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Systems of Linear Equations and Matrices

Systems of Linear Equations and Matrices. Introduction. We had to solve 2 simultaneous linear equations in order to find the Break-even point and the equilibrium point . These are two real world situations which call for solving a system of linear equations in two or more variables.

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Systems of Linear Equations and Matrices

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  1. Systems of Linear Equations and Matrices

  2. Introduction We had to solve 2 simultaneous linear equations in order to find the Break-even point and the equilibrium point . These are two real world situations which call for solving a system of linear equations in two or more variables. Nature of the solution system of linear equations. We know the graph of each equation in system ax + by = h …I cx + dy = k …II is a straight line in the plane, so that geometrically the solution to the system is the point of intersection of the two lines I and II. There are 3 particular cases (solutions) that may occur • Intersect at exactly one point, • Be parallel and coincident, or • Be parallel and distinct Lamar University 2

  3. A System of Equation with exactly one solution Consider the system 2x – y = 1 …I 3x + 2y = 12 …II Like we have done so many times before we take equation I and put it in this form y = 2x – 1 Substituting this expression in II we get 3x + 2( 2x- 1 ) = 12 3x + 4x – 2 = 12 7x = 14 x = 2 Substituting in I we get y = 2(2) – 1 = 3 Thus the point of intersection is (2,3) Lamar University 3

  4. System of equations with infinitely many solutions Consider the following system 2x – y = 1 …I 6x – 3y = 3 …II Solving the first equation in terms of x we get y = 2x – 1 Substituting this in II we get 6x – 3(2x – 1) = 3 6x – 6x + 3 = 3 0 = 0 This result simply tells us that the first equation is equivalent to the second. Geometrically the two equations in the system represent the same line and all solutions lie on the line. Lamar University 4

  5. A system of equations with no solution Consider the system 2x – y = 1 …I 6x – 3y = 12 …II The first equation is equivalent to y = 2x – 1 substituting this in equation II we get 6x – 3 ( 2x – 1 ) = 12 6x – 6x + 3 = 12 0 = 9 We all know this is not possible. Thus there is no solution in this case Lamar University 5

  6. Solution to Systems of Equations Just as a linear equation in two variables represents a straight line in the plane, it can be shown that a linear equation ax + by + cz = d ( a, b and c not simultaneously equal to zero) in three variables represents a plane in three dimensional space.The following equations represent a plane in space a1x + b1y + c1z = d1 a2x + b2y + c2z = d2 a3x + b3y + c3z = d3 As before these planes have one and only one solution,, infinitely many solutions or then no solution at all. Lamar University 6

  7. Problem The Ace Novelty company wishes to produces three types of souvenirs : types A, B , and C. To manufacture a type A souvenir requires 2 minutes on machine I, 1 minute on machine II, and 2 minutes on machine III. A type-B souvenir requires 1 minute on machine I, 3 minutes on machine II and 1 minute on machine III. A type – C souvenir requires 1 minute on machine I and 2 minutes on machine II and III. There are 3 hours available on machine I, 5 hours available on machine II and 4 hours available on machine III for processing the order. How many souvenirs of each type should Ace Novelty make in order to use all of the available time? Formulate but do not solve the problem. Lamar University 7

  8. Problem We have to determine the number of each of the three types of souvenirs to be made. Let x, y , z denote the respective numbers of type A, type B and type C souvenirs to be made. The total amount of time that machine I is used is given by 2x + y + z and must equal 180 minutes. This leads to the equation 2x + y + z = 180 (Time spent on machine I) Similarly x + 3y + 2z = 300 (Time spent on machine II) 2x + y + 2z = 240 (Time spent on machine III) Lamar University 8

  9. Problem Investments Kelly Fisher has a total of $30,000 invested in two municipal bonds that have yields of 8% and 10% interest per year, respectively. If the interest Kelly receives from the bonds in a year is $2640, how much does she have to invest in each bond Solution: Let the amount of money invested in the bonds yielding 8% be x dollars and the amount of money invested in in bonds yielding 10% be y dollars. Then x + y = 30,000. Also since the yield from both investments totals $ 2640, we have 0.08x + 0.10y = 2640 The solution of the problem can be found by solving the system of equations x + y = 30,000 0.08x + 0.10y = 2640 Lamar University 9

