linear algebra. session 1 - math.tamu.eduroquesol/math_304_fall... · a survey linear algebra is a...

Introduction

Linear Algebra. Session 1

Dr. Marco A Roque Sol

08/28/2018

Dr. Marco A Roque Sol Linear Algebra. Session 1

Introduction

Table of ContentsA SurveySistems of linear equationsGaussian Elimination. Gauss-Jordan Reduction

Table of contents

IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains


Introduction


Table of contents

Introduction

Systems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains


Introduction


Table of contents

IntroductionSystems of linear equations

Gaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains


Introduction


Table of contents

IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reduction

Matrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains


Introduction


Table of contents

IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebra

DeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains


Introduction


Table of contents

IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminants

Vector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains


Introduction


Table of contents

IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spaces

Linear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains


Introduction


Table of contents

IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independence

Basis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains


Introduction


Table of contents

IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimension

Coordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains


Introduction


Table of contents

IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basis

Linear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains


Introduction


Table of contents

IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformations

OrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains


Introduction


Table of contents

IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonality

Inner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains


Introduction


Table of contents

IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and norms

The Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains


Introduction


Table of contents

IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization process

Eigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains


Introduction


Table of contents

IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value Decomposition

Matrix exponentials, Diagonalization, and Markov Chains


Introduction


Table of contents

IntroductionSystems of linear equationsGaussian elimination, Gauss-Jordan reductionMatrices, matrix algebraDeterminantsVector spacesLinear independenceBasis and dimensionCoordinates, change of basisLinear transformationsOrthogonalityInner products and normsThe Gram-Schmidt orthogonalization processEigenvalues and eigenvectors. Singular Value DecompositionMatrix exponentials, Diagonalization, and Markov Chains


Introduction


A Survey

Linear Algebra

( Medical Imaging )


Introduction


A Survey

Linear Algebra

( Medical Imaging )Dr. Marco A Roque Sol Linear Algebra. Session 1

Introduction


A Survey

Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.

Here are some concrete areas where we can find applications ofLinear Algebra.

Abstract Thinking

Chemistry

Coding theory

Coupled oscillations

Cryptography

Economics


Introduction


A Survey

Linear algebra is a cornerstone

in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.


Abstract Thinking

Chemistry

Coding theory


Cryptography

Economics


Introduction


A Survey

Linear algebra is a cornerstone in undergraduate mathematicaleducation.

It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.


Abstract Thinking

Chemistry

Coding theory


Cryptography

Economics


Introduction


A Survey

Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used

by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.


Abstract Thinking

Chemistry

Coding theory


Cryptography

Economics


Introduction


A Survey

Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence.

It hence, evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.


Abstract Thinking

Chemistry

Coding theory


Cryptography

Economics


Introduction


A Survey

Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence,

evolves naturally towardsabstraction. For most students, it is a first contact with modernmathematics.


Abstract Thinking

Chemistry

Coding theory


Cryptography

Economics


Introduction


A Survey

Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction.

For most students, it is a first contact with modernmathematics.


Abstract Thinking

Chemistry

Coding theory


Cryptography

Economics


Introduction


A Survey

Linear algebra is a cornerstone in undergraduate mathematicaleducation. It develops a general language used by all scientists andis interdisciplinary in essence. It hence, evolves naturally towardsabstraction. For most students,

it is a first contact with modernmathematics.


Abstract Thinking

Chemistry

Coding theory


Cryptography

Economics


Introduction


A Survey



Abstract Thinking

Chemistry

Coding theory


Cryptography

Economics


Introduction


A Survey


Here are some concrete areas

where we can find applications ofLinear Algebra.

Abstract Thinking

Chemistry

Coding theory


Cryptography

Economics


Introduction


A Survey


Here are some concrete areas where we can find applications

ofLinear Algebra.

Abstract Thinking

Chemistry

Coding theory


Cryptography

Economics


Introduction


A Survey



Abstract Thinking

Chemistry

Coding theory


Cryptography

Economics


Introduction


A Survey

Classical Electromagnetism.

Geophysics.

Elimination Theory.

Game Theory.

Genetics.

Geometry.

Graph theory.

Heat distribution.

Image compression.


Introduction


A Survey

Linear Programming.

Markov Chains.

Networks.

Sociology

The Fibonacci numbers.

Eigenstates.


Introduction


A Survey

Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2

A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).

For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1

And the graph of the lines is:


Introduction


A Survey

Definition (Linear Equation)

An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2





Introduction


A Survey

Definition (Linear Equation)An equation

of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2





Introduction


A Survey

Definition (Linear Equation)An equation of the form

ax + by = c (5x + 7y = 6) is called linearbecause its solution set is a straight line in R2





Introduction


A Survey

Definition (Linear Equation)An equation of the form ax + by = c

(5x + 7y = 6) is called linearbecause its solution set is a straight line in R2





Introduction


A Survey

Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6)

is called linearbecause its solution set is a straight line in R2





Introduction


A Survey

Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linear

because its solution set is a straight line in R2





Introduction


A Survey

Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause

its solution set is a straight line in R2





Introduction


A Survey

Definition (Linear Equation)An equation of the form ax + by = c (5x + 7y = 6) is called linearbecause its solution set is

a straight line in R2





Introduction


A Survey






Introduction


A Survey


A solution

of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).




Introduction


A Survey


A solution of the equation

is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).




Introduction


A Survey


A solution of the equation is a pair of numbers

(x0, y0) ∈ R2 suchthat ax0 + by0 = c (5x0 + 7y0 = 6).




Introduction


A Survey


A solution of the equation is a pair of numbers (x0, y0) ∈ R2

suchthat ax0 + by0 = c (5x0 + 7y0 = 6).




Introduction


A Survey


A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat

ax0 + by0 = c (5x0 + 7y0 = 6).




Introduction


A Survey


A solution of the equation is a pair of numbers (x0, y0) ∈ R2 suchthat ax0 + by0 = c

(5x0 + 7y0 = 6).




Introduction


A Survey






Introduction


A Survey



For example,

(1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1



Introduction


A Survey



For example, (1, 1/7) and

(−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1



Introduction


A Survey



For example, (1, 1/7) and (−1/5, 1),

are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1



Introduction


A Survey



For example, (1, 1/7) and (−1/5, 1), are solutions.

In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1



Introduction


A Survey



For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,

we can write the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1



Introduction


A Survey



For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write

the first solution, as x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1



Introduction


A Survey



For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as

x = 1, y = 1/5 and the secondone, as x = −1/5, y = 1



Introduction


A Survey



For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5

and the secondone, as x = −1/5, y = 1



Introduction


A Survey



For example, (1, 1/7) and (−1/5, 1), are solutions. In another way,we can write the first solution, as x = 1, y = 1/5 and the secondone, as

x = −1/5, y = 1



Introduction


A Survey






Introduction


Sistems of linear equations

General equation of a line

The General equation of a line is given by

ax0 + by0 = c

where x , y are variables and a, b, and c are constants (except forthe case a = b = 0 )

Definition

A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b

where a1, a2, · · · an and b are constants.

A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that

a1α1 + a2α2 + · · ·+ anαn = b


Introduction




The General equation

of a line is given by

ax0 + by0 = c


Definition




a1α1 + a2α2 + · · ·+ anαn = b


Introduction




The General equation of a line

is given by

ax0 + by0 = c


Definition




a1α1 + a2α2 + · · ·+ anαn = b


Introduction





ax0 + by0 = c


Definition




a1α1 + a2α2 + · · ·+ anαn = b


Introduction





ax0 + by0 = c

where x , y

are variables and a, b, and c are constants (except forthe case a = b = 0 )

Definition




a1α1 + a2α2 + · · ·+ anαn = b


Introduction





ax0 + by0 = c

where x , y are variables and

a, b, and c are constants (except forthe case a = b = 0 )

Definition




a1α1 + a2α2 + · · ·+ anαn = b


Introduction





ax0 + by0 = c

where x , y are variables and a, b, and c

are constants (except forthe case a = b = 0 )

Definition




a1α1 + a2α2 + · · ·+ anαn = b


Introduction





ax0 + by0 = c


Definition




a1α1 + a2α2 + · · ·+ anαn = b


Introduction





ax0 + by0 = c


Definition

A linear equation

in the variables x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b



a1α1 + a2α2 + · · ·+ anαn = b


Introduction





ax0 + by0 = c


Definition

A linear equation in the variables

x1, x2, · · · , xn, is an equation ofthe form: a1x1 + a2x2 + · · · anxn = b



a1α1 + a2α2 + · · ·+ anαn = b


Introduction





ax0 + by0 = c


Definition

A linear equation in the variables x1, x2, · · · , xn, is an equation ofthe form:

a1x1 + a2x2 + · · · anxn = b



a1α1 + a2α2 + · · ·+ anαn = b


Introduction





ax0 + by0 = c


Definition




a1α1 + a2α2 + · · ·+ anαn = b


Introduction





ax0 + by0 = c


Definition



A solution

of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn such that

a1α1 + a2α2 + · · ·+ anαn = b


Introduction





ax0 + by0 = c


Definition



A solution of the equation

is an array of numbersα1, α2, · · ·αn ∈ Rn such that

a1α1 + a2α2 + · · ·+ anαn = b


Introduction





ax0 + by0 = c


Definition



A solution of the equation is an array of numbersα1, α2, · · ·αn ∈ Rn

such that

a1α1 + a2α2 + · · ·+ anαn = b


Introduction





ax0 + by0 = c


Definition




a1α1 + a2α2 + · · ·+ anαn = b


Introduction





ax0 + by0 = c


Definition




a1α1 + a2α2 + · · ·+ anαn = bDr. Marco A Roque Sol Linear Algebra. Session 1

Introduction


System of linear equations


A System of linear equations is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm

Here x1, x2, · · · xn are variables and aij , bj are constants.

In many applications, we will find such a systems of equations andit is here, where all the study of Linear Algebra starts.


Introduction




A System of linear equations

is an expression of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm




Introduction




A System of linear equations is an expression

of the form:a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm




Introduction




A System of linear equations is an expression of the form:

a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm




Introduction





a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm




Introduction





a21x1 + a22x2 + · · ·+ a2nxn = b2

...am1x1 + am2x2 + · · ·+ amnxn = bm




Introduction





a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm




Introduction





a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm

Here

x1, x2, · · · xn are variables and aij , bj are constants.



Introduction





a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm

Here x1, x2, · · · xn

are variables and aij , bj are constants.



Introduction





a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm

Here x1, x2, · · · xn are variables and

aij , bj are constants.



Introduction





a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm

Here x1, x2, · · · xn are variables and aij , bj

are constants.



Introduction





a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm




Introduction





a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm


In many applications,

we will find such a systems of equations andit is here, where all the study of Linear Algebra starts.


Introduction





a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm


In many applications, we will find such a systems of equations

andit is here, where all the study of Linear Algebra starts.


Introduction





a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm


In many applications, we will find such a systems of equations andit is here,

where all the study of Linear Algebra starts.


Introduction





a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm


In many applications, we will find such a systems of equations andit is here, where all the study

of Linear Algebra starts.


Introduction





a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm


In many applications, we will find such a systems of equations andit is here, where all the study of Linear Algebra

starts.


Introduction





a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm




Introduction


Systems of linear equations

Example 1.1

Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and solving the system of two equations in twounknowns {

x − y = −22x + 3y = 6

⇔ ( Equivalent System){x = y −2

2x + 3y = 6

⇔ {x = y − 2

2(y − 2) + 3y = 6


Introduction



Example 1.1

Find the point of intersection

of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and solving the system of two equations in twounknowns {

x − y = −22x + 3y = 6


2x + 3y = 6

⇔ {x = y − 2

2(y − 2) + 3y = 6


Introduction



Example 1.1

Find the point of intersection of the lines x − y = −2 and

2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and solving the system of two equations in twounknowns {

x − y = −22x + 3y = 6


2x + 3y = 6

⇔ {x = y − 2

2(y − 2) + 3y = 6


Introduction



Example 1.1

Find the point of intersection of the lines x − y = −2 and2x + 3y = 6,

in R2. You can find the solution of this problem bysetting up and solving the system of two equations in twounknowns {

x − y = −22x + 3y = 6


2x + 3y = 6

⇔ {x = y − 2

2(y − 2) + 3y = 6


Introduction



Example 1.1

Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2.

