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Chapter 4 Systems of Linear Equations; Matrices Section Section Section Section Section Section Section 1 2 3 4 5 6 7 – – – – – – – Review: Sys of Linear Eg in Two Var Sys of Linear Eq and Aug Matr Gauss-Jordan Elimination Matrices: Basic Operations Inverse of a Square Matrix Matrix Eq and Sys of Linear Eq Leontief Input-Output Analysis Introduction Chapter 4 Systems of linear equations can be used to solve resource allocation problems in business and economics. Such systems can involve many equations in many variables. • After reviewing methods for solving two linear equations in two variables, we use matrices and matrix operations to develop procedures that are suitable for solving linear systems of any size. • We also discuss the Wassily Leontief application of matrices to economic planning of industrialized countries. Barnett/Ziegler/Byleen College Mathematics 12e 2 Chapter 4 Systems of Linear Equations; Matrices Section 1 Review: Systems of Linear Equations in Two Variables Learning Objectives for Section 4.1 Review: Systems of Linear Equations in Two Variables The student will be able to solve systems of linear equations in two variables by graphing The student will be able to solve these systems using substitution The student will be able to solve these systems using elimination by addition The student will be able to solve applications of linear systems. Barnett/Ziegler/Byleen College Mathematics 12e 4 Table of Content Systems of Linear Equations in Two Variables Graphing Substitution Elimination by Addition Applications Barnett/Ziegler/Byleen College Mathematics 12e 5 Terms solution solution set consistent / inconsistent system dependent / independent system substitution elimination by addition equilibrium price / quantity Barnett/Ziegler/Byleen College Mathematics 12e 6 Opening Example A restaurant serves two types of fish dinners- small for $5.99 and large for $8.99. One day, there were 134 total orders of fish, and the total receipts for these 134 orders was $1,024.66. How many small fish dinners and how many large fish dinners were ordered? x fish dinner-small y fish dinner-large x+ y = 134 5.99x + 8.99y = 1,024.66 Barnett/Ziegler/Byleen College Mathematics 12e 7 Opening Example x fish dinner-small y fish dinner-large x+ y = 134 5.99x + 8.99y = 1,024.66 Now we have a system of two linear equations in two variables. It is easy to find ordered pairs (x, y) that satisfies one or the other of these equations. To solve this system we have to find an ordered pair of real number that satisfies both equation at the same time. In general, we have the following definition: Barnett/Ziegler/Byleen College Mathematics 12e 8 Systems of Two Equations in Two Variables We are given the linear system ax + by = h cx + dy = k a, b, c, d, h, k real constants A solution is an ordered pair (x0, y0) that will satisfy each equation (make a true equation when substituted into that equation). The solution set is the set of all ordered pairs that satisfy both equations. In this section, we wish to find the solution set of a system of linear equations. To solve a system is to find its solution set. Barnett/Ziegler/Byleen College Mathematics 12e 9 Opening Example We will consider three methods of solving such systems: • Graphing • Susbtitution and • Elimination by addition Barnett/Ziegler/Byleen College Mathematics 12e 10 Solve by Graphing One method to find the solution of a system of linear equations is to graph each equation on a coordinate plane and to determine the point of intersection (if it exists). The drawback of this method is that it is not very accurate in most cases, but does give a general location of the point of intersection. Lets take a look at an example: Solve the following system by graphing: 3x + 5y = –9 x+ 4y = –10 Barnett/Ziegler/Byleen College Mathematics 12e 11 Solve by Graphing Solution 3x + 5y = –9 x+ 4y = –10 First line (intercept method) If x = 0, y= –9/5 If y = 0 , x = –3 Plot points and draw line (2, –3) Second line: Intercepts are (0, –5/2) , ( –10,0) From the graph we estimate that the point of intersection is (2,–3). Check: 3(2) + 5(–3)= –9 and 2 + 4(–3)= –10. Both check. Barnett/Ziegler/Byleen College Mathematics 12e 12 Another Example Now, you try one: Solve the system by graphing: 2x + 3 = y x + 2y = –4 Barnett/Ziegler/Byleen College Mathematics 12e 13 Another Example Solution Now, you try one: Solve the system by graphing: 2x + 3 = y x + 2y = –4 First line (intercept method) If x = 0, y= 3 If y = 0 , x = –3/2 Plot points and draw line Barnett/Ziegler/Byleen College Mathematics 12e Second line: Intercepts are (0, –2) , ( –4,0) From the graph we estimate that the point of intersection is (-2,–1). 