Finding Area Using a DeterminantDate: 05/25/2007 at 03:55:12 From: Richard Subject: How does the determinant of a 2x2 matrix give the area? I just can't see how this formula gives the area (determinant): M = | 1 1| |-1 1| det M = 1(1) - 1(-1) = 1 + 1 = 2, which is indeed the area of the parallelogram defined by the two vectors (1,-1) and (1,1). What are the relations in the formula? I've tried putting them on paper numerous times, but I just can't figure it out! Date: 05/25/2007 at 17:10:45 From: Doctor Fenton Subject: Re: How does the determinant of a 2x2 matrix give the area? Hi Richard, Thanks for writing to Dr. Math. This isn't a trivial result. This can be shown by some algebra and the dot product for vectors. If u = (x1,y1) and v = (x2,y2) are vectors, and ||u|| represents the length of u, etc., then the dot product of u and v, u.v, is defined by u.v = x1*x2 + y1*y2 . Using the Law of Cosines, it can be shown that u.v = ||u||*||v||*cos(@) where @ is the angle between u and v. The area of the parallelogram formed by u and v is given by ^--------------- /| / u / | / / |h / / | / /@ | / ----------------> v |<--->| L Area = h*||v|| since ||v|| is the length of the base, and h is the altitude as shown in the diagram. If L is the length of the projection of the vector u on the vector v as shown, then L ----- = cos(@) , or L = ||u||*cos(@) . ||u|| By Pythagoras, h^2 = ||u||^2 - L^2 = ||u||^2 - (||u||*cos(@))^2 (u.v)^2 = ||u||^2 - ||u||^2 * ----------------- ||u||^2 * ||v||^2 Multiplying by ||v||^2 gives (Area)^2 = ||u||^2 * ||v||^2 - (u.v)^2 = (x1^2 + y1^2)*(x2^2 + y2^2) - (x1*x2 + y1*y2)^2 If you simplify this expression, you can show that this equation can be written as (Area)^2 = (x1*y2 - x2*y1)^2 , and the right side is the square of det [x1 y1] [x2 y2] . If you have any questions, please write back and I will try to explain further. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/ Date: 05/28/2007 at 04:14:37 From: Richard Subject: Thank you (How does the determinant of a 2x2 matrix give the area?) Hey Dr. Fenton! Thanks a million! It all makes sense now! I keep forgetting to look at the geometry! |
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