Linear Independence - Math 218bfitzpat/teaching/218s20/lectures/... · Math 218 Brian D....

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Linear IndependenceMath 218

Brian D. Fitzpatrick

Duke University

March 3, 2020

MATH

Overview

Geometric MotivationCounting “Directions”

BackgroundLinear Combinations as Matrix MultiplicationAn Important Observation

Definitions and ExamplesDefinitionsExamples

The Linear Independence TestStatementExample

The “Pivot Columns” of a MatrixDefinitionExample


QuestionWhat does Span{ #»v } “look like”?

AnswerSpan{ #»v } consists of all “multiples” of { #»v }.

Span{ #»v }

2 · #»v3 · #»v

−2 · #»v−3 · #»v

#»v

Assuming #»v 6= #»

O , Span{ #»v } is the line containing #»v .




Span{ #»v }

2 · #»v3 · #»v

−2 · #»v−3 · #»v

#»v






Span{ #»v }

2 · #»v3 · #»v

−2 · #»v−3 · #»v

#»v






Span{ #»v }

2 · #»v

3 · #»v

−2 · #»v−3 · #»v

#»v






Span{ #»v }

2 · #»v3 · #»v

−2 · #»v−3 · #»v

#»v






Span{ #»v }

2 · #»v

3 · #»v

−2 · #»v−3 · #»v

#»v






Span{ #»v }

2 · #»v3 · #»v

−2 · #»v−3 · #»v

#»v






Span{ #»v }

2 · #»v3 · #»v

−2 · #»v

−3 · #»v

#»v






Span{ #»v }

2 · #»v3 · #»v

−2 · #»v−3 · #»v

#»v






Span{ #»v }

2 · #»v3 · #»v

−2 · #»v

−3 · #»v

#»v






Span{ #»v }

2 · #»v3 · #»v

−2 · #»v−3 · #»v

#»v






Span{ #»v }

2 · #»v3 · #»v

−2 · #»v−3 · #»v

#»v






Span{ #»v }

2 · #»v3 · #»v

−2 · #»v−3 · #»v

#»v




QuestionWhat does Span{ #»v 1,

#»v 2} “look like”?

AnswerSpan{ #»v 1,

#»v 2} consists of all “linear combinations” of { #»v 1,#»v 2}.

Span{#»v 1,#»v 2}

#»v 1

#»v 2

Assuming #»v 1 and #»v 2 are not parallel, Span{ #»v 1,#»v 2} is the plane

containing #»v 1 and #»v 2.




AnswerSpan{ #»v 1,


Span{#»v 1,#»v 2}

#»v 1

#»v 2






AnswerSpan{ #»v 1,


Span{#»v 1,#»v 2}

#»v 1

#»v 2






AnswerSpan{ #»v 1,


Span{#»v 1,#»v 2}

#»v 1

#»v 2






AnswerSpan{ #»v 1,


Span{#»v 1,#»v 2}

#»v 1

#»v 2






AnswerSpan{ #»v 1,


Span{#»v 1,#»v 2}

#»v 1

#»v 2






AnswerSpan{ #»v 1,


Span{#»v 1,#»v 2}

#»v 1

#»v 2






AnswerSpan{ #»v 1,


Span{#»v 1,#»v 2}

#»v 1

#»v 2






AnswerSpan{ #»v 1,


Span{#»v 1,#»v 2}

#»v 1

#»v 2






AnswerSpan{ #»v 1,


Span{#»v 1,#»v 2}

#»v 1

#»v 2




Span{ #»v }Defines a one-directional object (line) if #»v 6= #»

O .

Span{ #»v 1,#»v 2}

Defines a two-directional object (plane) if #»v 1 and #»v 2 are notparallel.

QuestionHow can we determine if Span{ #»v 1,

#»v 2, . . . ,#»v n} defines an

n-directional object?

AnswerSpan{ #»v 1,

#»v 2, . . . ,#»v n} defines an n-directional object if the list

{ #»v 1,#»v 2, . . . ,

#»v n} is linearly independent.



O .

