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Current time:0:00Total duration:21:37

Determinant and area of a parallelogram

Video transcript

I've got a two-by-two matrix here and let's just say it's entries are a b c and d and it's composed of two column vectors we've done this before let's call this first column v1 and let's call the second column v2 so we can rewrite it here we could say v1 is equal to AC and we kids write that V 2 is equal to B D and these are both members of r2 and just to have a nice visualization in our head let's graph these two let me draw my axes it's my vertical axis that's my horizontal axis and maybe V 1 looks something like this the 1 might look something like that so that is V 1 its horizontal component will be a its vertical coordinate if you view this is maybe a position vector or where you're going to however just how we're drawing it is C and then V 2 let's just say it looks something like this let's say that they're not the same vector so V 2 looks like that V 2 its horizontal coordinate is going to be B and it's vertical coordinate is going to be D now what we're going to concern ourselves with in this video is the parallelogram generated by these two guys what I mean by that is imagine that these two guys are position vectors that are specifying points on a parallelogram and then of course the zero or not of course but the origin is also another point on the parallelogram so what will be the last point on the parallelogram well you can imagine a parallelogram we already have two sides of it so the other two sides have to be parallel so one side will look like that it's parallel to v1 the way I've drawn it and then the other side will look like this that's our parallelogram the parallelogram generated by V 2 and V 1 we're going to concern ourselves with specifically is the area the area of the parallelogram parallelogram generated by V 1 vitu so how do we figure that out so in general if i have just any parallelogram this is kind of a tilted one but if I just have any parallelogram let me just draw any parallelogram right there the area the area is just equal to the base so that could be the base times the height times the height so it's equal to base I'll write capital B since we have a lowercase B their base times height that's what the area of a parallelogram would be now what are the base and the height in this situation so let me write this down the area of our parallelogram is equal to the base times the height and actually well let me just write it here so what is the base here the base here is going to be the length of this right of our vector V so this is our base so this right here is going to be the length of vector v1 the length of this orange vector right here and what's the height of this parallelogram going to be so we could drop a perpendicular here we could drop a perpendicular here and that the length of this line right here is going to be our height is going to be our height so how can we figure out that you know we know what v1 is so we can figure out the base pretty easily but how can we figure out how can we figure out the height well one thing we can do is if we can figure out this guy right here if we can figure out this guy we could use a Pythagorean theorem because the length of this vector plus squared plus the h plus h squared is going to be equal to the length of v2 squared so let's see if we can do that what is this guy what is this green guy right here well if you imagine a line let's imagine some line L so let's say L is a line is a line spanned by v1 spanned by v1 which means you take all of the multiples of v1 and all of the positions that they specify will create a set of points and that is my line l so you take all the multiples of v1 you're going every point along this line along a line that's like that now if we have L defined that way that line right there is a line see it let me do it a little bit better and it goes through v1 and it just keeps going over there what is this green line right there this green line that we're concerned with that's the projection that is the projection onto L of what well we have a perpendicular here you can imagine the light source coming down I don't know if that analogy helps you but it's kind of the shadow with v2 onto that line so it's a projection of v2 of your vector v2 onto L is this green line right there so if we want to figure out H we want to figure out H we can just use the Pythagorean theorem so we can say that the length of H squared or we could just write well I'm just writing H as the length I'm not even specifying as a vector so we could say that H squared a squared where that is the length of this line plus the length of this vector squared and the length of that vector squared is the length of the projection onto L of V to do that I'll do it in a different color so the length of the projection onto L of v2 squared all right we're just doing the Pythagorean theorem this squared plus this squared is going to equal that squared it's going to be equal to it's going to be equal to the length of v2 squared that's just Pythagorean theorem Pythagorean theorem nothing fancy there here so how can we simplify what we want to figure out we want to solve for H and actually just let's just solve for H squared for now because it will keep things a little bit simpler so we can say that H squared H squared is equal to this guy it's equal to the length of my vector v2 squared minus the length of the projection squared so minus I'll do that in purple minus the length of the projection the projection onto L of v2 squared remember this thing is just this thing right here we're just doing