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  1. Systems of Equationsand Matrices Chapter 5

  2. 5.1Systems of Equations in Two Variables Solve a system of two linear equations in two variables by graphing. Solve a system of two linear equations in two variables using the substitution and the elimination methods. Use systems of two linear equations to solve applied problems.

  3. Systems of Equations • A system of equations is composed of two or more equations considered simultaneously. Example: 5x y = 5 4x y = 3 This is a system of two linear equations in two variables. The solution set of this system consists of all ordered pairs that make both equations true. The ordered pair (2, 5) is a solution of this system.

  4. When we graph a system of linear equations, each point at which the equations intersect is a solution of both equations and therefore a solution of the system of equations. Let’s solve the previous system graphically. 5x y = 5 4x y = 3 Solution: We see that the graph intersects at the single point (2, 5), so this is the solution of the system of equations. Solving Systems of Equations Graphically

  5. Systems of Equations • If a system of equations has at least one solution, it is consistent. If the system has no solutions, it is inconsistent. • If a system of two linear equations in two variables has an infinite number of solutions, the equations are dependent. Otherwise, they are independent.

  6. Illustration of Graphs Graphs of linear equations may be related to each other in one of three ways.

  7. Substitution Method The substitution method is a technique that gives accurate results when solving systems of equations. It is most often used when a variable is alone on one side of an equation or when it is easy to solve for a variable. One equation is used to express one variable in terms of the other, then it is substituted in the other equation. Example: Let’s use this method to solve the previous system. 5x y = 5 4x y = 3

  8. Solve the first equation for y. 5x y = 5 y = 5x  5 Then we substitute 5x  5 for y in the second equation to give an equation in one variable. 4x  (5x  5) = 3 4x  5x + 5 = 3 x = 2 Now we use back-substitution and substitute 2 for x in either original equation. 4x  y = 3 4(2)  y = 3 8  y = 3 y = 5 We find the solution to the system of equations to be (2, 5), once again. Solution

  9. Elimination Method Using the elimination method, we eliminate one variable by adding the two equations. If the coefficients of a variable are opposites, that variable can be eliminated by simply adding the original equations. If the coefficients are not opposites, it is necessary to multiply one or both equations by suitable constants, before we add.

  10. Example Solve the system using the elimination method. 6x + 2y = 4 10x + 7y =  8 If we multiply the first equation by 5 and the second equation by 3, we will be able to eliminate the x variable. 30x + 10y = 20 Substituting: 6x + 2y = 4 30x  21y = 24 6x + 2(4) = 4 11y = 44 6x  8 = 4 y = 4 6x = 12 The solution is (2,  4). x = 2

  11. Solve the system. x  3y = 9 (1) 2x  6y = 3 (2) Solution: 2x + 6y = 18 Mult. (1) by 2 2x  6y = 3 0 = 21 There are no values of x and y in which 0 = 21. So this system has no solution. The graphs of the equations are of parallel lines. Solve the system. 9x + 6y = 48 (1) 3x + 2y = 16 (2) Solution: 9x + 6y = 48 9x 6y = 48 Mult. (2) by 3 0 = 0 When we obtain the equation 0 = 0, we know the equations are dependent. There are infinitely manysolutions. The graphs of the equations are identical. More Examples

  12. Application Ethan and Ian are twins. They have decided to save all of the money they earn, at their part-time jobs, to buy a car to share at college. One week, Ethan worked 8 hours and Ian worked 14 hours. Together they saved $256. The next week, Ethan worked 12 hours and Ian worked 16 hours and they earned $324. How much does each twin make per hour?

  13. Solution Letting E represent Ethan and I represent Ian, the following system can be obtained. 8E + 14I = 256 Mult by 12 96E + 168I = 3072 12E + 16I = 324 Mult by 8 96E  128I = 2592 40I = 480 I = 12 Solve for E. 8E + 14(12) = 256 8E = 88 E = 11 Ian makes $12 per hour while Ethan makes $11 per hour.

  14. Graphical Solution • y1 = and y2 =

  15. 5.2Systems of Equations in Three Variables Solve systems of linear equations in three variables. Use systems of three equations to solve applied problems. Model a situation using a quadratic function.