Span{ #»v 1,#»v 2}



#»v 2, . . . ,#»v n} defines an


AnswerSpan{ #»v 1,


{ #»v 1,#»v 2, . . . ,




O .

Span{ #»v 1,#»v 2}



#»v 2, . . . ,#»v n} defines an


AnswerSpan{ #»v 1,


{ #»v 1,#»v 2, . . . ,




O .

Span{ #»v 1,#»v 2}



#»v 2, . . . ,#»v n} defines an


AnswerSpan{ #»v 1,


{ #»v 1,#»v 2, . . . ,



Example

Consider the matrix A given by

A =

−5 −15 −10 −20−2 −6 −4 −8

2 6 4 8

Note that Col(A) ⊂ R3. Each column is a multiple of the firstcolumn.

3 · #»a 1

−4 · #»a 1

−2 · #»a 1

Col(A)#»a 1

This illustrates that Col(A) = Span{〈−5, −2, 2〉 }.


Example


A =

−5 −15 −10 −20−2 −6 −4 −8

2 6 4 8

Note that Col(A) ⊂ R3.

Each column is a multiple of the firstcolumn.

3 · #»a 1

−4 · #»a 1

−2 · #»a 1

Col(A)#»a 1



Example


A =

−5 −15 −10 −20−2 −6 −4 −8

2 6 4 8


3 · #»a 1

−4 · #»a 1

−2 · #»a 1

Col(A)#»a 1



Example


A =

−5 −15 −10 −20−2 −6 −4 −8

2 6 4 8


3 · #»a 1

−4 · #»a 1

−2 · #»a 1

Col(A)

#»a 1



Example


A =

−5 −15 −10 −20−2 −6 −4 −8

2 6 4 8


3 · #»a 1

−4 · #»a 1

−2 · #»a 1

Col(A)

#»a 1



Example


A =

−5 −15 −10 −20−2 −6 −4 −8

2 6 4 8


3 · #»a 1

−4 · #»a 1

−2 · #»a 1

Col(A)

#»a 1



Example


A =

−5 −15 −10 −20−2 −6 −4 −8

2 6 4 8


3 · #»a 1

−4 · #»a 1

−2 · #»a 1

Col(A)

#»a 1



Example


A =

−5 −15 −10 −20−2 −6 −4 −8

2 6 4 8


3 · #»a 1

−4 · #»a 1

−2 · #»a 1

Col(A)#»a 1



Example


A =

−5 −15 −10 −20−2 −6 −4 −8

2 6 4 8


3 · #»a 1

−4 · #»a 1

−2 · #»a 1

Col(A)#»a 1



Example


A =

1 2 7 6−8 −16 3 11

3 6 −2 −52 4 −9 −11


2 · #»a 1

#»a 3

− #»a 1 + #»a 3#»a 1

This means that Col(A) = Span{〈1, −8, 3, 2〉 , 〈7, 3, −2, −9〉 }.


Example


A =

1 2 7 6−8 −16 3 11

3 6 −2 −52 4 −9 −11


2 · #»a 1

#»a 3

− #»a 1 + #»a 3#»a 1



Example


A =

1 2 7 6−8 −16 3 11

3 6 −2 −52 4 −9 −11


2 · #»a 1

#»a 3

− #»a 1 + #»a 3

#»a 1



Example


A =

1 2 7 6−8 −16 3 11

3 6 −2 −52 4 −9 −11


2 · #»a 1

#»a 3

− #»a 1 + #»a 3

#»a 1



Example


A =

1 2 7 6−8 −16 3 11

3 6 −2 −52 4 −9 −11


2 · #»a 1

#»a 3

− #»a 1 + #»a 3

#»a 1



Example


A =

1 2 7 6−8 −16 3 11

3 6 −2 −52 4 −9 −11


2 · #»a 1

#»a 3

− #»a 1 + #»a 3#»a 1



Example


A =

1 2 7 6−8 −16 3 11

3 6 −2 −52 4 −9 −11


2 · #»a 1

#»a 3

− #»a 1 + #»a 3#»a 1


BackgroundLinear Combinations as Matrix Multiplication

ObservationLet { #»v 1,

#»v 2, . . . ,#»v n} be a list of vectors in Rm.

