Study Guides (390,000)
CA (150,000)
York (10,000)
MATH (500)
Midterm

MATH 1025 Study Guide - Midterm Guide: Linear Map, Gang Of YouthsExam


Department
Mathematics and Statistics
Course Code
MATH 1025
Professor
Asia Weiss
Study Guide
Midterm

This preview shows pages 1-3. to view the full 9 pages of the document.
YORK UNIVERSITY
Faculty of Science
Department of Mathematics and Statistics
MATH 1025 3.00 – Section N
Test #3 Solutions March 26, 2019
Carefully read and complete the information below in ink.
SOLUTIONS
INSTRUCTIONS:
1. This is a closed book, closed notes test, duration – 50 minutes.
2. Calculators, Internet connected devices or other aids are NOT permitted.
3. There are 6 questions on pages 2 – 8. Answer the all questions.
4. Your work must justify your answers. Show your work on the space provided. Write your
solutions only on pages 2 – 8. Check the backs of the sheets for questions.
5. There are 55 points on the test, but to get 100% you will need 50 points. Do the easiest
questions first. GOOD LUCK!
6. Remain seated until we collect all the test papers.
Copyright 2019 c
I. Raguimov

Only pages 1-3 are available for preview. Some parts have been intentionally blurred.

1. (8 pts) Let
H= span{
2
3
1
,
4
6
2
,
0
2
2
}.
Is Ha subspace of R3? Answer “YES” or “NO”, and justify your answer.
If “YES” find a basis for H. What is the dimension of H?
ANSWER: YES.
Every span of a set of 4-dimensional vectors is a subspace of R4?
Or alternatively, using the subspace axioms:
u H&v H=(u +v)H,
u H=(k⃗u)HkR.
Let
A=
2 4 0
3 6 2
1 2 2
.
Then H= col (A).
Reduce Ato RREF:
2 4 0
3 6 2
1 2 2
1 2 0
3 6 2
1 2 2
1 2 0
0 0 2
0 0 2
120
001
000
.
Since Ahas two pivots, rank A = 2.The positions of the pivots show that the first and third
columns of Aare linearly independent and form a basis for col (A).
Therefore, dim H = 2 and B={
2
3
1
,
0
2
2
}is a basis for H.
Or alternative, using a REF of the matrix.
Continues...
Copyright 2019 c
I. Raguimov
2

Only pages 1-3 are available for preview. Some parts have been intentionally blurred.

2. (7 + 6 pts) Let
A=
221 3
111 2
1 1 1 0
.
(a) Find the rank of Aas well as bases for column and row spaces of A.
ANSWER:
ANSWER:
Reduce Ato the RREF:
221 3
111 2
1 1 1 0
R1R2
111 2
221 3
1 1 1 0
(2)R1+R2,(1)R1+R3
111 2
0011
0 0 2 2
(2)R2+R3
111 2
0011
0000
(1)R2+Rq
11 0 1
0 0 1 1
0 0 0 0
.
Since the RREF of Ahas two pivots, rank A = r = 2.The positions of the pivots show
that the first and third columns of Aare linearly independent and form a basis for
col (A),that is, B1={
2
1
1
,
1
1
1
}is a basis for col (A).
The positions of the pivots also show that the first two rows of the RREF of Ais a basis
for row (A),that is, B2={[1,1,0,1],[0,0,1,1]}is a basis for row (A).
Or alternative using a REF of the matrix.
Continues...
Copyright 2019 c
I. Raguimov
3
You're Reading a Preview

Unlock to view full version