Linear Algebra (Aljabar Linier) Week 2 Universitas Multimedia Nusantara Serpong, Tangerang Dr. Ananda Kusuma Ph: 081338227031,

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Presentation transcript:

Linear Algebra (Aljabar Linier) Week 2 Universitas Multimedia Nusantara Serpong, Tangerang Dr. Ananda Kusuma Ph: ,

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 R 2 and R 3, ≤, show that this is equivalent to Cauchy-Schwarz Inequality

Systems of Linear Equations

Linear Equations Recall equation of a line in R 2 and a plane in R 3 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 R 2 Solve the following systems of linear equations

Example in R 3

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

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:

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 Sets and 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!