You can find the solution of this problem bysetting up and solving the system of two equations in twounknowns {

x − y = −22x + 3y = 6


2x + 3y = 6

⇔ {x = y − 2

2(y − 2) + 3y = 6


Introduction



Example 1.1

Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find

the solution of this problem bysetting up and solving the system of two equations in twounknowns {

x − y = −22x + 3y = 6


2x + 3y = 6

⇔ {x = y − 2

2(y − 2) + 3y = 6


Introduction



Example 1.1

Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution

of this problem bysetting up and solving the system of two equations in twounknowns {

x − y = −22x + 3y = 6


2x + 3y = 6

⇔ {x = y − 2

2(y − 2) + 3y = 6


Introduction



Example 1.1

Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem

bysetting up and solving the system of two equations in twounknowns {

x − y = −22x + 3y = 6


2x + 3y = 6

⇔ {x = y − 2

2(y − 2) + 3y = 6


Introduction



Example 1.1

Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and

solving the system of two equations in twounknowns {

x − y = −22x + 3y = 6


2x + 3y = 6

⇔ {x = y − 2

2(y − 2) + 3y = 6


Introduction



Example 1.1

Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and solving the system

of two equations in twounknowns {

x − y = −22x + 3y = 6


2x + 3y = 6

⇔ {x = y − 2

2(y − 2) + 3y = 6


Introduction



Example 1.1

Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and solving the system of two equations

in twounknowns {

x − y = −22x + 3y = 6


2x + 3y = 6

⇔ {x = y − 2

2(y − 2) + 3y = 6


Introduction



Example 1.1

Find the point of intersection of the lines x − y = −2 and2x + 3y = 6, in R2. You can find the solution of this problem bysetting up and solving the system of two equations in twounknowns

{x − y = −2

2x + 3y = 6


2x + 3y = 6

⇔ {x = y − 2

2(y − 2) + 3y = 6


Introduction



Example 1.1


x − y = −22x + 3y = 6


2x + 3y = 6

⇔ {x = y − 2

2(y − 2) + 3y = 6


Introduction



Example 1.1


x − y = −22x + 3y = 6

⇔ ( Equivalent System)

{x = y −2

2x + 3y = 6

⇔ {x = y − 2

2(y − 2) + 3y = 6


Introduction



Example 1.1


x − y = −22x + 3y = 6


2x + 3y = 6

⇔

{x = y − 2

2(y − 2) + 3y = 6


Introduction



Example 1.1


x − y = −22x + 3y = 6


2x + 3y = 6

⇔ {x = y − 2

2(y − 2) + 3y = 6


Introduction



⇔ {x = y − 2

5y = 10

⇔ {x = y − 2y = 2

⇔ {x = 0y = 2

Thus, the solution is given by the point (0, 2)


Introduction



⇔

{x = y − 2

5y = 10

⇔ {x = y − 2y = 2

⇔ {x = 0y = 2



Introduction



⇔ {x = y − 2

5y = 10

⇔ {x = y − 2y = 2

⇔ {x = 0y = 2



Introduction



⇔ {x = y − 2

5y = 10

⇔

{x = y − 2y = 2

⇔ {x = 0y = 2



Introduction



⇔ {x = y − 2

5y = 10

⇔ {x = y − 2y = 2

⇔

{x = 0y = 2



Introduction



⇔ {x = y − 2

5y = 10

⇔ {x = y − 2y = 2

⇔ {x = 0y = 2



Introduction



⇔ {x = y − 2

5y = 10

⇔ {x = y − 2y = 2

⇔ {x = 0y = 2

Thus,

the solution is given by the point (0, 2)


Introduction



⇔ {x = y − 2

5y = 10

⇔ {x = y − 2y = 2

⇔ {x = 0y = 2

Thus, the solution

is given by the point (0, 2)


Introduction



⇔ {x = y − 2

5y = 10

⇔ {x = y − 2y = 2

⇔ {x = 0y = 2

Thus, the solution is given by

the point (0, 2)


Introduction



⇔ {x = y − 2

5y = 10

⇔ {x = y − 2y = 2

⇔ {x = 0y = 2



Introduction



{x − y = −2

2x + 3y = 6x = 0; y = 2


Introduction



{x − y = −2

2x + 3y = 6

x = 0; y = 2


Introduction



{x − y = −2

2x + 3y = 6x = 0; y = 2


Introduction



In a similar way, we have

{2x + 3y = 22x + 3y = 6

inconsistent system (no solution)


Introduction




{2x + 3y = 22x + 3y = 6

inconsistent system

(no solution)


Introduction




{2x + 3y = 22x + 3y = 6

inconsistent system (no solution)


Introduction



{4x + 6y = 122x + 3y = 6

⇒ 2x+3y = 6 (infinitely many solutions)


Introduction



{4x + 6y = 122x + 3y = 6

⇒ 2x+3y = 6

(infinitely many solutions)


Introduction



{4x + 6y = 122x + 3y = 6

⇒ 2x+3y = 6 (infinitely many solutions)


Introduction



Solving systems of linear equations

Elimination Method

Algorithm:(1) pick a variable, solve one of the equations for it, and eliminateit from the other equations;

(2) put aside the equation used in the elimination, and return tostep (1).

The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.


Introduction




Elimination Method

Algorithm:

(1) pick a variable, solve one of the equations for it, and eliminateit from the other equations;




Introduction




Elimination Method

Algorithm:(1) pick a variable,

solve one of the equations for it, and eliminateit from the other equations;




Introduction




Elimination Method

Algorithm:(1) pick a variable, solve one of the equations for it, and

eliminateit from the other equations;




Introduction




Elimination Method

Algorithm:(1) pick a variable, solve one of the equations for it, and eliminateit

from the other equations;




Introduction




Elimination Method





Introduction




Elimination Method


(2) put aside the equation

used in the elimination, and return tostep (1).



Introduction




Elimination Method


(2) put aside the equation used in the elimination, and

return tostep (1).



Introduction




Elimination Method





Introduction




Elimination Method



The algorithm

reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.


Introduction




Elimination Method



The algorithm reduces the number of variables

(as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.


Introduction




Elimination Method



The algorithm reduces the number of variables (as well as thenumber of equations),

hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.


Introduction




Elimination Method



The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after

a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear how to complete solution.


Introduction




Elimination Method



The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops,

the system is simplified so that it shouldbe clear how to complete solution.


Introduction




Elimination Method



The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that

it shouldbe clear how to complete solution.


Introduction




Elimination Method



The algorithm reduces the number of variables (as well as thenumber of equations), hence it stops after a finite number of steps.After the algorithm stops, the system is simplified so that it shouldbe clear

how to complete solution.


Introduction




Elimination Method





Introduction



Example 1.2 x − y = 2

2x − y− z = 3x + y+ z = 6

Solve the 1st equation for x :x = y + 2

2x − y− z = 3x + y+ z = 6

⇔Eliminate x from the 2nd and 3rd equations:

x = y + 22(y + 2) − y− z = 3(y + 2) + y+ z = 6


Introduction



Example 1.2

x − y = 2

2x − y− z = 3x + y+ z = 6


2x − y− z = 3x + y+ z = 6


x = y + 22(y + 2) − y− z = 3(y + 2) + y+ z = 6


Introduction




2x − y− z = 3x + y+ z = 6


2x − y− z = 3x + y+ z = 6


x = y + 22(y + 2) − y− z = 3(y + 2) + y+ z = 6


Introduction




2x − y− z = 3x + y+ z = 6

Solve the 1st equation for x :

x = y + 2

2x − y− z = 3x + y+ z = 6


x = y + 22(y + 2) − y− z = 3(y + 2) + y+ z = 6


Introduction




2x − y− z = 3x + y+ z = 6


2x − y− z = 3x + y+ z = 6

⇔

Eliminate x from the 2nd and 3rd equations:x = y + 2

2(y + 2) − y− z = 3(y + 2) + y+ z = 6


Introduction




2x − y− z = 3x + y+ z = 6


2x − y− z = 3x + y+ z = 6


x = y + 2

2(y + 2) − y− z = 3(y + 2) + y+ z = 6


Introduction




2x − y− z = 3x + y+ z = 6


2x − y− z = 3x + y+ z = 6


x = y + 22(y + 2) − y− z = 3(y + 2) + y+ z = 6


Introduction



⇔

Simplify: x = y+ 2y− z = −12y + z = 4

In this way we have that, the whole system has been reduced to asystem (2nd and 3rd equations) of two linear equations in twovariables.

Solve the 2nd equation for y :x = y+ 2y = z− 12y + z = 4

⇔


Introduction



⇔Simplify:

x = y+ 2y− z = −12y + z = 4



⇔


Introduction



⇔Simplify:

x = y+ 2y− z = −12y + z = 4

In this way

we have that, the whole system has been reduced to asystem (2nd and 3rd equations) of two linear equations in twovariables.


⇔


Introduction



⇔Simplify:

x = y+ 2y− z = −12y + z = 4

In this way we have that,

the whole system has been reduced to asystem (2nd and 3rd equations) of two linear equations in twovariables.


⇔


Introduction



⇔Simplify:

x = y+ 2y− z = −12y + z = 4

In this way we have that, the whole system has been reduced

to asystem (2nd and 3rd equations) of two linear equations in twovariables.


⇔


Introduction



⇔Simplify:

x = y+ 2y− z = −12y + z = 4

In this way we have that, the whole system has been reduced to asystem (2nd and 3rd equations)

of two linear equations in twovariables.


⇔


Introduction



⇔Simplify:

x = y+ 2y− z = −12y + z = 4

In this way we have that, the whole system has been reduced to asystem (2nd and 3rd equations) of two linear equations

in twovariables.


⇔


Introduction



⇔Simplify:

x = y+ 2y− z = −12y + z = 4



⇔


Introduction



⇔Simplify:

x = y+ 2y− z = −12y + z = 4


Solve the 2nd equation for y :

x = y+ 2y = z− 12y + z = 4

⇔


Introduction



⇔Simplify:

x = y+ 2y− z = −12y + z = 4



⇔Dr. Marco A Roque Sol Linear Algebra. Session 1

Introduction



Eliminate y from the 3rd equation:x = y+ 2y = z+ 1

2(z − 1) + z = 4⇔

Simplify x = y+ 2y = z+ 13z = 6

⇔

Thus,the elimination process, has been completed. Now, thesystem is easily solved by back substitution.


Introduction



Eliminate y from the 3rd equation:

x = y+ 2y = z+ 1

2(z − 1) + z = 4⇔


⇔



Introduction




2(z − 1) + z = 4⇔


⇔



Introduction




2(z − 1) + z = 4⇔

Simplify

x = y+ 2y = z+ 13z = 6

⇔



Introduction




2(z − 1) + z = 4⇔


⇔



Introduction




2(z − 1) + z = 4⇔


⇔

Thus,

the elimination process, has been completed. Now, thesystem is easily solved by back substitution.


Introduction




2(z − 1) + z = 4⇔


⇔

Thus,the elimination process,

has been completed. Now, thesystem is easily solved by back substitution.


Introduction




2(z − 1) + z = 4⇔


⇔

Thus,the elimination process, has been completed.

Now, thesystem is easily solved by back substitution.


Introduction




2(z − 1) + z = 4⇔


⇔

Thus,the elimination process, has been completed. Now,

thesystem is easily solved by back substitution.


Introduction




2(z − 1) + z = 4⇔


⇔

Thus,the elimination process, has been completed. Now, thesystem

is easily solved by back substitution.


Introduction




2(z − 1) + z = 4⇔


⇔

Thus,the elimination process, has been completed. Now, thesystem is easily solved by

back substitution.


Introduction




2(z − 1) + z = 4⇔


⇔



Introduction



That is, we find z from the 3rd equation, then substitute it in the2nd equation and find y , then substitute y and z in the 1stequation and find x .

x = y + 2y = z + 1

z = 2

x = y + 2y = 1z = 2

x = 3y = 1z = 2

Finally, the System of linear equations:x − y = 2

2x − y− z = 3x + y+ z = 6

has as a solution, the point (x , y , z) = (3, 1, 2) .


Introduction



That is,

we find z from the 3rd equation, then substitute it in the2nd equation and find y , then substitute y and z in the 1stequation and find x .

x = y + 2y = z + 1

z = 2

x = y + 2y = 1z = 2

x = 3y = 1z = 2


2x − y− z = 3x + y+ z = 6



Introduction



That is, we find z

from the 3rd equation, then substitute it in the2nd equation and find y , then substitute y and z in the 1stequation and find x .

x = y + 2y = z + 1

z = 2

x = y + 2y = 1z = 2

x = 3y = 1z = 2


2x − y− z = 3x + y+ z = 6



Introduction



That is, we find z from the 3rd equation,

then substitute it in the2nd equation and find y , then substitute y and z in the 1stequation and find x .

x = y + 2y = z + 1

z = 2

x = y + 2y = 1z = 2

x = 3y = 1z = 2


2x − y− z = 3x + y+ z = 6



Introduction



That is, we find z from the 3rd equation, then substitute it

in the2nd equation and find y , then substitute y and z in the 1stequation and find x .

x = y + 2y = z + 1

z = 2

x = y + 2y = 1z = 2

x = 3y = 1z = 2


2x − y− z = 3x + y+ z = 6



Introduction



That is, we find z from the 3rd equation, then substitute it in the2nd equation and

find y , then substitute y and z in the 1stequation and find x .

x = y + 2y = z + 1

z = 2

x = y + 2y = 1z = 2

x = 3y = 1z = 2


2x − y− z = 3x + y+ z = 6



Introduction



That is, we find z from the 3rd equation, then substitute it in the2nd equation and find y ,

then substitute y and z in the 1stequation and find x .

x = y + 2y = z + 1

z = 2

x = y + 2y = 1z = 2

x = 3y = 1z = 2


2x − y− z = 3x + y+ z = 6



Introduction



That is, we find z from the 3rd equation, then substitute it in the2nd equation and find y , then substitute y and z

in the 1stequation and find x .

x = y + 2y = z + 1

z = 2

x = y + 2y = 1z = 2

x = 3y = 1z = 2


2x − y− z = 3x + y+ z = 6



Introduction



That is, we find z from the 3rd equation, then substitute it in the2nd equation and find y , then substitute y and z in the 1stequation and

find x .

x = y + 2y = z + 1

z = 2

x = y + 2y = 1z = 2

x = 3y = 1z = 2


2x − y− z = 3x + y+ z = 6



Introduction




x = y + 2y = z + 1

z = 2

x = y + 2y = 1z = 2

x = 3y = 1z = 2


2x − y− z = 3x + y+ z = 6



Introduction




x = y + 2y = z + 1

z = 2

x = y + 2y = 1z = 2

x = 3y = 1z = 2

Finally,

the System of linear equations:x − y = 2

2x − y− z = 3x + y+ z = 6



Introduction




x = y + 2y = z + 1

z = 2

x = y + 2y = 1z = 2

x = 3y = 1z = 2

Finally, the System of linear equations:

x − y = 2

2x − y− z = 3x + y+ z = 6



Introduction




x = y + 2y = z + 1

z = 2

x = y + 2y = 1z = 2

x = 3y = 1z = 2


2x − y− z = 3x + y+ z = 6



Introduction




x = y + 2y = z + 1

z = 2

x = y + 2y = 1z = 2

x = 3y = 1z = 2


2x − y− z = 3x + y+ z = 6

has as a solution,

the point (x , y , z) = (3, 1, 2) .


Introduction




x = y + 2y = z + 1

z = 2

x = y + 2y = 1z = 2

x = 3y = 1z = 2


2x − y− z = 3x + y+ z = 6

has as a solution, the point

(x , y , z) = (3, 1, 2) .


Introduction




x = y + 2y = z + 1

z = 2

x = y + 2y = 1z = 2

x = 3y = 1z = 2


2x − y− z = 3x + y+ z = 6

has as a solution, the point (x , y , z) = (3, 1, 2) .Dr. Marco A Roque Sol Linear Algebra. Session 1

Introduction


Gaussian Elimination. Gauss-Jordan Reduction

Gaussian elimination


Remember that a System of linear equations is an expression ofthe form:

a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm


A solution of the system is a common solution of all equationspresent in the system.

A system of linear equations can have one solution, infinitely manysolutions, or no solution at all.


