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three variables. An infinite number of solutions can result from several situations. The three planes could be the same, so that a solution to one equation will be the solution to the other two equations. All three equations could be different but they intersect on a line, which has infinite solutions. Or two of the equations could be the same and intersect the third on a line. Example 11.16 Finding the Solution to a Dependent System of Equations Find the solution to the given system of three equations in three variables. 2x + y−3z = 0 (1) 4x + 2y−6z = 0 (2) x − y + z = 0 (3) Solution First, we can multiply equation (1) by −2 and add it to equation (2). −4x−2y + 6z = 0 equation (1) multiplied by −2 4x + 2y−6z = 0 (2) ____________________________________________ 0 = 0 We do not need to proceed any further. The result we get is an identity, 0 = 0, which tells us that this system has an infinite number of solutions. There are other ways to begin to solve this system, such as multiplying equation (3) by −2, and adding it to equation (1). We then perform the same steps as above and find the same result, 0 = 0. When a system is dependent, we can find general expressions for the solutions. Adding equations (1) and (3), we have 1240 Chapter 11 Systems of Equations and Inequalities We then solve the resulting equation for z. 2x + y−3z = 0 x − y + z = 0 _____________ 3x−2z = 0 3x−2z = 0 z = 3 2 x We back-substitute the expression for z into one of the equations and solve for y. x⎞ ⎠ = 0 ⎛ 2x + y − 3 ⎝ 3 2 2x + − 2x x So the general solution is ⎛ x, 3 2 dependent on the value selected for x. ⎝x, 5 2 x⎞ ⎠. In this solution, x can be any real number. The values of y and z are Analysis As shown in Figure 11.18, two of the planes are the same and they intersect the third plane on a line. The solution set is infinite, as all points along the intersection line will satisfy all three
equations. Figure 11.18 Does the generic solution to a dependent system always have to be written in terms of x? No, you can write the generic solution in terms of any of the variables, but it is common to write it in terms of x and if needed x and y. 11.11 Solve the following system. x + y + z = 7 3x − 2y − z = 4 x + 6y + 5z = 24 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1241 Access these online resources for additional instruction and practice with systems of equations in three variables. • Ex 1: System of Three Equations with Three Unknowns Using Elimination (http://openstaxcollege.org/l/systhree) • Ex. 2: System of Three Equations with Three Unknowns Using Elimination (http://openstaxcollege.org/l/systhelim) 1242 Chapter 11 Systems of Equations and Inequalities 11.2 EXERCISES Verbal Can a linear system of three equations have exactly two 78. solutions? Explain why or why not 79. If a given ordered triple solves the system of equations, is that solution unique? If so, explain why. If not, give an example where it is not unique. 80. If a given ordered triple does not solve the system of equations, is there no solution? If so, explain why. If not, give an example. Using the method of addition, is there only one way to 81. solve the system? 82. Can you explain whether there can be only one method to solve a linear system of equations? If yes, give an example of such a system of equations. If not, explain why not. Algebraic For the following exercises, determine whether the ordered triple given is the solution to the system of equations. 83. 84. 85. 86. 2x−6y + 6z = −12 x + 4y + 5z = −1 −x + 2y + 3z = −1 and (0, 1, −1) 6x − y + 3z = 6 3x + 5y + 2z = 0 x + y = 0 and (3, −3, −5) 6x−7y + z = 2 −x − y + 3z = 4 2x + y − z = 1 and (
4, 2, −61 and (4, 4, −1) 87. −x − y + 2z = 3 5x + 8y−3z = 4 −x + 3y−5z = −5 and (4, 1, −7) the following exercises, For substitution. solve each system by 3x−4y + 2z = −15 2x + 4y + z = 16 2x + 3y + 5z = 20 88. 89. This content is available for free at https://cnx.org/content/col11758/1.5 5x−2y + 3z = 20 2x−4y−3z = −9 x + 6y−8z = 21 90. 91. 92. 5x + 2y + 4z = 9 −3x + 2y + z = 10 4x−3y + 5z = −3 4x−3y + 5z = 31 −x + 2y + 4z = 20 x + 5y−2z = −29 5x−2y + 3z = 4 −4x + 6y−7z = −1 3x + 2y − z = 4 93. 4x + 6y + 9z = 0 −5x + 2y−6z = 3 7x−4y + 3z = −3 For the following exercises, solve each system by Gaussian elimination. 2x − y + 3z = 17 −5x + 4y−2z = −46 2y + 5z = −7 5x−6y + 3z = 50 − x + 4y = 10 2x − z = 10 2x + 3y−6z = 1 −4x−6y + 12z = −2 x + 2y + 5z = 10 4x + 6y−2z = 8 6x + 9y−3z = 12 −2x−3y + z = −4 2x + 3y−4z = 5 −3x + 2y + z = 11 −x + 5y + 3z = 4 10x + 2y−14z = 8 −x−2y−4z = −1 −12x−6y + 6z = −12 94. 95. 96. 97. 98. 99. 100. Chapter 11 Systems of Equations and Inequalities 1243 x + y + z = 14 2y
+ 3z = −14 −16y−24z = −112 101. 102. 103. 104. 5x−3y + 4z = −1 −4x + 2y−3z = 0 −x + 5y + 7z = −11 x + y + z = 0 2x − y + 3z = 0 x − z = 0 3x + 2y−5z = 6 5x−4y + 3z = −12 4x + 5y−2z = 15 x + y + z = 0 2x − y + 3z = 0 x − z = 1 105. 3x − 1 2 y − z = − 1 2 4x + 106. 107. 6x−5y + 6z = 38 4x − 74 = − 13 10 z = − 7 20 z = − 5 4 108 10 2 1 2 1 4 1 8 109. 1108 z = −5 111 = − 23 3 z = 0 112 = 55 12 z = 5 3 113. 114. 115. 116. 117. 118. 119. 1 40 − 1 2 x + 3 12 x + 1 y + 1 80 60 16 3 8 z = 1 100 z = − 1 5 z = 3 20 0.1x−0.2y + 0.3z = 2 0.5x−0.1y + 0.4z = 8 0.7x−0.2y + 0.3z = 8 0.2x + 0.1y−0.3z = 0.2 0.8x + 0.4y−1.2z = 0.1 1.6x + 0.8y−2.4z = 0.2 1.1x + 0.7y−3.1z = −1.79 2.1x + 0.5y−1.6z = −0.13 0.5x + 0.4y−0.5z = −0.07 0.5x−0.5y + 0.5z = 10 0.2x−0.2y + 0.2z = 4 0.1x−0.1y + 0.1z = 2 0.1x + 0.2y + 0.3z = 0.37 0.1x−0.2y−0.3z = −0.27 0.5x−0.1y
