Math 2270 Spring 2004
Homework 18 Solutions
Section 6.3) 3, 4, 7, 11, 13, 14, 22, 25, 26, 31
(3) We find the vectors that make up the sides of the triangle.
"
5
7
#
−
"
#
4
3
=
"
1
4
#
"
10
1
#
−
"
#
=
a1
a2
#
4
3
"
6
−2
=
"
#
Then, to find the area of the triangle we use Fact 6.3.3.
1 1 6
2 4 −2
=
1
(−2 − 24) = −13
2
Taking the absolute value, we get an area of 13.
(4) We use the same procedure as in problem (3).
"
b1
b2
#
−
"
a1
a2
#
=
"
b1 − a1
b2 − a2
#
"
c1
c2
#
−
"
c1 − a1
c2 − a2
#
Then, to find the area of the triangle we use Fact 6.3.3.
1 b1 − a1 c1 − a1
2 b2 − a2 c2 − a2
=
=
1
(b1 c2 − a2 b1 − a1 c2 + a1 a2 − b2 c1 + a1 b2 + a2 c1 − a1 a2 )
2
1
[a1 (b2 − c2 ) − b1 (a2 − c2 ) + c1 (a2 − b2 )]
2
a b c 1
1
1 1
= a2 b2 c2 2 1 1 1 Let’s check and see if this formula works for the vectors given in problem (3).
5 4 10 1
1
1
7 3 1 = [5(3 − 1) − 4(7 − 1) + 10(7 − 3)] = [10 − 24 + 40] = 13
2
2
2
1 1 1 1
Math 2270 Spring 2004
Homework 18 Solutions
(7) There are several ways to approach this problem. For example, we can divide the shape
into four triangles (each with one vertex at the origin) and add their areas, or we can measure
the area of the entire region at once. Here we’ll use the four triangle approach along with
the formula from problem (4).
−7 −5 0 1
1
77
7 −6 0 = (42+35) =
A2 =
2
2
2
1 1 1 3 5 0 1
1
35
A3 =
−4 5 0 = (15 + 20) =
2
2
2
1 1 1 Therefore, the total area is
35 +
−7 5 0 1
1
7 5 0 = (−35−35) = −35
A1 =
2
2
1 1 1 −5 3 0
1
−6 −4 0
A4 =
2
1 1 1
77 35
+
+ 19 = 110
2
2
1
= (20 + 18) = 19
2
(11) We will construct a matrix for the transformation with respect to the basis {~v1 , ~v2 }.
(Notice that we know that ~v1 and ~v2 are linearly independent. If they were not, they would
have to be multiples of one another. Assume that ~v1 = k~v2 . Then, T (~v1 ) = T (k~v2 ) =
kT (~v2 ) = 4k~v2 = 4~v1 . We are given that T (~v1 ) = 3~v1 , so the vectors ~v1 and ~v2 cannot be
scalar multiples of one another. Therefore, they must be independent.)
B=
"
3 0
0 4
#
The determinant of this matrix is 12, so det(A)=12 as well.
(13) We use Fact 6.3.7.
AT A =
The area is
v
u
u
t
"
1 1 1 1
1 2 3 4
#





1
1
1
1
1
2
3
4





=
"
4 10
10 30
√
√
4 10 √
= 120 − 100 = 20 = 2 5
10 30 2
#
Math 2270 Spring 2004
Homework 18 Solutions
(14) Again, we use Fact 6.3.7.
 1
1 0 0 0 
 0
AT A = 
 1 1 1 1 
 0
1 2 3 4
0
1
1
1
1


1
2
3
4







1 1 1

=  1 4 10 

1 10 30
The area is
v
u
u
u
u
t
1 1 1 q
√
√
1 4 10 = (120 − 100) − (30 − 10) + (10 − 4) = 20 − 20 + 6 = 6
1 10 30 (22) We have
A=
"
3 7
4 11
#
~b =
"
1
3
#
We use Cramer’s rule (Fact 6.3.9).
1 7 = 11 − 21 = −10
det(A1 ) =
3 11 3 1 =9−4=5
det(A2 ) =
4 3 3 7 = 33 − 28 = 5
det(A) =
4 11 Then, x1 = −10/5 = −2 and x2 = 5/5 = 1.
(25) We use Fact 6.3.10.
The entries of adj(A) are:
a11
1 0 =1
=
0 1 a21 =
−
0 0 =0
2 1 1 0 1 det(A) = 0 1 0 = 1(1 − 2) = −1
2 0 1 a12 =
a22
−
0 1 =0
0 1 1 1 = −1
=
2 1 3
a13
0 1 = −1
=
1 0 a23 =
−
1 1 =0
0 0 Math 2270 Spring 2004
a31
Homework 18 Solutions
0 1 =
= −2
2 0 a32 =
Now, we construct the matrix A−1 .
A−1
−
1 0 =0
2 0 

a33
1 0 =
=1
0 1 

−1 0 1
1 0 −1
1 


0 = 0 1 0 
=

 0 −1
−1
2 0 −1
−2 0 1
(26) Yes, the entries of A−1 must be integers. By Fact 6.3.10, the entries of the inverse of
a matrix are found by calculating the determinants of all of the minors of the matrix and
then dividing them by the determinant of the original matrix. If all of the entries of A are
integers, the determinants of the minors will also be integers. If det(A) = 1, we will be
dividing integer values by one, leaving integer values.
(31) We start by considering the equilateral triangle sitting on the x−axis with sides of
length 2. The base of the triangle
√ is formed by the vector ~v1 = 2~e1 . The second vector, ~v2 ,
has x1 = 1 and x2 = 1 · tan π3 = 3. (Remember that π/3 radians is 60 degrees.) Now, we
must define a linear transformation that transforms this equilateral triangle into the right
triangle with sides of length 3, 4 and 5. We assume that T (~v1 ) = 3~e1 and T (~v2 ) = 4~e2 . We
must find a matrix A such that
A
"
2
0
#
=
"
3
0
#
and
A
"
√1
3
#
=
"
0
4
#
which is equivalent to solving the following system for the matrix A.
"
2 √1
0 3
#
"
#
3 0
A
=
0 4
"
#" √
"
#
#
"
#
√
√
1
1
3 0
3/2 − 3/2
3 −1
3 3 −3
√
A= √
= √
=
0
2
0
8
0
4/ 3
2 3 0 4
2 3
The determinant of the matrix A will determine how the area of the circle scales when it is
transformed into an
√ ellipse√by the√transformation T . We have
det(A) = 12/2 3 = 6/ 3 = 2 3.
All that remains is to find the area of the largest circle that can be inscribed into the
equilateral triangle. Using the symmetry√of an equilateral triangle we see that the radius
of the circle must be r = 1 · tan π6 = 1/ 3. Therefore, the area of the circle is π/3. The
√
transformation
√enlarges the area of the circle by 2 3, so the area of the ellipse must be
√
2π 3/3 = 2π/ 3.
4