  10. Problem Management Decisions: The management of Hartman Rent A Car has allocated $ 1.25 million to buy a fleet of new automobiles consisting of compact, intermediate and full size cars. Compacts costs $ 10,000 each, intermediate costs $ 15,000 each and full size costs $20,000 each. If Hartman purchases twice as many compacts as intermediate cars and the total number of cars to be purchased is 100 determine how many cars of each type will be purchased.(Assume the entire budget will be used) Solution: Let x, y and z denote the number of compact intermediate and full size cars respectively, to be purchased. The cost incurred in buying the the specified number of cars is 10000x+ 15000y + 20000z . Since the budget is 1.25 million we have the system 10,000 x + 15,000 y + 20,000z = 1,250,000 x – 2y = 0 x + y + z = 100 Lamar University 10

  11. Gauss Jordan Method The operations of the Gauss-Jordan method are: • Interchange any two equations. • Replace an equation by a nonzero constant multiple of itself. • Replace an equation by the sum of that equation and a constant multiple of any other equation. Lamar University 11

  12. Problem Lets apply the Gauss – Jordan method to this system 2x + 4y = 8 3x – 2y = 4 We begin by working with the first, or x, column. First, we transform the system into and equivalent system in which the coefficient of x in the first equation is 1 2x + 4y = 8 3x – 2y = 4 x + 2y = 4 3x - 2y = 4 Now we eliminate x from the second equation (3)x + (3)2y = (3)4 3x – 2y = 4 - 8y = - 8 Lamar University 12

  13. Problem Then we obtain the following equivalent system in which the coefficient of y in the second equation is 1 x + 2y = 4 y = 1 Now eliminate y in the first equation by multiplying the second equation by –2 and adding to the first equation x + 2y = 4 (-2)y = (-2)1 Adding the two we get x = 2 y = 1 Lamar University 13

  14. Problem Solve: 2x+ 4y + 6z = 22 3x + 8y + 5z = 27 - x + y + 2z = 2 First we Transform this system into an equivalent system in which the coefficient of x in the first equation is 1: 2x+ 4y + 6z = 22 …I 3x + 8y + 5z = 27 …II - x + y + 2z = 2 …III Multiply I by (1/2) we get x+ 2y + 3z = 11 3x + 8y + 5z = 27 - x + y + 2z = 2 Lamar University 14

  15. Problem Our next aim is to eliminate x in all equations except the first x + 2y + 3z = 11 …I 3x + 8y + 5z = 27 …II - x + y + 2z = 2 …III x + 2y + 3z = 11 2y – 4z = - 6 ( multiply I with –3 and add to II) - x + y + 2z = 2 x + 2y + 3z = 11 2y – 4z = - 6 3y + 5z = 13 (add I to II) Lamar University 15

  16. Problem The system is reduced to x + 2y + 3z = 11 …I 2y – 4z = - 6 …II 3y + 5z = 13 …III We reduce the II equation x + 2y + 3z = 11 …I y – 2z = - 3 …II (multiplying by (1/2) 3y + 5z = 13 …III We now eliminate y from all the equations except II Lamar University 16

  17. Problem x + 2y + 3z = 11 …I y – 2z = - 3 …II 3y + 5z = 13 …III We get x + + 7z = 17 (sum of I + (-2)times the II) y - 2z = -3 11z = 22 (sum of III + (-3) times II) Multiply III by (1/11) we get x + + 7z = 17 …I y - 2z = -3 …II z = 2 …III Lamar University 17

  18. Problem x + + 7z = 17 …I y - 2z = -3 …II z = 2 …III Eliminating z from all the equations except the III we get x = 3 ( sum of I and (-7) III) y = 1 ( sum of II and (2)III) z = 2 Thus the point (3,1,2) lie in the point of intersection of the three planes. Lamar University 18

  19. Augmented Matrices The system 2x+ 4y + 6z = 22 3x + 8y + 5z = 27 - x + y + 2z = 2 may be represented by the matrix Problem 2 4 6 22 3 8 5 27 -1 1 2 2 Augmented Matrix Coefficient Matrix Constant Matrix Lamar University 19

  20. Augmented Matrices Row- Reduced Form • Each row consisting entirely of zeros lies below any other row having non zero entries. • The first non zero entry in reach row is 1 ( called a leading1) • In any two successive (nonzero) rows, the leading 1 in the lower row lies to the right of the leading 1 in the upper row. • If a column contains a leading 1, then the other entries in that column are zeros. Lamar University 20