Introduction





Remember

that a System of linear equations is an expression ofthe form:

a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm





Introduction





Remember that a System of linear equations

is an expression ofthe form:

a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm





Introduction





Remember that a System of linear equations is an expression

ofthe form:

a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm





Introduction






a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm





Introduction






a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm

Here x1, x2, · · · xn

are variables and aij , bj are constants.




Introduction






a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm

Here x1, x2, · · · xn are variables and

aij , bj are constants.




Introduction






a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm





Introduction






a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm


A solution of the system

is a common solution of all equationspresent in the system.



Introduction






a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm


A solution of the system is a common solution of all equations

present in the system.



Introduction






a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm





Introduction






a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm



A system

of linear equations can have one solution, infinitely manysolutions, or no solution at all.


Introduction






a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm



A system of linear equations

can have one solution, infinitely manysolutions, or no solution at all.


Introduction






a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm



A system of linear equations can have

one solution, infinitely manysolutions, or no solution at all.


Introduction






a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm



A system of linear equations can have one solution,

infinitely manysolutions, or no solution at all.


Introduction






a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm



A system of linear equations can have one solution, infinitely manysolutions, or

no solution at all.


Introduction






a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm





Introduction



Using the Elimination method we had already shown the first caseand using it again we can illustrate the other two extra cases

Example 1.3 x + y− 2z = 1

y− z = 3−x + 4y− 3z = 14

Solve the 1st equation for x :⇔ ( Equivalent Systems)

x = −y+ 2z+ 1y − z = 3

−x + 4y− 3z = 14

Eliminate x from the 3rd equation :


Introduction



Using the Elimination method

we had already shown the first caseand using it again we can illustrate the other two extra cases


y− z = 3−x + 4y− 3z = 14


x = −y+ 2z+ 1y − z = 3

−x + 4y− 3z = 14



Introduction



Using the Elimination method we had already shown

the first caseand using it again we can illustrate the other two extra cases


y− z = 3−x + 4y− 3z = 14


x = −y+ 2z+ 1y − z = 3

−x + 4y− 3z = 14



Introduction



Using the Elimination method we had already shown the first caseand

using it again we can illustrate the other two extra cases


y− z = 3−x + 4y− 3z = 14


x = −y+ 2z+ 1y − z = 3

−x + 4y− 3z = 14



Introduction



Using the Elimination method we had already shown the first caseand using it again

we can illustrate the other two extra cases


y− z = 3−x + 4y− 3z = 14


x = −y+ 2z+ 1y − z = 3

−x + 4y− 3z = 14



Introduction



Using the Elimination method we had already shown the first caseand using it again we can illustrate

the other two extra cases


y− z = 3−x + 4y− 3z = 14


x = −y+ 2z+ 1y − z = 3

−x + 4y− 3z = 14



Introduction



Using the Elimination method we had already shown the first caseand using it again we can illustrate the other

two extra cases


y− z = 3−x + 4y− 3z = 14


x = −y+ 2z+ 1y − z = 3

−x + 4y− 3z = 14



Introduction





y− z = 3−x + 4y− 3z = 14


x = −y+ 2z+ 1y − z = 3

−x + 4y− 3z = 14



Introduction




Example 1.3

x + y− 2z = 1

y− z = 3−x + 4y− 3z = 14


x = −y+ 2z+ 1y − z = 3

−x + 4y− 3z = 14



Introduction





y− z = 3−x + 4y− 3z = 14


x = −y+ 2z+ 1y − z = 3

−x + 4y− 3z = 14



Introduction





y− z = 3−x + 4y− 3z = 14


⇔ ( Equivalent Systems)x = −y+ 2z+ 1

y − z = 3−x + 4y− 3z = 14



Introduction





y− z = 3−x + 4y− 3z = 14


x = −y+ 2z+ 1

y − z = 3−x + 4y− 3z = 14



Introduction





y− z = 3−x + 4y− 3z = 14


x = −y+ 2z+ 1y − z = 3

−x + 4y− 3z = 14



Introduction





y− z = 3−x + 4y− 3z = 14


x = −y+ 2z+ 1y − z = 3

−x + 4y− 3z = 14

Eliminate x from the 3rd equation :Dr. Marco A Roque Sol Linear Algebra. Session 1

Introduction



⇔ x = −y + 2z + 1

y − z = 3−(−y + 2z + 1) + 4y − 3z = 14

Simplify : x = −y + 2z + 1

y − z = 35y − 5z = 15

Solve the 2nd equation for yx = −y+ 2z+ 1y = z+ 35y − 5z = 15


Introduction



⇔

x = −y + 2z + 1

y − z = 3−(−y + 2z + 1) + 4y − 3z = 14


y − z = 35y − 5z = 15



Introduction



⇔ x = −y + 2z + 1

y − z = 3−(−y + 2z + 1) + 4y − 3z = 14


y − z = 35y − 5z = 15



Introduction



⇔ x = −y + 2z + 1

y − z = 3−(−y + 2z + 1) + 4y − 3z = 14

Simplify :

x = −y + 2z + 1

y − z = 35y − 5z = 15



Introduction



⇔ x = −y + 2z + 1

y − z = 3−(−y + 2z + 1) + 4y − 3z = 14


y − z = 35y − 5z = 15



Introduction



⇔ x = −y + 2z + 1

y − z = 3−(−y + 2z + 1) + 4y − 3z = 14


y − z = 35y − 5z = 15

Solve the 2nd equation for y

x = −y+ 2z+ 1y = z+ 35y − 5z = 15


Introduction



⇔ x = −y + 2z + 1

y − z = 3−(−y + 2z + 1) + 4y − 3z = 14


y − z = 35y − 5z = 15



Introduction



Eliminate y from the 3rd equation:x = −y+ 2z+ 1y = z+ 3

5(z + 3)− 5z = 15

⇔Simplify :

x = −y+ 2z+ 1y = z + 3

0 = 0

Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .


Introduction




x = −y+ 2z+ 1y = z+ 3

5(z + 3)− 5z = 15

⇔Simplify :

x = −y+ 2z+ 1y = z + 3

0 = 0



Introduction




5(z + 3)− 5z = 15

⇔Simplify :

x = −y+ 2z+ 1y = z + 3

0 = 0



Introduction




5(z + 3)− 5z = 15

⇔

Simplify : x = −y+ 2z+ 1y = z + 3

0 = 0



Introduction




5(z + 3)− 5z = 15

⇔Simplify :

x = −y+ 2z+ 1y = z + 3

0 = 0



Introduction




5(z + 3)− 5z = 15

⇔Simplify :

x = −y+ 2z+ 1y = z + 3

0 = 0

Now,

the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .


Introduction




5(z + 3)− 5z = 15

⇔Simplify :

x = −y+ 2z+ 1y = z + 3

0 = 0

Now, the elimination process is completed.

The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .


Introduction




5(z + 3)− 5z = 15

⇔Simplify :

x = −y+ 2z+ 1y = z + 3

0 = 0

Now, the elimination process is completed. The last equation

isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .


Introduction




5(z + 3)− 5z = 15

⇔Simplify :

x = −y+ 2z+ 1y = z + 3

0 = 0

Now, the elimination process is completed. The last equation isactually 0z = 0.

Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .


Introduction




5(z + 3)− 5z = 15

⇔Simplify :

x = −y+ 2z+ 1y = z + 3

0 = 0

Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z ,

is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .


Introduction




5(z + 3)− 5z = 15

⇔Simplify :

x = −y+ 2z+ 1y = z + 3

0 = 0

Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is,

it can beassigned an arbitrary value. Then, x and y are found by backsubstitution .


Introduction




5(z + 3)− 5z = 15

⇔Simplify :

x = −y+ 2z+ 1y = z + 3

0 = 0

Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned

an arbitrary value. Then, x and y are found by backsubstitution .


Introduction




5(z + 3)− 5z = 15

⇔Simplify :

x = −y+ 2z+ 1y = z + 3

0 = 0

Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value.

Then, x and y are found by backsubstitution .


Introduction




5(z + 3)− 5z = 15

⇔Simplify :

x = −y+ 2z+ 1y = z + 3

0 = 0

Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y

are found by backsubstitution .


Introduction




5(z + 3)− 5z = 15

⇔Simplify :

x = −y+ 2z+ 1y = z + 3

0 = 0

Now, the elimination process is completed. The last equation isactually 0z = 0. Hence z , is a free variable, that is, it can beassigned an arbitrary value. Then, x and y are found by

backsubstitution .


Introduction




5(z + 3)− 5z = 15

⇔Simplify :

x = −y+ 2z+ 1y = z + 3

0 = 0



Introduction



z = t (a parameter)

y = z + 3x = −y + 2z + 1

⇔ z = t

y = t + 3x = t − 2

Thus, the system x + y− 2z = 1

y − z = 3−x + 4y− 3z = 14


Introduction



z = t (a parameter)

y = z + 3x = −y + 2z + 1

⇔

z = t

y = t + 3x = t − 2


y − z = 3−x + 4y− 3z = 14


Introduction



z = t (a parameter)

y = z + 3x = −y + 2z + 1

⇔ z = t

y = t + 3x = t − 2


y − z = 3−x + 4y− 3z = 14


Introduction



z = t (a parameter)

y = z + 3x = −y + 2z + 1

⇔ z = t

y = t + 3x = t − 2

Thus, the system

x + y− 2z = 1

y − z = 3−x + 4y− 3z = 14


Introduction



z = t (a parameter)

y = z + 3x = −y + 2z + 1

⇔ z = t

y = t + 3x = t − 2


y − z = 3−x + 4y− 3z = 14


Introduction



has the General Solution

(x , y , z) = (t − 2, t + 3, t)

or in vector form ( vector equation of a line )

(x , y , z) = (−2, 3, 0) + t(1, 1, 1)

The set of all solutions is a straight line in R3 passing through thepoint P = (−2, 3, 0) in the direction v =< 1, 1, 1 > .


Introduction




(x , y , z) = (t − 2, t + 3, t)

or in vector form

( vector equation of a line )

(x , y , z) = (−2, 3, 0) + t(1, 1, 1)



Introduction




(x , y , z) = (t − 2, t + 3, t)


(x , y , z) = (−2, 3, 0) + t(1, 1, 1)



Introduction




(x , y , z) = (t − 2, t + 3, t)


(x , y , z) = (−2, 3, 0) + t(1, 1, 1)

The set of all solutions

is a straight line in R3 passing through thepoint P = (−2, 3, 0) in the direction v =< 1, 1, 1 > .


Introduction




(x , y , z) = (t − 2, t + 3, t)


(x , y , z) = (−2, 3, 0) + t(1, 1, 1)

The set of all solutions is a straight line

in R3 passing through thepoint P = (−2, 3, 0) in the direction v =< 1, 1, 1 > .


Introduction




(x , y , z) = (t − 2, t + 3, t)


(x , y , z) = (−2, 3, 0) + t(1, 1, 1)

The set of all solutions is a straight line in R3

passing through thepoint P = (−2, 3, 0) in the direction v =< 1, 1, 1 > .


Introduction




(x , y , z) = (t − 2, t + 3, t)


(x , y , z) = (−2, 3, 0) + t(1, 1, 1)

The set of all solutions is a straight line in R3 passing through thepoint

P = (−2, 3, 0) in the direction v =< 1, 1, 1 > .


Introduction




(x , y , z) = (t − 2, t + 3, t)


(x , y , z) = (−2, 3, 0) + t(1, 1, 1)

The set of all solutions is a straight line in R3 passing through thepoint P = (−2, 3, 0)

in the direction v =< 1, 1, 1 > .


Introduction




(x , y , z) = (t − 2, t + 3, t)


(x , y , z) = (−2, 3, 0) + t(1, 1, 1)

The set of all solutions is a straight line in R3 passing through thepoint P = (−2, 3, 0) in the direction

v =< 1, 1, 1 > .


Introduction




(x , y , z) = (t − 2, t + 3, t)


(x , y , z) = (−2, 3, 0) + t(1, 1, 1)



Introduction



Example 1.4

x + y− 2z = 1

y− z = 3−x + 4y− 3z = 1


⇔ (Equivalent System )x = −y + 2z + 1

y − z = 3−x + 4y − 3z = 1


Introduction



Example 1.4

x + y− 2z = 1

y− z = 3−x + 4y− 3z = 1


⇔ (Equivalent System )

x = −y + 2z + 1

y − z = 3−x + 4y − 3z = 1


Introduction



Example 1.4

x + y− 2z = 1

y− z = 3−x + 4y− 3z = 1


⇔ (Equivalent System )x = −y + 2z + 1

y − z = 3−x + 4y − 3z = 1


Introduction



Eliminate x from the 3rd equation:x = −y + 2z + 1

y − z = 3−(−y + 2z + 1) + 4y − 3z = 1

Simplify: x = −y + 2z + 1

y − z = 35y − 5z = 2

Solve the second equation for y :x = −y + 2z + 1

y = z + 35y − 5z = 2


Introduction



Eliminate x from the 3rd equation:

x = −y + 2z + 1

y − z = 3−(−y + 2z + 1) + 4y − 3z = 1


y − z = 35y − 5z = 2


y = z + 35y − 5z = 2


Introduction




y − z = 3−(−y + 2z + 1) + 4y − 3z = 1


y − z = 35y − 5z = 2


y = z + 35y − 5z = 2


Introduction




y − z = 3−(−y + 2z + 1) + 4y − 3z = 1

Simplify:

x = −y + 2z + 1

y − z = 35y − 5z = 2


y = z + 35y − 5z = 2


Introduction




y − z = 3−(−y + 2z + 1) + 4y − 3z = 1


y − z = 35y − 5z = 2


y = z + 35y − 5z = 2


Introduction




y − z = 3−(−y + 2z + 1) + 4y − 3z = 1


y − z = 35y − 5z = 2

Solve the second equation for y :

x = −y + 2z + 1

y = z + 35y − 5z = 2


Introduction




y − z = 3−(−y + 2z + 1) + 4y − 3z = 1


y − z = 35y − 5z = 2


y = z + 35y − 5z = 2


Introduction



Eliminate y from the 3rd equation:x = −y + 2z + 1

y = z + 35(z + 3)− 5z = 2


y = z + 315 = 2

Now, the elimination process is completed. The last equationactually is giving us a contradiction. Hence, there is no solution forthis system.