  16. Solving Systems of Equations in Three Variables • A linear equation in three variables is an equation equivalent to one of the form Ax + By + Cz = D. • A, B, C, and D are real numbers and A, B, and C are not all 0. • A solution of a system of three equations in three variables is an ordered triple that makes all three equations true. Example: The triple (4, 0, 3) is the solution of this system of equations. We can verify this by substituting 4 for x, 0 for y, and 3 for z in each equation. x  2y + 4z = 8 2x + 2y  z = 11 x + y  2z = 10

  17. Gaussian Elimination • An algebraic method used to solve systems in three variables. • The original system is transformed to an equivalent one of the form: Ax + By + Cz = D, Ey + Fz = G, Hz = K. Then the third equation is solved for z and back-substitution is used to find y and then x.

  18. Operations The following operations can be used to transform the original system to an equivalent system in the desired form. • Interchange any two equations. • Multiply both sides of one of the equations by a nonzero constant. • Add a nonzero multiple of one equation to another equation.

  19. Example Solution: Choose 1 variable to eliminate using 2 different pairs of equations. Let’s eliminate x from equations (2) and (3). x + 3y + 2z = 9 x y + 3z = 16 3x  4y + 2z = 28 x  3y  2z = 9 Mult. (1) by 1 x  y + 3z = 16 (2) 4y + z = 7 (4) 3x  9y  6z = 27 Mult. (1) by 3 3x  4y + 2z = 28 (3) 13y  4z = 1 (5)

  20. Example continued Now we have… x + 3y + 2z = 9 (1) 4y + z = 7 (4) 13y  4z = 1 (5) Next, we multiply equation (4) by 4 to make the z coefficient a multiple of the z coefficient in the equation below it. x + 3y + 2z = 9 (1) 16y + 4z = 28 (6) 13y  4z = 1 (5)

  21. Example continued Now, we add equations 5 and 6. 13y  4z = 1 (5) 16y + 4z = 28 (6) 29y = 29 Now, we have the system of equations: x + 3y + 2z = 9 (1) 13y  4z = 1 (5) 29y = 29 (7) Next, we solve equation (7) for y: 29y = 29 y = 1

  22. Example continued Then, we back-substitute 1 in equation (5) and solve for z. 13(1)  4z = 1 13  4z = 1 4z = 12 z = 3 Finally, we substitute 1 for y and 3 for z in equation (1) and solve for x: x + 3(1) + 2(3) = 9 x  3 + 6 = 9 x = 6 • The triple (6, 1, 3) is the solution of this system.

  23. Graphs • The graph of a linear equation in three variables is a plane. Thus the solution set of such a system is the intersection of three planes.

  24. Rush Hour Hours City Traffic Hours Highway Hours Total Fuel Used (gal) Week 1 2 9 3 15 Week 2 7 8 3 24 Week 3 6 18 6 34 Application A food service distributor conducted a study to predict fuel usage for new delivery routes, for a particular truck. Use the chart to find the rates of fuel in rush hour traffic, city traffic, and on the highway.

  25. Solution • Familiarize. We let x, y, and z represent the hours in rush hour traffic, city traffic, and highway, respectively. • Translate. We have three equations: 2x + 9y + 3z = 15 (1) 7x + 8y + 3z = 24 (2) 6x + 18y + 6z = 34 (3) • Carry Out. We will solve this equation by eliminating z from equations (2) and (3). 2x  9y  3z = 15 Mult. (1) by 1 7x + 8y + 3z = 24 (2) 5x  y = 9 (4)

  26. Solution continued Next, we can solve for x: 4x  18y  6z = 30 Mult. (1) by 2 6x + 18y + 6z = 34 (3) 2x = 4 x = 2 Next, we can solve for y by substituting 2 for x in equation (4): 5(2)  y = 9 y = 1 Finally, we can substitute 2 for x and 1 for y in equation (1) to solve for z: 2(2) + 9(1) + 3z = 15 4 + 9 + 3z = 15 z = Solving the system we get (2, 1, ).

  27. Solution continued Check: Substituting 2 for x, 1 for y, and for z, we see that the solution makes each of the three equations true. State: In rush hour traffic the distribution truck uses fuel at a rate of 2 gallons per hour. In city traffic, the same truck uses 1 gallon of fuel per hour. In highway traffic, the same truck used gallon of fuel per hour.

  28. 5.3Matrices and Systems of Equations Solve systems of equations using matrices.

  29. Matrices • A rectangular array of numbers is called a matrix (plural, matrices). Example: • The matrix shown above is an augmented matrix because it contains not only the coefficients but also the constant terms. • The matrix is called the coefficient matrix.