Every linearcombination

c1 · #»v 1 + c2 · #»v 2 + · · ·+ cn · #»v n =#»

b

is of the form A #»c =#»

b where

A =[

#»v 1#»v 2 · · · #»v n

]#»c =

c1c2...cn

#»

b =

b1b2...bm

Note that A is an m × n matrix.



#»v 2, . . . ,#»v n} be a list of vectors in Rm. Every linear

combination

c1 · #»v 1 + c2 · #»v 2 + · · ·+ cn · #»v n =#»

b


b where

A =[

#»v 1#»v 2 · · · #»v n

]#»c =

c1c2...cn

#»

b =

b1b2...bm





combination

c1 · #»v 1 + c2 · #»v 2 + · · ·+ cn · #»v n =#»

b


b where

A =[

#»v 1#»v 2 · · · #»v n

]#»c =

c1c2...cn

#»

b =

b1b2...bm





combination

c1 · #»v 1 + c2 · #»v 2 + · · ·+ cn · #»v n =#»

b


b where

A =[

#»v 1#»v 2 · · · #»v n

]#»c =

c1c2...cn

#»

b =

b1b2...bm



Example

The linear combination

c1 ·[

131

]+ c2 ·

[0−3

]+ c3 ·

[7−5

]+ c4 ·

[−5−3

]=

[33−11

]may be written as

[1 0 7 −5

31 −3 −5 −3

] c1c2c3c4

=

[33−11

]


Example

The linear combination

c1 ·[

131

]+ c2 ·

[0−3

]+ c3 ·

[7−5

]+ c4 ·

[−5−3

]=

[33−11

]may be written as

[1 0 7 −5

31 −3 −5 −3

] c1c2c3c4

=

[33−11

]

BackgroundAn Important Observation


#»v 2, . . . ,#»v n} be a list of vectors in Rm. The equation

c1 · #»v 1 + c2 · #»v 2 + · · ·+ cn · #»v n =#»

O (∗)

can always be solved by c1 = c2 = · · · = cn = 0.

This is called thetrivial linear combination.

QuestionWhen is (∗) solved by a nontrivial linear combination?




c1 · #»v 1 + c2 · #»v 2 + · · ·+ cn · #»v n =#»

O (∗)

can always be solved by c1 = c2 = · · · = cn = 0. This is called thetrivial linear combination.





c1 · #»v 1 + c2 · #»v 2 + · · ·+ cn · #»v n =#»

O (∗)

can always be solved by c1 = c2 = · · · = cn = 0. This is called thetrivial linear combination.


Definitions and ExamplesDefinitions

DefinitionA list of vectors { #»v 1,

#»v 2, . . . ,#»v n} in Rm is linearly independent if

the only solution to

c1 · #»v 1 + c2 · #»v 2 + · · ·+ cn · #»v n =#»

O

is the trivial solution c1 = c2 = · · · = cn = 0.

The list{ #»v 1,

#»v 2, . . . ,#»v n} is linearly dependent if it is not linearly

independent.

NoteThe list { #»v 1,

#»v 2, . . . ,#»v n} is linearly dependent if the equation

c1 · #»v 1 + c2 · #»v 2 + · · ·+ cn · #»v n =#»

O

has a nontrivial solution.





c1 · #»v 1 + c2 · #»v 2 + · · ·+ cn · #»v n =#»

O

is the trivial solution c1 = c2 = · · · = cn = 0. The list{ #»v 1,


independent.



c1 · #»v 1 + c2 · #»v 2 + · · ·+ cn · #»v n =#»

O






c1 · #»v 1 + c2 · #»v 2 + · · ·+ cn · #»v n =#»

O

is the trivial solution c1 = c2 = · · · = cn = 0. The list{ #»v 1,


independent.



c1 · #»v 1 + c2 · #»v 2 + · · ·+ cn · #»v n =#»

O


Definitions and ExamplesExamples

Example

Note that

(3)

10−4

+ (1)

−21

15

+ (1)

−1−1−3

+ (0)

−53

41

=

000

This means that columns of 1 −2 −1 −50 1 −1 3−4 15 −3 41

are linearly dependent.


