Chapter 6 Systems of Equations and Matrices Copyright

Chapter 6 Systems of Equations and Matrices Copyright © 2016, 2012 Pearson Education, Inc. 6 -1

Section 6. 1 6. 2 6. 3 6. 4 6. 5 6. 6 6. 7 6. 8 Systems of Equations in Two Variables Systems of Equations in Three Variables Matrices and Systems of Equations Matrix Operations Inverses of Matrices Determinants and Cramer’s Rule Systems of Inequalities and Linear Programming Partial Fractions Copyright © 2016, 2012 Pearson Education, Inc. 6 -2

6. 2 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. Copyright © 2016, 2012 Pearson Education, Inc. 6 -3

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, where 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, 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 2 y + 4 z = 8 2 x + 2 y z = 11 x + y 2 z = 10 Copyright © 2016, 2012 Pearson Education, Inc. 6 -4

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. Copyright © 2016, 2012 Pearson Education, Inc. 6 -5

Operations The following operations can be used to transform the original system to an equivalent system in the desired form. 1. 2. 3. 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. Copyright © 2016, 2012 Pearson Education, Inc. 6 -6

Example Solve the system x + 3 y + 2 z = 9 x y + 3 z = 16 3 x 4 y + 2 z = 28 Solution: Choose 1 variable to eliminate using 2 different pairs of equations. Let’s eliminate x from equations (2) and (3). Copyright © 2016, 2012 Pearson Education, Inc. 6 -7

Example x 3 y 2 z = 9 x y + 3 z = 16 4 y + z = 7 3 x 9 y 6 z = 27 3 x 4 y + 2 z = 28 13 y 4 z = 1 Mult. (1) by 1 (2) (4) Mult. (1) by 3 (3) (5) Copyright © 2016, 2012 Pearson Education, Inc. 6 -8

Example continued Now we have… x + 3 y + 2 z = 9 4 y + z = 7 13 y 4 z = 1 (1) (4) (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 + 3 y + 2 z = 9 (1) 16 y + 4 z = 28 (6) 13 y 4 z = 1 (5) Copyright © 2016, 2012 Pearson Education, Inc. 6 -9

Example continued Now, we add equations 5 and 6. 13 y 4 z = 1 (5) 16 y + 4 z = 28 (6) 29 y = 29 Now, we have the system of equations: x + 3 y + 2 z = 9 (1) 13 y 4 z = 1 (5) 29 y = 29 (7) Copyright © 2016, 2012 Pearson Education, Inc. 6 -10

Example continued Next, we solve equation (7) for y: 29 y = 29 y = 1 Then, we back-substitute 1 in equation (5) and solve for z. 13( 1) 4 z = 1 13 4 z = 12 z=3 Copyright © 2016, 2012 Pearson Education, Inc. 6 -11

Example continued 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. Copyright © 2016, 2012 Pearson Education, Inc. 6 -12

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. Copyright © 2016, 2012 Pearson Education, Inc. 6 -13

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 3 Rush Hours 2 7 6 City Traffic Hours 9 8 18 Copyright © 2016, 2012 Pearson Education, Inc. Highway Hours 3 3 6 Total Fuel Used (gal) 15 24 34 6 -14

Solution 1. 2. 3. Familiarize. We let x, y, and z represent the hours in rush hour traffic, city traffic, and highway, respectively. Translate. We have three equations: 2 x + 9 y + 3 z = 15 (1) 7 x + 8 y + 3 z = 24 (2) 6 x + 18 y + 6 z = 34 (3) Carry Out. We will solve this system by eliminating z from equations (2) and (3). 2 x 9 y 3 z = 15 Mult. (1) by 1 7 x + 8 y + 3 z = 24 (2) 5 x y = 9 (4) Copyright © 2016, 2012 Pearson Education, Inc. 6 -15

Solution continued Next, we can solve for x: 4 x 18 y 6 z = 30 Mult. (1) by 2 6 x + 18 y + 6 z = 34 (3) 2 x = 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) + 3 z = 15 4 + 9 + 3 z = 15 3 z = 2 z= Solving the system we get (2, 1, ). Copyright © 2016, 2012 Pearson Education, Inc. 6 -16

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. Copyright © 2016, 2012 Pearson Education, Inc. 6 -17

Mathematical Models and Applications Recall that when we model a situation using a linear function f(x) = mx + b, we need to know two data points in order to determine m and b. For a quadratic model, f(x) = ax 2 + bc + c we need three data points in order to determine a, b, and c. Copyright © 2016, 2012 Pearson Education, Inc. 6 -18

Example The table below lists sales of winter sports equipment, apparel, and accessories in November in three recent years. Use the data to find a quadratic function that gives the snow-sports sales as a function of the number of years after 2006. Then use the function to estimate snow-sports sales in 2009. Copyright © 2016, 2012 Pearson Education, Inc. 6 -19

Solution Let x = the number of years after 2006 s(x) = snow-sports sales x = 0 corresponds to 2006 x = 1 corresponds to 2007 x = 2 corresponds to 2008 Use three data points. (0, 296), (1, 423), and (2, 402) to find a, b, and c. Copyright © 2016, 2012 Pearson Education, Inc. 6 -20

Solution Now have a system of three equations in the variables a, b, and c. Solving the system, we get: Estimate sales in 2009, we find f(3). Copyright © 2016, 2012 Pearson Education, Inc. 6 -21