Class Notes
(808,385)
Canada
(493,173)
University of Toronto Scarborough
(30,809)
Mathematics
(837)
MATB24H3
(13)
Sophie Chrysostomou
(12)
Lecture
Lecture12.pdf
Unlock Document
University of Toronto Scarborough
Mathematics
MATB24H3
Sophie Chrysostomou
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 nbox 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 nbox is:
  ▯▯▯ 
▯ T
V olume = det(A A)
In case we have an mbox 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 1box in R : the 1box determine by te 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 1box is its length.
1
m
A 2box 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 2box 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
More
Less
Related notes for MATB24H3