Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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Area With Determinants - Concept

Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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One thing thatdeterminantsare useful for is in calculating thearea determinantof a parallelogram formed by 2 two-dimensional vectors. The matrix made from these two vectors has a determinant equal to the area of the parallelogram. Area determinants are quick and easy to solve if you know how to solve a 2x2 determinant. A parallelogram in three dimensions is found using thecross product.

One really neat application of determinants is that they can be used to find the area of a parallelogram.
Now suppose I create a parallelogram using these four points a, b, c and d. I've got the parallelogram graphed here and it turns out that if you pick any vertex of the parallelogram I'm going to pick a and look at the vectors ab and ad that you can use those vectors to form a determinant whose value is going to be the area. Now ab vector ab goes from point a to point b and so its components are going to be 9-1 which is 8 -4--2 -4+2 -2 and ad goes from point a to point d and so its components are going to be going to be 0-1 -1 and 3--2 5, so the way we we do this is we form a determinant using these these vectors as the rows of the matrix so the determinant is 8,-2 and -1,5.
Now I have to do one other thing. Remember when you when you switch rows of a determinant you change the sign of the determinant so it's possible to get a negative answer from this determinant so you should always take the absolute value. And I'm going to write it that way because if I wrote the absolute value with vertical lines it would be little confusing with our notation for the determinant but again you can enter these vectors in either order as long as you take the absolute value afterwards.
Okay so let's calculate this area so its absolute of 8 times 5 40 minus -2 times -1 2 so it's the absolute value of 38 which is 38 and so the area of this parallelogram is 38 so just remember that the procedure is, you pick a vertex on the parallelogram and then you have that vertex will share two vectors in this case ab and ad. If you use those the components of those vectors to form a determinant, calculate that determinant then take its absolute value, if we had reversed the order of these vectors we would have gotten -38 we would have needed that absolute value to get us an area which has to be positive so make sure you remember the absolute value but it's the absolute value of the determinant of those vectors.

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