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DM559 Linear and Integer Programming Lecture 2 Systems of Linear Equations Marco Chiarandini Department of Mathematics & Computer Science University of Southern Denmark Systems of Linear Equations Outline 1. Systems of Linear Equations


  1. DM559 Linear and Integer Programming Lecture 2 Systems of Linear Equations Marco Chiarandini Department of Mathematics & Computer Science University of Southern Denmark

  2. Systems of Linear Equations Outline 1. Systems of Linear Equations 2

  3. Systems of Linear Equations Outline 1. Systems of Linear Equations 3

  4. Systems of Linear Equations A Motivating Example You are organizing the next party at IMADA. In order to make it a memorable experience for everybody, you want to prepare 8 liters of a drink that contains 32% of alcohol. At the store you find two products that contain 50% alcohol and 10% alcohol, respectively. How much of each is needed? • Let x represent the amount of the 50% product needed. • Let y represent the amount of the 10% product needed. • The total amount of the mixture must be 8 liters. x + y = 8 • The amount of alcohol from each product in the end result must be 32% of 8 liters, or 0.32(8). 0 . 50 x + 0 . 10 y = 0 . 32 ( 8 ) • We have a system of two linear equations and two variables (aka, unknowns). � x + y = 8 0 . 50 x + 0 . 10 y = 0 . 32 ( 8 ) • How do you find a solution? • Is there always a solution? 4

  5. Systems of Linear Equations Another Motivating Example The picture shows a network of one-way streets with traffic flowing in the directions indicated. The flow rates along the streets are measured as the average number of vehicles per hour. Determine the flow rates in the inner branches of the network. Intersection In Flow Out Flow A: 750 + x 3 = 250 + x 4 B: 200 + x 4 = 300 + x 1 C: 200 + x 1 = 400 + x 2 D: 300 + x 2 = 400 + x 3  − x 3 + x 4 = 500   − x 1 + x 4 = 100  x 1 − x 2 = 200   x 2 − x 3 = 100  A system of linear equations in 4 unknowns 5

  6. Systems of Linear Equations Yet Another Motivating Example Polynomial Interpolation Given any n points in the xy -plane that have distinct x -coordinates, there is a unique polynomial of degree n − 1 or less whose graph passes through those points. The graph of the polynomial is the graph of the equation: y = a 0 + a 1 x + a 2 x 2 + · · · + a n − 1 x n − 1 The coordinates of the points ( x 1 , y 1 ) , ( x 2 , y 2 ) , . . . , ( x n , y n ) must satisfy:  a 0 + a 1 x 1 + a 2 x 2 1 + · · · + a n − 1 x n − 1 = y 1 1   2 + · · · + a n − 1 x n − 1 a 0 + a 1 x 2 + a 2 x 2 = y 2   2 . . . . . . . . . . . . . . . . . = .    a 0 + a 1 x n + a 2 x 2 n + · · · + a n − 1 x n − 1 = y n  n 6

  7. Systems of Linear Equations Find a cubic polynomial that passes through the points: ( 1 , 3 ) ( 2 , − 2 ) ( 3 , − 5 ) ( 4 , 0 ) y = a 0 + a 1 x + a 2 x 2 + a 3 x 3  a 0 + a 1 + a 2 + a 3 = 3   a 0 + a 1 2 + a 2 4 + a 3 8 = − 2  a 0 + a 1 3 + a 2 9 + a 3 27 = − 5   a 0 + a 1 4 + a 2 16 + a 3 64 = 0   1 1 1 1     3  a 0 1 2 4 8 − 2 a 1        =       1 3 9 27 − 5 a 2      1 4 16 64 0 a 3 7

  8. Systems of Linear Equations Systems of Linear Equations Definition (Linear Equation) A linear equation in the n variables x 1 , x 2 , . . . , x n is an equation that can be written in the form: a 1 x 1 + a 2 x 2 + · · · + a n x n = b a 1 , a 2 , . . . , a n , b ∈ R , ∃ i : a i � = 0 Definition (System of linear equations, aka linear system) A system of m linear equations in n unknowns x 1 , x 2 , . . . , x n is a set of m equations of the form a 11 x 1 + a 12 x 2 + · · · + a 1 n x n = b 1 a 21 x 1 + a 22 x 2 + · · · + a 2 n x n = b 2 . . The numbers a ij are known as . . . . the coefficients of the system. a m 1 x 1 + a m 2 x 2 + · · · + a mn x n = b m We say that s 1 , s 2 , . . . , s n is a solution of the system if all m equations hold true when x 1 = s 1 , x 2 = s 2 , . . . , x n = s n 8

  9. Systems of Linear Equations Examples x 1 + x 2 + x 3 + x 4 + x 5 = 3 2 x 1 + x 2 + x 3 + x 4 + 2 x 5 = 4 x 1 − x 2 − x 3 + x 4 + x 5 = 5 + x 4 + x 5 = 4 x 1 has solution x 1 = − 1 , x 2 = − 2 , x 3 = 1 , x 4 = 3 , x 5 = 2 . Is it the only one? x 1 + x 2 + x 3 + x 4 + x 5 = 3 2 x 1 + x 2 + x 3 + x 4 + 2 x 5 = 4 x 1 − x 2 − x 3 + x 4 + x 5 = 5 + x 4 + x 5 = 6 x 1 has no solutions 9

