Kenyon College
paquind@kenyon.edu
Math 224
Homework 9 Solutions
Section 6.1. Note: The form of the answers for the problems in this section is
not unique. I have written basis vectors using integers, when possible, but many
of your answers will likely con
Kenyon College
Dana Paquin
paquind@kenyon.edu
Math 224
Homework 3 Solutions
Section 1.6
1.6 #4: Let S = cfw_[x, y ] | x, y 0. Since [1, 1] is in S , but 1 [1, 1] = [1, 1] is not in S ,
we see that S is not closed under scalar multiplication, so S is not a
Kenyon College
Dana Paquin
paquind@kenyon.edu
Math 224
Homework 6 Solutions
Section 4.4
4.4 #2
1
0
0
1
A = 1 1 .
2
1
1 3
Using Maple, we compute det AT A = 80 = 4 5. So the volume of the 2-box
in R5 determined by the given vectors is 4 5 .
4.4 #8
1 3 4
A
Kenyon College
Dana Paquin
paquind@kenyon.edu
Math 224
Homework 7 Solutions
Section 5.1
5.1 #2 p() = det(AI ) = (+3)(2), so the eigenvalues of A are 1 = 3 and 2 = 2 .
Next, we compute the eigenvectors of A.
(A (3)I ) =
10 5
10 5
,
1 1/2
. Since the second
Kenyon College
Dana Paquin
paquind@kenyon.edu
Math 224
Homework 11 Solutions
Section 3.2
3.2 #2: The set is NOT a subspace of P since it is not closed under vector addition. For
example, p(x) = x4 + x3 and q (x) = x4 are both in the set, but p(x)+ q (x) =
Kenyon College
paquind@kenyon.edu
Math 224
Homework 10 Solutions
Section 6.3
6.3 #2: Let
3/5 0 4/5
A = 4/5 0 3/5 .
0
01
Then
AT A = I,
so A is orthogonal and
A 1
3/5 4/5 0
0
1 .
= AT = 0
4/5 3/5 0
6.3 #20: Suppose that A is orthogonal. Then AT is also ort
Kenyon College
Dana Paquin
paquind@kenyon.edu
Math 224
Homework 8 Solutions
Section 5.2
5.2 #2 The characteristic polynomial is
p() = det(A I ) = ( 2)( 5),
so the eigenvalues of A are 1 = 2 and 2 = 5. The eigenvectors corresponding
2r
, r = 0. The eigenve
Kenyon College
Dana Paquin
paquind@kenyon.edu
Math 224
Homework 1 Solutions
Section 1.1
1.1 #8: 4(3u + 2v 5w) = 4([3, 9, 6] + [8, 0, 2] [15, 5, 10]) = [80, 56, -72]
1.1 #26: All c R . Since [3, 5] and [6, 11] are not parallel, every vector [1, c] forms
th