−0.3z = −0.03 0.5x−0.5y−0.3z = 0.13 0.4x−0.1y−0.3z = 0.11 0.2x−0.8y−0.9z = −0.32 120. 0.5x + 0.2y−0.3z = 1 0.4x−0.6y + 0.7z = 0.8 0.3x−0.1y−0.9z = 0.6 1244 121. 122. 0.3x + 0.3y + 0.5z = 0.6 0.4x + 0.4y + 0.4z = 1.8 0.4x + 0.2y + 0.1z = 1.6 0.8x + 0.8y + 0.8z = 2.4 0.3x−0.5y + 0.2z = 0 0.1x + 0.2y + 0.3z = 0.6 Extensions For the following exercises, solve the system for x, y, and z. 123−1 2 x−2 3 y−−3 3 = 2 3 = 0 + + + 124. 5x−3y − z + 1 2 = 1 2 + 2z = −3 y−9 2 −4y + z = 4 6x + x + 8 2 125. 126. 127. x + 4 7 x−− 12 −−5 2 y−3 2 + − − + z−− 4x + 3y−2z = 11 0.02x + 0.015y−0.01z = 0.065 = 1 Real-World Applications Three even numbers sum up to 108. The smaller is 128. half the larger and the middle number is 3 4 the larger. What are the three numbers? Three numbers sum up to 147. The smallest number is 129. half the middle number, which is half the largest number. What are the three numbers? Chapter 11 Systems of Equations and Inequalities attendance. There were 400 people total. There were twice as many parents as grandparents, and 50 more children than parents. How many children, parents, and grandparents were in attendance? An animal shelter has a total of 350 animals 131. comprised of cats, dogs, and rabbits. If the number of rabbits is 5 less than one
-half the number of cats, and there are 20 more cats than dogs, how many of each animal are at the shelter? Your roommate, Sarah, offered to buy groceries for 132. you and your other roommate. The total bill was $82. She forgot to save the individual receipts but remembered that your groceries were $0.05 cheaper than half of her groceries, and that your other roommate’s groceries were $2.10 more than your groceries. How much was each of your share of the groceries? Your roommate, John, offered to buy household 133. supplies for you and your other roommate. You live near the border of three states, each of which has a different sales tax. The total amount of money spent was $100.75. Your supplies were bought with 5% tax, John’s with 8% tax, and your third roommate’s with 9% sales tax. The total amount of money spent without taxes is $93.50. If your supplies before tax were $1 more than half of what your third roommate’s supplies were before tax, how much did each of you spend? Give your answer both with and without taxes. Three coworkers work for the same employer. Their 134. jobs are warehouse manager, office manager, and truck driver. The sum of the annual salaries of the warehouse manager and office manager is $82,000. The office manager makes $4,000 more than the truck driver annually. The annual salaries of the warehouse manager and the truck driver total $78,000. What is the annual salary of each of the co-workers? At a carnival, $2,914.25 in receipts were taken at the 135. end of the day. The cost of a child’s ticket was $20.50, an adult ticket was $29.75, and a senior citizen ticket was $15.25. There were twice as many senior citizens as adults in attendance, and 20 more children than senior citizens. How many children, adult, and senior citizen tickets were sold? A local band sells out for their concert. They sell all 136. 1,175 tickets for a total purse of $28,112.50. The tickets were priced at $20 for student tickets, $22.50 for children, and $29 for adult tickets. If the band sold twice as many adult as children tickets, how many of each type was sold? In a bag, a child has 325 coins worth $19.50. There
137. were three types of coins: pennies, nickels, and dimes. If the bag contained the same number of nickels as dimes, how many of each type of coin was in the bag? 130. consisting of children, parents, and grandparents, At a family reunion, there were only blood relatives, in 138. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1245 production, and Saudi Arabia produced 2% more oil than Russia. What percent of the world oil production did Saudi Arabia, the United States, and Russia produce?[2] The top three sources of oil imports for the United 144. States in the same year were Saudi Arabia, Mexico, and Canada. The three top countries accounted for 47% of oil imports. The United States imported 1.8% more from Saudi Arabia than they did from Mexico, and 1.7% more from Saudi Arabia than they did from Canada. What percent of imports were from these three the United States oil countries?[3] The top three oil producers in the United States in a 145. certain year are the Gulf of Mexico, Texas, and Alaska. The three regions were responsible for 64% of the United States oil production. The Gulf of Mexico and Texas combined for 47% of oil production. Texas produced 3% more than Alaska. What percent of United States oil production came from these regions?[4] At one time, in the United States, 398 species of 146. animals were on the endangered species list. The top groups were mammals, birds, and fish, which comprised 55% of the endangered species. Birds accounted for 0.7% more than fish, and fish accounted for 1.5% more than mammals. What percent of the endangered species came from mammals, birds, and fish? Meat consumption in the United States can be broken 147. into three categories: red meat, poultry, and fish. If fish makes up 4% less than one-quarter of poultry consumption, and red meat consumption is 18.2% higher than poultry of meat consumption, what consumption?[5] percentages the are Last year, at Haven’s Pond Car Dealership, for a particular model of BMW, Jeep, and Toyota, one could purchase all three cars for a total of $140,000. This year, due to inflation, the same cars would cost $151,830. The cost of the