14 Example 2 Solve the following systems by graphing: (A) x – 2y = 2 x+ y= 5 (B) x + 2y = -4 2x + 4y = 8 Barnett/Ziegler/Byleen College Mathematics 12e (C) 2x + 4y = 8 x + 2y = 4 15 Example 2 - Solution (B) Lines are parallel (each has slope -1/2) – no solutions. x + 2y = -4 2x + 4y = 8 (4,1) (A) Interception at one point – exactly one solution. x – 2y = 2 x+ y= 5 (C) Lines coincide – infinite number of solutions. 2x + 4y = 8 x + 2y = 4 Barnett/Ziegler/Byleen College Mathematics 12e 16 Basic Terms A consistent linear system is one that has one or more solutions. • If a consistent system has exactly one solution (unique solution), it is said to be independent. An independent system will occur when the two lines have different slopes. • If a consistent system has more than one solution, it is said to be dependent. A dependent system will occur when the two lines have the same slope and the same y intercept. In other words, the graphs of the lines will coincide. There will be an infinite number of points of intersection. Two systems of equations are equivalent if they have the same solution set. Barnett/Ziegler/Byleen College Mathematics 12e 17 Basic Terms An inconsistent linear system is one that has no solutions. This will occur when two lines have the same slope but different y intercepts. In this case, the lines will be parallel and will never intersect. Referring to the three systems in the Example 2, the system in part (A) is consistent and independent with the unique solution x = 4, y = 1. The system in part (B) is inconsistent. And the system in part (C) is consistent and dependent with an infinitive number of solutions. Barnett/Ziegler/Byleen College Mathematics 12e 18 Example 2 - Solution (B) Lines are parallel (each has slope -1/2) – no solutions. x + 2y = -4 2x + 4y = 8 inconsistent (4,1) (A) Interception at one point – exactly one solution. x – 2y = 2 x+ y= 5 consistent and independent (C) Lines coincide – infinite number of solutions. 2x + 4y = 8 x + 2y = 4 consistent and dependent Barnett/Ziegler/Byleen College Mathematics 12e 19 Basic Terms Example: Determine if the system is consistent, independent, dependent or inconsistent: 2x – 5y = 6 –4x + 10y = –1 Barnett/Ziegler/Byleen College Mathematics 12e 20 Solution of Example Solve each equation for y to obtain the slope intercept form of the equation: 2𝑥 − 5𝑦 = 6 −4𝑥 + 10𝑦 = −1 −5𝑦 = −2𝑥 + 6 10𝑦 = 4𝑥 − 1 4 1 5𝑦 = 2𝑥 − 6 𝑦= 𝑥− 𝑦= 2 𝑥 5 − 6 5 𝑦= 10 2 𝑥 5 − 10 1 10 Since each equation has the same slope but different y intercepts, they will not intersect. This is an inconsistent system. Barnett/Ziegler/Byleen College Mathematics 12e 21 Basic Terms By graphing a system of two linear equations in two variables, we gain useful information about the solution set of the system. In general, any two lines in a coordinate plane must intersect in exactly one point, be parallel, or coincide (have identical graphs). The example 2 illustrate the only three possible types of solutions for systems of two linear equations in two variables. This is summarized as follows… Barnett/Ziegler/Byleen College Mathematics 12e 22 Possible Solutions to a Linear System A linear system ax + by = h cx + dy = k must have A. Exactly one solution or B. No solution Consistent and independent Inconsistent or C. Infinitely many solutions There are no other possibilities. Barnett/Ziegler/Byleen College Mathematics 12e Consistent and dependent 23 Method of Substitution Although the method of graphing is intuitive, it is not very accurate in most cases, unless done by calculator. There is another method that is 100% accurate. It is called the method of substitution. This method is an algebraic one. It works well when the coefficients of x or y are either 1 or –1. For example, let’s solve the previous system 2x + 3 = y x + 2y = -4 using the method of substitution. Barnett/Ziegler/Byleen College Mathematics 12e 24 Method of Substitution (continued) The steps for this method are as follows: 1) Solve one of the equations for either x or y. 2) Substitute that result into the other equation to obtain an equation in a single variable (either x or y). 