  21. Augmented Matrices Row operations • Interchange any two rows • Replace any row by a non zero constant multiple of itself • Replace any row by the sum of that row and a constant multiple of any other row Lamar University 21

  22. Augmented Matrices Notations for Row Operations Operation 1 : Ri Rj to mean: interchange row i with row j Operation 2 : cRi to mean : replace row i with c times row i. Operation 3 : Ri + aRj to mean : Replace row i with the sum of row i and a times row j . Lamar University 22

  23. Problem Using the notation just introduced Can also be solved as 1/3 R1 R2- 2 R1 R2- 2 R1 R1- R2 Lamar University 23

  24. Gauss Jordan Elimination Method • Write the Augmented matrix corresponding to the linear system. • Interchange rows (Operation 1), if necessary, to obtain an augmented matrix in which the first entry in the first row is nonzero. Then pivot the matrix about this entry. • Interchange the second row with any row below it, if necessary, to obtain an augmented matrix in which the second entry in the second row is nonzero. Pivot the matrix about this entry. • Continue till the final matrix is in the row-reduced form Lamar University 24

  25. Problem Solve the system of linear equations given by 3x – 2y + 8 z = 9 - 2x + 2y + z = 3 x + 2y – 3z = 8 Using Gauss-Jordan Method we obtain the following sequence R1+ R2 R2+ 2R1 R3 - R1 Lamar University 25

  26. Problem R2 R3 R3- 2R2 1/2R2 1/31R3 R1- 9R3 R3+6R3 Lamar University 26

  27. Problem The Ace Novelty company wishes to produces three types of souvenirs : types A, B , and C. To manufacture a type A souvenir requires 2 minutes on machine I, 1 minute on machine II, and 2 minutes on machine III. A type-B souvenir requires 1 minute on machine I, 3 minutes on machine II and 1 minute on machine III. A type – C souvenir requires 1 minute on machine I and 2 minutes on machine II and III. There are 3 hours available on machine I, 5 hours available on machine II and 4 hours available on machine III for processing the order. How many souvenirs of each type should Ace Novelty make in order to use all of the available time? Formulate but do not solve the problem. Lamar University 27

  28. Problem We have seen the system is : 2x + y + z = 180 x + 3y + 2z = 300 2x + y + 2z = 240 R1 R2 R2- 2R1 R3- 2R1 Lamar University 28

  29. Problem Thus x = 36, y = 48 and z = 60; that is Ace Novelty should make 36 type A souvenirs, 48 type B souvenirs, and 60 type C souvenirs in order to use all available machine time. -1/5R2 R1- 3R2 R1- 1/5R3 R3- 5R2 R2- 3/5R3 Lamar University 29

  30. Assignment Exercise 2.1 1, 3, 5, 10, 15, 17, 25 Exercise 2.2 1,4, 6, 8, 19,23, 27,30, 37, 44 Excersice 2.3 1,4,9,15,20,30,38 Homework due on 9th September 2002 Lamar University 30

  31. Solutions of Linear Equations(Infinite solutions) Solve the system of linear equations given by x + 2y – 3z = - 2 3x – y – 2z = 1 2x + 3y – 5z = - 3 Using Gauss Jordan elimination R2- 3R1 R3- 2R1 Lamar University 31

  32. Problem -1/7R2 The last augmented matrix is in row-reduced form. Interpreting it as a system of linear equations gives x – z = 0 y – z = - 1 R2- 2R2 R3 + R2 Lamar University 32

  33. Problem A system of two equations in the three variables x,y, and z. Let’s now single out one variable – say z and solve for x and y in terms of it we obtain x = z y = z – 1 If we assign particular value to z –say z = 0 we obtain x = 0 and y = -1 giving (0,- 1, 0) as the solution by setting the value of z = 1 the solution will be (1,0,1) Thus we see that this system has infinite solutions. Lamar University 33

  34. Problem (No Solution) Solve the system of linear equations given by x + y + z = 1 3x – y – z = 4 x + 5y + 5z =1 Solution: We notice that the third row in the last matrix reads 0x + 0y + 0z = - 1 that is 0 = -1 we can safely say that the system is inconsistent and has no solution R2- 3R1 R3+ R1 R3- R1 Lamar University 34