BMW increased by 8%, the Jeep by 5%, and the Toyota by 12%. If the price of last year’s Jeep was $7,000 less than the price of last year’s BMW, what was the price of each of the three cars last year? A recent college graduate took advantage of his 139. business education and invested in three investments immediately after graduating. He invested $80,500 into three accounts, one that paid 4% simple interest, one that % simple interest, and one that paid 21 paid 31 % simple 2 8 interest. He earned $2,670 interest at the end of one year. If the amount of the money invested in the second account was four times the amount invested in the third account, how much was invested in each account? You inherit one million dollars. You invest it all in 140. three accounts for one year. The first account pays 3% compounded annually, the second account pays 4% compounded annually, and the third account pays 2% compounded annually. After one year, you earn $34,000 in interest. If you invest four times the money into the account that pays 3% compared to 2%, how much did you invest in each account? You inherit one hundred thousand dollars. You invest 141. it all in three accounts for one year. The first account pays 4% compounded annually, the second account pays 3% compounded annually, and the third account pays 2% compounded annually. After one year, you earn $3,650 in interest. If you invest five times the money in the account that pays 4% compared to 3%, how much did you invest in each account? The top three countries in oil consumption in a certain 142. year are as follows: the United States, Japan, and China. In millions of barrels per day, the three top countries consumed 39.8% of the world’s consumed oil. The United States consumed 0.7% more than four times China’s consumption. The United States consumed 5% more than triple Japan’s consumption. What percent of the world oil consumption did the United States, Japan, and China consume?[1] The top three countries in oil production in the same 143. year are Saudi Arabia, the United States, and Russia. In millions of barrels per day, the top three countries produced 31.4% of the world’s produced oil. Saudi Arabia and the the world’s United States combined for 22.1% of 1. “Oil reserves
, production and consumption in 2001,” accessed April 6, 2014, http://scaruffi.com/politics/oil.html. 2. “Oil reserves, production and consumption in 2001,” accessed April 6, 2014, http://scaruffi.com/politics/oil.html. 3. “Oil reserves, production and consumption in 2001,” accessed April 6, 2014, http://scaruffi.com/politics/oil.html. 4. “USA: The coming global oil crisis,” accessed April 6, 2014, http://www.oilcrisis.com/us/. 5. “The United States Meat Industry at a Glance,” accessed April 6, 2014, http://www.meatami.com/ht/d/sp/i/47465/pid/ 47465. 1246 Chapter 11 Systems of Equations and Inequalities 11.3 \| Systems of Nonlinear Equations and Inequalities: Two Variables Learning Objectives In this section, you will: 11.3.1 Solve a system of nonlinear equations using substitution. 11.3.2 Solve a system of nonlinear equations using elimination. 11.3.3 Graph a nonlinear inequality. 11.3.4 Graph a system of nonlinear inequalities. Halley’s Comet (Figure 11.19) orbits the sun about once every 75 years. Its path can be considered to be a very elongated ellipse. Other comets follow similar paths in space. These orbital paths can be studied using systems of equations. These systems, however, are different from the ones we considered in the previous section because the equations are not linear. Figure 11.19 Halley’s Comet (credit: "NASA Blueshift"/Flickr) In this section, we will consider the intersection of a parabola and a line, a circle and a line, and a circle and an ellipse. The methods for solving systems of nonlinear equations are similar to those for linear equations. Solving a System of Nonlinear Equations Using Substitution A system of nonlinear equations is a system of two or more equations in two or more variables containing at least one equation that is not linear. Recall that a linear equation can take the form Ax + By + C = 0. Any equation that cannot be written in this form in nonlinear. The substitution method we used for linear systems is the same method we will
use for nonlinear systems. We solve one equation for one variable and then substitute the result into the second equation to solve for another variable, and so on. There is, however, a variation in the possible outcomes. Intersection of a Parabola and a Line There are three possible types of solutions for a system of nonlinear equations involving a parabola and a line. Possible Types of Solutions for Points of Intersection of a Parabola and a Line Figure 11.20 illustrates possible solution sets for a system of equations involving a parabola and a line. • No solution. The line will never intersect the parabola. • One solution. The line is tangent to the parabola and intersects the parabola at exactly one point. • Two solutions. The line crosses on the inside of the parabola and intersects the parabola at two points. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1247 Figure 11.20 Given a system of equations containing a line and a parabola, find the solution. 1. Solve the linear equation for one of the variables. 