**Unformatted text preview: **LAST (Family) NAME:
FIRST (Given) NAME:
Email address: @mail.utoronto.ca STUDENT NUMBER: University of Toronto Fall 2019/2020 MAT223 Midterm 1 Solutions Duration: 110 minutes
Aids Allowed: None Faculty of Arts and Sciences Exam Reminders:
• Fill out your name, student number, and email address at the top of this page.
• Do not begin writing the exam until instructed to do so.
• As a student, you help create a fair and inclusive writing environment; unauthorized aids are prohibited and using
one may result in you being charged with an academic offence.
• Turn off and place all cell phones, smart watches, electronic devices, and unauthorized study materials in your
bag under your desk. These devices may not be left in your pockets.
• If you are feeling ill and unable to finish your exam, please bring it to the attention of an Exam Facilitator.
• In the event of a fire alarm, do not check cell phones or other electronic devices unless authorized to do so. Special Instructions:
• Write legibly and darkly.
• Cross out any work you do not wish to have scored, and clearly indicate if there is work on another page you want
scored.
• Show all of your work. Unsupported answers may not earn credit. Exam Format and Grading Scheme: Answers must be written on the examination paper.
Question: 1 2 3 4 5 6 7 8 9 10 Total Points: 10 4 10 6 4 8 10 4 7 5 68 You must hand in all examination materials after the exam 1. Complete the following sentences with a mathematically correct definition. No marks will be awarded
for a “close” but incorrect definition.
(a) (2 points) The span of the vectors ~v1 , ~v2 , ~v3 is
the set {~x : ~x = a~v1 + b~v2 + c~v3 for some a, b, c ∈ R}. (b) (2 points) The vectors ~u and ~v are orthogonal if
~u · ~v = 0. (c) (2 points) The vectors ~v1 , ~v2 , ~v3 are linearly dependent if
there are scalars a, b, c, not all zero, such that ~0 = a~v1 + b~v2 + c~v3 . (d) (2 points) The vector w
~ is a convex linear combination of the vectors ~v1 , ~v2 , ~v3 if
there are scalars a, b, c ≥ 0 satisfying a + b + c = 1 and w
~ = a~v1 + b~v2 + c~v3 . (e) (2 points) The vector ~u ∈ R3 is a unit vector if
k~uk = 1. + 2x3 = −1 x1
2. Consider the system x1 + x2 + 5x3 = 0 . x1 − x2 − x3 = −2
In this problem you may use the fact 1 1
rref
1 that 0
2 −1
1 0 2 −1
1
5
0 = 0 1 3 1 .
−1 −1 −2
0 0 0 0 (a) (2 points) Write the complete solution to the system. Express your answer in vector form. −2
−1
~x = t −3 + 1 1
0 (b) (2 points) Are there values of a, b, and c that make the system x1
+ 2x3 x + x + 5x
1
2
3 x1 − x 2 − x3 x1 + ax2 + bx3 = −1
=0
= −2
=c inconsistent? If so, give an example of such values. If not, explain why not.
a = 0, b = 2, c = 0 3. For each of the following, give an example if possible. Otherwise, explain why it is impossible.
(
ax + y = 0
(a) (2 points) Numbers a, b ∈ R so that the system
is inconsistent.
bx + y = 0
Impossible. x = y = 0 is a solution to the system regardless of the values of a and b. (b) (2 points) Three linearly independent vectors in R2 .
Impossible. Every subset of a linearly independent set is linearly independent. Two linearly
independent vectors in R2 span R2 . Thus, if there were three linearly independent vectors in R2 ,
one would necessarily be in the span of the other two, which is a contradiction. (c) (2 points) Sets A, B ⊆ R2 so that span(A) = span(A ∪ B).
A = B = {} (d) (2 points) Vectors ~a, ~b, ~c ∈ R3 that are mutually orthogonal (i.e., every pair of vectors is orthogonal).
~a = ~e1 , ~b = ~e2 , ~c = ~e3
2
(e) (2 points) Vectors ~u, ~v ∈ R such that w
~=
is not a linear combination of ~u and ~v .
1
~u = ~v = ~0
2
2
−2
4. (6 points) Let ~u =
and ~v =
. Philbert and Tiffany are discussing whether ~0 can be written
2
−2
as a convex linear combination of ~u and ~v .
Philbert thinks yes: Since ~0 is on the line segment connecting ~u and ~v , it must be that ~0 is a
convex linear combination of ~u and ~v .
However, Tiffany disagrees: Since ~0 = 1~u + 1~v and 1 + 1 > 1, it cannot be that ~0 is a convex
linear combination of ~u and ~v .
Explain to Philbert and Tiffany, using complete English sentences, whether ~0 is a convex linear combination of ~u and ~v . Your explanation must (i) include relevant definitions, and (ii) point out where
Philbert’s and Tiffany’s reasoning is correct/incorrect.
By definition, ~0 is a convex linear combination of ~u and ~v if there exist scalars a, b ≥ 0 with a + b = 1
and
~0 = a~u + b~v .
In this situation, we see
~0 = 1 ~u + 1 ~v .