3) Solve the equation for that variable. 4) Substitute this value into any convenient equation to obtain the value of the remaining variable. Barnett/Ziegler/Byleen College Mathematics 12e 25 Example (continued) 2x + 3 = y x + 2y = –4 x + 2(2x + 3) = –4 x + 4x + 6 = –4 5x + 6 = –4 5x = –10 x = –2 Barnett/Ziegler/Byleen College Mathematics 12e Substitute y from first equation into second equation Solve the resulting equation After we find x = –2, then from the first equation, we have 2(–2) + 3 = y or y = –1. Our solution is (–2, –1) 26 Another Example Solve the system using substitution: 3x – 2y = 7 y = 2x – 3 Barnett/Ziegler/Byleen College Mathematics 12e 27 Another Example Solution Solve the system using substitution: 3x – 2y = 7 y = 2x – 3 Solution: 3x 2 y 7 y 2x 3 3x 2(2 x 3) 7 3x 4 x 6 7 x 1 x 1 y 2 1 3 y 5 Barnett/Ziegler/Byleen College Mathematics 12e 28 Elimination by Addition The method of substitution is not preferable if none of the coefficients of x and y are 1 or –1. For example, substitution is not the preferred method for the system below: 2x – 7y = 3 –5x + 3y = 7 A better method is elimination by addition. The following operations can be used to produce equivalent systems: • Two equations can be interchanged. • An equation can be multiplied by a non-zero constant. • A constant multiple of one equation and is added to another equation. Barnett/Ziegler/Byleen College Mathematics 12e 29 Elimination by Addition Example For our system, we will seek to eliminate the x variable. The coefficients are 2 and –5. Our goal is to obtain coefficients of x that are additive inverses of each other. We can accomplish this by multiplying the first equation by 5, and the second equation by 2. Next, we can add the two equations to eliminate the x-variable. Solve for y Substitute y value into original equation and solve for x Write solution as an ordered pair Barnett/Ziegler/Byleen College Mathematics 12e 5x 3y 7 5(2x 7 y) 5(3) 2(5x 3y){ 2(7) 10x 35y 15 10x 6 y 14 0x 29 y 29 2x 7 y 3 y 1 2x 7(1) { 3 2x 7 3 2x 4 x 2 (2,1) 30 Elimination by Addition Another Example Solve 2x – 5y = 6 –4x + 10y = –1 Barnett/Ziegler/Byleen College Mathematics 12e 31 Elimination by Addition Another Example 2x – 5y = 6 –4x + 10y = –1 Solution: 1. Eliminate x by multiplying equation 1 by 2 . 2. Add two equations 3. Upon adding the equations, both variables are eliminated producing the false equation 0 = 11 Solve Barnett/Ziegler/Byleen College Mathematics 12e 2x 5 y 6 4 x 10 y 1 4 x 10 y 12 4 x 10 y 1 0 11 4. Conclusion: If a false equation arises, the system is inconsistent and there is no solution. 32 Elimination by Addition Additional Considerations Let see what happens in the elimination process when a system has either no solution or infinitely many solutions. Barnett/Ziegler/Byleen College Mathematics 12e 33 Elimination by Addition Additional Considerations Consider the following system: 2x – 6y = -3 x + 3y = 2 Multiplying the second equation by -2 and adding, we obtain: 2x – 6y = -3 -2x – 6y = 2 0 = -7 not possible We have obtained a contradiction. The initial assumption that the system has solutions is false. The system has no solutions and the graphs of the equations are parallel lines. Barnett/Ziegler/Byleen College Mathematics 12e 34 Elimination by Addition Additional Considerations Now consider the system: x – ½y = 4 -2x + y = -8 Multiplying the top equation by 2 and adding, we obtain: 2x – y = 8 -2x + y = -8 0 = 0 This result implies that the equations are equivalent; that is, their graphs coincide and the system is dependent. Barnett/Ziegler/Byleen College Mathematics 12e 35 Elimination by Addition Additional Considerations If we let x = k, where k is any real number, and solve either equation for y, we obtain y = 2k – 8. So (k, 2k – 8) is a solution to the system for any real number k. The variable k is called a parameter and replacing k with a real number produces a particular solution to the system. For example, some particular solutions to the system are: k = -1 k = -1 k = -1 k = -1 (-1, -10) (2, -4) (5, 2) (9.4, 10.8) Barnett/Ziegler/Byleen