Introduction




x = −y + 2z + 1

y = z + 35(z + 3)− 5z = 2


y = z + 315 = 2



Introduction




y = z + 35(z + 3)− 5z = 2


y = z + 315 = 2



Introduction




y = z + 35(z + 3)− 5z = 2

Simplify:

x = −y + 2z + 1

y = z + 315 = 2



Introduction




y = z + 35(z + 3)− 5z = 2


y = z + 315 = 2



Introduction




y = z + 35(z + 3)− 5z = 2


y = z + 315 = 2

Now,

the elimination process is completed. The last equationactually is giving us a contradiction. Hence, there is no solution forthis system.


Introduction




y = z + 35(z + 3)− 5z = 2


y = z + 315 = 2

Now, the elimination process is completed.

The last equationactually is giving us a contradiction. Hence, there is no solution forthis system.


Introduction




y = z + 35(z + 3)− 5z = 2


y = z + 315 = 2

Now, the elimination process is completed. The last equation

actually is giving us a contradiction. Hence, there is no solution forthis system.


Introduction




y = z + 35(z + 3)− 5z = 2


y = z + 315 = 2

Now, the elimination process is completed. The last equationactually is giving us

a contradiction. Hence, there is no solution forthis system.


Introduction




y = z + 35(z + 3)− 5z = 2


y = z + 315 = 2



Introduction




y− z = 3−x + 4y− 3z = 1


φ = empty set


Introduction



Thus, the system

x + y− 2z = 1

y− z = 3−x + 4y− 3z = 1


φ = empty set


Introduction




y− z = 3−x + 4y− 3z = 1


φ = empty set


Introduction




Gaussian elimination is a modification of the elimination methodthat allows only so-called elementary operations .

Elementary operations for systems of linear equations:

(1) to multiply an equation by a nonzero scalar;

(2) to add an equation multiplied by a scalar to another equation;

(3) to interchange any two equations.


Introduction





is a modification of the elimination methodthat allows only so-called elementary operations .






Introduction




Gaussian elimination is a modification of

the elimination methodthat allows only so-called elementary operations .






Introduction




Gaussian elimination is a modification of the elimination method

that allows only so-called elementary operations .






Introduction




Gaussian elimination is a modification of the elimination methodthat allows only

so-called elementary operations .






Introduction










Introduction





Elementary operations

for systems of linear equations:





Introduction





Elementary operations for systems of

linear equations:





Introduction










Introduction






(1) to multiply

an equation by a nonzero scalar;




Introduction






(1) to multiply an equation

by a nonzero scalar;




Introduction










Introduction







(2) to add

an equation multiplied by a scalar to another equation;



Introduction







(2) to add an equation

multiplied by a scalar to another equation;



Introduction







(2) to add an equation multiplied by a scalar

to another equation;



Introduction










Introduction








(3) to interchange

any two equations.


Introduction










Introduction



The next result, is very important, because it ensures that thisoperations will not modify the solution set of the original system.

Theorem

(i) Applying elementary operations to a system of linear equationsdoes not change the solution set of the system.

(ii) Any elementary operation can be undone by anotherelementary operation.


Introduction



The next result,

is very important, because it ensures that thisoperations will not modify the solution set of the original system.

Theorem




Introduction



The next result, is very important,

because it ensures that thisoperations will not modify the solution set of the original system.

Theorem




Introduction



The next result, is very important, because it ensures

that thisoperations will not modify the solution set of the original system.

Theorem




Introduction



The next result, is very important, because it ensures that thisoperations

will not modify the solution set of the original system.

Theorem




Introduction



The next result, is very important, because it ensures that thisoperations will not

modify the solution set of the original system.

Theorem




Introduction



The next result, is very important, because it ensures that thisoperations will not modify the solution set

of the original system.

Theorem




Introduction




Theorem




Introduction




Theorem

(i) Applying

elementary operations to a system of linear equationsdoes not change the solution set of the system.



Introduction




Theorem

(i) Applying elementary operations

to a system of linear equationsdoes not change the solution set of the system.



Introduction




Theorem

(i) Applying elementary operations to a system

of linear equationsdoes not change the solution set of the system.



Introduction




Theorem

(i) Applying elementary operations to a system of linear equations

does not change the solution set of the system.



Introduction




Theorem

(i) Applying elementary operations to a system of linear equationsdoes not change

the solution set of the system.



Introduction




Theorem

(i) Applying elementary operations to a system of linear equationsdoes not change the solution set

of the system.



Introduction




Theorem




Introduction




Theorem


(ii) Any

elementary operation can be undone by anotherelementary operation.


Introduction




Theorem


(ii) Any elementary operation

can be undone by anotherelementary operation.


Introduction




Theorem


(ii) Any elementary operation can be undone

by anotherelementary operation.


Introduction




Theorem


(ii) Any elementary operation can be undone by another

elementary operation.


Introduction




Theorem




Introduction



Operation 1: Multiply the ith equation by r 6= 0

a11x1 + a12x2 + · · ·+ a1nxn = b1...

ai1x1 + ai2x2 + · · ·+ ainxn = bi...

am1x1 + am2x2 + · · ·+ amnxn = bm

⇒

a11x1 + a12x2 + · · ·+ a1nxn = b1...

(rai1)x1 + (rai2)x2 + · · ·+ (rain)xn = rbi...

am1x1 + am2x2 + · · ·+ amnxn = bm

To undo the operation, multiply the ith equation by r−1.


Introduction



Operation 1: Multiply

the ith equation by r 6= 0

a11x1 + a12x2 + · · ·+ a1nxn = b1...

ai1x1 + ai2x2 + · · ·+ ainxn = bi...

am1x1 + am2x2 + · · ·+ amnxn = bm

⇒

a11x1 + a12x2 + · · ·+ a1nxn = b1...


am1x1 + am2x2 + · · ·+ amnxn = bm



Introduction



Operation 1: Multiply the ith equation

by r 6= 0

a11x1 + a12x2 + · · ·+ a1nxn = b1...

ai1x1 + ai2x2 + · · ·+ ainxn = bi...

am1x1 + am2x2 + · · ·+ amnxn = bm

⇒

a11x1 + a12x2 + · · ·+ a1nxn = b1...


am1x1 + am2x2 + · · ·+ amnxn = bm



Introduction



Operation 1: Multiply the ith equation by

r 6= 0

a11x1 + a12x2 + · · ·+ a1nxn = b1...

ai1x1 + ai2x2 + · · ·+ ainxn = bi...

am1x1 + am2x2 + · · ·+ amnxn = bm

⇒

a11x1 + a12x2 + · · ·+ a1nxn = b1...


am1x1 + am2x2 + · · ·+ amnxn = bm



Introduction




a11x1 + a12x2 + · · ·+ a1nxn = b1...

ai1x1 + ai2x2 + · · ·+ ainxn = bi...

am1x1 + am2x2 + · · ·+ amnxn = bm

⇒

a11x1 + a12x2 + · · ·+ a1nxn = b1...


am1x1 + am2x2 + · · ·+ amnxn = bm



Introduction




a11x1 + a12x2 + · · ·+ a1nxn = b1...

ai1x1 + ai2x2 + · · ·+ ainxn = bi...

am1x1 + am2x2 + · · ·+ amnxn = bm

⇒

a11x1 + a12x2 + · · ·+ a1nxn = b1...


am1x1 + am2x2 + · · ·+ amnxn = bm

To undo

the operation, multiply the ith equation by r−1.


Introduction




a11x1 + a12x2 + · · ·+ a1nxn = b1...

ai1x1 + ai2x2 + · · ·+ ainxn = bi...

am1x1 + am2x2 + · · ·+ amnxn = bm

⇒

a11x1 + a12x2 + · · ·+ a1nxn = b1...


am1x1 + am2x2 + · · ·+ amnxn = bm

To undo the operation,

multiply the ith equation by r−1.


Introduction




a11x1 + a12x2 + · · ·+ a1nxn = b1...

ai1x1 + ai2x2 + · · ·+ ainxn = bi...

am1x1 + am2x2 + · · ·+ amnxn = bm

⇒

a11x1 + a12x2 + · · ·+ a1nxn = b1...


am1x1 + am2x2 + · · ·+ amnxn = bm

To undo the operation, multiply the ith equation

by r−1.


Introduction




a11x1 + a12x2 + · · ·+ a1nxn = b1...

ai1x1 + ai2x2 + · · ·+ ainxn = bi...

am1x1 + am2x2 + · · ·+ amnxn = bm

⇒

a11x1 + a12x2 + · · ·+ a1nxn = b1...


am1x1 + am2x2 + · · ·+ amnxn = bm

To undo the operation, multiply the ith equation by

r−1.


Introduction




a11x1 + a12x2 + · · ·+ a1nxn = b1...

ai1x1 + ai2x2 + · · ·+ ainxn = bi...

am1x1 + am2x2 + · · ·+ amnxn = bm

⇒

a11x1 + a12x2 + · · ·+ a1nxn = b1...


am1x1 + am2x2 + · · ·+ amnxn = bm

To undo the operation, multiply the ith equation by r−1.Dr. Marco A Roque Sol Linear Algebra. Session 1

Introduction



Operation 2: Add r times the ith equation to the jth equation

...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...(aj1 + rai1)x1 + (aj2 + rai2)x2 + · · ·+ (ajn + rain)xn = bj + rbi

...To undo the operation, add −r times the ith equationto the jth equation .


Introduction



Operation 2: Add

r times the ith equation to the jth equation

...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...ai1x1 + ai2x2 + · · ·+ ainxn = bi




Introduction



Operation 2: Add r times

the ith equation to the jth equation

...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...ai1x1 + ai2x2 + · · ·+ ainxn = bi




Introduction



Operation 2: Add r times the ith equation

to the jth equation

...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...ai1x1 + ai2x2 + · · ·+ ainxn = bi




Introduction




...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...ai1x1 + ai2x2 + · · ·+ ainxn = bi




Introduction




...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...ai1x1 + ai2x2 + · · ·+ ainxn = bi


...

To undo the operation, add −r times the ith equationto the jth equation .


Introduction




...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...ai1x1 + ai2x2 + · · ·+ ainxn = bi


...To undo

the operation, add −r times the ith equationto the jth equation .


Introduction




...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...ai1x1 + ai2x2 + · · ·+ ainxn = bi


...To undo the operation,

add −r times the ith equationto the jth equation .


Introduction




...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...ai1x1 + ai2x2 + · · ·+ ainxn = bi


...To undo the operation, add −r times

the ith equationto the jth equation .


Introduction




...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...ai1x1 + ai2x2 + · · ·+ ainxn = bi


...To undo the operation, add −r times the ith equation

to the jth equation .


Introduction




...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...ai1x1 + ai2x2 + · · ·+ ainxn = bi




Introduction



Operation 3: interchange the ith and jth equations.

...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...

To undo the operation, apply it once more .


Introduction



Operation 3: interchange

the ith and jth equations.

...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...



Introduction



Operation 3: interchange the ith and

jth equations.

...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...



Introduction




...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...



Introduction




...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...

To undo

the operation, apply it once more .


Introduction




...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...

To undo the operation,

apply it once more .


Introduction




...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...

To undo the operation, apply it

once more .


Introduction




...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...

⇒

...aj1x1 + aj2x2 + · · ·+ ajnxn = bj

...ai1x1 + ai2x2 + · · ·+ ainxn = bi

...

To undo the operation, apply it once more .Dr. Marco A Roque Sol Linear Algebra. Session 1

Introduction




2x − y − z = 3x + y + z = 6

Add −2 times the 1st equation to the 2nd equation:x − y = 2

y − z = −1x + y + z = 6

R2 := R2 − 2 ∗ R1

Add −1 times the 1st equation to the 3rd equation:x − y = 2

y − z = −12y + z = 4

R3 := (−1) ∗ R1 + R3


Introduction



Example 1.5

x − y = 2

2x − y − z = 3x + y + z = 6


y − z = −1x + y + z = 6

R2 := R2 − 2 ∗ R1


y − z = −12y + z = 4

R3 := (−1) ∗ R1 + R3


Introduction




2x − y − z = 3x + y + z = 6


y − z = −1x + y + z = 6

R2 := R2 − 2 ∗ R1


y − z = −12y + z = 4

R3 := (−1) ∗ R1 + R3


Introduction




2x − y − z = 3x + y + z = 6

Add

−2 times the 1st equation to the 2nd equation:x − y = 2

y − z = −1x + y + z = 6

R2 := R2 − 2 ∗ R1


y − z = −12y + z = 4

R3 := (−1) ∗ R1 + R3


Introduction




2x − y − z = 3x + y + z = 6

Add −2 times

the 1st equation to the 2nd equation:x − y = 2

y − z = −1x + y + z = 6

R2 := R2 − 2 ∗ R1


y − z = −12y + z = 4

R3 := (−1) ∗ R1 + R3


Introduction




2x − y − z = 3x + y + z = 6

Add −2 times the 1st equation to

the 2nd equation:x − y = 2

y − z = −1x + y + z = 6

R2 := R2 − 2 ∗ R1


y − z = −12y + z = 4

R3 := (−1) ∗ R1 + R3


Introduction




2x − y − z = 3x + y + z = 6

Add −2 times the 1st equation to the 2nd equation:

x − y = 2

y − z = −1x + y + z = 6

R2 := R2 − 2 ∗ R1


y − z = −12y + z = 4

R3 := (−1) ∗ R1 + R3


Introduction




2x − y − z = 3x + y + z = 6


y − z = −1x + y + z = 6

R2 := R2 − 2 ∗ R1


y − z = −12y + z = 4

R3 := (−1) ∗ R1 + R3


Introduction




2x − y − z = 3x + y + z = 6


y − z = −1x + y + z = 6

R2 := R2 − 2 ∗ R1

Add

−1 times the 1st equation to the 3rd equation:x − y = 2

y − z = −12y + z = 4

R3 := (−1) ∗ R1 + R3


Introduction




2x − y − z = 3x + y + z = 6


y − z = −1x + y + z = 6

R2 := R2 − 2 ∗ R1

Add −1 times

the 1st equation to the 3rd equation:x − y = 2

y − z = −12y + z = 4

R3 := (−1) ∗ R1 + R3


Introduction




2x − y − z = 3x + y + z = 6


y − z = −1x + y + z = 6

R2 := R2 − 2 ∗ R1

Add −1 times the 1st equation to

the 3rd equation:x − y = 2

y − z = −12y + z = 4

R3 := (−1) ∗ R1 + R3


Introduction




2x − y − z = 3x + y + z = 6


y − z = −1x + y + z = 6

R2 := R2 − 2 ∗ R1

Add −1 times the 1st equation to the 3rd equation:

x − y = 2

y − z = −12y + z = 4

R3 := (−1) ∗ R1 + R3


Introduction




2x − y − z = 3x + y + z = 6


y − z = −1x + y + z = 6

R2 := R2 − 2 ∗ R1


y − z = −12y + z = 4

R3 := (−1) ∗ R1 + R3


Introduction



Add −2 times the 2nd equation to the 3rd equation:x − y = 2

y − z = −13z = 6

R3 := (−2) ∗ R2 + R3

Note: At this point, the elimination process is completed, and wecan solve the system by back substitution. However, we cancontinue with elementary operations

