Linear Algebra(Aljabar Linier)

Week 2

Universitas Multimedia NusantaraSerpong, Tangerang

Dr. Ananda Kusumae-mail: ananda_kusuma@yahoo.com

Ph: 081338227031, 081908058069

Agenda

• Review on Vectors– Exercises

• Systems of Linear Equations– Introduction– Direct Methods

• Matrices and echelon form• Gaussian Elimination• Gauss-Jordan Elimination

– Spanning Sets and Linear Independence– Iterative Methods

• Jacobi’s• Gauss-Seidel

– Applications– Exercises

Vectors: Exercises

• Using dot/inner product, compute angle between two vectors v and d

• Find projection of v onto d, i.e.• Given that B=(1,0,2) and line through the point

A=(3,1,1), with direction vector d=[-1,1,0], compute the distance from B to line

• In R2 and R3, ≤ , show that this is equivalent to Cauchy-Schwarz Inequality

v

Systems of Linear Equations

Linear Equations

Recall equation of a line in R2 and a plane in R3 from last week’s lecture

Which ones are linear?

A system of linear equations

• A system of linear equations is a finite set of linear equations, each with the same variables.

• A solution of a system of linear equations is a vector that is simultaneously a solution of each equation in the system

• The solution set of a system of linear equations is the set of all solutions of the system

• A system of linear equations with real coefficients has either:

Example in R2

• Solve the following systems of linear equations

Example in R3

Homogeneous Linear Systems

• A homogeneous system cannot have no solution. It will have either a unique solution (namely the zero or trivial solution) or infinitely many solutions

• A homogenous system of m linear equations with n variables, where m < n, then the system has infinitely many solutions

Solving a system of linear equations

• Two linear systems are called equivalent if they have the same solution sets

• Example: which one is easier to solve?

• The approach to solving a system of linear equations is to transform the given system into an equivalent one that is easier to solve

– Triangular structure and use back-substitution to solve– Develop strategy for transforming a given system in an equivalent one

Example

• Solve the system

• Hint: find triangular structure and use back-substitution• Utilize matrix useful in real-life applications when the systems

are large or have coefficients that are not nice

Augmented matrix of the system

Coefficient matrix bA |

Example

• The solution is [3,-1,2]

Direct Methods for Solving Linear Systems

Introduction

• Based on the idea of reducing the augmented matrix of the given system to a form that can then be solved by backsubtitution– Direct leads directly to the solution (if one exists) in a finite

number of steps– In solving a linear system, it will not always possible to reduce the

coeffient matrix to triangular form, but we can always achieve a staircase pattern in the nonzero entries of the final matrix

Row Reduction: Convert a matrix to echelon form

• Notation

• Exercise: reduce the following matrix to echelon form

• Remember that row echelon form of a matrix is not unique– Doing different sequences of row operations can give different row echelon forms

Row Equivalent

• Elementary row operations are reversible– What is the elementary row operation that undoes , ,– Example:

ji RR ikR ji kRR

Gaussian Elimination

• Examples:– Solve the following

Rank of a matrix

Gauss-Jordan Elimination

• Modification of Gaussian Elimination simplify back substitution phase

Examples

• Check the following for reduced row echelon form

• Solve the following using Gauss-Jordan Elimination

Spanning Setsand Linear Independence

Introduction (1)

• Examples:

Introduction (2)

Spanning Sets

• Examples:

Linear Independence

Example:- Determine whether the following set of vectors are linearly independent

Some theorems: linear dependence

The End

To be continued next week

Thank you for your attention!