Example

Note that

(3)

10−4

+ (1)

−21

15

+ (1)

−1−1−3

+ (0)

−53

41

=

000

This means that columns of 1 −2 −1 −5

0 1 −1 3−4 15 −3 41

are linearly dependent.


Example

Determine if the list { #»v 1,#»v 2,

#»v 3} is linearly independent where

#»v 1 = 〈−1, 1, 1, 2〉 #»v 2 = 〈1, −2, −1, −3〉 #»v 3 = 〈1, 0, 0, 2〉

To determine if { #»v 1,#»v 2,

#»v 3} is linearly independent, we consider

c1 · #»v 1 + c2 · #»v 2 + c3 · #»v 3 =#»

O

This gives the system−1 1 1 0

1 −2 0 01 −1 0 02 −3 2 0

1 0 0 00 1 0 00 0 1 00 0 0 0

Thus c1 = c2 = c3 = 0. Hence { #»v 1,

#»v 2,#»v 3} is linearly

independent.


Example



#»v 1 = 〈−1, 1, 1, 2〉 #»v 2 = 〈1, −2, −1, −3〉 #»v 3 = 〈1, 0, 0, 2〉



c1 · #»v 1 + c2 · #»v 2 + c3 · #»v 3 =#»

O


1 −2 0 01 −1 0 02 −3 2 0

1 0 0 00 1 0 00 0 1 00 0 0 0

Thus c1 = c2 = c3 = 0. Hence { #»v 1,


independent.


Example



#»v 1 = 〈−1, 1, 1, 2〉 #»v 2 = 〈1, −2, −1, −3〉 #»v 3 = 〈1, 0, 0, 2〉



c1 · #»v 1 + c2 · #»v 2 + c3 · #»v 3 =#»

O


1 −2 0 01 −1 0 02 −3 2 0

1 0 0 00 1 0 00 0 1 00 0 0 0

Thus c1 = c2 = c3 = 0. Hence { #»v 1,


independent.


Example



#»v 1 = 〈−1, 1, 1, 2〉 #»v 2 = 〈1, −2, −1, −3〉 #»v 3 = 〈1, 0, 0, 2〉



c1 · #»v 1 + c2 · #»v 2 + c3 · #»v 3 =#»

O


1 −2 0 01 −1 0 02 −3 2 0

1 0 0 00 1 0 00 0 1 00 0 0 0

Thus c1 = c2 = c3 = 0. Hence { #»v 1,#»v 2,

#»v 3} is linearlyindependent.


Example



#»v 1 = 〈−1, 1, 1, 2〉 #»v 2 = 〈1, −2, −1, −3〉 #»v 3 = 〈1, 0, 0, 2〉



c1 · #»v 1 + c2 · #»v 2 + c3 · #»v 3 =#»

O


1 −2 0 01 −1 0 02 −3 2 0

1 0 0 00 1 0 00 0 1 00 0 0 0

Thus c1 = c2 = c3 = 0.

Hence { #»v 1,#»v 2,

#»v 3} is linearlyindependent.


Example



#»v 1 = 〈−1, 1, 1, 2〉 #»v 2 = 〈1, −2, −1, −3〉 #»v 3 = 〈1, 0, 0, 2〉



c1 · #»v 1 + c2 · #»v 2 + c3 · #»v 3 =#»

O


1 −2 0 01 −1 0 02 −3 2 0

1 0 0 00 1 0 00 0 1 00 0 0 0

Thus c1 = c2 = c3 = 0. Hence { #»v 1,


independent.


Example

Suppose that { #»v 1,#»v 2,

#»v 3,#»v 4} is linearly independent. Show that

{ #»v 1 − #»v 4,#»v 2 − #»v 4,

#»v 3 − #»v 4} is linearly independent.

