2. 1 Consider a system of linear equations.  The variables, or unknowns, are referred to as x 1, x 2, …, x n while the a ij ’s and b j ’s are constants.  A set of such equations is called a linear system of m equations in n variables. A solution to a linear set of m equations in n unknowns is a set of values for the unknowns that satisfies each of the system’s m equations. A solution to a linear set of m equations in n unknowns is a set of values for the unknowns that satisfies each of the system’s m equations. 2.2 Matrices and Systems of Linear Equations (p.20)

3. 2 Example 5: Solution to Linear System (p.21) Show that a solution to the linear system and that is not a solution to the linear system. Show that a solution to the linear system and that is not a solution to the linear system.

4. 3 Example 5 Solution To show that is a solution, x 1 =1 and x 2 =2 must be substituted in both equations. The equations must be satisfied. The vector is not a solution, because x 1 =3 and x 2 =1 fail to satisfy 2x 1 -x 2 =0

5. 4 Matrices can simplify and compactly represent a system of linear equations. (p.21) This system of linear equations may be written as Ax=b and is called it’s matrix representation. This system of linear equations may be written as Ax=b and is called it’s matrix representation. A : ( 係數矩陣 ) A : coefficent matrix ( 係數矩陣 ) X : ( 變數矩陣 ) X : variable matrix ( 變數矩陣 ) b : ( 常數矩陣 ) b : constant matrix ( 常數矩陣 ) : ( 擴增矩陣 ) A|b : augmented matrix ( 擴增矩陣 )

6. 5 Row equivalent ( 列同義 ) & Elementary matrix ( 基本矩陣 )

7. 6 交 換 乘常數 相 加

8. 7 Elementary row operations( 基本列運算 ) Elementary row operations (ERO) transforms a given matrix A into a new matrix A’ via one of the following operations:  Type 1 ERO –A’ is obtained by multiplying any row of A by a nonzero scalar. ( 乘上一非零定數 )  Type 2 ERO – Multiply any row of A (say, row i) by a nonzero scalar c. For some j ≠ i, let row j of A’ = c*(row i of A) + row j of A and the other rows of A’ be the same as the rows of A. ( 乘上一非零定數 + 另一列 )  Type 3 ERO – Interchange any two rows of A. ( 任二列交換 ) (p.22)

9. 8 Reduced Row Echelon Form( 簡約列梯陣 )

10. 9 。 矩陣之基本列運算 (EROs) 可求得一方陣的逆方 陣 ( ) 。 。 簡約列梯陣可解線性方程組。 ，使所得矩 陣適合某一特殊形式 ( 同義 ) 。 基本列運算將矩陣變形成另一矩陣，使所得矩 陣適合某一特殊形式 ( 同義 ) 。 原始矩陣與所得矩陣並無相等關係 原始矩陣與所得矩陣並無相等關係 基本列運算 (EROs) 之功能

11. 10 Example :

12. 11

13. 12 2.3 2.3 The Gauss-Jordan Method (p.22) Also Gauss-Jordan Elimination( 高斯 - 約旦消去法 ). Using the Gauss-Jordan method, it can be shown that any system of linear equations must satisfy one of the following three cases:  Case 1. The system has no solution.  Case 2. The system has a unique solution.  Case 3. The system has an infinite number of solutions. The Gauss-Jordan method is important because many of the manipulations used in this method are used when solving linear programming problems by the simplex algorithm.

14. 13 The Gauss-Jordan Method(continued) The Gauss-Jordan method solves a linear equation system by utilizing EROs in a systematic fashion.

15. 14 Case1 ： a unique solution P.24 The steps to using the Gauss-Jordan method The augmented matrix representation is A|b =

16. 15 Step 1 Multiply row 1 by ½. This Type 1 ERO yields Step 2 Replace row 2 of A1|b1 by -2(row 1 A1|b1) + row 2 of A1|b1. The result of this Type 2 ERO is A 1 |b 1 = A 2 |b 2 =

17. 16 Step 3 Replace row 3 of A 2 |b 2 by -1(row 1 of A 2 |b 2 ) + row 3 of A 2 |b 2 The result of this Type 2 ERO is The first column has now been transformed into A 3 |b 3 =

18. 17 Step 4 Multiply row 2 of A 3 |b 3 by -1/3. The result of this Type 1 ERO is Step 5 Replace row 1 of A 4 |b 4 by -1(row 2 of A 4 |b 4 ) + row 1 of A 4 |b 4. The result of this Type 2 ERO is A 4 |b 4 = A 5 |b 5 =

19. 18 Step 6 Place row 3 of A5|b5 by 2(row 2 of A5|b5) + row 3 of A5|b5. The result of this Type 2 ERO is Column 2 has now been transformed into A 6 |b 6 =

20. 19 Step 7 Multiply row 3 of A 6 |b 6 by 6/5. The result of this Type 1 ERO is Step 8 Replace row 1 of A 7 |b 7 by -5/6(row 3 of A 7 |b 7 )+A 7 |b 7. The result of this Type 2 ERO is A 7 |b 7 = A 8 |b 8 =

21. 20 Step 9 Replace row 2 of A 8 |b 8 by 1/3(row 3 of A 8 |b 8 )+ row 2 of A 8 |b 8. The result of this Type 2 ERO is A 9 |b 9 represents the system of equations and thus the unique solution A 9 |b 9 =

22. 21 Case2 ： no solution Example 6 (p.28)

23. 22 Case3 ： an infinite number of solutions Example 7 (p.28)

24. 23 Exercise : Solving linear system by Gauss-Jordan Elimination

25. 24

26. 25 原擴增矩陣經有限次之基本列運算後，同義於一上三角矩陣， 故由後代法 (backward-substitution) 解得 x 1,x 2,x 3 。

27. 26 若繼續矩陣之基本列運算，使其係數矩陣同義於一單位矩陣， 則可直接求得 x 1,x 2,x 3 ，而不必使用後代法的運算步驟，繼續 矩陣之基本列運算。

28. 27

29. 28 After the Gauss Jordan method has been applied to any linear system, a variable that appears with a coefficient of 1 in a single equation and a coefficient of 0 in all other equations is called a basic variable ( 基本變數 BV). Any variable that is not a basic variable is called a nonbasic variable ( 非基本變數 NBV). Basic Variables (p.30)

30. 29 p.30 No solution Unique solution BV={x1,x2,x3}, NBV is empty. Infinite number of solutions BV={x1,x2,x3} NBV={x4,x5}

31. 30 Summary of Gauss-Jordan Method Does A' | b ' have a row [ 0, 0,..., 0 | c ] (c 0 ) ? A x = b has no solution. Find BV and NBV. Is NBV empty? A x = b has a unique solution. A x = b has an infinite number of solutions. YesNo Yes No

32. 31 Exercise ： 方程組 ，求所有 a 之值。 (1)no solution (2)a unique solution (3) infinite number of solutions (3)a=3 (2) 除了 a=3,-3 以外 (1) a=-3