Multiply the 3rd equation by 1/3:x − y = 2

y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction



Add

−2 times the 2nd equation to the 3rd equation:x − y = 2

y − z = −13z = 6

R3 := (−2) ∗ R2 + R3



y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction



Add −2 times

the 2nd equation to the 3rd equation:x − y = 2

y − z = −13z = 6

R3 := (−2) ∗ R2 + R3



y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction



Add −2 times the 2nd equation to

the 3rd equation:x − y = 2

y − z = −13z = 6

R3 := (−2) ∗ R2 + R3



y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction



Add −2 times the 2nd equation to the 3rd equation:

x − y = 2

y − z = −13z = 6

R3 := (−2) ∗ R2 + R3



y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction




y − z = −13z = 6

R3 := (−2) ∗ R2 + R3



y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction




y − z = −13z = 6

R3 := (−2) ∗ R2 + R3

Note:

At this point, the elimination process is completed, and wecan solve the system by back substitution. However, we cancontinue with elementary operations


y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction




y − z = −13z = 6

R3 := (−2) ∗ R2 + R3

Note: At this point,

the elimination process is completed, and wecan solve the system by back substitution. However, we cancontinue with elementary operations


y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction




y − z = −13z = 6

R3 := (−2) ∗ R2 + R3

Note: At this point, the elimination process

is completed, and wecan solve the system by back substitution. However, we cancontinue with elementary operations


y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction




y − z = −13z = 6

R3 := (−2) ∗ R2 + R3

Note: At this point, the elimination process is completed, and

wecan solve the system by back substitution. However, we cancontinue with elementary operations


y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction




y − z = −13z = 6

R3 := (−2) ∗ R2 + R3

Note: At this point, the elimination process is completed, and wecan solve

the system by back substitution. However, we cancontinue with elementary operations


y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction




y − z = −13z = 6

R3 := (−2) ∗ R2 + R3

Note: At this point, the elimination process is completed, and wecan solve the system by

back substitution. However, we cancontinue with elementary operations


y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction




y − z = −13z = 6

R3 := (−2) ∗ R2 + R3

Note: At this point, the elimination process is completed, and wecan solve the system by back substitution.

However, we cancontinue with elementary operations


y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction




y − z = −13z = 6

R3 := (−2) ∗ R2 + R3

Note: At this point, the elimination process is completed, and wecan solve the system by back substitution. However,

we cancontinue with elementary operations


y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction




y − z = −13z = 6

R3 := (−2) ∗ R2 + R3

Note: At this point, the elimination process is completed, and wecan solve the system by back substitution. However, we cancontinue

with elementary operations


y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction




y − z = −13z = 6

R3 := (−2) ∗ R2 + R3



y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction




y − z = −13z = 6

R3 := (−2) ∗ R2 + R3


Multiply

the 3rd equation by 1/3:x − y = 2

y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction




y − z = −13z = 6

R3 := (−2) ∗ R2 + R3


Multiply the 3rd equation by 1/3:

x − y = 2

y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction




y − z = −13z = 6

R3 := (−2) ∗ R2 + R3



y − z = −1z = 2

R3 := (1/3) ∗ R3


Introduction



Add the 3rd equation to the 2nd equation:x − y = 2

y = 1z = 2

R2 := R2 + R3

Add the 2nd equation to the 1st equation:x = 3

y = 1z = 2

R1 := R2 + R1

Thus, the system x − y = 2

2x − y − z = 3x + y + z = 6

has as the solution set, the point (3, 1, 2)


Introduction



Add

the 3rd equation to the 2nd equation:x − y = 2

y = 1z = 2

R2 := R2 + R3


y = 1z = 2

R1 := R2 + R1


2x − y − z = 3x + y + z = 6



Introduction



Add the 3rd equation to

the 2nd equation:x − y = 2

y = 1z = 2

R2 := R2 + R3


y = 1z = 2

R1 := R2 + R1


2x − y − z = 3x + y + z = 6



Introduction



Add the 3rd equation to the 2nd equation:

x − y = 2

y = 1z = 2

R2 := R2 + R3


y = 1z = 2

R1 := R2 + R1


2x − y − z = 3x + y + z = 6



Introduction




y = 1z = 2

R2 := R2 + R3


y = 1z = 2

R1 := R2 + R1


2x − y − z = 3x + y + z = 6



Introduction




y = 1z = 2

R2 := R2 + R3

Add

the 2nd equation to the 1st equation:x = 3

y = 1z = 2

R1 := R2 + R1


2x − y − z = 3x + y + z = 6



Introduction




y = 1z = 2

R2 := R2 + R3

Add the 2nd equation to

the 1st equation:x = 3

y = 1z = 2

R1 := R2 + R1


2x − y − z = 3x + y + z = 6



Introduction




y = 1z = 2

R2 := R2 + R3

Add the 2nd equation to the 1st equation:

x = 3

y = 1z = 2

R1 := R2 + R1


2x − y − z = 3x + y + z = 6



Introduction




y = 1z = 2

R2 := R2 + R3


y = 1z = 2

R1 := R2 + R1


2x − y − z = 3x + y + z = 6



Introduction




y = 1z = 2

R2 := R2 + R3


y = 1z = 2

R1 := R2 + R1

Thus,

the system x − y = 2

2x − y − z = 3x + y + z = 6



Introduction




y = 1z = 2

R2 := R2 + R3


y = 1z = 2

R1 := R2 + R1

Thus, the system

x − y = 2

2x − y − z = 3x + y + z = 6



Introduction




y = 1z = 2

R2 := R2 + R3


y = 1z = 2

R1 := R2 + R1


2x − y − z = 3x + y + z = 6



Introduction




y = 1z = 2

R2 := R2 + R3


y = 1z = 2

R1 := R2 + R1


2x − y − z = 3x + y + z = 6

has

as the solution set, the point (3, 1, 2)


Introduction




y = 1z = 2

R2 := R2 + R3


y = 1z = 2

R1 := R2 + R1


2x − y − z = 3x + y + z = 6

has as the solution set,

the point (3, 1, 2)


Introduction




y = 1z = 2

R2 := R2 + R3


y = 1z = 2

R1 := R2 + R1


2x − y − z = 3x + y + z = 6

has as the solution set, the point (3, 1, 2)Dr. Marco A Roque Sol Linear Algebra. Session 1

Introduction


Row echelon form. Gauss-Jordan Reduction

Matrices

Definition. A matrix is a rectangular array of numbers.

The dimension of a matrix is given by

dimensions = (number of rows) X ( number of columns)

Thus we have

n × n : Square matrixn × 1 : Column vector1× n : Row vector


Introduction



Matrices




Thus we have

n × n : Square matrix

n × 1 : Column vector1× n : Row vector


Introduction



Matrices




Thus we have

n × n : Square matrixn × 1 : Column vector

1× n : Row vector


Introduction



Matrices




Thus we have

n × n : Square matrixn × 1 : Column vector1× n : Row vector


Introduction



Examples of matrices are:

2 7−1 03 3

(3× 2)

(2 7 0−1 1 5

)(2× 3)

358

(3× 1)

(2 4 9

)(1× 3)

(−2 01 5

)(2× 2)


Introduction




2 7−1 03 3

(3× 2)

(2 7 0−1 1 5

)

(2× 3)

358

(3× 1)

(2 4 9

)(1× 3)

(−2 01 5

)(2× 2)


Introduction




2 7−1 03 3

(3× 2)

(2 7 0−1 1 5

)(2× 3)

358

(3× 1)

(2 4 9

)(1× 3)

(−2 01 5

)(2× 2)


Introduction




2 7−1 03 3

(3× 2)

(2 7 0−1 1 5

)(2× 3)

358

(3× 1)

(2 4 9

)

(1× 3)

(−2 01 5

)(2× 2)


Introduction




2 7−1 03 3

(3× 2)

(2 7 0−1 1 5

)(2× 3)

358

(3× 1)

(2 4 9

)(1× 3)

(−2 01 5

)(2× 2)


Introduction




2 7−1 03 3

(3× 2)

(2 7 0−1 1 5

)(2× 3)

358

(3× 1)

(2 4 9

)(1× 3)

(−2 01 5

)

(2× 2)


Introduction




2 7−1 03 3

(3× 2)

(2 7 0−1 1 5

)(2× 3)

358

(3× 1)

(2 4 9

)(1× 3)

(−2 01 5

)(2× 2)


Introduction



From a system of linear equations:a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm

We can find the coefficient matrix and column vector of theright-hand sides:

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

b1

b2...

bm


Introduction



From a system of linear equations:

a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm


a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

b1

b2...

bm


Introduction




a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm


a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

b1

b2...

bm


Introduction




a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm

We can find

the coefficient matrix and column vector of theright-hand sides:

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

b1

b2...

bm


Introduction




a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm

We can find the coefficient matrix and

column vector of theright-hand sides:

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

b1

b2...

bm


Introduction




a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm

We can find the coefficient matrix and column vector

of theright-hand sides:

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

b1

b2...

bm


Introduction




a21x1 + a22x2 + · · ·+ a2nxn = b2...

am1x1 + am2x2 + · · ·+ amnxn = bm


a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

b1

b2...

bm


Introduction



and also associated to the linear system we have the Augmentedmatrix:

a11 a12 · · · a1n b1

a21 a22 · · · a2n b2...

am1 am2 · · · amn bm

Now,remember that using Gaussian elimination, the solution of asystem of linear equations splits into two parts:

(A) Elimination and(B) Back substitution.


Introduction



and also

associated to the linear system we have the Augmentedmatrix:

a11 a12 · · · a1n b1

a21 a22 · · · a2n b2...





Introduction



and also associated to the linear system

we have the Augmentedmatrix:

a11 a12 · · · a1n b1

a21 a22 · · · a2n b2...





Introduction




a11 a12 · · · a1n b1

a21 a22 · · · a2n b2...





Introduction




a11 a12 · · · a1n b1

a21 a22 · · · a2n b2...


Now,

remember that using Gaussian elimination, the solution of asystem of linear equations splits into two parts:



Introduction




a11 a12 · · · a1n b1

a21 a22 · · · a2n b2...


Now,remember that using Gaussian elimination,

the solution of asystem of linear equations splits into two parts:



Introduction




a11 a12 · · · a1n b1

a21 a22 · · · a2n b2...


Now,remember that using Gaussian elimination, the solution of asystem of linear equations

splits into two parts:



Introduction




a11 a12 · · · a1n b1

a21 a22 · · · a2n b2...





Introduction




a11 a12 · · · a1n b1

a21 a22 · · · a2n b2...



(A) Elimination and

(B) Back substitution.


Introduction




a11 a12 · · · a1n b1

a21 a22 · · · a2n b2...





Introduction



Both parts can be done by applying a finite number of elementaryoperations:

(1) to multiply a row by a nonzero scalar;

(2) to add the ith row multiplied by some r ∈ R to the jth row;

(3) to interchange two rows.

Notation


Introduction



Both parts

can be done by applying a finite number of elementaryoperations:




Notation


Introduction



Both parts can be done

by applying a finite number of elementaryoperations:




Notation


Introduction



Both parts can be done by applying a finite number

of elementaryoperations:




Notation


Introduction



Both parts can be done by applying a finite number of elementaryoperations:




Notation


Introduction



Augmented matrix:

a11 a12 · · · a1n b1

a21 a22 · · · a2n b2...


=

v1v2...

vm

where vi = (ai1 ai1 ai1 · · · ai1, bi ) is a row vector.


Introduction



Augmented matrix:

a11 a12 · · · a1n b1

a21 a22 · · · a2n b2...


=

v1v2...

vm

where

vi = (ai1 ai1 ai1 · · · ai1, bi ) is a row vector.


Introduction



Augmented matrix:

a11 a12 · · · a1n b1

a21 a22 · · · a2n b2...


=

v1v2...

vm

where vi = (ai1 ai1 ai1 · · · ai1, bi )

is a row vector.


Introduction



Augmented matrix:

a11 a12 · · · a1n b1

a21 a22 · · · a2n b2...


=

v1v2...

vm

where vi = (ai1 ai1 ai1 · · · ai1, bi ) is a row vector.


Introduction



Elementary row operations on Matrices

Operation 1: To multiply the ith row by r 6= 0

v1...

vi...

vm

⇒

v1...

rvi...

vm


Introduction




Operation 1:

To multiply the ith row by r 6= 0

v1...

vi...

vm

⇒

v1...

rvi...

vm


Introduction




Operation 1: To multiply the ith row by r 6= 0

v1...

vi...

vm

⇒

v1...

rvi...

vm


Introduction



Operation 2: To add the ith row multiplied by r to the jth row

v1...

vi...

vj...

vm

⇒

v1...

vi...

vj + rvi...

vm


Introduction



Operation 2:

To add the ith row multiplied by r to the jth row

v1...

vi...

vj...

vm

⇒

v1...

vi...

vj + rvi...

vm


Introduction



Operation 2: To add the ith row multiplied by r to the jth row

v1...

vi...

vj...

vm

⇒

v1...

vi...

vj + rvi...

vm


Introduction



Operation 3: To interchange the ith row with the jth row

v1...

vi...

vj...

vm

⇒

v1...

vj...

vi...

vm


Introduction



Operation 3:

To interchange the ith row with the jth row

v1...

vi...

vj...

vm

⇒

v1...

vj...

vi...

vm


Introduction



Operation 3: To interchange the ith row with the jth row

v1...

vi...

vj...

vm

⇒

v1...

vj...

vi...

vm


Introduction



Row echelon form

Definition. Leading entry of a matrix is the first nonzero entry ina row.