SolutionSuppose that

c1 · ( #»v 1 − #»v 4) + c2 · ( #»v 2 − #»v 4) + c3 · ( #»v 3 − #»v 4) =#»

O

Then

c1 · #»v 1 + c2 · #»v 2 + c3 · #»v 3 + (−c1 − c2 − c3) · #»v 4 =#»

O (∗)

Since { #»v 1,#»v 2,

#»v 3,#»v 4} is linearly independent, each coefficient in

(∗) must be zero. In particular, c1 = c2 = c3 = 0.


Example



{ #»v 1 − #»v 4,#»v 2 − #»v 4,



c1 · ( #»v 1 − #»v 4) + c2 · ( #»v 2 − #»v 4) + c3 · ( #»v 3 − #»v 4) =#»

O

Then

c1 · #»v 1 + c2 · #»v 2 + c3 · #»v 3 + (−c1 − c2 − c3) · #»v 4 =#»

O (∗)

Since { #»v 1,#»v 2,




Example



{ #»v 1 − #»v 4,#»v 2 − #»v 4,



c1 · ( #»v 1 − #»v 4) + c2 · ( #»v 2 − #»v 4) + c3 · ( #»v 3 − #»v 4) =#»

O

Then

c1 · #»v 1 + c2 · #»v 2 + c3 · #»v 3 + (−c1 − c2 − c3) · #»v 4 =#»

O (∗)

Since { #»v 1,#»v 2,




Example



{ #»v 1 − #»v 4,#»v 2 − #»v 4,



c1 · ( #»v 1 − #»v 4) + c2 · ( #»v 2 − #»v 4) + c3 · ( #»v 3 − #»v 4) =#»

O

Then

c1 · #»v 1 + c2 · #»v 2 + c3 · #»v 3 + (−c1 − c2 − c3) · #»v 4 =#»

O (∗)

Since { #»v 1,#»v 2,


(∗) must be zero.

In particular, c1 = c2 = c3 = 0.


Example



{ #»v 1 − #»v 4,#»v 2 − #»v 4,



c1 · ( #»v 1 − #»v 4) + c2 · ( #»v 2 − #»v 4) + c3 · ( #»v 3 − #»v 4) =#»

O

Then

c1 · #»v 1 + c2 · #»v 2 + c3 · #»v 3 + (−c1 − c2 − c3) · #»v 4 =#»

O (∗)

Since { #»v 1,#»v 2,



The Linear Independence TestStatement

NoteThe equation

c1 · #»v 1 + c2 · #»v 2 + · · ·+ ck · #»v k =#»

O

is given by the augmented matrix[#»v 1

#»v 2 · · · #»v k#»

O]

So, { #»v 1,#»v 2, . . . ,

#»v k} is linearly independent if and only if

A =[

#»v 1#»v 2 · · · #»v k

]has full column rank.


NoteThe equation

c1 · #»v 1 + c2 · #»v 2 + · · ·+ ck · #»v k =#»

O


#»v 2 · · · #»v k#»

O]

So, { #»v 1,#»v 2, . . . ,


A =[

#»v 1#»v 2 · · · #»v k

]has

full column rank.


NoteThe equation

c1 · #»v 1 + c2 · #»v 2 + · · ·+ ck · #»v k =#»

O


#»v 2 · · · #»v k#»

O]

So, { #»v 1,#»v 2, . . . ,


A =[

#»v 1#»v 2 · · · #»v k



Theorem (The Linear Independence Test)

The list { #»v 1,#»v 2, . . . ,

#»v k} is linearly independent if and only if[#»v 1

#»v 2 · · · #»v k


The Linear Independence TestExample

Example

Determine if the list

{〈−14, −5, −5, −3〉 , 〈−53, −19, −18, −13〉 , 〈78, 28, 26, 20〉 }

is linearly independent.

SolutionNote that

rref

−14 −53 78−5 −19 28−5 −18 26−3 −13 20

=

1 0 20 1 −20 0 00 0 0

Since rank = 2 < # columns, the list is linearly dependent.


