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1. 資訊科學數學13 :Solutions of Linear Systems 陳光琦助理教授 (Kuang-Chi Chen) chichen6@mail.tcu.edu.tw

2. Solutions of Linear Systems of Equations 1.6 Solutions of Linear Systems of Equations

3. Row Echelon Form • Definition – Row echelon form (r.e.f.) An mn matrix A is said to be in row echelon form if (a) All zero rows, if there are any, appear at the bottom of the matrix (b) The first nonzero entry from the left of a nonzero row is a 1; a leading one of the row (c) For each nonzero row, the leading one appears to the right and below any leading one’s in preceding rows

4. Reduced Row Echelon Form • Definition – Reduced row echelon form An mn matrix A is said to be in reduced row echelon form if (a) A is in row echelon form (b) If a column contains a leading one, then all other entries in that column are zero (列梯形式; 簡化之列梯形式)

5. Example 1 - in row echelon form • E.g. 1

6. E.g. 2 – not reduced row echelon form • E.g. 2 , , , Nonzero element above leading 1 in row 2

7. Three Basic Types of Elementary Row Operations • Type 1 – Interchange row i and row j are interchanged • Type 2 – Multiply row i = row i times c • Type 3 – Add Add d times row r of A to row s of A row s = row s + d row r

8. Example 3 E1 ⇒ • E.g. 3  E2 E3

9. Row Equivalence • Definition – Row Equivalence An mn matrix A is said to be row equivalence to an mn matrix B if B can be obtained by applying a finite sequence of elementary row operations to the matrix A .

10. Example 4 E3 ⇒ • E.g. 4 E1 E2 ⇒

11. Theorem 1.5 • Theorem 1.5 Every mn matrix is row equivalent to a matrix in row echelon form .

12. Pivot column Pivot E.g. 5 - Procedure of Row Echelon Form • E.g. 5 Step 1 – Find the pivotal column Step 2 – Identify the pivot in the pivotal column

13. (cont’d) • E.g. 5 Step 3 – Interchange if necessary so that the pivot is in the 1st row Step 4 – Multiply so that the pivot equals to 1 pivot

14. (cont’d) • E.g. 5 Step 5 – Make all entries in the pivot column, except the entry where the pivot was located, equal to zero

15. (cont’d) • E.g. 5 Step 6 – Ignore the first row and repeat ⇒ ⇒ ……… ⇒

16. Example 6 • Example 6

17. Remark • Remark - There may be more than one matrix in row echelon form that is row equivalent to a given matrix. - A matrix in row echelon form (r.e.f.) that is row equivalent to A is called “a row echelon form of A”.

18. Theorem 1.6 • Theorem 1.6 - Every mn matrix is row equivalent to a unique matrix in reduced row echelon form.

19. Theorem 1.7 • Theorem 1.7 Let Ax = b and Cx = d be two linear systems each of m equations in n unknowns. If the augmented matrices [A|b] and [C|d] of these systems are row equivalent, then both linear systems have the same solutions.

20. Corollary 1.1 • Corollary 1.1 If A and C are row equivalent mn matrices, then the linear system Ax = 0 and Cx = 0 have exactly the same solutions.

21. Gauss-Jordan Reduction Procedure • The Gauss-Jordan reduction procedure Step 1. Form the augmented matrix [A|b] Step 2. Obtain the reduced row echelon form [C|d] of the augmented matrix [A|b] by using elementary row operations Step 3. For each nonzero row of [C|d], solve the corresponding equation. (augmented matrix 擴增矩陣)

22. Gauss Elimination Procedure • The Gauss elimination procedure Step 1. Form the augmented matrix [A|b] Step 2. Obtain a row echelon form [C|d] of the augmented matrix [A|b] by using elementary row operations Step 3. Solving the linear system corresponding to [C|d], by back substitution (後代入法).

