Math 2341, Solutions to Homework No. 1
1. Let f : X V , g : X V and h : X V be functions from X to V , and let a, b K. Now
O : X V is given by O(x) = 0 for all x X. For all x X, we have
(f + (g + h)(x) = f (x) + (g + h)(x) = f (x) + (g(x) + h(x)
= (f (x)
S OLUTIONS TO M ATH 1341 , T EST 2 ( YELLOW
EXAM )
1. B is correct.
2. The correct answer is C.
A3 = 2I = det(A)3 = (2)3 det(I )
= det(A)3 = (2)3
= detA = 2
(by Theorem 3 in Section 2.1)
(by Theorem 1 in Section 2.2)
3. C is the correct answer: if x = 0 a
S OLUTIONS TO M ATH 1341 , T EST 1 ( WHITE EXAM )
1. The vector n = (2, 3, 5) is normal to P , so we may use n as a direction vector
for L. Since the line L passes through the point (2, 1, 1) and has direction vector
(2, 3, 5),
(x, y, z ) = (2, 1, 1) + t(
April 28, 2004
Professeur: A. Sebbar
Mathematical Methods III MAT 1303
FINAL EXAM
This is a closed book exam. Non-graphing, standard scientic calculators are permitted
Problem 1 [3pts]: Find the tangent line to the graph of the function f (x) = 2x3 at
x =
S OLUTIONS TO M ATH 1341 , T EST 1 ( YELLOW EXAM )
1. The vector n = (2, 1, 1) is normal to P , so we may use n as a direction vector for L.
Since the line L passes through the point (2, 3, 5) and has direction vector (2, 1, 1),
(x, y, z ) = (2, 3, 5) + t
S OLUTIONS TO M ATH 1341 , T EST 2 ( WHITE EXAM )
1. F is correct.
2. The correct answer is G.
A3 = 3I = det(A)3 = (3)3 det(I )
= det(A)3 = (3)3
= detA = 3
(by Theorem 3 in Section 2.1)
(by Theorem 1 in Section 2.2)
3. G is the correct answer: if x = 2 an
S OLUTIONS TO M ATH 1341 , T EST 3 ( WHITE EXAM )
1. U does not contain the zero vector 0 = 0 0 0 0
T
R4 , and is therefore not a subspace.
The set V is not closed under addition: For example X = 1 1 0
lie in V , but their sum X + Y = 0 2 0
T
T
and Y = 1
S OLUTIONS TO M ATH 1341 , T EST 3 ( YELLOW
EXAM )
1. The set U is not closed under addition: For example X = 0 1 1
T
and Y = 0 1 1
T
T
lie in U , but their sum X + Y = 0 2 0 does not. The set V is a subspace, since it is
closed under scalar multiplicatio