Example


{〈−14, −5, −5, −3〉 , 〈−53, −19, −18, −13〉 , 〈78, 28, 26, 20〉 }


SolutionNote that

rref

−14 −53 78−5 −19 28−5 −18 26−3 −13 20

=

1 0 20 1 −20 0 00 0 0



Example


{〈−14, −5, −5, −3〉 , 〈−53, −19, −18, −13〉 , 〈78, 28, 26, 20〉 }


SolutionNote that

rref

−14 −53 78−5 −19 28−5 −18 26−3 −13 20

=

1 0 20 1 −20 0 00 0 0



Example


{〈−1, −5, −1, 1〉 , 〈7, −14, −1, −2〉 , 〈33, −31, 2, −13〉 }


SolutionNote that

rref

−1 7 33−5 −14 −31−1 −1 2

1 −2 −13

=

1 0 00 1 00 0 10 0 0

Since rank = 3 = # columns, the list is linearly independent.


Example


{〈−1, −5, −1, 1〉 , 〈7, −14, −1, −2〉 , 〈33, −31, 2, −13〉 }


SolutionNote that

rref

−1 7 33−5 −14 −31−1 −1 2

1 −2 −13

=

1 0 00 1 00 0 10 0 0



Example


{〈−1, −5, −1, 1〉 , 〈7, −14, −1, −2〉 , 〈33, −31, 2, −13〉 }


SolutionNote that

rref

−1 7 33−5 −14 −31−1 −1 2

1 −2 −13

=

1 0 00 1 00 0 10 0 0



Example

Determine if the list 1

5−5

,

−2−914

,

−6−29

35

,

526−23


SolutionThe matrix

A =

1 −2 −6 55 −9 −29 26−5 14 35 −23

satisfies rank(A) ≤ 3. Since rank(A) 6= # columns = 4, the list isnot linearly independent.


Example


5−5

,

−2−914

,

−6−29

35

,

526−23


SolutionThe matrix

A =

1 −2 −6 55 −9 −29 26−5 14 35 −23

satisfies rank(A) ≤

3. Since rank(A) 6= # columns = 4, the list isnot linearly independent.


Example


5−5

,

−2−914

,

−6−29

35

,

526−23


SolutionThe matrix

A =

1 −2 −6 55 −9 −29 26−5 14 35 −23

satisfies rank(A) ≤ 3.

Since rank(A) 6= # columns = 4, the list isnot linearly independent.


Example


5−5

,

−2−914

,

−6−29

35

,

526−23


SolutionThe matrix

A =

1 −2 −6 55 −9 −29 26−5 14 35 −23

satisfies rank(A) ≤ 3. Since rank(A) 6= # columns = 4, the list isnot linearly independent.

The “Pivot Columns” of a MatrixDefinition

DefinitionThe pivot columns of a matrix A are the columns of A thatcorrespond to pivot columns in rref(A).

Example

Consider the calculation

rref

A3 −9 7−2 6 2

1 −3 −613 −39 0

=

1 −3 00 0 10 0 00 0 0

The pivot columns of A are 〈3, −2, 1, 13〉 and 〈7, 2, −6, 0〉 .



Example


rref

A3 −9 7−2 6 2

1 −3 −613 −39 0

=

1 −3 00 0 10 0 00 0 0




Example


rref

A3 −9 7−2 6 2

1 −3 −613 −39 0

=

1 −3 00 0 10 0 00 0 0



TheoremThe pivot columns of a matrix are linearly independent.

TheoremLet { #»v 1,

#»v 2, . . . ,#»v d} be the pivot columns of A. Then

Col(A) = Col([

#»v 1#»v 2 · · · #»v d

])

In particular, every column of A is a linear combination of the pivotcolumns of A.


TheoremThe pivot columns of a matrix are linearly independent.

TheoremLet { #»v 1,

#»v 2, . . . ,#»v d} be the pivot columns of A. Then

Col(A) = Col([

#»v 1#»v 2 · · · #»v d

])

In particular, every column of A is a linear combination of the pivotcolumns of A.