The goal of the Gaussian elimination is to convert the augmentedmatrix into row echelon form

1) Leading entries shift to the right as we go from the first row tothe last one;

2) Each leading entry is equal to 1.


Introduction



Row echelon form

Definition. Leading entry

of a matrix is the first nonzero entry ina row.





Introduction



Row echelon form

Definition. Leading entry of a matrix is

the first nonzero entry ina row.





Introduction



Row echelon form






Introduction



Row echelon form


The goal of

the Gaussian elimination is to convert the augmentedmatrix into row echelon form




Introduction



Row echelon form


The goal of the Gaussian elimination

is to convert the augmentedmatrix into row echelon form




Introduction



Row echelon form


The goal of the Gaussian elimination is to convert

the augmentedmatrix into row echelon form




Introduction



Row echelon form


The goal of the Gaussian elimination is to convert the augmentedmatrix

into row echelon form




Introduction



Row echelon form






Introduction



Row echelon form



1) Leading entries

shift to the right as we go from the first row tothe last one;



Introduction



Row echelon form



1) Leading entries shift to the right

as we go from the first row tothe last one;



Introduction



Row echelon form



1) Leading entries shift to the right as we go from

the first row tothe last one;



Introduction



Row echelon form



1) Leading entries shift to the right as we go from the first row to

the last one;



Introduction



Row echelon form






Introduction



Row echelon form




2) Each

leading entry is equal to 1.


Introduction



Row echelon form




2) Each leading entry

is equal to 1.


Introduction



Row echelon form




2) Each leading entry is equal to

1.


Introduction



Row echelon form






Introduction



Thus, we have this example in the echelon form

1 −1 4 1 7 9 6 1 3 9 −50 1 3 −2 7 1 5 3 3 4 −10 0 0 0 1 5 6 −3 1 2 10 0 0 0 0 0 1 8 1 7 20 0 0 0 0 0 0 1 −2 7 −30 0 0 0 0 0 0 0 1 −2 90 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0


Introduction



Thus,

we have this example in the echelon form

1 −1 4 1 7 9 6 1 3 9 −50 1 3 −2 7 1 5 3 3 4 −10 0 0 0 1 5 6 −3 1 2 10 0 0 0 0 0 1 8 1 7 20 0 0 0 0 0 0 1 −2 7 −30 0 0 0 0 0 0 0 1 −2 90 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0


Introduction



Thus, we have this example

in the echelon form

1 −1 4 1 7 9 6 1 3 9 −50 1 3 −2 7 1 5 3 3 4 −10 0 0 0 1 5 6 −3 1 2 10 0 0 0 0 0 1 8 1 7 20 0 0 0 0 0 0 1 −2 7 −30 0 0 0 0 0 0 0 1 −2 90 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0


Introduction



Thus, we have this example in the echelon form

1 −1 4 1 7 9 6 1 3 9 −50 1 3 −2 7 1 5 3 3 4 −10 0 0 0 1 5 6 −3 1 2 10 0 0 0 0 0 1 8 1 7 20 0 0 0 0 0 0 1 −2 7 −30 0 0 0 0 0 0 0 1 −2 90 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0


Introduction



Echelon FormationDr. Marco A Roque Sol Linear Algebra. Session 1

Introduction



Row echelon form

General augmented matrix in row echelon form

� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗

� � ∗ ∗� ∗

�

1) Leading entries are boxed (all equal to 1);

2) All the entries below, the (imaginary) staircase line are zero;


Introduction



Row echelon form


� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗

� � ∗ ∗� ∗

�

1) Leading entries

are boxed (all equal to 1);



Introduction



Row echelon form


� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗

� � ∗ ∗� ∗

�

1) Leading entries are boxed

(all equal to 1);



Introduction



Row echelon form


� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗

� � ∗ ∗� ∗

�




Introduction



Row echelon form


� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗

� � ∗ ∗� ∗

�


2) All

the entries below, the (imaginary) staircase line are zero;


Introduction



Row echelon form


� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗

� � ∗ ∗� ∗

�


2) All the entries below,

the (imaginary) staircase line are zero;


Introduction



Row echelon form


� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗

� � ∗ ∗� ∗

�


2) All the entries below, the (imaginary)

staircase line are zero;


Introduction



Row echelon form


� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗

� � ∗ ∗� ∗

�


2) All the entries below, the (imaginary) staircase line

are zero;


Introduction



Row echelon form


� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗

� � ∗ ∗� ∗

�




Introduction



3) Each step of the (imaginary) staircase has height 1;

4) Each circle marks a column without a leading entry thatcorresponds to a free variable.

Strict triangular form

Is a particular case of row echelon form that can occur for systemsof n variables :


Introduction



3) Each step

of the (imaginary) staircase has height 1;





Introduction



3) Each step of the

(imaginary) staircase has height 1;





Introduction



3) Each step of the (imaginary) staircase

has height 1;





Introduction



3) Each step of the (imaginary) staircase has height

1;





Introduction








Introduction






Is a particular case

of row echelon form that can occur for systemsof n variables :


Introduction






Is a particular case of row echelon form

that can occur for systemsof n variables :


Introduction








Introduction



� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗ ∗� ∗ ∗ ∗

� ∗ ∗� ∗

1) No zero rows.

2) No free variables.


Introduction



� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗ ∗� ∗ ∗ ∗

� ∗ ∗� ∗

1) No zero rows.

2) No free variables.Dr. Marco A Roque Sol Linear Algebra. Session 1

Introduction



Consistency check

The original system of linear equations is consistent if there is noleading entry in the rightmost column of the augmented matrix inrow echelon form.

Augmented matrix of an inconsistent system

� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗

� � ∗ ∗� ∗

�


Introduction



Consistency check

The original system

of linear equations is consistent if there is noleading entry in the rightmost column of the augmented matrix inrow echelon form.


� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗

� � ∗ ∗� ∗

�


Introduction



Consistency check

The original system of linear equations

is consistent if there is noleading entry in the rightmost column of the augmented matrix inrow echelon form.


� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗

� � ∗ ∗� ∗

�


Introduction



Consistency check

The original system of linear equations is consistent

if there is noleading entry in the rightmost column of the augmented matrix inrow echelon form.


� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗

� � ∗ ∗� ∗

�


Introduction



Consistency check

The original system of linear equations is consistent if there is noleading entry

in the rightmost column of the augmented matrix inrow echelon form.


� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗

� � ∗ ∗� ∗

�


Introduction



Consistency check

The original system of linear equations is consistent if there is noleading entry in the rightmost column

of the augmented matrix inrow echelon form.


� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗

� � ∗ ∗� ∗

�


Introduction



Consistency check

The original system of linear equations is consistent if there is noleading entry in the rightmost column of the augmented matrix

inrow echelon form.


� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗

� � ∗ ∗� ∗

�


Introduction



Consistency check

The original system of linear equations is consistent if there is noleading entry in the rightmost column of the augmented matrix inrow echelon form.


� ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗

� � ∗ ∗� ∗

�


Introduction



The goal of the Gauss-Jordan reduction is to convert theaugmented matrix into reduced row echelon form :

1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗

1 � ∗ ∗1 ∗ ∗

1 � ∗1 ∗

1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circlecorresponds to a free variable.


Introduction



The goal

of the Gauss-Jordan reduction is to convert theaugmented matrix into reduced row echelon form :

1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗

1 � ∗ ∗1 ∗ ∗

1 � ∗1 ∗



Introduction



The goal of the Gauss-Jordan reduction

is to convert theaugmented matrix into reduced row echelon form :

1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗

1 � ∗ ∗1 ∗ ∗

1 � ∗1 ∗



Introduction



The goal of the Gauss-Jordan reduction is to convert

theaugmented matrix into reduced row echelon form :

1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗

1 � ∗ ∗1 ∗ ∗

1 � ∗1 ∗



Introduction



The goal of the Gauss-Jordan reduction is to convert theaugmented matrix

into reduced row echelon form :

1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗

1 � ∗ ∗1 ∗ ∗

1 � ∗1 ∗



Introduction




1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗

1 � ∗ ∗1 ∗ ∗

1 � ∗1 ∗



Introduction




1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗

1 � ∗ ∗1 ∗ ∗

1 � ∗1 ∗

1) All entries

below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circlecorresponds to a free variable.


Introduction




1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗

1 � ∗ ∗1 ∗ ∗

1 � ∗1 ∗

1) All entries below the staircase line

are zero; 2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circlecorresponds to a free variable.


Introduction




1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗

1 � ∗ ∗1 ∗ ∗

1 � ∗1 ∗

1) All entries below the staircase line are zero;

2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circlecorresponds to a free variable.


Introduction




1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗

1 � ∗ ∗1 ∗ ∗

1 � ∗1 ∗

1) All entries below the staircase line are zero; 2) Each boxed entry

is 1, the other entries in its column are zero; 3) Each circlecorresponds to a free variable.


Introduction




1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗

1 � ∗ ∗1 ∗ ∗

1 � ∗1 ∗

1) All entries below the staircase line are zero; 2) Each boxed entryis 1,

the other entries in its column are zero; 3) Each circlecorresponds to a free variable.


Introduction




1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗

1 � ∗ ∗1 ∗ ∗

1 � ∗1 ∗

1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries

in its column are zero; 3) Each circlecorresponds to a free variable.


Introduction




1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗

1 � ∗ ∗1 ∗ ∗

1 � ∗1 ∗

1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column

are zero; 3) Each circlecorresponds to a free variable.


Introduction




1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗

1 � ∗ ∗1 ∗ ∗

1 � ∗1 ∗

1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column are zero;

3) Each circlecorresponds to a free variable.


Introduction




1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗

1 � ∗ ∗1 ∗ ∗

1 � ∗1 ∗

1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circle

corresponds to a free variable.


Introduction




1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗

1 � ∗ ∗1 ∗ ∗

1 � ∗1 ∗

1) All entries below the staircase line are zero; 2) Each boxed entryis 1, the other entries in its column are zero; 3) Each circlecorresponds to

a free variable.


Introduction




1 ∗ ∗ ∗ ∗ ∗1 � � ∗ ∗ ∗

1 � ∗ ∗1 ∗ ∗

1 � ∗1 ∗



Introduction



Example 1.6From a previous example, we have

x − y = 1

2x − y− z = 3x + y+ z = 6

⇒

1 −1 0 12 −1 −1 31 1 1 6

Row echelon form (also strict triangular):x − y = 1

y − z = 1x + y+ z = 1

⇒

1 −1 0 1

0 1 −1 1

0 0 1 1


Introduction



Example 1.6

From a previous example, we have

x − y = 1

2x − y− z = 3x + y+ z = 6

⇒

1 −1 0 12 −1 −1 31 1 1 6


y − z = 1x + y+ z = 1

⇒

1 −1 0 1

0 1 −1 1

0 0 1 1


Introduction



Example 1.6From a previous example,

we have

x − y = 1

2x − y− z = 3x + y+ z = 6

⇒

1 −1 0 12 −1 −1 31 1 1 6


y − z = 1x + y+ z = 1

⇒

1 −1 0 1

0 1 −1 1

0 0 1 1


Introduction




x − y = 1

2x − y− z = 3x + y+ z = 6

⇒

1 −1 0 12 −1 −1 31 1 1 6


y − z = 1x + y+ z = 1

⇒

1 −1 0 1

0 1 −1 1

0 0 1 1


Introduction




x − y = 1

2x − y− z = 3x + y+ z = 6

⇒

1 −1 0 12 −1 −1 31 1 1 6

Row echelon form (also strict triangular):

x − y = 1

y − z = 1x + y+ z = 1

⇒

1 −1 0 1

0 1 −1 1

0 0 1 1


Introduction




x − y = 1

2x − y− z = 3x + y+ z = 6

⇒

1 −1 0 12 −1 −1 31 1 1 6


y − z = 1x + y+ z = 1

⇒

1 −1 0 1

0 1 −1 1

0 0 1 1


Introduction



Reduced row echelon form :x = 3

y = 1z = 2

⇒

1 0 0 3

0 1 0 1

0 0 1 2

Example 1.7

x + y− 2z = 1

y− z = 3−x + 4y− 3z = 6

⇒

1 1 −2 10 1 −1 3−1 4 −3 1


Introduction



Reduced row echelon form :

x = 3

y = 1z = 2

⇒

1 0 0 3

0 1 0 1

0 0 1 2

Example 1.7

x + y− 2z = 1

y− z = 3−x + 4y− 3z = 6

⇒

1 1 −2 10 1 −1 3−1 4 −3 1


Introduction



Reduced row echelon form :x = 3

y = 1z = 2

⇒

1 0 0 3

0 1 0 1

0 0 1 2

Example 1.7

x + y− 2z = 1

y− z = 3−x + 4y− 3z = 6

⇒

1 1 −2 10 1 −1 3−1 4 −3 1


Introduction



Row echelon form:

x + y− 2z = 1

y− z = 30 = 1

⇒

1 1 −2 1

0 1 −1 3

0 0 0 1

Reduced row echelon form:x + − z = 0

y− z = 00 = 1

⇒

1 0 −1 0

0 1 −1 0

0 0 0 1

Inconsistent system


Introduction



How to solve a system of linear equations

1) Order the variables.

2) Write down the augmented matrix of the system.

3) Convert the matrix to row echelon form

4) Check for consistency.

5) Convert the matrix to reduced row echelon form.

6) Write down the system corresponding to the reduced rowechelon form.


Introduction





2) Write down

the augmented matrix of the system.






Introduction





2) Write down the augmented matrix

of the system.






Introduction











Introduction






3) Convert the matrix

to row echelon form





Introduction











Introduction








5) Convert the matrix

to reduced row echelon form.



Introduction








5) Convert the matrix to reduced row

echelon form.



Introduction











Introduction









6) Write down

the system corresponding to the reduced rowechelon form.


Introduction









6) Write down the system corresponding

to the reduced rowechelon form.


Introduction











Introduction



7) Determine leading and free variables.

8) Rewrite the system so that the leading variables are on the leftwhile everything else is on the right.

9) Assign parameters to the free variables and write down thegeneral solution in parametric form.

Example 1.7 {x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10

In this case the variables are x1, x2, x3, x4. and the Augmentedmatrix is :


Introduction




8) Rewrite the system

so that the leading variables are on the leftwhile everything else is on the right.


Example 1.7 {x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10



Introduction




8) Rewrite the system so that the leading variables

are on the leftwhile everything else is on the right.