  30. Matrices continued • The rows of a matrix are horizontal. • The columns of a matrix are vertical. • The matrix shown has 2 rows and 3 columns. • A matrix with m rows and n columns is said to be of orderm n. • When m = n the matrix is said to be square.

  31. Gaussian Elimination with Matrices • Row-Equivalent Operations 1. Interchange any two rows. 2. Multiply each entry in a row by the same nonzero constant. 3. Add a nonzero multiple of one row to another row.

  32. Example Solve the following system: First, we write the augmented matrix, writing 0 for the missing y-term in the last equation. Our goal is to find a row-equivalent matrix of the form .

  33. New row 1 = row 2 New row 2 = row 1 Example continued We multiply the first row by 2 and add it to the second row. We also multiply the first row by 4 and add it to the third row.

  34. Example continued We multiply the second row by 1/5 to get a 1 in the second row, second column. We multiply the second row by 12 and add it to the third row. Now, we can write the system of equations that corresponds to the last matrix above:

  35. Example continued • We back-substitute 3 for z in equation (2) and solve for y. • Next, we back-substitute 1 for y and 3 for z in equation (1) and solve for x. • The triple (2, 1, 3) checks in the original system of equations, so it is the solution.

  36. Row-Echelon Form 1. If a row does not consist entirely of 0’s, then the first nonzero element in the row is a 1 (called a leading 1). 2. For any two successive nonzero rows, the leading 1 in the lower row is farther to the right than the leading 1 in the higher row. 3. All the rows consisting entirely of 0’s are at the bottom of the matrix. If a fourth property is also satisfied, a matrix is said to be in reduced row-echelon form: 4. Each column that contains a leading 1 has 0’s everywhere else.

  37. Example • Which of the following matrices are in row-echelon form? a) b) c) d) Matrices (a) and (d) satisfy the row-echelon criteria. In (b) the first nonzero element is not 1. In (c), the row consisting entirely of 0’s is not at the bottom of the matrix.

  38. Gauss-Jordan Elimination • We perform row-equivalent operations on a matrix to obtain a row-equivalent matrix in row-echelon form. We continue to apply these operations until we have a matrix in reduced row-echelon form. Example: Use Gauss-Jordan elimination to solve the system of equations from the previous example; we had obtained the matrix .

  39. Gauss-Jordan Elimination continued • We continue to perform row-equivalent operations until we have a matrix in reduced row-echelon form. • Next, we multiply the second row by 3 and add it to the first row.

  40. Gauss-Jordan Elimination continued • Writing the system of equations that corresponds to this matrix, we have • We can actually read the solution, (2, 1, 3), directly from the last column of the reduced row-echelon matrix.

  41. Special Systems • When a row consists entirely of 0’s, the equations are dependent and the system is equivalent. • When we obtain a row whose only nonzero entry occurs in the last column, we have an inconsistent system of equations. For example, in the matrix • the last row corresponds to the false equation 0 = 9, so we know the original system has no solution.

  42. 5.4Matrix Operations Add, subtract, and multiply matrices when possible. Write a matrix equation equivalent to a system of equations.

  43. Matrices • A capital letter is generally used to name a matrix, and lower-case letters with double subscripts generally denote its entries. • For example, a23 read “a sub two three,” indicates the entry in the second row and the third column. • Two matrices are equal if they have the same order and corresponding entries are equal.

  44. Matrix Addition and Subtraction • To add or subtract matrices, we add or subtract their corresponding entries. • Addition and Subtraction of Matrices Given two m n matrices A= [aij] and B = [bij], their sum is A + B = [aij + bij] and their difference is A  B = [aij  bij].

  45. Example • Find A + Bfor each of the following. a) b)

  46. Example continued • We have a pair of 2  2 matrices in part (a) and a pair of 3  2 matrices in part (b). Since each pair has the same order we can add their corresponding entries.

  47. Find C  D for each of the following. a) Since the order of each matrix is 3  2, we can subtract corresponding entries. b) Since the matrices do not have the same order, we cannot subtract them. Example

  48. Scalar Multiplication • When we find the product of a number and a matrix, we obtain a scalar product. • The scalar product of a number k and a matrix A is the matrix denoted kA, obtained by multiplying each entry of A by the number k. The number k is called a scalar.

  49. Example • Find 4A and (2)Afor . Solution: