glad to answer them as soon as I can. If you like this video, please click the like button, subscribe. If you wanna see more videos like this, or you wanna get updates about future videos and other things related to this channel. And best of luck in your journey through linear algebra. I hope to see you next time. Bye, everyone.