School

York UniversityDepartment

Mathematics and StatisticsCourse Code

MATH 1025Professor

Asia WeissStudy Guide

MidtermThis

**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)∈H∀k∈R.

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=

2−2−1 3

1−1−1 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:

2−2−1 3

1−1−1 2

−1 1 −1 0

R1↔R2

−→

1−1−1 2

2−2−1 3

−1 1 −1 0

(−2)R1+R2,(1)R1+R3

−→

1−1−1 2

001−1

0 0 −2 2

(2)R2+R3

−→

1−1−1 2

001−1

0000

(1)R2+Rq

−→

1−1 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

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