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## Linear algebra

### Course: Linear algebra > Unit 2

Lesson 6: More determinant depth- Determinant when row multiplied by scalar
- (correction) scalar multiplication of row
- Determinant when row is added
- Duplicate row determinant
- Determinant after row operations
- Upper triangular determinant
- Simpler 4x4 determinant
- Determinant and area of a parallelogram
- Determinant as scaling factor

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# Determinant and area of a parallelogram

Realizing that the determinant of a 2x2 matrix is equal to the area of the parallelogram defined by the column vectors of the matrix. Created by Sal Khan.

## Want to join the conversation?

- This is kind of off topic but can we also use cross product to calculate the area of the parallelogram? like v1 cross v2? If so, they would be different method to achieve the same answer?(9 votes)
- Yes, you can. The cross product is used to do this is the last few videos in the calculus playlist. A good way to see why is to consider the alternate way of calculating the cross product.(6 votes)

- Does this work for any kind of area or only for parallelograms?(4 votes)
- Half of the parallelogram is the triangle created by v1 and v2 so you can find the area of a triangle as being the absolute value of half of the determinant.(5 votes)

- Does this extend to higher dimensional vectors?(6 votes)
- I think it at least applies to 3 dimensions:

http://en.wikipedia.org/wiki/Parallelepiped#Volume

And apparently for higher dimensions, the volume is equal to the determinant of the Gram matrix (but I don't yet understand the Gram matrix very well)(3 votes)

- that was really neat and beautiful(5 votes)
- Hi, this might be kind of weird question out of the blue. But would taking the determinate of a vector be somewhat similar to taking the integral of a function (where both of them seeks for the area)?(2 votes)
- That is a good line of thought.

This is what i think....Having given it a thought.

A vector has magnitude and direction. It has a zero determinant.

Given a linear function above the x-axis(for simplity), the integral of the function is the area under the graph.

This linear function can also be thought of as

line kV1(scaler multiples of some vector v1) through the origin or

Vo + kV1 where the scaler multiples start off from a vector Vo.

This line can be seen as a matrix-vector product

T = AX taking an input Rn and mapping it to R1.

Meaning it takes a vector in Rn and squishes it to a line.

Now finding the determinant of A(the transformation matrix) is 0. det(A).

That is, the determinant of the transformation matrix is 0 and the determinant of the line (if viewed as a long vector) is also zero.

Nonetheless, the area below the line may not be zero but the determinant will always be zero.

The case gets 🤢 if the function is not linear. Probably some crazy curve.

The integral is still the area under the curve.

But what is the determinant? This is not a linear transformation mind you, and not a square matrix(n,n)... because it maps from some Rn to R1. And so no determinant. This transformation matrix A takes a vector in Rn and maps it onto a point. HA!

So

T = AX

where A is 1 by n matrix and X is n by 1 vector in Rn.

T = A(1,n) * X(n, 1)

Happy Learning_HL(2 votes)

- how do I use the parallelogram rule to construct the resultants for vectors(2 votes)
- Sketch the two vectors head to tail and draw a line from the tail of the first to the head of the second.(2 votes)

- At14:25Sal says that V1.V1 is in the numerator, and also in the denominator, so they cancel out, but then V1.V1 is still in the final equation he uses. Shouldn`t V1.V1 have canceled out, and should we not just be left with v2.v2 - (v2.v1) ^2?(2 votes)
- solve quadrilateral abcd vertices a(4,4),b(2,0),c(-4,-2) and d(-2,2) prove that abcd is a parallelogram(2 votes)
- Why does sals Height line at a 90 degree angle with V2? Shouldn't it be at a 45 degree angle?(2 votes)
- Never mind the shape is at an angle itself.(1 vote)

- If we take a rectangle with one angle at the origin, which is a special case of a parallelogram, having its base x and its height y, its area is xy which is the same as the determinate | x 0 |

| 0 y | .

Is it possible from this to construct a proof for all parallelograms?(2 votes)