Example 1.7 {x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10



Introduction




8) Rewrite the system so that the leading variables are on the left

while everything else is on the right.


Example 1.7 {x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10



Introduction




8) Rewrite the system so that the leading variables are on the leftwhile everything else

is on the right.


Example 1.7 {x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10



Introduction






Example 1.7 {x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10



Introduction





9) Assign parameters

to the free variables and write down thegeneral solution in parametric form.

Example 1.7 {x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10



Introduction





9) Assign parameters to the free variables and

write down thegeneral solution in parametric form.

Example 1.7 {x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10



Introduction





9) Assign parameters to the free variables and write down thegeneral solution

in parametric form.

Example 1.7 {x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10



Introduction






Example 1.7 {x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10



Introduction






Example 1.7

{x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10



Introduction






Example 1.7 {x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10



Introduction






Example 1.7 {x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10

In this case

the variables are x1, x2, x3, x4. and the Augmentedmatrix is :


Introduction






Example 1.7 {x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10

In this case the variables are

x1, x2, x3, x4. and the Augmentedmatrix is :


Introduction






Example 1.7 {x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10

In this case the variables are x1,

x2, x3, x4. and the Augmentedmatrix is :


Introduction






Example 1.7 {x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10

In this case the variables are x1, x2,

x3, x4. and the Augmentedmatrix is :


Introduction






Example 1.7 {x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10

In this case the variables are x1, x2, x3,

x4. and the Augmentedmatrix is :


Introduction






Example 1.7 {x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10

In this case the variables are x1, x2, x3, x4.

and the Augmentedmatrix is :


Introduction






Example 1.7 {x2 + 2x3 + 3x4 = 6

x1 + 2x2 + 3x3 + 4x4 = 10



Introduction



(0 1 2 3 61 2 3 4 10

)To get it into row echelon form, we exchange the two rows:(

1 2 3 4 100 1 2 3 6

)Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(

1 0 −1 −2 −2

0 1 2 3 6

)The leading variables are x1 and x2 ; hence x3 and x4 are freevariables


Introduction



(0 1 2 3 61 2 3 4 10

)

To get it into row echelon form, we exchange the two rows:(1 2 3 4 100 1 2 3 6


1 0 −1 −2 −2

0 1 2 3 6



Introduction



(0 1 2 3 61 2 3 4 10

)To get it

into row echelon form, we exchange the two rows:(1 2 3 4 100 1 2 3 6


1 0 −1 −2 −2

0 1 2 3 6



Introduction



(0 1 2 3 61 2 3 4 10

)To get it into row echelon form,

we exchange the two rows:(1 2 3 4 100 1 2 3 6


1 0 −1 −2 −2

0 1 2 3 6



Introduction



(0 1 2 3 61 2 3 4 10

)To get it into row echelon form, we exchange

the two rows:(1 2 3 4 100 1 2 3 6


1 0 −1 −2 −2

0 1 2 3 6



Introduction



(0 1 2 3 61 2 3 4 10

)To get it into row echelon form, we exchange the two rows:

(1 2 3 4 100 1 2 3 6


1 0 −1 −2 −2

0 1 2 3 6



Introduction



(0 1 2 3 61 2 3 4 10


1 2 3 4 100 1 2 3 6

)

Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(

1 0 −1 −2 −2

0 1 2 3 6



Introduction



(0 1 2 3 61 2 3 4 10


1 2 3 4 100 1 2 3 6

)Consistency check

is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(

1 0 −1 −2 −2

0 1 2 3 6



Introduction



(0 1 2 3 61 2 3 4 10


1 2 3 4 100 1 2 3 6

)Consistency check is passed.

To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:(

1 0 −1 −2 −2

0 1 2 3 6



Introduction



(0 1 2 3 61 2 3 4 10


1 2 3 4 100 1 2 3 6

)Consistency check is passed. To convert

into reduced row echelonform, add −2 times the 2nd row to the 1st row:(

1 0 −1 −2 −2

0 1 2 3 6



Introduction



(0 1 2 3 61 2 3 4 10


1 2 3 4 100 1 2 3 6

)Consistency check is passed. To convert into reduced row echelonform,

add −2 times the 2nd row to the 1st row:(1 0 −1 −2 −2

0 1 2 3 6



Introduction



(0 1 2 3 61 2 3 4 10


1 2 3 4 100 1 2 3 6

)Consistency check is passed. To convert into reduced row echelonform, add −2 times

the 2nd row to the 1st row:(1 0 −1 −2 −2

0 1 2 3 6



Introduction



(0 1 2 3 61 2 3 4 10


1 2 3 4 100 1 2 3 6

)Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row

to the 1st row:(1 0 −1 −2 −2

0 1 2 3 6



Introduction



(0 1 2 3 61 2 3 4 10


1 2 3 4 100 1 2 3 6

)Consistency check is passed. To convert into reduced row echelonform, add −2 times the 2nd row to the 1st row:

(1 0 −1 −2 −2

0 1 2 3 6



Introduction



(0 1 2 3 61 2 3 4 10


1 2 3 4 100 1 2 3 6


1 0 −1 −2 −2

0 1 2 3 6

)

The leading variables are x1 and x2 ; hence x3 and x4 are freevariables


Introduction



(0 1 2 3 61 2 3 4 10


1 2 3 4 100 1 2 3 6


1 0 −1 −2 −2

0 1 2 3 6

)The leading variables

are x1 and x2 ; hence x3 and x4 are freevariables


Introduction



(0 1 2 3 61 2 3 4 10


1 2 3 4 100 1 2 3 6


1 0 −1 −2 −2

0 1 2 3 6

)The leading variables are x1 and

x2 ; hence x3 and x4 are freevariables


Introduction



(0 1 2 3 61 2 3 4 10


1 2 3 4 100 1 2 3 6


1 0 −1 −2 −2

0 1 2 3 6

)The leading variables are x1 and x2 ; hence

x3 and x4 are freevariables


Introduction



(0 1 2 3 61 2 3 4 10


1 2 3 4 100 1 2 3 6


1 0 −1 −2 −2

0 1 2 3 6

)The leading variables are x1 and x2 ; hence x3 and

x4 are freevariables


Introduction



(0 1 2 3 61 2 3 4 10


1 2 3 4 100 1 2 3 6


1 0 −1 −2 −2

0 1 2 3 6

)The leading variables are x1 and x2 ; hence x3 and x4

are freevariables


Introduction



(0 1 2 3 61 2 3 4 10


1 2 3 4 100 1 2 3 6


1 0 −1 −2 −2

0 1 2 3 6



Introduction



Back to the system:{x1 − x3 − 2x4 = −1x2 + 2x3 + 3x4 = 6

⇒{

x1 = x3 + 2x4 − 2x2 = −2x3 − 3x4 + 6

and the general solution is given byx1 = t + 2s − 2x2 = −2t − 3s + 6x3 = tx4 = s

(t, s ∈ R)

In vector form ( vector equation of a plane in the space ),

(x1, x2, x3, x4) = (−2, 6, 0, 0) + t(1,−2, 1, 0) + s(2,−3, 0, 1)


Introduction



Back to the system:

{x1 − x3 − 2x4 = −1x2 + 2x3 + 3x4 = 6

⇒{

x1 = x3 + 2x4 − 2x2 = −2x3 − 3x4 + 6


(t, s ∈ R)


(x1, x2, x3, x4) = (−2, 6, 0, 0) + t(1,−2, 1, 0) + s(2,−3, 0, 1)


Introduction




⇒

{x1 = x3 + 2x4 − 2

x2 = −2x3 − 3x4 + 6


(t, s ∈ R)


(x1, x2, x3, x4) = (−2, 6, 0, 0) + t(1,−2, 1, 0) + s(2,−3, 0, 1)


Introduction




⇒{

x1 = x3 + 2x4 − 2x2 = −2x3 − 3x4 + 6


(t, s ∈ R)


(x1, x2, x3, x4) = (−2, 6, 0, 0) + t(1,−2, 1, 0) + s(2,−3, 0, 1)


Introduction




⇒{

x1 = x3 + 2x4 − 2x2 = −2x3 − 3x4 + 6

and the general solution is given by

x1 = t + 2s − 2x2 = −2t − 3s + 6x3 = tx4 = s

(t, s ∈ R)


(x1, x2, x3, x4) = (−2, 6, 0, 0) + t(1,−2, 1, 0) + s(2,−3, 0, 1)


Introduction




⇒{

x1 = x3 + 2x4 − 2x2 = −2x3 − 3x4 + 6


(t, s ∈ R)


(x1, x2, x3, x4) = (−2, 6, 0, 0) + t(1,−2, 1, 0) + s(2,−3, 0, 1)


Introduction




⇒{

x1 = x3 + 2x4 − 2x2 = −2x3 − 3x4 + 6


(t, s ∈ R)

In vector form

( vector equation of a plane in the space ),

(x1, x2, x3, x4) = (−2, 6, 0, 0) + t(1,−2, 1, 0) + s(2,−3, 0, 1)


Introduction




⇒{

x1 = x3 + 2x4 − 2x2 = −2x3 − 3x4 + 6


(t, s ∈ R)


(x1, x2, x3, x4) = (−2, 6, 0, 0) + t(1,−2, 1, 0) + s(2,−3, 0, 1)


Introduction



Example 1.8 y + 3z = 0

x + y − 2z = 0x + 2y + az = 0

a ∈ R

The system is homogeneous (all right-hand sides are zeros).Therefore it is consistent ( x = y = z = 0 is a solution )The augmented matrix is: 0 1 3 0

1 1 2 01 2 a 0


Introduction



Example 1.8

y + 3z = 0

x + y − 2z = 0x + 2y + az = 0

a ∈ R


1 1 2 01 2 a 0


Introduction




x + y − 2z = 0x + 2y + az = 0

a ∈ R


1 1 2 01 2 a 0


Introduction




x + y − 2z = 0x + 2y + az = 0

a ∈ R

The system

is homogeneous (all right-hand sides are zeros).Therefore it is consistent ( x = y = z = 0 is a solution )The augmented matrix is: 0 1 3 0

1 1 2 01 2 a 0


Introduction




x + y − 2z = 0x + 2y + az = 0

a ∈ R

The system is homogeneous

(all right-hand sides are zeros).Therefore it is consistent ( x = y = z = 0 is a solution )The augmented matrix is: 0 1 3 0

1 1 2 01 2 a 0


Introduction




x + y − 2z = 0x + 2y + az = 0

a ∈ R

The system is homogeneous (all right-hand sides are zeros).

Therefore it is consistent ( x = y = z = 0 is a solution )The augmented matrix is: 0 1 3 0

1 1 2 01 2 a 0


Introduction




x + y − 2z = 0x + 2y + az = 0

a ∈ R

The system is homogeneous (all right-hand sides are zeros).Therefore

it is consistent ( x = y = z = 0 is a solution )The augmented matrix is: 0 1 3 0

1 1 2 01 2 a 0


Introduction




x + y − 2z = 0x + 2y + az = 0

a ∈ R

The system is homogeneous (all right-hand sides are zeros).Therefore it is consistent

( x = y = z = 0 is a solution )The augmented matrix is: 0 1 3 0

1 1 2 01 2 a 0


Introduction




x + y − 2z = 0x + 2y + az = 0

a ∈ R

The system is homogeneous (all right-hand sides are zeros).Therefore it is consistent ( x = y = z = 0 is a solution )

The augmented matrix is: 0 1 3 01 1 2 01 2 a 0


Introduction




x + y − 2z = 0x + 2y + az = 0

a ∈ R

The system is homogeneous (all right-hand sides are zeros).Therefore it is consistent ( x = y = z = 0 is a solution )The augmented matrix is:

0 1 3 01 1 2 01 2 a 0


Introduction




x + y − 2z = 0x + 2y + az = 0

a ∈ R


1 1 2 01 2 a 0


Introduction



Since the 1st row cannot serve as a pivotal one, we interchange itwith the 2nd row: 0 1 3 0

1 1 −2 01 2 a 0

⇒

1 1 −2 00 1 3 01 2 a 0

Now we can start the elimination. First subtract the 1st row fromthe 3rd row: 1 1 −2 0

0 1 3 01 2 a 0

⇒

1 1 −2 00 1 3 00 1 a + 2 0


Introduction



Since

the 1st row cannot serve as a pivotal one, we interchange itwith the 2nd row: 0 1 3 0

1 1 −2 01 2 a 0

⇒

1 1 −2 00 1 3 01 2 a 0


0 1 3 01 2 a 0

⇒

1 1 −2 00 1 3 00 1 a + 2 0


Introduction



Since the 1st row

cannot serve as a pivotal one, we interchange itwith the 2nd row: 0 1 3 0

1 1 −2 01 2 a 0

⇒

1 1 −2 00 1 3 01 2 a 0


0 1 3 01 2 a 0

⇒

1 1 −2 00 1 3 00 1 a + 2 0


Introduction



Since the 1st row cannot serve

as a pivotal one, we interchange itwith the 2nd row: 0 1 3 0

1 1 −2 01 2 a 0

⇒

1 1 −2 00 1 3 01 2 a 0


0 1 3 01 2 a 0

⇒

1 1 −2 00 1 3 00 1 a + 2 0


Introduction



Since the 1st row cannot serve as a pivotal one,

we interchange itwith the 2nd row: 0 1 3 0

1 1 −2 01 2 a 0

⇒

1 1 −2 00 1 3 01 2 a 0


0 1 3 01 2 a 0

⇒

1 1 −2 00 1 3 00 1 a + 2 0


Introduction



Since the 1st row cannot serve as a pivotal one, we interchange it

with the 2nd row: 0 1 3 01 1 −2 01 2 a 0

⇒

1 1 −2 00 1 3 01 2 a 0


0 1 3 01 2 a 0

⇒

1 1 −2 00 1 3 00 1 a + 2 0


Introduction



Since the 1st row cannot serve as a pivotal one, we interchange itwith the 2nd row:

0 1 3 01 1 −2 01 2 a 0

⇒

1 1 −2 00 1 3 01 2 a 0


0 1 3 01 2 a 0

⇒

1 1 −2 00 1 3 00 1 a + 2 0


Introduction




1 1 −2 01 2 a 0

⇒

1 1 −2 00 1 3 01 2 a 0


0 1 3 01 2 a 0

⇒

1 1 −2 00 1 3 00 1 a + 2 0


Introduction




1 1 −2 01 2 a 0

⇒

1 1 −2 00 1 3 01 2 a 0

Now

we can start the elimination. First subtract the 1st row fromthe 3rd row: 1 1 −2 0

0 1 3 01 2 a 0

⇒

1 1 −2 00 1 3 00 1 a + 2 0


Introduction




1 1 −2 01 2 a 0

⇒

1 1 −2 00 1 3 01 2 a 0

Now we can start the elimination.

