Linear equations

Linear eq

1. ELECTRICAL-1 MATHS GROUP ID NUMBER :- 31,32,33,34,35,36

2. Systems of Linear Equations How to: solve by graphing, substitution, linear combinations, and special types of linear systems

3. What is a Linear System, Anyways? A linear system includes two, or more, equations, and each includes two or more variables. When two equations are used to model a problem, it is called a linear system .

4. Introduction to Systems of Linear Equations a 1 , a 2 , a 3 , … , a n , b : real number a 1 : leading coefficient x 1 : leading variable a linear equation in n variables:

5. a system of m linear equations in n variables: 5 Consistent: A system of linear equations has at least one solution . Inconsistent: A system of linear equations has no solution . Elementary Linear Algebra: Section 1.1, p.4

6. Finding a Solution by Graphing Since our chances of guessing the right coordinates to try for a solution are not that high, we’ll be more successful if we try a different technique. Since a solution of a system of equations is a solution common to both equations, it would also be a point common to the graphs of both equations. So to find the solution of a system of 2 linear equations, graph the equations and see where the lines intersect.

7. Finding a Solution by Graphing Solve the following system of equations by graphing. 2 x – y = 6 and x + 3 y = 10 x y First, graph 2 x – y = 6. (0, -6) (3, 0) (6, 6) Second, graph x + 3 y = 10 (1, 3) (-2, 4) (-5, 5) The lines APPEAR to intersect at (4, 2). (4, 2) Example Continued.

8. Finding a Solution by Graphing Although the solution to the system of equations appears to be (4, 2), you still need to check the answer by substituting x = 4 and y = 2 into the two equations. First equation, 2( 4 ) – 2 = 8 – 2 = 6 true Second equation, 4 + 3( 2 ) = 4 + 6 = 10 true The point (4, 2) checks, so it is the solution of the system. Example continued

9. Finding a Solution by Graphing Solve the following system of equations by graphing. – x + 3 y = 6 and 3 x – 9 y = 9 x y First, graph – x + 3 y = 6 (-6, 0) (0, 2) (6, 4) Second, graph 3x-9y=9 (0, -1) (6, 1) (3, 0) The lines APPEAR to be parallel. Example Continued.

10. Finding a Solution by Graphing Although the lines appear to be parallel, you still need to check that they have the same slope. You can do this by solving for y . First equation, – x + 3 y = 6 3 y = x + 6 (add x to both sides) y = x + 2 (divide both sides by 3) Second equation, 3 x – 9 y = 9 – 9 y = – 3 x + 9 (subtract 3 x from both sides) y = x – 1 (divide both sides by –9) Example continued Both lines have a slope of , so they are parallel and do not intersect. Hence, there is no solution to the system

11. SLOPE Slope is the ratio of the vertical rise to the horizontal run between any two points on a line. Usually referred to as the rise over run. Slope triangle between two points. Notice that the slope triangle can be drawn two different ways. Rise is -10 because we went down Run is -6 because we went to the left Rise is 10 because we went up Run is 6 because we went to the right Another way to find slope

12. FORMULA FOR FINDING SLOPE The formula is used when you know two points of a line. EXAMPLE

13. Find the slope of the line between the two points (-4, 8) and (10, -4) If it helps label the points. Then use the formula

14. Matrix equation m  n matrix: 14 (4) For a square matrix, the entries a 11 , a 22 , …, a nn are called the main diagonal entries. (3) If , then the matrix is called square of order n . Notes: (1) Every entry a ij in a matrix is a number. (2) A matrix with m rows and n columns is said to be of size m  n . Elementary Linear Algebra: Section 1.2, p.14

15. Matrix form: Coefficient matrix:

16. The Substitution Method Another method (beside getting lucky with trial and error or graphing the equations) that can be used to solve systems of equations is called the substitution method . You solve one equation for one of the variables, then substitute the new form of the equation into the other equation for the solved variable.

17. The Substitution Method Solve the following system using the substitution method. 3 x – y = 6 and – 4 x + 2 y = –8 Solving the first equation for y , 3 x – y = 6 – y = –3 x + 6 (subtract 3 x from both sides) y = 3 x – 6 (multiply both sides by – 1) Substitute this value for y in the second equation. –4 x + 2y = –8 –4 x + 2(3 x – 6) = –8 (replace y with result from first equation) –4 x + 6 x – 12 = –8 (use the distributive property) 2 x – 12 = –8 (simplify the left side) 2 x = 4 (add 12 to both sides) x = 2 (divide both sides by 2) Example Continued.

