Linear Algebra I & Mathematics Tutorial 1bNagoya University, G30 Program

Fall 2020Instructor: Henrik Bachmann

Homework 1: Linear systems

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Deadline: 18th October, 2020

Exercise 1. (1 Point) Solve the Exercises 2 & 3 and write the solutions down by hand (paper, tablet) orby computer (Latex). Create one pdf-file (for example, by using a scanner app on your phone) and sendit before the deadline ends (any time on 18th October is fine) to henrik.bachmann@math.nagoya-u.ac.jp.Use the following format as a filename: ”Familyname Firstname LA1 HW1.pdf”.

A linear system is said to be on row-reduced echelon form if the following three conditions aresatisfied:

(i) The first (that it, the leftmost) variable in each equation has coe�cient 1.

(ii) If xi is the first variable in one of the equations, then it does not occur in any other equation in thesystem.

(iii) If xi is the first variable in one equation, then the equations below it do not contain any of thevariables x1, x2, . . . , xi�1.

Exercise 2. (2+2+2+1+1 = 8 Points) Which of the following linear systems are on row-reduced echelonform? For those that are not, find an equivalent system (i.e. one which has the same solutions) that ison row-reduced echelon form. For each system, find all solutions.

i)

⇢x1 + x2 + x3 + 2x4 = 0

x2 � x4 = 0

ii)

8<

:

x1 + 4x2 + 7x3 = 12x1 + 5x2 + 8x3 = 23x1 + 6x2 + 10x3 = 1

iii) x1 + 2x2 + 3x3 + 4x4 = 5

iv)

8<

:

x1 = 2x2 = 0x3 = 2

v)

⇢x1 + 3x2 = 1

3x1 + 9x2 = 2

Exercise 3. (5 Points) Decide for which real numbers a 2 R the following linear system has solutions.Give all the solutions in these cases.

8<

:

2x1 + 12x2 + 7x3 = 12a+ 72x1 + 4x2 + 2x3 = 12ax1 + 10x2 + 6x3 = 7a+ 8

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Version: October 3, 2020

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Solution

qThis is net on r ref

should be 0 since Xs is the first non zerovariable in the second row

ÖHref

Solutions pivot X ti 3tz fort ttRvar.XztzX3tfürtz

0 This is not on rrefO

First non zero coefficients are not I

X 14 2 7 3 1 X 14 2 7 3 12X 15 2 8 3 2 3 2 6 3 0

IX 6 2 10 3 1 6 2 11 3 2

X 14 2 7 3 1 µxi zx.ro IIII6 2 11 3 2

EE EErref

Solution X I

pijart.frz 43 2

This is on rref

Solutions pivots X 5 2T 3T 4Tvar

Ht fort ttRX3tztIax ts

This is on rref

Solution X 2

pijat.frzOX32

oThis is netonrref

is not 0

9

D

This linear system has no Solutions

We first bring this lin.snstemonrref

2X t 12 2 7 3 12 7 1 10 2 6 3 7 812g 2X 112 2 7 3 12 72

I 7 8 2 4 2 3 ka

X 1 10 2 6 3 7 8 1 10 2 6 3 7 8

Ifa 8 5 3 4 90 2 2

The last equation can just be satisfied if2a ZO.ie.a

9 1 1 10 2 6 3 7 8 1 In8 2 5 9 7 6 to I

t.io

ß X 110 2 6 3 1 X fXztIg Xz 15 3 5

Pivot Ittf rrefSolutions XE F It for TER

free I X twar