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1、管理數學 Chapter 2: System of Linear Equations,XX. XX, XXX by XXXX,Agenda,Linear Systems as Mathematical Models Linear Systems Having One or No Solutions Linear Systems Having Many Solutions,Linear - Line,a1 x + a2 y = b or a1 x1 + a2 x2 + an xn = b,2x + 3y = 4 x2 + y2 = 1 x1 - x2 + x3 - x4 = 6 z = 5 -
2、3x + y/2 sin x + ey = 1 xy = 2 7x1 + 3x2 + 9/x3 + 2x4 = 1 x1 + 2x2 + 3x3 + nxn = 1,Which one is linear?,Example 1,A firm produces bargain and deluxe TV sets by buying the components, assembling them, and testing the sets before shipping.,Resources,The bargain set requires 3 hours to assemble and 1 h
3、our to test. The deluxe set requires 4 hours to assemble and 2 hours to test. The firm has 390 hours for assembly and 170 hours for testing each week.,Question,Use a system of linear equations to model the number of each type of TV set that the company can produce each week while using all of its av
4、ailable labor.,Problem Formulation,Define decision variables (unit of scale) Define the linear relation between variables (write the linear equations),Example 2,A dietitian is to combine a total of 5 servings of cream of mushroom soup, tuna, and green beans, among other ingredients, in making a cass
5、erole.,Ingredient Nutritions,Each serving of soup has 15 calories and 1 gram of protein, each serving of tuna has 160 calories and 12 gram of protein, and each serving of green beans has 20 calories and 1 gram of protein.,Question,If these three foods are to furnish 380 calories and 27 grams of prot
6、ein in casserole, how many servings of each should be used?,Example 3,A retailer has warehouses in Lima and Canton, from which two storesone in Tiffin and one in Danvilleplace orders for bicycles. Tiffin orders 38 and Danville orders 46.,Limitations and Question,Each warehouse has enough to supply a
7、ll orders but twice as many are to be shipped from Lima to Danville as from Canton to Tiffin. Write the linear equations.,Agenda,Linear Systems as Mathematical Models Linear Systems Having One or No Solutions Linear Systems Having Many Solutions,Solving a System of Linear Equations,Problem formulati
8、on: variable definition and equations Algorithms or formula Interpretation of solutions,System of 2 Linear Equations,x1 + x2 + x3 = 2 -(1) 2x1 + 3x2 + x3 = 3 -(2) x1 - x2 - 2x3 = -6 -(3),(1),(2),(3),System of 3 Linear Equations,A System of Linear Equations ( A Linear System ),A finite collection of
9、linear equations a11 x1 + a12 x2 + a1n xn = b1 a21 x1 + a22 x2 + a2n xn = b2 am1 x1 + am2 x2 + amn xn = bm,A Solution,To a equation: a1x1 + a2x2 + + anxn = b ( t1, t2, tn ) To a linear system : A solution to each of linear equation simultaneously ps. Solution Set,Elementary Transformations,1. Interc
10、hange the position of two equations. 2. Multiply both sides of an equation by a nonzero constant. 3. Add a multiple of one equation to another equation.,x1 + x2 + x3 = 2 -(1) 2x1 + 3x2 + x3 = 3 -(2) x1 - x2 - 2x3 = -6 -(3),Interchange (1) and (3). Multiply (2) by 1/2. Add a -1 multiple (2) to (1).,C