2
2
Since 1
2 ≥ 0 and 1
2 + 1
2 = 1, it must be that ~0 is a convex linear combination of ~u and ~v . In this situation, Philbert’s intuitions is correct (all points on the line segment connecting ~u and ~v are
convex linear combinations of ~u and ~v ), but Tiffany’s is wrong. Tiffany showed that not all solutions
to
~0 = a~u + b~v
come from convex combinations. However, for ~0 to be a convex combination of ~u and ~v , we only need
one solution to the above equation where the coefficients satisfy a, b ≥ 0 and a + b = 1. 1
1
3
5. Let P ⊆ R3 be the plane given in vector form by ~x = t 0 + s 1 + 1.
1
1
3
(a) (2 points) Express P as a set (e.g., using set-builder notation). 1
1
3 P = ~x ∈ R3 : ~x = t 0 + s 1 + 1 for some t, s ∈ R 1
1
3
or x 3 y ∈R :x−z =0
P= z (b) (2 points) If possible, express P as a span. Otherwise, explain why it is impossible. 0
1
1 Since 0 ∈ P, we know P can also be written in vector form as ~x = t 0 + s 1 and so
0
1
1 1 1 0 , 1
P = span 1
1
1 ~
−2
−1
6. Let ~a =
,b=
, and ~c =
, and let S = {~x ∈ R2 : k~xk = 1}.
2
−1
−2
Draw the following subsets of R2 .
(a) (2 points) A vector ~u so that ~u and ~a are linearly dependent. (c) (2 points) span({~b}) (b) (2 points) S + {~a, ~c} (d) (2 points) W ⊆ R2 consisting of two vectors
such that W ∩ S consists of one vector. 7. For the following questions, circle IDENTICAL if the two sets are equal, INTERSECT if the sets
intersect but are not identical, or DISJOINT if they do not intersect at any point or the sets cannot
be compared. No explanation is needed.
2
1
−2
−1
(a) (2 points) The lines given in vector form by ~x = t
+
and ~x = t
+
−1
−1
1
0
IDENTICAL INTERSECT DISJOINT 2
(b) (2 points)
The line in R with equation x + y = 2 and the line given in vector form by ~x =
1
2
t
+
.
1
0 IDENTICAL INTERSECT (
2x − y + z = 0
(c) (2 points) The line in R3 given by the system
x=z
IDENTICAL INTERSECT DISJOINT and the z-axis.
DISJOINT 2 3
(d) (2 points) The plane in R with equation x + y − z = 4 and the span of 0 . 2
IDENTICAL INTERSECT DISJOINT (e) (2 points) The xy-plane in R3 and Q = R2 .
IDENTICAL INTERSECT DISJOINT 8. Consider the plane A and the lines B and C given in vector form by
A z }| {
1
1
~x = t 1 + s 2
1
1 B z }| {
3
3
~x = t −3 + 7
3
3 C z }| {
10
0
~x = t 20 + 10
30
0 (a) (2 points) Write down a system of linear equations that you could solve to determine whether or
not A ∩ B ∩ C is non-empty. Do not solve the system. x1 + x2 − 3x3 x1 + 2x2 + 3x3 x1 + x2 − 3x3 3x3 − 10x4 − 3x3 − 20x4 3x3 − 30x4 x1 + x2
− 10x4 x1 + 2x2
− 20x4 x1 + x2
− 30x4 =3
=7
=3
= −3
=3
= −3
=0
= 10
=0 (b) (2 points) Write down the augmented matrix corresponding to your system from part (a). 1
1 1 0 0 0 1 1
1 1 −3 0
3
2 3
0
7 1 −3 0
3
0 3 −10 −3 0 −3 −20 3 0 3 −30 −3 1 0 −10 0 2 0 −20 10 1 0 −30 0 1
9. Let X ⊆ R3 be the set of all vectors orthogonal to ~u = 2 .
−2
(a) (2 points) Write X using set-builder notation.
X = {~x ∈ R3 : ~x · ~u = 0} (b) (2 points) Give three different examples of vectors in X. −2
2
~0, 1 , 0
0
1 1 (c) (3 points) Find a vector ~v ∈ X which is also orthogonal to 1 and which satisfies k~v k = 6.
0
(You do not need to simplify your answer.) −2
p By inspection, we see that ~v must be a multiple of 2 . This vector has a length of (−2)2 + 22 + 12 =
1
3, so we need to double its length. Thus, −4 ~v = ± 4 .
2 10. In this question, you will work with a new definition.
Let ~u, ~v ∈ R2 be vectors. The symplectic span of ~u and ~v is the set
~x ∈ R2 : ~x = a~u + b~v for some a, b ≥ 0 such that a + b ≤ 1
1
−1
Define X to be the symplectic span of
and
.
1
1
(a) (2 points) Draw X.
1
(b) (3 points) Let ~x ∈ X. What is the largest possible value of ~x ·
? Justify your answer.
0
From the definition, we know
1
−1
a−b
~x = a
+b
=
,
1
1
a+b
and so
1
~x ·
=a−b
0 for suitable a and b.
Since a, b ≥ 0 and a + b≤ 1, the largest that a − b could be occurs when a = 1 and b = 0.
1
Therefore, the largest ~x ·
could be is 1.
0 YOU MUST SUBMIT THIS PAGE.
If you would like work on this page scored, then clearly indicate to which question the work belongs and
indicate on the page containing the original question that there is work on this page to score. YOU MUST SUBMIT THIS PAGE.
If you would like work on this page scored, then clearly indicate to which question the work belongs and
indicate on the page containing the original question that there is work on this page to score. YOU MUST SUBMIT THIS PAGE.
If you would like work on this page scored, then clearly indicate to which question the work belongs and
indicate on the page containing the original question that there is work on this page to score. ...

View
Full Document