2. Substitute the expression obtained in step one into the parabola equation. 3. Solve for the remaining variable. 4. Check your solutions in both equations. Example 11.17 Solving a System of Nonlinear Equations Representing a Parabola and a Line Solve the system of equations. x − y = −1 y = x2 + 1 Solution Solve the first equation for x and then substitute the resulting expression into the second equation. x − y = −1 x = y−1 y = x2 + 1 y = (y−1)2 + 1 Solve for x. Substitute expression for x. Expand the equation and set it equal to zero. y = (y−1)2 = (y2 −2y + 1) + 1 = y2 −2y + 2 0 = y2 −3y + 2 = (y−2)(y−1) 1248 Chapter 11 Systems of Equations and Inequalities Solving for y gives y = 2 and y = 1. Next, substitute each value for y into the first equation to solve for x. Always substitute the value into the linear equation to check for extraneous solutions. x − y = −1 x − (2) = −
1 x = 1 x − (1) = −1 x = 0 The solutions are (1, 2) and (0, 1), which can be verified by substituting these (x, y) values into both of the original equations. See Figure 11.21. Figure 11.21 Could we have substituted values for y into the second equation to solve for x in Example 11.17? Yes, but because x is squared in the second equation this could give us extraneous solutions for x. For y = 1 This gives us the same value as in the solution. For y = 2 y = x2 + 1 y = x2 + 1 x2 = 0 x = ± 0 = 0 y = x2 + 1 2 = x2 + 1 x2 = 1 x = ± 1 = ± 1 Notice that −1 is an extraneous solution. 11.12 Solve the given system of equations by substitution. 3x − y = −2 2x2 − y = 0 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1249 Intersection of a Circle and a Line Just as with a parabola and a line, there are three possible outcomes when solving a system of equations representing a circle and a line. Possible Types of Solutions for the Points of Intersection of a Circle and a Line Figure 11.22 illustrates possible solution sets for a system of equations involving a circle and a line. • No solution. The line does not intersect the circle. • One solution. The line is tangent to the circle and intersects the circle at exactly one point. • Two solutions. The line crosses the circle and intersects it at two points. Figure 11.22 Given a system of equations containing a line and a circle, find the solution. 1. Solve the linear equation for one of the variables. 2. Substitute the expression obtained in step one into the equation for the circle. 3. Solve for the remaining variable. 4. Check your solutions in both equations. Example 11.18 Finding the Intersection of a Circle and a Line by Substitution Find the intersection of the given circle and the given line by substitution. x2 + y2 = 5 y = 3x−5 Solution One of the equations has already been solved for y. We will substitute y = 3x−5 into the equation for the circle. x2 + (3x−5)
2 = 5 x2 + 9x2 −30x + 25 = 5 10x2 −30x + 20 = 0 1250 Chapter 11 Systems of Equations and Inequalities Now, we factor and solve for x. 10(x2 − 3x + 2) = 0 10(x − 2)(x − 1) = 0 x = 2 x = 1 Substitute the two x-values into the original linear equation to solve for y. y = 3(2)−5 = 1 y = 3(1)−5 = −2 The line intersects the circle at (2, 1) and (1, −2), which can be verified by substituting these (x, y) values into both of the original equations. See Figure 11.23. Figure 11.23 11.13 Solve the system of nonlinear equations. x2 + y2 = 10 x−3y = −10 Solving a System of Nonlinear Equations Using Elimination We have seen that substitution is often the preferred method when a system of equations includes a linear equation and a nonlinear equation. However, when both equations in the system have like variables of the second degree, solving them using elimination by addition is often easier than substitution. Generally, elimination is a far simpler method when the system involves only two equations in two variables (a two-by-two system), rather than a three-by-three system, as there are fewer steps. As an example, we will investigate the possible types of solutions when solving a system of equations representing a circle and an ellipse. Possible Types of Solutions for the Points of Intersection of a Circle and an Ellipse Figure 11.24 illustrates possible solution sets for a system of equations involving a circle and an ellipse. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1251 • No solution. The circle and ellipse do not intersect. One shape is inside the other or the circle and the ellipse are a distance away from the other. • One solution. The circle and ellipse are tangent to each other, and intersect at exactly one point. • Two solutions. The circle and the ellipse intersect at two points. • Three solutions. The circle and the ellipse intersect at three points. • Four solutions. The circle and the ellipse intersect at four points. Figure 11