the Pythagorean theorem so let's see if we can simplify this or write it in terms that we understand so v2 is the length of a vector squared this is just equal to let me write it this way this is just equal to v2 dot v2 you take a vector u dotted with itself and you get the length of that vector squared we saw that many many videos ago and then what is this guy going to be equal to well the projection I'll do it over here the projection onto L of v2 is going to be equal to v2 dot the spy spanning vector which is v1 so its v2 dot v1 over the spanning vector dotted with itself v1 dot v1 we saw this several videos ago when we learned about projections and this is just a number this is just a number right there and then it's going to be times the spanning vector itself so times v1 that is what the projection is so it's going to be this minus the projection the length of the projection squared so what is the length of the projection squared well that's this guy dotted with himself this guy dotted with himself let me write it this way let me take it let me take it step by step so this is going to be minus - I want to make sure I can still see that up there so I don't have to redo write it minus the projection is going to be it looks a little complicated but hopefully things will simplify v2 dot v1 over v1 dot v1 times switch colors times the vector this is all just going to end up being a number remember you take the dot products you get numbers times the vector v1 and we're going to take the length of that whole thing we're going to take the length of that whole thing squared and all of this is going to be equal to H squared once again just the Pythagorean theorem this or this squared which is the height squared is equal to your hypotenuse squared this is your hypotenuse squared minus the other side squared this is the other side squared looks a little complicated with just the projection of this guy onto that right there so let's see if we can simplify this a little bit we're just going to have to break out some algebra or it let's see what we can do here well actually not algebra some linear algebra so what is this guy well this guy is just the dot product of this with itself so this is just equal to we know I mean any vector if you take the square of its length it's just that vector dotted with itself so this thing if we taking the square of this guy's length it's just equal to this guy dotted with themselves so let me write everything over again so we get h squared H squared is equal to v2 dot v2 and then - this guy dotted with himself so - v2 dot v1 over v1 dot v1 times the vector v1 dotted with itself dotted with it dotted with v2 dot v1 remember this green part is just a number over v1 dot v1 times v1 and what is this equal to let's just simplify this these are just scalar quantities and we saw that the dot product is associative with respect to scalar scalar quantity so we can just change the order here so this is going to be equal to the scalar quantity times itself so as we could say this is equal to v2 v2 dot v1 let me write it this way v2 dot v1 that's going to be and we're going to multiply the numerator times itself v2 dot v1 and then all of that over v1 dot v1 times v1 dot v1 remember I'm just taking these two terms and multiplying them by each other I'm just switching the order because and then we know that the scalars can kind of be taken out times these two guys dot each other times v1 dot v1 that's what this simplifies to now this is now a number we had vectors here but when you take the dot product you just get a number and this number is the same as this number so we can simplify it a little bit so we can cross those two guys out and then we are left with we are left with that our height squared our height squared is equal to it's equal to this expression times itself right we have it times itself twice so it's equal to well let me start over here it's equal to v2 dot v2 v2 dot v2 minus this guy times itself so v2 dot v1 squared all of that over just one of these guys v1 v1 dot v1 that is what the height squared is now we have the height squared we could take the square root if we just want to solve for the height but to keep our math simple to keep our math simple we know that area is equal to base times height let's just tell you what the area squared is equal to because then both of these terms will get squared so we know that the so if the area is equal to base times height so at the beginning of the video then the area squared is going to be equal to these two guys squared it's going to be equal to bass squared times height squared now what is the base squared let me do it like this so the base squared we already saw the base of our parallelogram is the length the base of our entire parallelogram remember this is the parallelogram in question the base is the length of vector v1 now what is the base squared the base squared is going to be the length of vector v1 squared or another way of writing that is v1 dot v1 and we already know what the height squared is it's this expression right there right let me write that down the height squared is the height squared right there so what is our area squared going to be our area squared our area squared let me go down here more space our area squared is going to be equal to our Bay squared which is v1 dot v1 times our height squared times times let me write it like this times let me know I was using the wrong color times this guy over here v2 dot v2 minus v2 dot v1 squared over v1 dot v1 now what is this simplify to well if this is just a number these are all just numbers so if I multiply if I distribute this out this is equal to what this times