The “Pivot Columns” of a MatrixExample

Example


rref

1 −3 4 −3 −11−2 6 −8 7 28

5 −15 20 −11 −31

=

1 −3 4 0 70 0 0 1 60 0 0 0 0

This shows that

Col

1 −3 4 −3 −11−2 6 −8 7 28

5 −15 20 −11 −31

= Col

1 −3−2 7

5 −11

We also have

#»a 2 = − 3 · #»a 1#»a 3 = 4 · #»a 1

#»a 5 = 7 · #»a 1 + 6 · #»a 4


Example


rref

1 −3 4 −3 −11−2 6 −8 7 28

5 −15 20 −11 −31

=

1 −3 4 0 70 0 0 1 60 0 0 0 0

This shows that

Col

1 −3 4 −3 −11−2 6 −8 7 28

5 −15 20 −11 −31

= Col

1 −3−2 7

5 −11

We also have

#»a 2 = − 3 · #»a 1#»a 3 = 4 · #»a 1

#»a 5 = 7 · #»a 1 + 6 · #»a 4


Example


rref

1 −3 4 −3 −11−2 6 −8 7 28

5 −15 20 −11 −31

=

1 −3 4 0 70 0 0 1 60 0 0 0 0

This shows that

Col

1 −3 4 −3 −11−2 6 −8 7 28

5 −15 20 −11 −31

= Col

1 −3−2 7

5 −11

We also have

#»a 2 =

− 3 · #»a 1#»a 3 = 4 · #»a 1

#»a 5 = 7 · #»a 1 + 6 · #»a 4


Example


rref

1 −3 4 −3 −11−2 6 −8 7 28

5 −15 20 −11 −31

=

1 −3 4 0 70 0 0 1 60 0 0 0 0

This shows that

Col

1 −3 4 −3 −11−2 6 −8 7 28

5 −15 20 −11 −31

= Col

1 −3−2 7

5 −11

We also have

#»a 2 = − 3 · #»a 1

#»a 3 = 4 · #»a 1#»a 5 = 7 · #»a 1 + 6 · #»a 4


Example


rref

1 −3 4 −3 −11−2 6 −8 7 28

5 −15 20 −11 −31

=

1 −3 4 0 70 0 0 1 60 0 0 0 0

This shows that

Col

1 −3 4 −3 −11−2 6 −8 7 28

5 −15 20 −11 −31

= Col

1 −3−2 7

5 −11

We also have

#»a 2 = − 3 · #»a 1#»a 3 =

4 · #»a 1#»a 5 = 7 · #»a 1 + 6 · #»a 4


Example


rref

1 −3 4 −3 −11−2 6 −8 7 28

5 −15 20 −11 −31

=

1 −3 4 0 70 0 0 1 60 0 0 0 0

This shows that

Col

1 −3 4 −3 −11−2 6 −8 7 28

5 −15 20 −11 −31

= Col

1 −3−2 7

5 −11

We also have

#»a 2 = − 3 · #»a 1#»a 3 = 4 · #»a 1

#»a 5 = 7 · #»a 1 + 6 · #»a 4


Example


rref

1 −3 4 −3 −11−2 6 −8 7 28

5 −15 20 −11 −31

=

1 −3 4 0 70 0 0 1 60 0 0 0 0

This shows that

Col

1 −3 4 −3 −11−2 6 −8 7 28

5 −15 20 −11 −31

= Col

1 −3−2 7

5 −11

We also have

#»a 2 = − 3 · #»a 1#»a 3 = 4 · #»a 1

#»a 5 =

7 · #»a 1 + 6 · #»a 4


Example


rref

1 −3 4 −3 −11−2 6 −8 7 28

5 −15 20 −11 −31

=

1 −3 4 0 70 0 0 1 60 0 0 0 0

This shows that

Col

1 −3 4 −3 −11−2 6 −8 7 28

5 −15 20 −11 −31

= Col

1 −3−2 7

5 −11

We also have

#»a 2 = − 3 · #»a 1#»a 3 = 4 · #»a 1

#»a 5 = 7 · #»a 1 + 6 · #»a 4

Date post:	22-Aug-2020
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