First subtract the 1st row fromthe 3rd row: 1 1 −2 0

0 1 3 01 2 a 0

⇒

1 1 −2 00 1 3 00 1 a + 2 0


Introduction




1 1 −2 01 2 a 0

⇒

1 1 −2 00 1 3 01 2 a 0

Now we can start the elimination. First

subtract the 1st row fromthe 3rd row: 1 1 −2 0

0 1 3 01 2 a 0

⇒

1 1 −2 00 1 3 00 1 a + 2 0


Introduction




1 1 −2 01 2 a 0

⇒

1 1 −2 00 1 3 01 2 a 0

Now we can start the elimination. First subtract the 1st row

fromthe 3rd row: 1 1 −2 0

0 1 3 01 2 a 0

⇒

1 1 −2 00 1 3 00 1 a + 2 0


Introduction




1 1 −2 01 2 a 0

⇒

1 1 −2 00 1 3 01 2 a 0

Now we can start the elimination. First subtract the 1st row fromthe 3rd row:

1 1 −2 00 1 3 01 2 a 0

⇒

1 1 −2 00 1 3 00 1 a + 2 0


Introduction




1 1 −2 01 2 a 0

⇒

1 1 −2 00 1 3 01 2 a 0


0 1 3 01 2 a 0

⇒

1 1 −2 00 1 3 00 1 a + 2 0


Introduction



The 2nd row is our new pivotal row. Subtract the 2nd row fromthe 3rd row: 1 1 −2 0

0 1 3 00 1 a + 2 0

⇒

1 1 −2 00 1 3 00 0 a− 1 0

At this point row reduction splits into two cases.Case 1: a 6= 1. In this case, multiply the 3rd row by (a− 1)−1 1 1 −2 0

0 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 1 0


Introduction



The 2nd row

is our new pivotal row. Subtract the 2nd row fromthe 3rd row: 1 1 −2 0

0 1 3 00 1 a + 2 0

⇒

1 1 −2 00 1 3 00 0 a− 1 0


0 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 1 0


Introduction



The 2nd row is our new pivotal row.

Subtract the 2nd row fromthe 3rd row: 1 1 −2 0

0 1 3 00 1 a + 2 0

⇒

1 1 −2 00 1 3 00 0 a− 1 0


0 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 1 0


Introduction



The 2nd row is our new pivotal row. Subtract the 2nd row

fromthe 3rd row: 1 1 −2 0

0 1 3 00 1 a + 2 0

⇒

1 1 −2 00 1 3 00 0 a− 1 0


0 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 1 0


Introduction



The 2nd row is our new pivotal row. Subtract the 2nd row fromthe 3rd row:

1 1 −2 00 1 3 00 1 a + 2 0

⇒

1 1 −2 00 1 3 00 0 a− 1 0


0 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 1 0


Introduction




0 1 3 00 1 a + 2 0

⇒

1 1 −2 00 1 3 00 0 a− 1 0


0 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 1 0


Introduction




0 1 3 00 1 a + 2 0

⇒

1 1 −2 00 1 3 00 0 a− 1 0

At this point

row reduction splits into two cases.Case 1: a 6= 1. In this case, multiply the 3rd row by (a− 1)−1 1 1 −2 0

0 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 1 0


Introduction




0 1 3 00 1 a + 2 0

⇒

1 1 −2 00 1 3 00 0 a− 1 0

At this point row reduction splits

into two cases.Case 1: a 6= 1. In this case, multiply the 3rd row by (a− 1)−1 1 1 −2 0

0 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 1 0


Introduction




0 1 3 00 1 a + 2 0

⇒

1 1 −2 00 1 3 00 0 a− 1 0

At this point row reduction splits into two cases.

Case 1: a 6= 1. In this case, multiply the 3rd row by (a− 1)−1 1 1 −2 00 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 1 0


Introduction




0 1 3 00 1 a + 2 0

⇒

1 1 −2 00 1 3 00 0 a− 1 0

At this point row reduction splits into two cases.Case 1: a 6= 1. In this case, multiply the 3rd row by (a− 1)−1

1 1 −2 00 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 1 0


Introduction




0 1 3 00 1 a + 2 0

⇒

1 1 −2 00 1 3 00 0 a− 1 0


0 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 1 0


Introduction



The matrix is converted into row echelon form. We proceedtowards reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0

0 1 3 00 0 1 0

⇒

1 1 −2 00 1 0 00 0 1 0

Add 2 times the 3rd row to the 1st row: 1 1 −2 0

0 1 0 00 0 1 0

⇒

1 1 0 00 1 0 00 0 1 0


Introduction



The matrix

is converted into row echelon form. We proceedtowards reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0

0 1 3 00 0 1 0

⇒

1 1 −2 00 1 0 00 0 1 0


0 1 0 00 0 1 0

⇒

1 1 0 00 1 0 00 0 1 0


Introduction



The matrix is converted

into row echelon form. We proceedtowards reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0

0 1 3 00 0 1 0

⇒

1 1 −2 00 1 0 00 0 1 0


0 1 0 00 0 1 0

⇒

1 1 0 00 1 0 00 0 1 0


Introduction



The matrix is converted into row echelon form.

We proceedtowards reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0

0 1 3 00 0 1 0

⇒

1 1 −2 00 1 0 00 0 1 0


0 1 0 00 0 1 0

⇒

1 1 0 00 1 0 00 0 1 0


Introduction



The matrix is converted into row echelon form. We proceedtowards

reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0

0 1 3 00 0 1 0

⇒

1 1 −2 00 1 0 00 0 1 0


0 1 0 00 0 1 0

⇒

1 1 0 00 1 0 00 0 1 0


Introduction



The matrix is converted into row echelon form. We proceedtowards reduced row echelon form.

Subtract 3 times the 3rd rowfrom the 2nd row: 1 1 −2 0

0 1 3 00 0 1 0

⇒

1 1 −2 00 1 0 00 0 1 0


0 1 0 00 0 1 0

⇒

1 1 0 00 1 0 00 0 1 0


Introduction



The matrix is converted into row echelon form. We proceedtowards reduced row echelon form. Subtract 3 times the 3rd rowfrom the 2nd row:

1 1 −2 00 1 3 00 0 1 0

⇒

1 1 −2 00 1 0 00 0 1 0


0 1 0 00 0 1 0

⇒

1 1 0 00 1 0 00 0 1 0


Introduction




0 1 3 00 0 1 0

⇒

1 1 −2 00 1 0 00 0 1 0


0 1 0 00 0 1 0

⇒

1 1 0 00 1 0 00 0 1 0


Introduction




0 1 3 00 0 1 0

⇒

1 1 −2 00 1 0 00 0 1 0

Add 2 times the 3rd row to the 1st row: 1 1 −2 00 1 0 00 0 1 0

⇒

1 1 0 00 1 0 00 0 1 0


Introduction




0 1 3 00 0 1 0

⇒

1 1 −2 00 1 0 00 0 1 0

Add 2 times

the 3rd row to the 1st row: 1 1 −2 00 1 0 00 0 1 0

⇒

1 1 0 00 1 0 00 0 1 0


Introduction




0 1 3 00 0 1 0

⇒

1 1 −2 00 1 0 00 0 1 0

Add 2 times the 3rd row

to the 1st row: 1 1 −2 00 1 0 00 0 1 0

⇒

1 1 0 00 1 0 00 0 1 0


Introduction




0 1 3 00 0 1 0

⇒

1 1 −2 00 1 0 00 0 1 0

Add 2 times the 3rd row to the 1st row:

1 1 −2 00 1 0 00 0 1 0

⇒

1 1 0 00 1 0 00 0 1 0


Introduction




0 1 3 00 0 1 0

⇒

1 1 −2 00 1 0 00 0 1 0


0 1 0 00 0 1 0

⇒

1 1 0 00 1 0 00 0 1 0


Introduction



Finally,subtract the 2nd row from the 1st row: 1 1 0 00 1 0 00 0 1 0

⇒

1 0 0 00 1 0 00 0 1 0

Thus x = y = z = 0 is the only solution


Introduction



Finally,

subtract the 2nd row from the 1st row: 1 1 0 00 1 0 00 0 1 0

⇒

1 0 0 00 1 0 00 0 1 0



Introduction



Finally,subtract the 2nd row

from the 1st row: 1 1 0 00 1 0 00 0 1 0

⇒

1 0 0 00 1 0 00 0 1 0



Introduction



Finally,subtract the 2nd row from the 1st row:

1 1 0 00 1 0 00 0 1 0

⇒

1 0 0 00 1 0 00 0 1 0



Introduction




⇒

1 0 0 00 1 0 00 0 1 0



Introduction




⇒

1 0 0 00 1 0 00 0 1 0

Thus x = y = z = 0

is the only solution


Introduction




⇒

1 0 0 00 1 0 00 0 1 0



Introduction



Case 2: a = 1. In this case, the matrix is already in row echelonform: 1 1 −2 0

0 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 0 0

To get reduced row echelon form, subtract the 2nd row from the1st row: 1 1 −2 0

0 1 3 00 0 0 0

⇒

1 0 −5 00 1 3 00 0 0 0


Introduction



Case 2: a = 1. In this case, the matrix is already in row echelonform:

1 1 −2 00 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 0 0


0 1 3 00 0 0 0

⇒

1 0 −5 00 1 3 00 0 0 0


Introduction




0 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 0 0


0 1 3 00 0 0 0

⇒

1 0 −5 00 1 3 00 0 0 0


Introduction




0 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 0 0

To get

reduced row echelon form, subtract the 2nd row from the1st row: 1 1 −2 0

0 1 3 00 0 0 0

⇒

1 0 −5 00 1 3 00 0 0 0


Introduction




0 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 0 0

To get reduced row echelon form,

subtract the 2nd row from the1st row: 1 1 −2 0

0 1 3 00 0 0 0

⇒

1 0 −5 00 1 3 00 0 0 0


Introduction




0 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 0 0

To get reduced row echelon form, subtract the 2nd row

from the1st row: 1 1 −2 0

0 1 3 00 0 0 0

⇒

1 0 −5 00 1 3 00 0 0 0


Introduction




0 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 0 0

To get reduced row echelon form, subtract the 2nd row from the1st row:

1 1 −2 00 1 3 00 0 0 0

⇒

1 0 −5 00 1 3 00 0 0 0


Introduction




0 1 3 00 0 a− 1 0

⇒

1 1 −2 00 1 3 00 0 0 0


0 1 3 00 0 0 0

⇒

1 0 −5 00 1 3 00 0 0 0


Introduction



Then, z(= t) is a free variable, and the solution is given by{x − 5z = 0y + 3z = 0

⇒{

x = 5z = 5ty = −3z = −3t

Thus, the System of linear equations:x + 3z = 0

x + y − 2z = 0x + 2y + az = 0

has as a solution:

if a 6= 1 then (x , y , z) = (0, 0, 0).if a = 1 then (x , y , z) = t(5,−3, 1) t ∈ R.


Introduction



Then,

z(= t) is a free variable, and the solution is given by{x − 5z = 0y + 3z = 0

⇒{

x = 5z = 5ty = −3z = −3t


x + y − 2z = 0x + 2y + az = 0

has as a solution:



Introduction



Then, z(= t) is a free variable, and

the solution is given by{x − 5z = 0y + 3z = 0

⇒{

x = 5z = 5ty = −3z = −3t


x + y − 2z = 0x + 2y + az = 0

has as a solution:



Introduction



Then, z(= t) is a free variable, and the solution is given by

{x − 5z = 0y + 3z = 0

⇒{

x = 5z = 5ty = −3z = −3t


x + y − 2z = 0x + 2y + az = 0

has as a solution:



Introduction




⇒

{x = 5z = 5ty = −3z = −3t


x + y − 2z = 0x + 2y + az = 0

has as a solution:



Introduction




⇒{

x = 5z = 5ty = −3z = −3t

Thus, the System of linear equations:

x + 3z = 0

x + y − 2z = 0x + 2y + az = 0

has as a solution:



Introduction




⇒{

x = 5z = 5ty = −3z = −3t


x + y − 2z = 0x + 2y + az = 0

has as a solution:



Introduction




⇒{

x = 5z = 5ty = −3z = −3t


x + y − 2z = 0x + 2y + az = 0

has as a solution:

if a 6= 1

then (x , y , z) = (0, 0, 0).if a = 1 then (x , y , z) = t(5,−3, 1) t ∈ R.


Introduction




⇒{

x = 5z = 5ty = −3z = −3t


x + y − 2z = 0x + 2y + az = 0

has as a solution:

if a 6= 1 then

(x , y , z) = (0, 0, 0).if a = 1 then (x , y , z) = t(5,−3, 1) t ∈ R.


Introduction




⇒{

x = 5z = 5ty = −3z = −3t


x + y − 2z = 0x + 2y + az = 0

has as a solution:

if a 6= 1 then (x , y , z) = (0, 0, 0).

if a = 1 then (x , y , z) = t(5,−3, 1) t ∈ R.


Introduction




⇒{

x = 5z = 5ty = −3z = −3t


x + y − 2z = 0x + 2y + az = 0

has as a solution:

if a 6= 1 then (x , y , z) = (0, 0, 0).if a = 1

then (x , y , z) = t(5,−3, 1) t ∈ R.


Introduction




⇒{

x = 5z = 5ty = −3z = −3t


x + y − 2z = 0x + 2y + az = 0

has as a solution:

if a 6= 1 then (x , y , z) = (0, 0, 0).if a = 1 then

(x , y , z) = t(5,−3, 1) t ∈ R.


Introduction




⇒{

x = 5z = 5ty = −3z = −3t


x + y − 2z = 0x + 2y + az = 0

has as a solution:

if a 6= 1 then (x , y , z) = (0, 0, 0).if a = 1 then (x , y , z) = t(5,−3, 1)

t ∈ R.


Introduction




⇒{

x = 5z = 5ty = −3z = −3t


x + y − 2z = 0x + 2y + az = 0

has as a solution:



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