18. The Substitution Method Substitute x = 2 into the first equation solved for y . y = 3 x – 6 = 3(2) – 6 = 6 – 6 = 0 Our computations have produced the point (2, 0). Check the point in the original equations. First equation, 3 x – y = 6 3( 2 ) – = 6 true Second equation, –4 x + 2 y = –8 –4( 2 ) + 2( ) = –8 true The solution of the system is (2, 0). Example continued

19. The Substitution Method Solving a System of Linear Equations by the Substitution Method Solve one of the equations for a variable. Substitute the expression from step 1 into the other equation. Solve the new equation. Substitute the value found in step 3 into either equation containing both variables. Check the proposed solution in the original equations.

20. The Substitution Method Solve the following system of equations using the substitution method. y = 2 x – 5 and 8 x – 4 y = 20 Since the first equation is already solved for y , substitute this value into the second equation. 8 x – 4 y = 20 8 x – 4(2 x – 5) = 20 (replace y with result from first equation) 8 x – 8 x + 20 = 20 (use distributive property) 20 = 20 (simplify left side) Example Continued.

21. The Substitution Method When you get a result, like the one on the previous slide, that is obviously true for any value of the replacements for the variables, this indicates that the two equations actually represent the same line. There are an infinite number of solutions for this system. Any solution of one equation would automatically be a solution of the other equation. This represents a consistent system and the linear equations are dependent equations. Example continued

22. The Substitution Method Solve the following system of equations using the substitution method . 3 x – y = 4 and 6 x – 2 y = 4 Solve the first equation for y. 3 x – y = 4 – y = –3 x + 4 (subtract 3x from both sides) y = 3 x – 4 (multiply both sides by –1) Substitute this value for y into the second equation. 6 x – 2 y = 4 6 x – 2(3 x – 4) = 4 (replace y with the result from the first equation) 6 x – 6 x + 8 = 4 (use distributive property) 8 = 4 (simplify the left side) Example Continued.

23. The Substitution Method When you get a result, like the one on the previous slide, that is never true for any value of the replacements for the variables, this indicates that the two equations actually are parallel and never intersect. There is no solution to this system. This represents an inconsistent system, even though the linear equations are independent. Example continued

24. Solving systems of linear equations by addition

25. The Elimination Method Another method that can be used to solve systems of equations is called the addition or elimination method . You multiply both equations by numbers that will allow you to combine the two equations and eliminate one of the variables.

26. The Elimination Method Solve the following system of equations using the elimination method. 6 x – 3 y = –3 and 4 x + 5 y = –9 Multiply both sides of the first equation by 5 and the second equation by 3. First equation, 5(6 x – 3 y ) = 5(–3) 30 x – 15 y = –15 (use the distributive property) Second equation, 3(4 x + 5 y ) = 3(–9) 12 x + 15 y = –27 (use the distributive property) Example Continued.

27. The Elimination Method Combine the two resulting equations (eliminating the variable y ). 30 x – 15 y = –15 12 x + 15 y = –27 42 x = –42 x = –1 (divide both sides by 42) Example continued Continued.

28. The Elimination Method Substitute the value for x into one of the original equations. 6 x – 3 y = –3 6( –1 ) – 3 y = –3 (replace the x value in the first equation) –6 – 3 y = –3 (simplify the left side) –3 y = –3 + 6 = 3 (add 6 to both sides and simplify) y = –1 (divide both sides by – 3) Our computations have produced the point (–1, –1). Example continued Continued.

29. The Elimination Method Check the point in the original equations. First equation, 6 x – 3 y = –3 6( –1 ) – 3(– 1 ) = –3 true Second equation, 4 x + 5 y = –9 4( –1 ) + 5( –1 ) = –9 true The solution of the system is (–1, –1). Example continued

30. The cryptographic method Use of matrix 1 2 0 3 To obtain the Hill cipher for the obain text message I AM HIDING Example

31. The C ryptographic method Solution If we group the plaintext into pairs and add the dummy letter G to fill out the last pair we obtain IA MH ID IN GG Example Continued

32. The cryptographic method 91 13 8 94 9 14 77 To cipher the paintext , we form the matrix product 1 2 9 11 = 0 3 1 3 Example Continued or equivalently

33. The cryptographic method Example Continued Which from , yields ciphertext KC To encipher the pair MH, we form the product 1 2 13 29 = 0 3 8 24

34. The cryptographic method Example Continued Whenever an integer greater than 25 occurs, it will be by the remainder that results when this integer is divided by 26 1 2 9 17 = 0 3 4 12 1 2 9 37 11 = or 0 3 14 42 16 1 2 7 21 = 0 3 7 21

35. The cryptographic method Example Continued The entire ciphertext message is KC CX QL KP UU which would usually be transmitted as a single string Without spaces KCCXQLKPUU