1. Solving System Of Linear Equations 线性方程组 二元一次方程组

2. Solve the following system of linear equations: 1. 2(2) (2’) -(1) Elimination 加减消元法 Substitution 代入消元法

3. 2. Gaussian Elimination 消元法 Alternative z = 1, y = 1, x = 4

4. Coefficient Matrix 系数矩阵 Augmented Matrix 增广矩阵 z = 1, y = 1, x = 4 R 2  – 2R 1 + R 2 R 3  – 3R 1 + R 2 R 3  –R 2 +R 3 R 3  0.5R 3 Alternative – PracticeAlgorithm Interchanging 2 rows: R i  R j Multiply a row by a constant: R i  kR i Add to a row a multiple of another row :R i  kR j +R i Elementary row transformations

5. R 1  – 2R 2 + R 1 R 1  – 3R 3 + R 1 R 2  – 2R 3 +R 2 ALGORITHM The matrix can be simplified further Computer Simulation x = 4, y = 1, z = 1

6. In general, how many solution(s) can a system of linear equations have? PRACTICE 1. 3. 2. SolutionBACK

7. SOLUTION 1. 3. z=, x = (1+ )/2, y = 4x – 3 z = 2 – where  IR BACK 0z=2 no solution 2.

8. Further points of discussion 1. Solving linear equations in 3 variables Computer program 2. Step by Step Illustration All steps shown http://www1.minn.net/~dchristo/GElim/GElim.htmlhttp://www1.minn.net/~dchristo/GElim/GElim.html With final score http://wims.unice.fr/wims/wims.cgi?lang=cn&cmd=new&module=U1%2Falgebra%2Fvisgauss.cn&type=system&size=3&field=Q http://wims.unice.fr/wims/wims.cgi?lang=cn&cmd=new&module=U1%2Falgebra%2Fvisgauss.cn&type=system&size=3&field=Q 2. What are the limitations of the above computer programs? http://www.mkaz.com/math/js_lalg3.html 1.How can the ALGORITHM described above be applied to 4 linear equations in 4 variables? n linear equations in n variables? Websites http://cos.cumt.edu.cn/math/shuxuekejian/xianxingdaishu/charp1/124.htm http://cos.cumt.edu.cn/math/shuxuekejian/xianxingdaishu/charp1/124.htm http://ws1.hkcampus.net/~ws1-kcy/al_pmath.html

9. Approach: (1) top down strategy (2) bottom up strategy Amy, Ben and Calvin play a game as follows. The player who loses each round must give each of the other players as much money as the player has at that time. In round 1, Amy loses and gives Ben and Calvin as much money as they each have. In round 2, Ben loses, and gives Amy and Calvin as much money as they each then have. Calvin loses in round 3 and gives Amy and Ben as much money as they each have. They decide to quit at this point and discover that they each have $24. How much money did they each start with? A Mathematical Problem

10. x – y – z = 6, 3y – x – z = 12 and 7z – x – y = 24; from which x, y and z can be solved. x=39, y=21, z=12 Method 1 – Top down strategy Let Amy, Ben and Calvin each had $x, $y and $z initially. AmyBenCalvin At start$x$y$z After round 1 After round 2 After round 3 x – y – z2y2z 3y – x – z 2(x – y – z) 4z 4(x – y – z) 2(3y – x – z) 7z – x – y NEXT

11. Instead of considering how much money Amy, Ben and Calvin had originally, work backwards from the moment when each of them has $24 each. Complete the following table and see how easily you can reach exactly the same conclusion as in method 1. Method 2 – Bottom up strategy AmyBenCalvin At start$x$y$z After round 1 After round 2 After round 3 BACK