# MATH 1025 Study Guide - Final Guide: Elementary Matrix, Parallelogram Law, Gaussian EliminationExam

by OC2218391

School

York UniversityDepartment

Mathematics and StatisticsCourse Code

MATH 1025Professor

Asia WeissStudy Guide

FinalThis

**preview**shows pages 1-2. to view the full**7 pages of the document.**YORK UNIVERSITY

Faculty of Science

Department of Mathematics and Statistics

MATH 1025 3.00 – Section N

Test #1 –February 01, 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 2–7. Answer the all questions.

4. Your work must justify your answers. Show your work on the space provided. Write your

solutions only on pages 2–7. Check the backs of the sheets for questions.

5. Do the easiest questions ﬁrst, GOOD LUCK!

6. Remain seated until we collect all the test papers.

Copyright 2019 c

I. Raguimov

Only pages 1-2 are available for preview. Some parts have been intentionally blurred.

1. (4 + 5 pts)

(a) Consider the triangle with vertices A, B and C. If Eis the midpoint of the side BC,

show that −→

AE =1

2(−−→

AB +−→

AC).

ANSWER:

−→

AE =−→

AC +−−→

CE =−→

AC +1

2−−→

CB.

So, −→

AE =1

2−→

AC +1

2(−→

AC +−−→

CB) = 1

2−→

AC +1

2−−→

AB.

∴−→

AE =1

2(−−→

AB +−→

AC).

OR EQUIVALENT using the Parallelogram Law.

(b) let −→

u=

3

1

−7

and −→

v=

1

0

5

.Write ~u =~uk+~u⊥,where ~ukis parallel to ~v and ~u⊥

is orthogonal to ~v.

ANSWER:

~uk=−−−−−→

proj ~v ~u =~u ·~v

k~vk2~v =(3)(1) + (1)(0) + (−7)(5)

(√12+ 02+ 52)2

1

0

5

=−16

13

1

0

5

=

−16

13

0

−80

13

.

Then

~u⊥=~u −~uk=

3

1

−7

−

−16

13

0

−80

13

=

55

13

1

−11

13

.

∴~u =

−16

13

0

−80

13

+

55

13

1

−11

13

.

Continues...

Copyright 2018 c

I. Raguimov

2

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