.24 Example 11.19 Solving a System of Nonlinear Equations Representing a Circle and an Ellipse Solve the system of nonlinear equations. x2 + y2 = 26 (1) 3x2 + 25y2 = 100 (2) Solution Let’s begin by multiplying equation (1) by −3, and adding it to equation (2). ( − 3)(x2 + y2) = ( − 3)(26) − 3x2 − 3y2 = − 78 3x2 + 25y2 = 100 22y2 = 22 After we add the two equations together, we solve for y. Substitute y = ± 1 into one of the equations and solve for x. y2 = 1 y = ± 1 = ± 1 1252 Chapter 11 Systems of Equations and Inequalities x2 + (1)2 = 26 x2 + 1 = 26 x2 = 25 x = ± 25 = ± 5 x2 + (−1)2 = 26 x2 + 1 = 26 x2 = 25 = ± 5 There are four solutions: (5, 1), (−5, 1), (5, −1), and (−5, −1). See Figure 11.25. Figure 11.25 11.14 Find the solution set for the given system of nonlinear equations. 4x2 + y2 = 13 x2 + y2 = 10 Graphing a Nonlinear Inequality All of the equations in the systems that we have encountered so far have involved equalities, but we may also encounter systems that involve inequalities. We have already learned to graph linear inequalities by graphing the corresponding equation, and then shading the region represented by the inequality symbol. Now, we will follow similar steps to graph a nonlinear inequality so that we can learn to solve systems of nonlinear inequalities. A nonlinear inequality is an inequality containing a nonlinear expression. Graphing a nonlinear inequality is much like graphing a linear inequality. Recall that when the inequality is greater than, y > a, or less than, y < a, the inequality is greater than or equal to, y ≥ a, or less than or equal to, y ≤ a, graphs will create regions in the plane, and we will test each region for a solution. If one point in the region works, the whole region works. That is the region we shade. See Figure 11.26. the graph is drawn with a dashed line. When the graph is drawn with
a solid line. The This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1253 Figure 11.26 (a) an example of y > a; (b) an example of y ≥ a; (c) an example of y < a; (d) an example of y ≤ a Given an inequality bounded by a parabola, sketch a graph. 1. Graph the parabola as if it were an equation. This is the boundary for the region that is the solution set. 2. 3. If the boundary is included in the region (the operator is ≤ or ≥ ), the parabola is graphed as a solid line. If the boundary is not included in the region (the operator is < or >), the parabola is graphed as a dashed line. 4. Test a point in one of the regions to determine whether it satisfies the inequality statement. If the statement is true, the solution set is the region including the point. If the statement is false, the solution set is the region on the other side of the boundary line. 5. Shade the region representing the solution set. Example 11.20 Graphing an Inequality for a Parabola Graph the inequality y > x2 + 1. Solution First, graph the corresponding equation y = x2 + 1. Since y > x2 + 1 has a greater than symbol, we draw the graph with a dashed line. Then we choose points to test both inside and outside the parabola. Let’s test the points (0, 2) and (2, 0). One point is clearly inside the parabola and the other point is clearly outside. 1254 Chapter 11 Systems of Equations and Inequalities y > x2 + 1 2 > (0)2 + 1 2 > 1 True 0 > (2)2 + 1 0 > 5 False The graph is shown in Figure 11.27. We can see that the solution set consists of all points inside the parabola, but not on the graph itself. Figure 11.27 Graphing a System of Nonlinear Inequalities Now that we have learned to graph nonlinear inequalities, we can learn how to graph systems of nonlinear inequalities. A system of nonlinear inequalities is a system of two or more inequalities in two or more variables containing at least one inequality that is not linear. Graphing a system
of nonlinear inequalities is similar to graphing a system of linear inequalities. The difference is that our graph may result in more shaded regions that represent a solution than we find in a system of linear inequalities. The solution to a nonlinear system of inequalities is the region of the graph where the shaded regions of the graph of each inequality overlap, or where the regions intersect, called the feasible region. Given a system of nonlinear inequalities, sketch a graph. 1. Find the intersection points by solving the corresponding system of nonlinear equations. 2. Graph the nonlinear equations. 3. Find the shaded regions of each inequality. 4. Identify the feasible region as the intersection of the shaded regions of each inequality or the set of points common to each inequality. Example 11.21 Graphing a System of Inequalities Graph the given system of inequalities. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1255 x2 − y ≤ 0 2x2 + y ≤ 12 Solution These two equations are clearly parabolas. We can find the points of intersection by the elimination process: Add both equations and the variable y will be eliminated. Then we solve for x. x2 − y = 0 2x2 + y = 12 ____________ 3x2 = 12 x2 = 4 x = ± 2 Substitute the x-values into one of the equations and solve for y. x2 − y = 0 (2) (−2) The two points of intersection are (2, 4) and (−2, 4). Notice that the equations can be rewritten as follows. x2 − y ≤ 0 x2 ≤ y y ≥ x2 2x2 + y ≤ 12 y ≤ −2x2 + 12 Graph each inequality. See Figure 11.28. The feasible region is the region between the two equations bounded by 2x2 + y ≤ 12 on the top and x2 − y ≤ 0 on the bottom. 1256 Chapter 11 Systems of Equations and Inequalities Figure 11.28 11.15 Graph the given system of inequalities. y ≥ x2 − 1 x − y ≥ − 1 Access these online resources for additional instruction and practice with nonlinear equations. • Solve a System of Nonlinear Equations Using Substitution (http://openstaxcollege.org/l/ nonlinsub) • Solve a System of Nonlinear Equations Using