this is equal to V 1 let me write color-coded v1 dot v1 times this guy times v2 dot v2 and then when I multiply this guy times that guy what happens well I have this guy in the numerator and that guy in the denominator so they cancel out so I'm just left with minus v2 v2 dot v1 squared now let's remind ourselves what these two vectors were v1 was the vector AC and v2 is the vector BD let me rewrite it down here so we have it to work with so v1 was equal to the vector AC and v2 is equal to the vector B D so what is v1 dot v1 what is v1 dot v1 that is equal to that is equal to a dot a which is or a times a squared plus C squared that's this right there and that's what's v2 dot v2 so we're going to have that times v2 dot v2 v2 dot v2 is b squared plus d squared and then we're going to have minus v2 dot v1 squared so what's v2 dot v1 it is it's B times a plus D times C or a times B plus we're just dotting these two guys so it's a B plus C D and then we're squaring it we're squaring in let's see what this simplifies to hopefully it simplifies to something let me switch colors so if we just multiply this out let me write it here our area squared is equal to a squared times B squared a squared B squared a squared times D squared plus a squared times D squared plus C squared times B squared plus C squared times B squared plus C squared times D squared plus C squared times D squared and then minus this guy squared what is that going to be equal to a B squared is a squared B squared and then I have I'm going to multiply these guys times each other twice so it's going to be plus two ABCD and then you're going to have a plus C squared d squared I just soiled this out that's the best way you could kind of think about it and then if I distribute this negative sign what do I have let me rewrite everything it's equal to a squared B squared plus a squared d squared plus C squared B squared plus C squared d squared minus a squared B squared minus 2 a b c d minus c squared d squared all I did is I distributed the minus sign now it looks like some things will simplify nicely and I remember all of this was equal to our area squared we have an a B squared we have a minus a B squared they cancel out we have a CD a minus C d squared and a CD squared so they cancel out so all we're left with is that the area of our parallelogram squared is equal to a squared d squared minus 2 ABCD plus C C squared plus C squared BC where'd C squared B squared now this might look a little bit bizarre to you but you might well if you made a substitution right here if you said that X is equal to ad and if you said Y is equal to CB then what does this become this is equal to this is equal to x squared minus two times XY plus y squared right hopefully you recognize this and this is just the same thing as X minus y squared so if we this was our substitutions we made I did this so you can visualize this a little bit better so we have our area squared is equal to X minus y squared or ad a t minus C B or L let me write it be C squared that's what the area of our parallelogram squared is and if you want to you know if you don't quite understand what I did here I just made these substitutions so you can recognize it better but just to understand that this is the same thing as this if you want you can just multiply this guy out and you'll get that right there but what is this what is this what is this thing right here it's the determinant this is the determinant of our original matrix of our original matrix up here well I call that matrix a and then I use the I used a again for area so let me write it this way area squared let me write it like this area area squared area squared is equal to ad minus BC squared this is so this is the area this is are all area these A's are all area but what is this this is the determinant of my matrix that is the determinant of that is the determinant of my matrix a my original matrix that I started the problem with which is equal to the determinant of a b c d right the determinant of this is ad minus bc by definition so your area this is exciting the area squared so the area of your parallelogram squared is equal to the determinant of the matrix whose column vectors construct that parallelogram is equal to the determinant of your matrix squared or if you take the square root of both sides you get the area is equal to the absolute value of the determinant of a which is a pretty neat outcome especially considering how much hairy algebra we had to go through let's go back all the way over here go back to the drawing so if we want to figure out the area of this parallelogram right here that is defined or that is created by the two column vectors of a matrix we literally just have to find the determinant of the matrix the area of this the area is equal to the absolute value of the determinant of a and you have to do that because this might be negative if you guess you could imagine if you kind of swapped these guys around if you swap some of the rows this guy would be negative but you can't have a negative area and it wouldn't really change the definition it really wouldn't change what span if you switch V 1 and V 2 you're still spanning the same parallelogram you just might get the negative of the determinant but that is a really neat outcome and you know when you first learn determinants in in school I mean we we learned them the first motivation for determinants was this idea of well you know when we take the inverse of a 2x2 this thing shows up in the denominator we call that determinant but now there's this other interpretation here that is actually the area of the parallelogram created by the column vectors of this matrix