Solving System Of Linear Equations 线性方程组 4x 5y 7 2x 4y 5 二元一次方程组 a11x1 a12 x 2 ...
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Solving System Of Linear Equations
线性方程组
4 2 x x 5 4 y y 二元一次方程组 7 5 a a 11 x 1 21 x 1 a a 12 x 22 x 2 2 ...
a 1 n x n ...
a 2 n x n b 1 b 2 a n 1 x 1 a n 2 x 2 ...
a nn x n b n
Solve the following system of linear equations:
1.
4 2 x x 5 y 4 y 7 5
2(2)
4 4 x x 8 5 y y 7 10
(2’) -(1)
4 x 5 3 y y 7 3 4 x 5 y y 1 7
Elimination
加减消元法 x 1 / 2 y 1 Substitution 代入消元法
2.
2 3 x x x 7 2 5 y y y 3 8 z z 13 z 9 21 32
Gaussian Elimination
消元法 2 3 x x x 7 2 5 y y y 3 8 z z 13 z 9 21 32 x 2 y 3 z y y 2 4 z z 3 5 9 x 2 y 3 z y 2 2 z z 2 3 9 x 2 y y 2 z 3 z z 1 3 9 z = 1, y = 1, x = 4
Alternative –
2 3 x x x 7 2 y y 5 y 3 8 z z 13 z 9 21 32 Elementary row transformations •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 1 2 3 2 5 7 3 8 13 R 2 R 3 – 2R 1 + R 2 – 3R 1 + R 2 1 0 0 9 21 32 2 1 1 3 2 4
Coefficient Matrix Augmented Matrix 9 3 5 R 3 –R 2 +R 3 1 0 0
系数矩阵 增广矩阵 2 3 9 R 3 0.5R
3 1 2 3 9 1 0 2 2 x 2 y y 3 2 2 2 z z 3 z 2 3 9 0 0 1 0 2 1 3 1 z = 1, y = 1, x = 4
ALGORITHM
1 2 3 2 5 7 3 8 13 9 21 32 1 0 0 2 1 1 3 2 4 9 3 5 1 0 0 2 1 0 3 2 2 9 3 2 1 0 0 2 1 0 3 2 1 9 3 1 R 1 R 2 – 3R 3 + R 1 – 2R 3 +R 2 1 0 0 2 1 0 0 0 1 6 1 1 R 1 – 2R 2 + R 1 1 0 0 0 1 0 0 0 1 4 1 1 x = 4, y = 1, z = 1 The matrix can be simplified further
1.
PRACTICE
7 5 x x x 3 6 y y y 3 z 17 22 z z 1 1 14 2.
3.
7 4 x 11 x x 2 3 y y y 5 3 z 8 z z 3 6 11 6 8 4 x x x 5 3 y y y z 3 z 2 z 10 0 13
In general, how many solution(s) can a system of linear equations have?
1.
SOLUTION
1 5 7 1 6 3 3 17 22 1 1 1 14 0 0 1 1 4 3 2 1 1 1 7 4 0 0 1 1 0 3 2 9 1 4 9 x y 3 z 1 y 2 z 4 z 1 z 1 , y 2 , x 6 2.
7 4 11 2 1 3 5 3 8 6 1 3 11 4 1 0 1 0 1 3 1 0 3 2 1 4 0 0 1 0 1 3 0 0 3 2
0z=2 no solution 3.
8 3 1 10 20 0 4 6 4 x 1 5 2 x y 3 2 z 3 0 13 1 4 26 z 0 1 0 z= , 10 3 13 10 2 0 13 4 2 0 1 0 1 3 1 1 2 0 1 4 0 0 1 0 1 3 0 x = (1+ )/2, y = 4x – 3 z = 2 – where IR 1 0 0
Computer program
1. Solving linear equations in 3 variables http://www.mkaz.com/math/js_lalg3.html
2. Step by Step Illustration All steps shown http://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
Further points of discussion
1. How can the
ALGORITHM
described above be applied to 4 linear equations in 4 variables? n linear equations in n variables?
2. What are the limitations of the above computer programs?
Websites
http://cos.cumt.edu.cn/math/shuxuekejian/xianxingdaishu/charp1/124.htm
http://ws1.hkcampus.net/~ws1-kcy/al_pmath.html
A Mathematical Problem
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?
Approach: (1) top down strategy (2) bottom up strategy
Method 1 – Top down strategy
Let Amy, Ben and Calvin each had $x, $y and $z initially.
Amy Ben Calvin At start After round 1 After round 2 After round 3 $x x – y – z 2(x – y – z) 4(x – y – z) $y 2y 3y – x – z 2(3y – x – z) $z 2z 4z 7z – x – y 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 2 – Bottom up strategy
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. Amy $x Ben $y Calvin $z At start After round 1 After round 2 After round 3