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# Math Problem: Linear Algebra Assignment Paper #4 (Math Problem Sample)

Instructions:

THE TASK IS AN ASSIGNMENT ON LINEAR ALGEBRA.

source..Content:

Linear Algebra: Assignment 4

University Affiliation

Student Name

Course Name

Date

Introduction

Assignment 4 of the linear algebra module deals with elementary vector operations with applications of algebra methods to solve linear systems of equations. Methods, properties and theorems are adopted from Howard Anton & Chris Rorres book, Elementary Linear Algebra (11th Edition). The solutions are herein indicated stepwise and in order of questions as they appear in the original assignment file.

Question 1(a)

u=(1,2,-3,5,0)

v=(0,4,-1,1,2)

w=(7,1,-4,-2,3)

* v+w=0,4,-1,1,2+7,1,-4,-2,3

=0+7,4+,1,-1-4,1-2,2+3

=7,5,-5,-1.5

* 3(2u-v=3(21,2,-3,5,0-(0,4,-1,1,2))

=3(2,4,-6,10,0-(0,4,-1,1,2))

=3(2-0,4-4,-6+1,10-1,0-2)

=3(2,0,-5,9,-2))

=(6,0,-15,27,-6)

* 3u-v)-(2u+4w)=

=31,2,-3,5,0-0,4,-1,1,2-21,2,-3,5,0+7,1,-4,-2,3

=(3,6,-9,15,0-(0,4,-1,1,2))-(2,4,-6,10,0+(28,4,-16,-8,12))

=3-0,6-4,-9+1,15-1,0-2-2+28,4+4,-6-16,10-8,0+12

=3,2,-8,14,-2-30,8,-22,2,12

=(-28,-6,-30,12,-14)

* 12(w-5v+2u)

=127,1,-4,-2,3-50,4,-1,1,2+21,2,-3,5,0+0,4,-1,1,2

=127,1,-4,-2,3-0,20, -5,5,10+2,4,-6,10,0+(0,4-1,1,2)

=127-0+2,1-20+4,-4+5-6,-2-5+10,3-10+0+0,4,-1,1,12

=129,-15,-5,3-7+(0,4,-1,1,2)

=92,-152,-52,32,-72+0,4-1,1,2

=(92,-72,-72,52,-32)

Question 1(b)

Given:

c1-1,0,2+c12,2,-2+c31,-2,1=(-6,12,4)

-c1+2c2+c3=-6 â€¦i

2c2-2c3=12 â€¦ii

2c1-2c2+c3=4 â€¦(iii)

Solve eqn i and eqn (iii) to eliminate the variable c1

-2c1+4c2+2c3=-12

2c1-2c2+c3=4

Adding the two equations, the following resultant equation is obtained:

2c2+3c3=-8 â€¦(iv)

Solving equation (ii) and equation (iv):

2c2-2c3=12

2c2+3c3=-8

âŸ¹5c3=-20

c3=-4

âˆ´ c2-c3=6 âŸ¹ c2=2

Also:

c1=6+2c2+c3

c1=6+4-4

c1=6

Question 2(a)

* Given:

u=(1,2-3,0)

v=5,1,2,-2

âˆ´d=1-52+2-12+-3-22+(0--21)2

=16+1+25+4

46

and

cosÎ¸=u.vuv=1*5+2*1+-3*2+0*212+22+(-3)2+02*52+12+22+(-2)2

=5+2-6+01434=11434

Î¸=cos-1(11434)=87.37

Hence Î¸<900

* u=(0,1,1,1,2)

v=2,1,0,-1,3

âˆ´d=4+0+1+4+1=10

cosÎ¸=0+1+0-1+61+1+1+44+1+1+9=6715

Î¸=cos-167=54.150

Î¸ is an acute angle since Î¸<900

Question 2(b)

According to Cauchy-Schwarz inequality

u.vâ‰¤uv

* u=(4,1,1)

v=1,2,3

u.v=4*1+1*2+1*3

=9

u.v=9

u=16+1+1=18

v=1+4+9=15

u.v=18*15=16.43

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