Main content

## Precalculus

### Course: Precalculus > Unit 7

Lesson 7: Matrices as transformations of the plane- Matrices as transformations of the plane
- Working with matrices as transformations of the plane
- Intro to determinant notation and computation
- Interpreting determinants in terms of area
- Finding area of figure after transformation using determinant
- Understand matrices as transformations of the plane
- Proof: Matrix determinant gives area of image of unit square under mapping
- Matrices as transformations
- Matrix from visual representation of transformation

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Interpreting determinants in terms of area

The determinant of a 2X2 matrix tells us what the area of the image of a unit square would be under the matrix transformation. This, in turn, allows us to tell what the area of the image of any figure would be under the transformation. Created by Sal Khan.

## Want to join the conversation?

- Does this concept applies as well to higher dimensions? I.e, determinant of 3x3 matrix as a volume of parallelepiped? Does it generalize to higher dimensions? Determinant of 4x4 matrix is ?? of 4D solid?(4 votes)
- Yes, that's correct. The determinant of a 1x1 matrix gives the length of a segment, of a 2x2 the area of a parallelogram, of a 3x3 the volume of a parallelepiped, and of an nxn the hypervolume of an n-dimensional parallelogram.(9 votes)

- What videos should I watch to realize how to get the area of triangles or parallelograms by using vectors on the coordinate plane?(4 votes)
- I don't know of any video specifically addressing it, but hopefully my answer is helpful nonetheless as I had a similar question and decided to solve it by hand. As we know, the area of a parallelogram is basically the base times the height.

It's simple enough to establish the base; use the pythagorean theorem on one of the vectors (preferably the longer).

The height is trickier. You need to know trigonometry and the unit circle for that - particularly sine/cos/tan (and their inverse) in order to calculate the angle between two vectors. Note that the matrix conveniently provides all the details needed to do the arctan operations! So you take the arctan of the vector with the greater angle [1,2] and subtract the arctan of the vector with the smaller angle [3,1] and you're left with the angle between the two vectors.

Once you have the angle, you simply need to set the vector that isn't being used as the base as the hypotenuse (again using the pythagorean theorem to calculate its length). You can then solve for the height by multiplying sine of the angle times the hypotenuse.

Multiply the base and the new-found height and you should then have the area of the parallelogram.

EDIT, this may be useful: https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:matrices/x9e81a4f98389efdf:matrices-as-transformations/v/proof-matrix-determinant-gives-area-of-image-of-unit-square-under-mapping(8 votes)

- How does |A| equal the absolute value of the determinant of A? I thought in the last video he said that in the context of matrices absolute value signs mean the determinant. Does it mean absolute value as well?(1 vote)
- How does |A| equal the absolute value of the determinant of A? No, they are two different definitions.

The meaning of || will depend on the context.(3 votes)

## Video transcript

- [Instructor] So I have
a two by two matrix here and we could view it as
having two column vectors. So the first column can
define this vector three, one, which I've depicted in blue here. And then that second
column, you can view it as telling us that we have
another vector, one, two, which I have depicted in this pink color. Now, an interesting
interpretation of the determinant of this two by two matrix is that the absolute value of the determinant is the area of the parallelogram defined by these two vectors. What I mean by the parallelogram defined by these two vectors? Well, imagine taking this
bottom vector and shifting it so it's tail is at the
head of this pink vector. So it would look like this, hand draw it. Looks something like that. And then if you were to
take this pink vector and copy it and shift
it up into the right. So its tail is at the head
of the original blue vector. It's going to look something like that. And so you can see, you
can use that technique to take any two vectors
in the coordinate plane. And they will define a parallelogram. And it turns out that the area of this parallelogram is going to be equal to the absolute value of the determinant of this matrix here. So what's that going to be? Well, we know of figure
out the determinant. It is three times two, which is six. Minus one times one, which is
one, which is equal to five. And of course the absolute
value of five is five. Now that's pretty cool in and of itself. We figured out one
interpretation of a determinant which will be useful as we build up our understanding of matrices. But another interpretation
is to say, all right, what if A is a transformation matrix, and I'm just rewriting it. So we know what a
transformation matrix does. It tells us what to do with
the unit vectors, so to speak. So for example, I have
this vector right over here which is the vector one, zero. We know that a transformation
matrix says, all right take that one, zero vector and turn it into the three, one vector. So turn that into that right over there. And we know we have the
other, or another unit vector. Let's call this, this right over here is the zero, one vector. Goes zero in the X direction,
one in the Y direction. And the transformation matrix says, hey, turn that into the one, two vector. But then you can think about it, it's also not just transforming
the individual vectors. It's also scaling up the
area defined by the vectors. So the area defined by these two original, we could say unit vectors,
we can see it's one by one. It's that area right over there. So this transformation is taking us from an area of one to an area, five. It's scaling it up by a factor of five. Now that's reasonably interesting just for this one unit square. But because of that
it'll actually scale up the area of any figure. Let's say had a figure like this. So this type of oval circle
thing, it has some area. If you apply this transformation matrix it will look something like this. I'm just approximating it. It would look something like that. So this will tell us that this bigger blob has five times the area
of this original blue blob because the bigger blob is the image. Once we've transformed the smaller blob by this transformation matrix. So if I were to tell you that the area of this smaller
circle is let's say 0.6 but then we were to
apply the transformation. And someone were to ask you, what is the area of this bigger blob? Which is the image of the smaller circle after the transformation? Well, you take 0.6, multiply
it by the absolute value of the determinant of the
transformation matrix. You'd multiply it by five. So 0.6 times five would
be three square units. And a hint at the reason
why this works is, is that any region on the coordinate plane can be represented as a series of squares. And then if you apply the transformation you're really just transforming each of those individual squares. So the scaling up of the area
would be the same scale you do to any one of those smaller squares.