## Video transcript

I've got a 2 by 2 matrix here,
and let's just say its 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 here. We can say v1 one is equal to
ac, and we could write that v2 is equal to bd. 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. That's my vertical axis. That's my horizontal axis. And maybe v1 looks something
like this. v1 might look something
like that. So that is v1. It's horizontal component will
be a, its vertical coordinant -- give you this as maybe a
position vector, or just how we're drawing it, is c. And then v2, let's just say it
looks something like this. Let's say that they're
not the same vector. So v2 looks like that. v2, its horizontal coordinate
is going to b, and its 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 -- or not of course but, the origin is also
another point in 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 look like that,
parallel to v1 the way I've drawn it, and the other side
will look like this. That's our parallelogram. The parallelogram generated
by v2 and v1. What we're going to concern
ourselves with specifically is the area of the parallelogram
generated by v1 and v2. So how do we figure that out? 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 is just equal to the base-- so
that could be the base-- times the height. So it's equal to base -- I'll
write capital B since we have a lowercase b there--
base times height. That's what the area of a
parallelogram would be. Now what are the base and the
height in this situation? 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 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? We could drop a perpendicular
here, and that, the length of this line right here, 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 the height? Well, one thing we can do is, if
we can figure out this guy right here, we could use the
Pythagorean theorem. Because the length of this
vector squared, 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
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 to get every point along this line. Just like that. Now if we have l defined that
way-- that line right there is l, I don't know if
you can 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 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 of 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 can just use the Pythagorean theorem. So we can say that the length
of H squared-- well I'm just writing H as the length,
I'm not even specifying it as a vector. So we could say that H 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 v2. I'll do that 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 the
length of v2 squared. That's just the Pythagorean
theorem. Nothing fancy there. So how can we simplify? We want to solve for H. And actually, let's just solve
for H squared for now because it'll keep things a little
bit simpler. So we can say that H squared is
equal to this guy, is 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 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 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, you dot it
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 spanning vector,
which is v1. So it's 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
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 length of the projection squared. What is the length of the
projection squared? Well that's this guy dotted
with himself. Let me write it this way, let
me take it step by step. So this is going to be minus--
I'm want to make sure I can still see that up there so I
don't have to rewrite it. 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 dot products, you get numbers--
times the vector v1. And 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, but
it was just a projection of this guy on to 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 let s 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 are taking
the square of this guy's length, it's just
equal to this guy dotted with himself. Let me write everything
over again. So we get H squared is equal to
v2 dot v2, and then minus this guy dotted with himself. So minus v2 dot v1 over v1 dot
v1, times the vector v1, dotted with itself. 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 quantities, so we can just
change the order here. So this is going to be
equal to the scalar quantity times itself. So we could say this is
equal to 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,
and then we know that the scalars can 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 a dot product, you just get a number. And this number is the
same as this number. So we can simplify
a little bit. So we can cross those two guys
out, and then we are left with that our height squared
is equal to this expression times itself. We have it times itself twice,
so it's equal to-- let me start over here. It's equal to v2 dot v2 minus
this guy times itself. So v2 dot v1 squared, all of
that over just one of these guys. 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, we
know that area is equal to base times height. Let's just say what the area
squared is equal to. Because then both of these
terms will get squared. So if the area is equal to base
times height-- we saw that 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 base
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
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. 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-- let me go
down here where I'll have more space-- our area squared is
going to be equal to our base squared, which is v1 dot v1
times our height squared. Let with me write
it like this. No, 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 does this
simplified to? Well, 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 v1--
let me color code it-- v1 dot v1 times this guy
times v2 dot v2. And then when I multiplied
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 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 bd. So what is v1 dot v1? That is equal to a dot
a, a times a, a squared plus c squared. That's this, right there. And that's what? v2 dot v2. So we're going to have
that times v2 dot v2. v2 dot v2 is v squared
plus d squared. And then we're going to have
minus v2 dot v1 squared. So what's v2 dot v1? It's b times a, plus d times c,
or a times b plus -- we're just dotting these two guys. So it's ab plus cd, and then
we're squaring it. And 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 times d squared,
plus c squared times b squared, plus c squared
times d squared. And then minus this
guy squared. What is that going
to be equal to? ab squared is a squared,
b squared. And then I'm going to multiply
these guys times each other twice, so that's going
to be plus 2abcd. And then you're going to have
a plus c squared, d squared. I just foiled this out, that's
the best way you could 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 2abcd, minus c squared, d squared. All I did is, I distributed
the minus sign. Now it looks like some things
will simplify nicely. And now remember, all this is
equal to our area squared. We have a ab squared, we have
a minus ab squared. They cancel out. We have a minus cd 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 2abcd plus c squared b squared. Now this might look a little bit
bizarre to you, but 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 x
squared minus 2 times xy plus y squared. Right? Hopefully you recognize this. And this is just the same thing
as x minus y squared. So, if this is our substitutions
we made-- I did this just so you can visualize
this a little bit better. So we have our area squared is
equal to x minus y squared or ad minus cb, or let me
write it, bc squared. That's what the area of our
parallelogram squared is. And if you don't quite
understand what I did here, I just made these substitutions
so you can recognize it better. But just 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 thing right here? It's the determinant. This is the determinant of
our original matrix. Well, I called that matrix A
and then I used A again for area, so let me write
it this way. Area squared -- let me
write it like this. Area squared is equal to
ad minus bc squared. So this is area, these
A's are all area. But what is this? This is the determinant
of my matrix. That is the determinant of my
matrix A, my original matrix that I started the problem with,
which is equal to the determinant of abcd. Right? The determinant of this is ad
minus bc, by definition. So your area-- this
is exciting! 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 is equal to
the absolute value of the determinant of A. And you have to do that because this might be negative. You can imagine if you swapped
these guys around, if you swapped 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 spanned. If you switched v1 and v2,
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
learned determinants in school-- I mean, we learned
the first motivation for a determinant was this idea of
when we take the inverse of a 2 by 2, this thing shows up in
the denominator and we call that the determinant. But now there's this other
interpretation here. There's actually the area of the
parallelogram created by the column vectors
of this matrix.