Elimination (http://openstaxcollege.org/l/ nonlinelim) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1257 11.3 EXERCISES Verbal 148. Explain whether a system of two nonlinear equations can have exactly two solutions. What about exactly three? If not, explain why not. If so, give an example of such a system, in graph form, and explain why your choice gives two or three answers. 149. When graphing an inequality, explain why we only need to test one point to determine whether an entire region is the solution? 150. When you graph a system of inequalities, will there always be a feasible region? If so, explain why. If not, give an example of a graph of inequalities that does not have a feasible region. Why does it not have a feasible region? If you graph a revenue and cost function, explain how 151. to determine in what regions there is profit. 152. If you perform your break-even analysis and there is more than one solution, explain how you would determine which x-values are profit and which are not. Algebraic For the following exercises, solve the system of nonlinear equations using substitution. 153. x + y = 4 x2 + y2 = 9 154. y = x−3 x2 + y2 = 9 155. 156. y = x x2 + y2 = 9 y = − x x2 + y2 = 9 157. x = 2 x2 − y2 = 9 For the following exercises, solve the system of nonlinear equations using elimination. 4x2 −9y2 = 36 4x2 + 9y2 = 36 x2 + y2 = 25 x2 − y2 = 1 158. 159. 160. 2x2 + 4y2 = 4 2x2 −4y2 = 25x−10 161. 162. y2 − x2 = 9 3x2 + 2y2 = 8 x2 + y2 + 1 16 y = 2x2 = 2500 For the following exercises, use any method to solve the system of nonlinear equations. 163. −2x2 + y = −5 6x − y = 9 164. −x2 + y = 2 − x + y = 2 165. 166. 167. 168. 169. 170. x2 + y2
= 1 y = 20x2 −1 x2 + y2 = 1 y = − x2 2x3 − x2 = y y = 1 2 − x 9x2 + 25y2 = 225 (x−6)2 + y2 = 1 x4 − x2 = y x2 + y = 0 2x3 − x2 = y x2 + y = 0 For the following exercises, use any method to solve the nonlinear system. 171. x2 + y2 = 9 y = 3 − x2 172. x2 − y2 = 9 x = 3 1258 173. 174. 175. 176. 177. 178. 179. 180. 181. 182. 183. 184. x2 − y2 = 9 y = 3 x2 − y2 = 9 x − y = 0 −x2 + y = 2 −4x + y = −1 −x2 + y = 2 2y = − x x2 + y2 = 25 x2 − y2 = 36 x2 + y2 = 1 y2 = x2 16x2 −9y2 + 144 = 0 y2 + x2 = 16 3x2 − y2 = 12 (x−1)2 + y2 = 1 3x2 − y2 = 12 (x−1)2 + y2 = 4 3x2 − y2 = 12 x2 + y2 = 16 x2 − y2 − 6x − 4y − 11 = 0 − x2 + y2 = 5 x2 + y2 −6y = 7 x2 + y = 1 185. x2 + y2 = 6 xy = 1 Graphical For the following exercises, graph the inequality. 186. x2 + y < 9 187. x2 + y2 < 4 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities the following exercises, graph the system of For inequalities. Label all points of intersection. 188. x2 + y < 1 y > 2x 189. x2 + y < −5 y > 5x + 10 190. 191. 192. x2 + y2 < 25 3x2 − y2 > 12 x2 − y2 > −4 x2 + y2 < 12 x2 + 3y2 > 16 3x2 − y2 < 1 Extensions For the following exercises,
graph the inequality. 193. y ≥ e x y ≤ ln(x) + 5 194. y ≤ − log(x) y ≤ e x For the following exercises, find the solutions to the nonlinear equations with two variables. 195. 196. 197. 198. 4 x2 + 1 x2 − 2 5 y2 = 24 y2 + 4 = 0 6 x2 − 1 x2 − 6 y2 = 8 y2 = 1 8 1 x2 − xy + y2 −2 = 0 x + 3y = 4 x2 − xy−2y2 −6 = 0 x2 + y2 = 1 199. x2 + 4xy−2y2 −6 = 0 x = y + 2 Chapter 11 Systems of Equations and Inequalities 1259 Technology the following exercises, system of For inequalities. Use a calculator to graph the system to confirm the answer. solve the 200. xy < 1 y > x 201. x2 + y < 3 y > 2x Real-World Applications For the following exercises, construct a system of nonlinear equations to describe the given behavior, then solve for the requested solutions. Two numbers add up to 300. One number is twice the 202. square of the other number. What are the numbers? The squares of two numbers add to 360. The second 203. number is half the value of the first number squared. What are the numbers? A laptop company has discovered their cost and 204. revenue functions for each day: C(x) = 3x2 −10x + 200 and R(x) = −2x2 + 100x + 50. If they want to make a profit, what is the range of laptops per day that they should produce? Round to the nearest number which would generate profit. 205. revenue A cell phone company has the following cost and functions: C(x) = 8x2 −600x + 21,500 and R(x) = −3x2 + 480x. What is the range of cell phones they should produce each day so there is profit? Round to the nearest number that generates profit. 1260 Chapter 11 Systems of Equations and Inequalities 11.4 \| Partial Fractions Learning Objectives In this section, you will: 11.4.1 Decompose P( x ) Q( x ), where Q( x ) has only nonrepeated linear factors. 11.4.2 Decompose P( x ) Q
( x ), where Q( x ) has repeated linear factors. 11.4.3 Decompose P( x ) Q( x ), where Q( x ) has a nonrepeated irreducible quadratic factor. 11.4.4 Decompose P( x ) Q( x ), where Q( x ) has a repeated irreducible quadratic factor. Earlier in this chapter, we studied systems of two equations in two variables, systems of three equations in three variables, and nonlinear systems. Here we introduce another way that systems of equations can be utilized—the decomposition of rational expressions. Fractions can be complicated; adding a variable in the denominator makes them even more so. The methods studied in this section will help simplify the concept of a rational expression. Decomposing P(x) Q(x) Where Q(x) Has Only Nonrepeated Linear Factors Recall the algebra regarding adding and subtracting rational expressions. These operations depend on finding a common denominator so that we can write the sum or difference as a single, simplified rational expression. In this section, we will look at partial fraction decomposition, which is the undoing of the procedure to add or subtract rational expressions. In other words, it is a return from the single simplified rational expression to the original expressions, called the partial fractions. For example, suppose we add the following fractions: We would first need to find a common denominator, (x + 2)(x−3). Next, we would write each expression with this common denominator and find the sum of the terms. 