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MATB24H3 (13)
Lecture

# Lecture12.pdf

5 Pages
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School
University of Toronto Scarborough
Department
Mathematics
Course
MATB24H3
Professor
Sophie Chrysostomou
Semester
Winter

Description
University of Toronto at Scarborough Department of Computer & Mathematical Sciences MAT B24S Fall 2011 MAT B24 Lecture 12 Section 4.4 DEFINITION: Let a , a , ▯▯▯ ,a be n linearly independent vectors in m 1 2 n m R for n ≤ m. We deﬁne the n-box in R determined by these vectos to be the set of all vectors x satisfying x = t 1 1 + t 2 2 + ▯▯▯ + tn n for 0 ≤ i ≤ 1 and i = 1,2,▯▯▯ ,n.   | | ▯▯▯ | If A = a1 a2 ▯▯▯ a k , then the ”volume” of the n-box is: | | ▯▯▯ | ▯ T V olume = det(A A) In case we have an m-box in R , then its volume would be |det(A)|. Note: (1)A is an m × n matrix. We cannot get the determinant of a matrix which is not a square matrix! (2) A A is a square matrix. (3) A A is a symmetric matrix. (4) If m = n, then: ▯ ▯ ▯ ▯ det(A A) = det(A )det(A) = det(A)det(A) = (det(A)) = |det(A)|. A 1-box in R : the 1-box determine by te vecto1 a is the line segment | from the origin to the tip 1f a . Let A =1  | 1 It’s ”volume” is: ▯ ▯   ▯ ▯ ▯ ▯ | ▯ det(A A) = ▯▯ − a 1 −  a1 = √ a1▯ 1 = ▯a1▯ = ▯a1▯, | the length of a . Thus the length of a 1-box is its length. 1 m A 2-box in R : It is a parallelogram with a vertex at the origin. Let us see what the expression for its ”volume” gives us here.   | | Let: A = a1 a2  Then: | | ▯ ▯  ▯ ▯   ▯ ▯ − a 1 − | | det(AA) = ▯ det  a1 a2  − a 2 − | | For simpliﬁcation, we will use the square of this expression: We will need that the angle betwe1n a an2 a is given in the expression: cosθ = a1▯ 2 or a ▯ a = cosθ ▯a ▯▯a ▯ (∗) ▯a1▯▯a2▯ 1 2 1 2 and that: ▯h▯ = ▯a 1sinθ from the parallelogram abov(∗∗) 2    ▯ ▯ | | T  − a 1 −   det(A A) = det − a 2 − a1 a2 | | ▯ ▯ ▯a1▯ a1 a 1 a2 ▯ = ▯ ▯ a2▯ a1 a 2 a2 = (a ▯1a 1(a 2 a2) − (2 ▯ 1 )1a ▯2a ) 2 2 2 = ▯a ▯1▯a ▯2− (▯a ▯1a ▯2osθ) from (*) 2 2 2 2 2 = ▯a ▯1▯a ▯2− ▯a ▯1▯a ▯2cos θ 2 2 2 = ▯a ▯1▯a ▯2(1 − cos θ) 2 2 2 = ▯a ▯1▯a ▯2sin θ = ▯a ▯ ▯h▯ 2 from (**) 2 = (▯a ▯ ▯h▯) 2 = Area 2 2 2 3 So, the volume of a 2-box in R or R is its area! EXAMPLE: Find the ”volume” of the 3- box in R determined by the vectors: 1 = [1,2,0,−1,−2], a 2 = [1,1,0,1,0], 3 = [0,0,1,1,1].   1 1 0    2 1 0  solution. : First we ﬁnd A : A =  0
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