CHAPTER 9: Systems of Equations and Matrices

Transcription

1 MAT 171 Precalculus Algebra Dr. Claude Moore Cape Fear Community College CHAPTER 9: Systems of Equations and Matrices 9.1 Systems of Equations in Two Variables 9.2 Systems of Equations in Three Variables 9. Matrices and Systems of Equations Khan Academy is free and has terrific videos. I suggest you check it out. The algebra section is at Systems 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. To model a quadratic function with the TI, see use of the TI Calculator on the Important Links webpage and click on Modeling Statistical Modeling with TI 8 Calculator linear, quadratic, and other regressions. ( 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 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, ) is the solution of this system of equations. We can verify this by substituting 4 for x, 0 for y, and for z in each equation. x 2y + 4z = 8 2x + 2y z = 11 x + y 2z = 10 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 backsubstitution is used to find y and then x. 1

2 Legal Operations The following operations can be used to transform the original system to an equivalent system in the desired form. 1. Interchange any two equations. 2. Multiply both sides of one of the equations by a nonzero constant.. Add a nonzero multiple of one equation to another equation. Example Solve the system (1) x + y + 2z = 9 (2) x y + z = 1 () x 4y + 2z = 28 Solution: Choose 1 variable to eliminate using 2 different pairs of equations. Let s eliminate x from equations (2) and (). Example Solve the system (1) x + y + 2z = 9 (2) x y + z = 1 () x 4y + 2z = 28 x y 2z = 9 Mult. (1) by 1 x y + z = 1 (2) 4y + z = 7 (4) x 9y z = 27 Mult. (1) by x 4y + 2z = 28 () 1y 4z = 1 (5) Now we have x + y + 2z = 9 (1) 4y + z = 7 (4) 1y 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 + y + 2z = 9 (1) 1y + 4z = 28 () 1y 4z = 1 (5) 2

3 Now, we add equations (5) and (). 1y 4z = 1 (5) 1y + 4z = 28 () 29y = 29 Now, we have the system of equations: x + y + 2z = 9 (1) 1y 4z = 1 (5) 29y = 29 (7) Next, we solve equation (7) for y: 29y = 29 y = 1 Then, we back substitute 1 in equation (5) and solve for z. 1( 1) 4z = 1 1 4z = 1 4z = 12 z = Finally, we substitute 1 for y and for z in equation (1) and solve for x: x + ( 1) + 2() = 9 x + = 9 x = 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. The triple (, 1, ) is the solution of this system.

4 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. Week 1 Week 2 Week Week 1 Week 2 Week Solution 1. Familiarize. We let x, y, and z represent the hours in rush hour traffic, city traffic, and highway, respectively. 2. Translate. We have three equations: 2x + 9y + z = 15 (1) 7x + 8y + z = 24 (2) x + 18y + z = 4 (). Carry Out. We will solve this equation by eliminating z from equations (2) and (). 2x 9y z = 15 Mult. (1) by 1 7x + 8y + z = 24 (2) 5x y = 9 (4) Solution continued Next, we can solve for x: 4x 18y z = 0 Mult. (1) by 2 x + 18y + z = 4 () 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) + z = z = 15 z = 2 z = Solving the system we get (2, 1, ). Solution continued 4. Check: Substituting 2 for x, 1 for y, and for z, we see that the solution makes each of the three equations true. 5. 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. 4

5 742/2. Solve the system of equations. x + y + z = 4 2x + y + 2z = x 2y + z = 0 742/8. Solve the system of equations. x + 2y z = 4 4x y + z = 8 5x y = 0 742/10. Solve the system of equations. x + y + 4z = 1 x + 4y + 5z = x + 8y + 11z = 2 74/1. Solve the system of equations. w + x y + z = 0 w + 2x + 2y + z = 5 w + x + y z = 4 2w + x + y z = 7 The last row represents 0 = 1 and yields no common solutions. 5

6 744/2. e Commerce. computerwarehouse.com charges $ for shipping orders up to 10 lb, $ 5 for orders from 10 lb up to 15 lb, and $ 7.50 for orders of 15 lb or more. One day shipping charges for 150 orders totaled $ 80. The number of orders under 10 lb was three times the number of orders weighing 15 lb or more. Find the number of packages shipped at each rate. 74/8. Preprimary School Enrollment. The table below lists the number of children age to 5 enrolled in nursery school and kindergarten, in millions, in various years. a) Use a graphing calculator to fit a quadratic function to the data, where x is the number of years after b) Use the function found in part (a) to estimate the number of children enrolled in preprimary school in 200. See next page for more calculator results. See Modeling tutorial for calculator at STAT.htm#algebra Set up the calculator as indicated in the Modeling section of the TI Tutorials. The quadratic graph passes very close to all points. Part (b) of problem: 2ND GRAPH to display the TABLE and scroll down to where X is 11 to see Y1 = 8.2.