I'm a little confused on which vectors are being mapped here. To my understanding, there's no position vectors (coming from the origin) that make up the rectangle shown. The rectangle is a shift variation of 7(1,0) + 5(0,1). Doesn't the transformation matrix turn two vectors into the weighted sums of the vectors in the matrix? Is the proper way to think about it: the transformation matrix maps (1,0) -> (5,4) and (0,1) -> (9,8) which will create a parallelogram with an area of 140 when you take the weight sum? Or is there more complex calculations underlying the transformation? Or a better way to think about it - each point on the original rectangle can be represented by a vector (a, b), each of which is transformed by the matrix shown. These will give new points (a', b') and all points (a', b') will be the new shape with an area of 140? I guess this would be the image that's being scaled?

Your final paragraph is correct. The fact that the sides of the rectangle could be thought of as vectors is irrelevant, and the same area-scaling property holds for areas with curved boundaries.

Why do you use the absolute value of determinant here?

You are multiplying the determinant by the area, and you cannot have a negative area. If you want the shape to be smaller, it would be multiplied by a fraction, such as 1/4, not a negative number.

Main content

Course: Precalculus > Unit 7

Lesson 7: Matrices as transformations of the plane

Finding area of figure after transformation using determinant

Name: Finding area of figure after transformation using determinant
Uploaded: 2021-03-10T10:02:23Z
Description: In this worked example, Sal finds the area of the image of a rectangle after a transformation defined by a given matrix. This is done by finding the area of the pre-image and multiplying by the matrix's determinant.

Google Classroom

In this worked example, Sal finds the area of the image of a rectangle after a transformation defined by a given matrix. This is done by finding the area of the pre-image and multiplying by the matrix's determinant. Created by Sal Khan.

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Star Birds
Posted 7 months ago. Direct link to Star Birds's post “I get the way to find the...”
I get the way to find the area, it's quite easy. But I don't fully understand why we need to muliply the determinant at
0:55
. In this situation, what does the determinant tell us? Thanks in advance!
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(2 votes)
Answer
- bob
  Posted 4 months ago. Direct link to bob's post “The determinant is basica...”
  The determinant is basically a scale factor, we are scaling the image/shape by the determinant
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  (0 votes)
Brian Ye
Posted a month ago. Direct link to Brian Ye's post “I'm a little confused on ...”
I'm a little confused on which vectors are being mapped here. To my understanding, there's no position vectors (coming from the origin) that make up the rectangle shown. The rectangle is a shift variation of 7(1,0) + 5(0,1). Doesn't the transformation matrix turn two vectors into the weighted sums of the vectors in the matrix?

Is the proper way to think about it: the transformation matrix maps (1,0) -> (5,4) and (0,1) -> (9,8) which will create a parallelogram with an area of 140 when you take the weight sum? Or is there more complex calculations underlying the transformation?

Or a better way to think about it - each point on the original rectangle can be represented by a vector (a, b), each of which is transformed by the matrix shown. These will give new points (a', b') and all points (a', b') will be the new shape with an area of 140? I guess this would be the image that's being scaled?
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(1 vote)
Answer
- kubleeka
  Posted a month ago. Direct link to kubleeka's post “Your final paragraph is c...”
  Your final paragraph is correct. The fact that the sides of the rectangle could be thought of as vectors is irrelevant, and the same area-scaling property holds for areas with curved boundaries.
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  (2 votes)
TheChessWizard2005
Posted a year ago. Direct link to TheChessWizard2005's post “Why do you use the absolu...”
Why do you use the absolute value of determinant here?
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(1 vote)
Answer
- ayili
  Posted a year ago. Direct link to ayili's post “You are multiplying the d...”
  You are multiplying the determinant by the area, and you cannot have a negative area. If you want the shape to be smaller, it would be multiplied by a fraction, such as 1/4, not a negative number.
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  (2 votes)

Video transcript

- [Tutor] We're told to consider this matrix transformation or this is a matrix that you can view, represents a transformation on the entire coordinate plane. And then they tell us that the transformation is performed on the following rectangle. So this is the rectangle before the transformation and they say, what is the area of the image of the rectangle under this transformation? So the image of the rectangle is what the rectangle becomes after the transformation. So pause this video and see if you can answer that before we work through it on our own. All right, so the main thing to realize is, if we have a matrix transformation or a transformation matrix like this if we take the absolute value of its determinant, that value tells us how much that transformation scales up areas of figures. So let's just do that, let's evaluate the absolute value of the determinant here. So the absolute value of the determinant would be the absolute value of 5 times 8, 5 times 8 minus 9 times 4, 9 times 4. Remember for a 2 by 2 matrix, the determinant is just this times this minus this times that. And so that's going to be the absolute value of 40 minus 36 which is just the absolute value of 4 which is just going to be equal to 4. So this tells us that this transformation will scale up area by a factor of 4. So what's the area before the transformation? Well, we can see that this is, let's see, it's 5 units tall and it is 7 units wide. So this has an area of 35 square units, pre transformation. So post transformation, we just multiply it by the absolute value of the determinant to get, let's see, 4 times 30 is 120 plus 4 times 5 is another 20. So this is going to get us to 140 square units and we're done.