2 x−3 + −1 ⎝ x + 2 x − 3 2x + 4 − x + 3 (x + 2)(x − 3) = x + 7 x2 − x − 6 Partial fraction decomposition is the reverse of this procedure. We would start with the solution and rewrite (decompose) it as the sum of two fractions. x + 7 x2 − x−6 Simplifie sum = 2 x−3 + −1 x + 2 Partial fraction decomposition We will investigate rational expressions with linear factors and quadratic factors in the denominator where the degree of the numerator is less than the degree of the denominator. Regardless of the type of expression we are decomposing, the first and most important thing to do is factor the denominator. When the denominator of the simplified expression contains distinct linear factors, it is likely that each of the
original rational expressions, which were added or subtracted, had one of the linear factors as the denominator. In other words, using the example above, the factors of x2 − x−6 are (x−3)(x + 2), the denominators of the decomposed rational expression. So we will rewrite the simplified form as the sum of individual fractions and use a variable for each numerator. Then, we will solve for each numerator using one of several methods available for partial fraction decomposition. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1261 Partial Fraction Decomposition of P(x) Q(x) : Q(x) Has Nonrepeated Linear Factors The partial fraction decomposition of P(x) Q(x) than the degree of Q(x) is when Q(x) has nonrepeated linear factors and the degree of P(x) is less P(x) Q(x) = A1 ⎝a1 x + b1 ⎛ ⎞ ⎠ + A2 ⎝a2 x + b2 ⎛ ⎞ ⎠ + A3 ⎝a3 x + b3 ⎛ ⎞ ⎠ + ⋅ ⋅ ⋅ + An ⎝an x + bn ⎛. ⎞ ⎠ Given a rational expression with distinct linear factors in the denominator, decompose it. 1. Use a variable for the original numerators, usually A, B, or C, depending on the number of factors, placing each variable over a single factor. For the purpose of this definition, we use An for each numerator A1 ⎝a1 x + b1 2. Multiply both sides of the equation by the common denominator to eliminate fractions. A2 ⎝a2 x + b2 An ⎝an x + bn P(x) Q(x. Expand the right side of the equation and collect like terms. 4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators. Example 11.22 Decomposing a Rational Function with Distinct Linear Factors Decompose the given rational expression with distinct linear factors. 3x (x +
2)(x−1) Solution We will separate the denominator factors and give each numerator a symbolic label, like A, B, or C. 3x (x + 2)(x−1) = A (x + 2) + B (x−1) Multiply both sides of the equation by the common denominator to eliminate the fractions: ⎡ (x + 2)(x−1) ⎣ 3x (x + 2)(x−1) ⎡ ⎤ ⎦ = (x + 2)(x−1) ⎣ A (x + 2) ⎤ ⎦ + (x + 2)(x−1) ⎡ ⎣ B (x−1) ⎤ ⎦ The resulting equation is Expand the right side of the equation and collect like terms. 3x = A(x−1) + B(x + 2) 3x = Ax − A + Bx + 2B 3x = (A + B)x − A + 2B Set up a system of equations associating corresponding coefficients + 2B 1262 Chapter 11 Systems of Equations and Inequalities Add the two equations and solve for B + 2B 3 = 0 + 3B 1 = B Substitute B = 1 into one of the original equations in the system. 3 = A + 1 2 = A Thus, the partial fraction decomposition is 3x (x + 2)(x−1) = 2 (x + 2) + 1 (x−1) Another method to use to solve for A or B is by considering the equation that resulted from eliminating the fractions and substituting a value for x that will make either the A- or B-term equal 0. If we let x = 1, the A- term becomes 0 and we can simply solve for B. 3x = A(x − 1) + B(x + 2) 3(1) = A[(1) − 1] + B[(1) + 2] 3 = 0 + 3B 1 = B Next, either substitute B = 1 into the equation and solve for A, or make the B-term 0 by substituting x = −2 into the equation. 3x = A(x − 1) + B(x + 2) 3( − 2) = A[( − 2) − 1] + B[( − 2) + 2] − 6 = − 3A
+ 0 = A −6 −3 2 = A We obtain the same values for A and B using either method, so the decompositions are the same using either method. 3x (x + 2)(x−1) = 2 (x + 2) + 1 (x−1) Although this method is not seen very often in textbooks, we present it here as an alternative that may make some partial fraction decompositions easier. It is known as the Heaviside method, named after Charles Heaviside, a pioneer in the study of electronics. 11.16 Find the partial fraction decomposition of the following expression. x (x−3)(x−2) Decomposing P(x) Q(x) Where Q(x) Has Repeated Linear Factors Some fractions we may come across are special cases that we can decompose into partial fractions with repeated linear factors. We must remember that we account for repeated factors by writing each factor in increasing powers. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1263 Partial Fraction Decomposition of P(x) Q(x) : Q(x) Has Repeated Linear Factors The partial fraction decomposition of P(x) Q(x) of P(x) is less than the degree of Q(x), is, when Q(x) has a repeated linear factor occurring n times and the degree P(x) Q(x) = A1 (ax + b) + A2 (ax + b)2 + A3 (ax + b)3 + ⋅ ⋅ ⋅ + An (ax + b) n Write the denominator powers in increasing order. Given a rational expression with repeated linear factors, decompose it. 1. Use a variable like A, B, or C for the numerators and account for increasing powers of the denominators. P(x) Q(x) = A1 (ax + b) + A2 (ax + b)2 +... + An (ax + b) n 2. Multiply both sides of the equation by the common denominator to eliminate fractions. 3. Expand the right side of the equation and collect like terms. 4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators. Example