College Mathematics 12e 36 Application A man walks at a rate of 3 miles per hour and jogs at a rate of 5 miles per hour. He walks and jogs a total distance of 3.5 miles in 0.9 hours. How long does the man jog? Barnett/Ziegler/Byleen College Mathematics 12e 37 Application A man walks at a rate of 3 miles per hour and jogs at a rate of 5 miles per hour. He walks and jogs a total distance of 3.5 miles in 0.9 hours. How long does the man jog? Solution: Let x represent the amount of time spent walking and y represent the amount of time spent jogging. Since the total time spent walking and jogging is 0.9 hours, we have the equation x + y = 0.9. We are given the total distance traveled as 3.5 miles. Since Distance = Rate x Time, distance walking = 3x and distance jogging = 5y. Then total distance is 3x + 5y= 3.5. Barnett/Ziegler/Byleen College Mathematics 12e 38 Application (continued) We can solve the system using Solution: substitution. x y 0.9 1. Solve the first equation for y y 0.9 x 2. Substitute this expression into 3 x 5 y 3.5 the second equation. 3 x 5(0.9 x) 3.5 3. Solve second equation for x 3 x 4.5 5 x 3.5 4. Find the y value by substituting this x value back into the first 2 x 1 equation. x 0.5 5. Answer the question: Time spent 0.5 y 0.9 jogging is 0.4 hours. y 0.4 Barnett/Ziegler/Byleen College Mathematics 12e 39 Supply and Demand The quantity of a product that people are willing to buy during some period of time depends on its price. Generally, the higher the price, the less the demand; the lower the price, the greater the demand. Similarly, the quantity of a product that a supplier is willing to sell during some period of time also depends on the price. Generally, a supplier will be willing to supply more of a product at higher prices and less of a product at lower prices. The simplest supply and demand model is a linear model where the graphs of a demand equation and a supply equation are straight lines. Barnett/Ziegler/Byleen College Mathematics 12e 40 Supply and Demand (continued) In supply and demand problems we are usually interested in finding the price at which supply will equal demand. This is called the equilibrium price, and the quantity sold at that price is called the equilibrium quantity. If we graph the the supply equation and the demand equation on the same axis, the point where the two lines intersect is called the equilibrium point. Its horizontal coordinate is the value of the equilibrium quantity, and its vertical coordinate is the value of the equilibrium price. Barnett/Ziegler/Byleen College Mathematics 12e 41 Supply and Demand Example Example: Suppose that the supply equation for long-life light bulbs is given by p = 1.04 q – 7.03, and that the demand equation for the bulbs is p = –0.81q + 7.5 where q is in thousands of cases. Find the equilibrium price and quantity, and graph the two equations in the same coordinate system. Barnett/Ziegler/Byleen College Mathematics 12e 42 Supply and Demand (Example continued) If we graph the two equations on a graphing calculator or manually, or solve by substitution, and find the intersection point, we see the graph below. Demand Curve 1.04 q – 7.03 = –0.81q + 7.5 1.85 q = 14.53 q = 7.854 p = 1.04 q – 7.03 Supply Curve p = 1.04 (7.854) – 7.03 = 1.14 Thus the equilibrium point is (7.854, 1.14), the equilibrium price is $1.14 per bulb, and the equilibrium quantity is 7,854 cases. Barnett/Ziegler/Byleen College Mathematics 12e 43 Now, Solve the Opening Example A restaurant serves two types of fish dinners- small for $5.99 each and large for $8.99. One day, there were 134 total orders of fish, and the total receipts for these 134 orders was $1024.66. How many small dinners and how many large dinners were ordered? x fish dinner-small y fish dinner-large x+ y = 134 5.99x + 8.99y = 1,024.66 Barnett/Ziegler/Byleen College Mathematics 12e 44 Solution Multiply the first equation by -8.99 and add to the second equation: (-8.99) (x + y) = 134(-8.99) -8.99x - 8.99y = -1,204.66 5.99x + 8.99y = -1,024.66 -3x + 0 = -180.00 x = 60 x + y = 134 60 + y = 134 y = 134 – 60 = 74 Answer: 60 small orders and 74 large orders Barnett/Ziegler/Byleen College Mathematics 12e 45 Chapter 4 Systems of Linear Equations; Matrices Section 1 Review: Systems of Linear Equations in Two Variables END Last Update: Abril 9/2013