  35. System with more equations than variables The system is x+ 2y = 4 x - 2y = 0 4x + 3y = 12 Solution: We see that the last row of the reduced matrix implies that 0=1, which is not possible, so we conclude that the given system has no solution. R2- R1 -1/4R2 R1- 2R2 R2- 4R1 R3+ 5R2 Lamar University 35

  36. System with more variables than equations The system is x + 2y – 3z + w = - 2 3x – y – 2z – 4w = 1 2x + 3y –5z + w = -3 R2- 3R1 -1/7R2 R3- 2R1 R1- 2R2 R3+ R2 Lamar University 36

  37. Problem The last matrix is in row reduced form. Observe that the given system is equivalent to the following system x – z – w = 0 y – z + w = - 1 of two equations with four variables. We can solve for the two of the variables in terms of the other two. Let z = s and w = t, we find that x = s + t y = s – t – 1 z = s w = t The solutions may b written in the form(s + t, s – t – 1, s, t) where s and t are real numbers Lamar University 37

  38. Application The figure shows the flow of downtown traffic in a certain city during rush hour on a typical weekday. The arrows indicate the direction of traffic flow on each one-way road, and the average number of vehicles entering and leaving each intersection per hour appears beside each road. Fifth and Sixth Avenues can each handle up to 2000 vehicles per hour without causing congestion, whereas maximum capacity of each of the two streets is 1000 vehicles per hour. The flow is controlled by traffic lights installed at each of the four intersections. 5th St. 4th St. 300 500 1200 800 5th Ave. x1 x4 x2 1300 1400 x3 6th Ave. 400 700 Lamar University 38

  39. Application • Write a general expression involving the rates of flow - x1 x2 x3 x4 – and suggest two possible flow patterns that will ensure that there is no traffic congestion. To avoid congestion, all traffic entering an intersection must leave the intersection. Applying this condition we obtain the following equations. 1500 = x1 + x4 1300 = x1 + x2 1800 = x2 + x3 2000 = x3 + x4 or x1 + x4 = 1500 x1 + x2 = 1300 x2 + x3 = 1800 x3 + x4 = = 2000 Lamar University 39

  40. Application Using Gauss - Jordan R2- R1 R4- R3 R3- R2 Lamar University 40

  41. Application The last matrix is in a row reduced form and is equivalent to a system of three linear variables in the four variables. They can be expressed as x4= t the system can be represented as x1 = 1500- t x2 = -200 + t x3 = 2000- t x4 = t For a meaningful solution, 200 ≤ t ≤ 1500 Let t = 300 then x1 = 1200, x2 = 100, x3 = 1700, x4 = 300 Lamar University 41

  42. Matrix Definition: Rectangular array of elements R * C Rows: Horizontal line of elements: a11 a12 a13 Column: Vertical line of elements a11 a12 a13 Element in third row and second column: A = Element designated as aij A= a32 Lamar University 42

  43. Matrix Rows matrix 1 2 5 7 Column matrix 3 4 7 9 Order is related to the number of rows and columns: Row is always listed first: (R * C) = Order A = R * C = (2*5) R * C = (4*2) B= Lamar University 43

  44. Equality of matrices Two matrices are said to be equal if they have the same size and their corresponding entries are equal for example, = Also ≠ Lamar University 44

  45. Addition & Subtraction Take for example Add 2 matrices + = Subtract 2 matrices - = Lamar University 45

  46. Laws for Matrix Addition If A,B and C are matrices of the same size, then • A + B = B + A (Commutative law) • (A + B ) + C = A + (B + C ) ( Associative law) The commutative law for matrix addition states that the order in which matrix addition is performed is immaterial. The associative law states that when adding three matrices together , we may add A and B and then add the result to C or then we can add A to the sum of B and C. A zero matrix is one in which all the entries is zero. Lamar University 46

  47. Transpose of a matrix A matrix got by interchanging the rows and column of a given matrix A is called a Transpose of A and is denoted by AT. For example: A = AT = Lamar University 47

  48. Scalar Multiplication • A matrix A may be may be multiplied by a real number, called a scalar in the context of matrix algebra. The scalar product, denoted by cA, is a matrix obtained by multiplying each entry of A by c. For example, the scalar product of the matrix A = And the scalar 3 is 3A = 3 = Lamar University 48

  49. Problem • Given A = B = Find the matrix X that will satisfy the matrix equation 2X + B = 3A Soln: We can say that 2X = 3A – B = 3 - = - = Lamar University 49

  50. Problem We have 2X = Or X = ½ = Lamar University 50

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