23. Example 8 • E.g. 8 Solve the linear system by Gauss-Jordan reduction - Step 1

24. (cont’d) • E.g. 8 - Solve the linear system by Gauss-Jordan reduction - Step 2

25. (cont’d) • E.g. 8 - Solve the linear system by Gauss-Jordan reduction - Step 3 x = 2 y = -1 z = 3

26. Example 9 • Example 9 - Solve the linear system by Gauss-Jordan reduction x + y + 2z – 5w = 3 2x + 5y – z – 9w = -3 2x + y – z + 3w = -11 x – 3y + 2z + 7w = -5

27. (cont’d) • Example 9 - Step 1 - Step 2

28. (cont’d) • E.g. 9 - Step 3 leading variables a free variable

29. Example 10 • Example 10 - Solve the linear system by Gauss-Jordan reduction x1 + 2x2 – 3x4 + x5 = 2 x1 + 2x2 + x3 – 3x4 + x5 + 2x6 = 3 x1 + 2x2 – 3x4 + 2x5 + x6 = 4 3x1 + 6x2 + x3 – 9x4 + 4x5 + 3x6 = 9

30. (cont’d) • Example 10 - Step 1 - Step 2

31. (cont’d) • Example 10 - Step 3 leading variables free variables

32. Example 11 • Example 11 - Solve the linear system by Gauss elimination x + 2y + 3z = 9 2x – y + z = 8 3x – z = 3

33. (cont’d) • Example 11 - Step 1 - Step 2

34. (cont’d) • Example 11 - Step 3 - By back substitution

35. Example 12 • Example 12 - Solve the linear system by Gauss elimination x + 2y + 3z + 4w = 5 x + 3y + 5z + 7w = 11 x – z – w = -6

36. (cont’d) • Example 12 - Step 1 - Step 2 - Step 3 ⇒ 0x + 0y + 0z + 0w = 1 ⇒ No solutions !!

37. Consistent and Inconsistent • Consistent and inconsistent - Consistent: Linear systems with at least one solution - Inconsistent: Linear systems with no solutions

38. Homogeneous Systems • A system of linear equations is said to be homogeneous if all the constant terms are zeros. a11x1 + a12x2 + … + a1nxn = 0 a21x1 + a22x2 + … + a2nxn = 0 … am1x1 + am2x2 + … + amnxn = 0 ⇒ Ax = 0 Thus, a homogeneous system always has the solution x1 = x2 = … = xn = 0 → the trivial solution

39. Example 13 • Example 13

40. Example 14 • Example 14

41. Theorem 1.8 • Theorem 1.8 A homogeneous system of m equations in n unknowns has a non-trivial solution if m < n, that is, if the number of unknowns exceeds the number of equations. namely, a homogeneous system has more variables than equations has many solutions. (a homogeneous system齊次系統; non-trivial solution 非零解)

42. Example 15 - A Homogeneous System • E.g. 15

43. (cont’d) • If let

44. A Homogeneous System Example x = xp + xh xp is a particular solution to the given system Axp = b , where b = [3 -3 -11 -5]T xh is a solution to the associated homogeneous system Axh = 0 .

45. Polynomial Interpolation • Polynomial Interpolation - The general form y = an – 1xn – 1 + an – 2xn – 2 + … + a1x + a0 E.g. n = 3, y = a2x2 + a1x + a0 Given three distinct points (x1 , y1), (x2 , y2), (x3 , y3), we have y1 = a2x12 + a1x1 + a0 y2 = a2x22 + a1x2 + a0 y3 = a2x32 + a1x3 + a0

46. Example 16 • Example 16 - Find the quadratic polynomial that interpolates the points (1, 3), (2, 4), (3, 7)

47. Example 17 – Temperature Distribution T1 = (260 – 100 + T2 + T3 )/4 or 4T1 – T2 – T3 = 160 T2 = (T1 + 100 + 40 + T4 )/4 or -T1 + 4T2 – T4 = 140 T3 = (60 + T1 + T4 + 0)/4 or -T1 + 4T3 – T4 = 60 T4 = (T2 + T3 + 40 + 0)/4 or -T2 – T3 + 4T4 = 40 ⇒ ⇒ T1 = 65, T2 = 60, T3 = 40, T4 = 35 .