Chapter 1

1. Chapter 1 Systems of Linear Equations Gareth Williams J & B, 滄海書局

2. 1.1 Matrices and Systems of Linear Equations (雞兔同籠) • 1.2 Gauss-Jordan Elimination • 1.3 Curve Fitting, Electrical Networks, and Traffic Flow

3. Figure 1.3 Many solution 4x – 2y = 6 6x – 3y = 9 Both equation have the same graph. Any point on the graph is a solution. Many solutions. Figure 1.2 No solution –2x + 3y = 3 –4x + 2y = 2 Lines are parallel.No point of intersection. No solutions. 1.1 Matrices and Systems of Linear Equations Figure 1.1 Unique solution x + 3y = 9 –6x + y = –4 Lines intersect at (3, 2) Unique solution, x = 3, y = 2.

4. Definition A matrix is a rectangular array of numbers. The numbers in the array are called the elements of the matrix. • Unique solution

5. No solutions • Many solutions

6. Submatrix • Rows and Columns

7. Identity Matrices • Size and Type • Location

8. matrix of coefficient and augmented matrix

9. Elementary Transformation Interchange two equations. Multiply both sides of an equation by a nonzero constant. Add a multiple of one equation to another equation. Elementary Row Operation Interchange two rows of a matrix. Multiply the elements of a row by a nonzero constant. Add a multiple of the elements of one row to the corresponding elements of another row.

10. Solution Analogous Matrix Method Augmented matrix: Equation Method Initial system: Eq2+(–2)Eq1 R2+(–2)R1 Eq3+(–1)Eq1 R3+(–1)R1 Example 1 Solving the following system of linear equation.

11. Eq1+(–1)Eq2 Eq1+(–1)Eq2 Eq3+(2)Eq2 Eq3+(2)Eq2 (–15)Eq3 (–15)R3 Eq1+(–2)Eq3 Eq2+Eq3 The solution is The solution is R1+(–2)R3 R2+R3

12. Example 2 Solving the following system of linear equation. Solution

13. Example 2

14. Example 3 Solving the system Solution

15. Example 3

16. Example 4 Solving the following three systems of linear equation, all of which have the same matrix of coefficients. Solution

17. Example 4

18. 1-2 Gauss-Jordan Elimination • Definition • A matrix is in reduced echelon form if • Any rows consisting entirely of zeros are grouped at the bottom of the matrix. • The first nonzero element of each other row is 1. This element is called a leading 1. • The leading 1 of each after the first is positioned to the right of the leading 1 of the previous row. • All other elements in a column that contains a leading 1 are zero.

19. Example In Reduced Echelon Form Not in Reduced Echelon Form

20. Example 1 Use the method of Gauss-Jordan elimination to find reduced echelon form of the following matrix. Solution

21. Example 1

22. Example 2 Solve, if possible, the system of equations Solution

23. The general solution to the system is Example 2

24. Example 3 Solve the system of equations Solution

25. Example 3

26. Example 4 Solve the system of equations Solution

27. Example 4

28. Example 5 This example illustrates a system that has no solution. Let us try to solve the system Solution

29. Example 5

30. Theorem 1.2 A system of homogeneous linear equations that has more variables than equations has many solutions. Homogeneous System of linear Equations Theorem 1.1 A system of homogeneous linear equations in n variables always has the solution x1 = 0, x2 = 0. …, xn = 0. This solution is called the trivial solution.

31. Figure 1.5 1-3 Curve Fitting, Electrical Networks, and Traffic Flow Curve Fitting The following problem occurs in many different branches of science. A set of data points is given and it is necessary to find a polynomial whose graph passes through the points. The points are often measurements in an experiment. The x-coordinates are called base points. It can be shown that if the base points are all distinct, then a unique polynomial of degree n – 1 (or less) Can be fitted to the points

32. Let the polynomial be , and substituting x = 1, y = 6; x = 2, y = 3; x = 3, y = 2 , we have Example 1 Determine the equation of the polynomial of degree 2 whose graph passes through the points (1, 6), (2, 3), (3, 2). Solution

33. Example 1

34. Figure 1.6 Example 1 We get a0 = , a1 = , a2 = . The parabola that passes through these points is y = – x + x2. Such as Figure, 1.6.

35. Electrical Network Analysis • Kirchhoff’s Laws • Junctions: All the current flowing into a junction must flow out of it. • Paths: The sum of the IR terms (I denotes current, R resistance) in any direction around a closed path is equal to the voltage in the path in that direction.

36. Figure 1.7 Solution Junction Junction B: I1 + I2 = I3 Junction C: I3 = I1 + I2 This two equations result in a single linear equation Example 2 Determine the currents through each branch of this network.

37. Path The problem thus reduces to solving the following system of three linear equations in three variables. Example 2 Using the method of Gauss-Jordan elimination, we get

38. Example 2 The currents are I1 = , I2 = , I3 = . The solution is unique.

39. Figure 1.8 Solution Junction Junction B: I1 + I2 = I3 Junction C: I3 = I1 + I2 giving I1 + I2 – I3 = 0. Example 3 Determinate the currents through the various branches of the electrical network in Figure 1.8.

40. Example 3 Path We have three equations in the three variables, I1, I2, and I3. Solving these equations, we get I1 = , I2 = , I3 = (amps).

41. Figure 1.9 Traffic Flow

42. Junction A Junction B Junction C Junction D Junction E Junction F

43. We get, Junction A Junction B Junction C Junction D Junction E Junction F Perform road work on x7. Because x3 0, the minimum flow 200 in the branch x7 can be obtained by making x3= 0 (closing DE to traffic)