  10. Systems of Linear Equations Matrices Definition (Matrix) A matrix is a rectangular array of numbers or symbols. It can be written as   a 12 · · · a 1 n a 11 a 21 a 22 · · · a 2 n   . . .  ...  . . .   . . .   a m 1 a m 2 · · · a mn • We denote this array by a single letter A or by ( a ij ) and • we say that A has m rows and n columns, or that it is an m × n matrix. • The size of A is m × n . • The number a ij is called the ( i , j ) entry or scalar. 10

  11. Systems of Linear Equations Vectors • An n × 1 matrix is a column vector, or simply a vector:   v 1 v 2   v =  .  .   .   v n The numbers v 1 , v 2 , . . . are known as the components (or entries) of v . • A row vector is a 1 × n matrix • We write vectors in lower boldcase type (writing by hand we can either underline them or add an arrow over v ). 11

  12. Systems of Linear Equations Definition (Coefficient Matrix) The matrix A = ( a ij ) , whose ( i , j ) entry is the coefficient a ij of the system of linear equations is called the coefficient matrix.   a 12 · · · a 1 n a 11 a 22 · · · a 2 n a 21   A = . . .  ...  . . .   . . .   a m 1 a m 2 · · · a mn Let x = [ x 1 , x 2 , . . . , x n ] T then n × 1 n × 1 m × n     · · · a 11 x 1 + a 12 x 2 + · · · + a 1 n x n a 11 a 12 a 1 n x 1   · · · a 21 x 1 + a 22 x 2 + · · · + a 2 n x n a 21 a 22 a 2 n x 2        = . . . . . .     ... . . .  . . .      . . . . . .      x n a m 1 x 1 + a m 2 x 2 + · · · + a mn x n a m 1 a m 2 · · · a mn 12 hence, the linear system can be written also as A x = b

  13. Systems of Linear Equations Row operations How do we find solutions? x 1 + x 2 + x 3 = 3 R1: R2: 2 x 1 + x 2 + x 3 = 4 R3: x 1 − x 2 + 2 x 3 = 5 Eliminate one of the variables from two of the equations R1’=R1: x 1 + x 2 + x 3 = 3 R2’=R2-2*R1: − x 2 − x 3 = − 2 R3’=R3: x 1 − x 2 + 2 x 3 = 5 x 1 + x 2 + x 3 = 3 R1’=R1: − x 2 − x 3 = − 2 R2’=R2: − 2 x 2 + x 3 = 2 R3’=R3-R1: We can now eliminate one of the variables in the last two equations to obtain the solution 13

  14. Systems of Linear Equations Row operations that do not alter solutions: RO1: multiply both sides of an equation by a non-zero constant RO2: interchange two equations RO3: add a multiple of one equation to another These operations only act on the coefficients of the system For a system A x = b :   1 1 1 3 � � A b = 2 1 1 4   1 − 1 2 5 14

  15. Systems of Linear Equations Problem Statement Given the system of linear equations: x 1 + x 2 + x 3 = 3 R1: R2: 2 x 1 + x 2 + x 3 = 4 x 1 − x 2 + 2 x 3 = 5 R3: Find whether it has any solution and in case characterize the solutions. 15

  16. Systems of Linear Equations Augmented Matrix Definition (Augmented Matrix and Elementary Row Operations) For a system of linear equations A x = b with a 11 a 12 · · · a 1 n    x 1   b 1  a 22 · · · a 2 n a 21 x 2 b 2       A = . . . x = b =  ...   .   .  . . . . .       . . . . .       a m 1 a m 2 · · · a mn x n b m the augmented matrix of the system is and the elementary row operations are: • RO1: multiply a row by a non-zero constant a 11 a 12 · · · a 1 n b 1   • RO2: interchange two rows a 21 a 22 · · · a 2 n b 2   � � A b = . . . .  ...  . . . . • RO3: add a multiple of one row to another   . . . .   a m 1 a m 2 · · · a mn b m They modify the linear system into an equivalent system (same solutions) 16

  17. Systems of Linear Equations Gaussian Elimination: Example Let’s consider the system A x = b with:   1 1 1 3   � � A b = 2 1 1 4     1 − 1 2 5 1. Left most column that is not all zeros 5. Cover up the top row and apply steps 1. to It is column 1 4. again 2. A non-zero entry at the top of this column  1 1 1 3  It is the one on the top 0 − 1 − 1 − 2   3. Make the entry 1 0 − 2 1 2 It is already 1 4. Make all entries below the leading one zero: R1’=R1  1 1 1 3  R2’=R2-2R1 0 − 1 − 1 − 2   R3’=R3-R1 0 − 2 1 2 17

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