11、ontinuous Operations,x1 + x2 + x3 = 2 -(1) x1 + 1.5x2 + 0.5x3 = 3 -(2) x1 - x2 - 2x3 = -6 -(3),A,Result of operations,x1 + x2 + x3 = 2 -(1) 2x1 + 3x2 + x3 = 3 -(2) x1 - x2 - 2x3 = -6 -(3),A,The Objective,Solve the problem by elementary transformation,x1 + x2 + x3 = 2 -(1) 2x1 + 3x2+ x3 = 3 -(2) x1 -
12、 x2 - 2x3= -6 -(3),Add a -2 multiple (1) to (2). Add a -1 multiple (1) to (3).,Iteration 1,x1 + x2 + x3 = 2 -(1) x2 - x3 = -1 -(2) - 2x2 -3x3 = -8 -(3),Add a -1 multiple (2) to (1). Add a 2 multiple (2) to (3).,Iteration 2,x1 + 2x3= 3 -(1) x2 - x3 = -1 -(2) -5x3 = -10 -(3),Multiply (2) by -1/5. Add
13、a -2 multiple (3) to (1). Add a 1 multiple (3) to (2).,Final Iteration,x1 = -1 -(1) x2 = 1 -(2) x3 = 2 -(3),Final answer: (x1, x2, x3) = (-1, 1, 2),Difficulty,It is very hard to carry variables, xis, through the calculation process when applying elementary transformation,2 3 -4 7 5 -1,7 1 0 5 -8 3,3
14、 5 6 0 -2 5 8 9 12,8 9 12,3 5 0 -2 8 9,5 -2 9,3 5 0 -2,Matrix,Transfer to be : AX = B,Matrix Notation,a11x1 + a12x2 + a1nxn = b1 a21x1 + a22x2 + a2nxn = b2 am1x1 + am2x2 + amnxn = bm,x1 + x2 + x3 = 2 -(1) 2x1 + 3x2 + x3 = 3 -(2) x1 - x2 - 2x3 = -6 -(3),Example 1,Transfer (I) to matrix format.,(I),x1
15、 x2 x3,X =,1 3 -2,B =,1 -1 -2 2 -3 -5 -1 3 5,A =,Matrix Representation,AX = B,3x1 + 2x2 - 5x3 = 7 x1 - 8x2 + 4x3 = 9 2x1 + 6x2 - 7x3 = -2,Transfer (I) to matrix format.,(I),Example 2,Matrix of Coefficients,Coefficients of the system or Matrix A,Augmented Matrix,Coefficients and RHS, or A | B .,Reduc
16、ed Echelon Form,1. Any rows with all zeros are at the bottom. 2. Leading 1. 3. Leading 1 to the right. 4. All other elements in a leading 1 column are zeros.,Examples I,Examples II,Elementary Row Operations,1. Interchange two rows 2. Multiply the elements of a row by a nonzero constant 3. Add a mult
17、iple of the elements of one row to the corresponding elements of another row.,x1 + x2 + x3 = 2 -(1) 2x1 + 3x2 + x3 = 3 -(2) x1 - x2 - 2x3 = -6 -(3),Continuous Operations,1 1 1 2 2 3 1 3 1 -1 -2 -6,Interchange r.1 and r.3. Multiply r.2 by 1/2. Add a -1 multiple r.2 to r.3.,Result,1 1 1 2 2 3 1 3 1 -1
18、 -2 -6,Through Elementary Row operations,The Objective,Equivalent Systems,Suppose that A and B are both systems of linear equations. A and B are equivalent if they are related through elementary transformations. A and B has the same solution if they are equivalent.,Solving a system of linear equatio
19、ns,Gauss-Jordan Elimination Gauss Elimination,Gauss-Jordan Elimination,1. Write the augmented matrix. 2. Derive the reduced echelon form of the augmented matrix 3. Write the system of equations corresponding to the reduced echelon form.,x1 + x2 + x3 = 2 -(1) 2x1 + 3x2 + x3 = 3 -(2) x1 - x2 - 2x3 = -
20、6 -(3),1 1 1 2 2 3 1 3 1 -1 -2 -6,Perform J-G Elimination,1 1 1 2 2 3 1 3 1 -1 -2 -6,Pivoting Step 1,Pivoting Step 2,Pivoting Step 4,1 0 0 -1 0 1 0 1 0 0 1 2,Stop?,Reduced Echelon Form?,Write the linear equations: x1 = -1, x2 = 1, x3 = 2,Agenda,Linear Systems as Mathematical Models Linear Systems Ha
21、ving One or No Solutions Linear Systems Having Many Solutions,Conditions of Solution Sets,Consistent : A linear system with at least one solution Inconsistent : A linear System with no solution,Conditions of Solution Set,Empty-infeasible One-feasible with unique solutions Many-feasible with infinite many solutions,