11.23 Decomposing with Repeated Linear Factors Decompose the given rational expression with repeated linear factors. −x2 + 2x + 4 x3 −4x2 + 4x Solution The denominator factors are x(x−2)2. To allow for the repeated factor of (x−2), the decomposition will include three denominators: x, (x−2), and (x−2)2. Thus, −x2 + 2x + 4 x3 −4x2 + 4x = A x + B (x−2) + C (x−2)2 Next, we multiply both sides by the common denominator. x(x−2)2 ⎡ ⎢−x2 + 2x + 4 x(x−2)2 ⎣ ⎡ ⎢A x + B ⎣ − x2 + 2x + 4 = A(x−2)2 + Bx(x−2) + Cx C (x−2)2 ⎤ ⎥ = ⎦ (x−2) + ⎤ ⎥x(x−2)2 ⎦ On the right side of the equation, we expand and collect like terms. −x2 + 2x + 4 = A(x2 − 4x + 4) + B(x2 − 2x) + Cx = Ax2 − 4Ax + 4A + Bx2 − 2Bx + Cx = (A + B)x2 + ( − 4A − 2B + C)x + 4A 1264 Chapter 11 Systems of Equations and Inequalities Next, we compare the coefficients of both sides. This will give the system of equations in three variables: −x2 + 2x + 4 = (A + B)x2 + (−4A−2B + C)x + 4A A + B = −1 (1) −4A−2B + C = 2 (2) 4A = 4 (3) Solving for A, we have Substitute A = 1 into equation (1). 4A = 4 A = 1 A + B = −1 (1) + B = −1 B = −2 Then, to solve for C, substitute the values for A and B into equation (2). −4A−2B + C = 2 −4(1)−2(−2
) + C = 2 − Thus, −x2 + 2x + 4 x3 −4x2 + 4x = 1 x − 2 (x−2) + 2 (x−2)2 11.17 Find the partial fraction decomposition of the expression with repeated linear factors. 6x−11 (x−1)2, Where Q(x) Has a Nonrepeated Irreducible Decomposing P(x) Q(x) Quadratic Factor So far, we have performed partial fraction decomposition with expressions that have had linear factors in the denominator, and we applied numerators A, B, or C representing constants. Now we will look at an example where one of the factors in the denominator is a quadratic expression that does not factor. This is referred to as an irreducible quadratic factor. In cases like this, we use a linear numerator such as Ax + B, Bx + C, etc. Decomposition of P(x) Q(x) : Q(x) Has a Nonrepeated Irreducible Quadratic Factor The partial fraction decomposition of P(x) Q(x) such that Q(x) has a nonrepeated irreducible quadratic factor and the degree of P(x) is less than the degree of Q(x) is written as P(x) Q(x) = A1 x + B1 ⎝a1 x2 + b1 x + c1 ⎛ ⎞ ⎠ + A2 x + B2 ⎝a2 x2 + b2 x + c2 ⎛ ⎞ ⎠ + ⋅ ⋅ ⋅ + An x + Bn ⎛ ⎝an x2 + bn x + cn ⎞ ⎠ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1265 The decomposition may contain more rational expressions if there are linear factors. Each linear factor will have a different constant numerator: A, B, C, and so on. Given a rational expression where the factors of the denominator are distinct, irreducible quadratic factors, decompose it. 1. Use variables such as A, B, or C for the constant numerators over linear factors, and
linear expressions such as A1 x + B1, A2 x + B2, etc., for the numerators of each quadratic factor in the denominator. P(x) Q(x) = A ax + b + A1 x + B1 ⎝a1 x2 + b1 x + c1 ⎛ ⎞ ⎠ + A2 x + B2 ⎝a2 x2 + b2 x + c2 ⎛ ⎞ ⎠ + ⋅ ⋅ ⋅ + An x + Bn ⎛ ⎝an x2 + bn x + cn ⎞ ⎠ 2. Multiply both sides of the equation by the common denominator to eliminate fractions. 3. Expand the right side of the equation and collect like terms. 4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators. Example 11.24 Decomposing P(x) Q(x) Factor When Q(x) Contains a Nonrepeated Irreducible Quadratic Find a partial fraction decomposition of the given expression. 8x2 + 12x−20 ⎛ ⎞ ⎝x2 + x + 2 (x + 3) ⎠ Solution We have one linear factor and one irreducible quadratic factor in the denominator, so one numerator will be a constant and the other numerator will be a linear expression. Thus, 8x2 + 12x−20 ⎛ ⎞ ⎝x2 + x + 2 (x + 3) ⎠ = A (x + 3) + Bx + C ⎛ ⎞ ⎝x2 + x + 2 ⎠ We follow the same steps as in previous problems. First, clear the fractions by multiplying both sides of the equation by the common denominator. ⎡ ⎢ 8x2 + 12x − 20 (x + 3)(x2 + x + 2) (x + 3)(x2 + x + 2x + 3) ⎣ + Bx + C (x2 + x + 2) ⎤ ⎥(x + 3)(x2 + x + 2) ⎦ 8x2 + 12x − 20 = A(x2 +
x + 2) + (Bx + C)(x + 3) Notice we could easily solve for A by choosing a value for x that will make the Bx + C term equal 0. Let x = −3 and substitute it into the equation. 8x2 + 12x − 20 = A(x2 + x + 2) + (Bx + C)(x + 3) 8( − 3)2 + 12( − 3) − 20 = A(( − 3)2 + ( − 3) + 2) + (B( − 3) + C)(( − 3) + 3) 16 = 8A A = 2 1266 Chapter 11 Systems of Equations and Inequalities Now that we know the value of A, substitute it back into the equation. Then expand the right side and collect like terms. 8x2 + 12x−20 = 2(x2 + x + 2) + (Bx + C)(x + 3) 8x2 + 12x−20 = 2x2 + 2x + 4 + Bx2 + 3B + Cx + 3C 8x2 + 12x−20 = (2 + B)x2 + (2 + 3B + C)x + (4 + 3C) Setting the coefficients of terms on the right side equal to the coefficients of terms on the left side gives the system of equations. 2 + B = 8 (1) 2 + 3B + C = 12 (2) 4 + 3C = −20 (3) Solve for B using equation (1) and solve for C using equation (3). (1 + 3C = −20 (3) 3C = −24 C = −8 Thus, the partial fraction decomposition of the expression is 8x2 + 12x−20 ⎛ ⎞ ⎝x2 + x + 2 (x + 3) ⎠ = 2 (x + 3) + 6x−8 ⎛ ⎞ ⎝x2 + x + 2 ⎠ Could we have just set up a system of equations to solve Example 11.24? Yes, we could have solved it by setting up a system of equations without solving for A first. The expansion on the right would be: 8x2 + 12x−20 = Ax2 + Ax + 2A + Bx2 + 3B + Cx + 3C 8x2 + 12x−20 =
(A + B)x2 + (A + 3B + C)x + (2A + 3C) So the system of equations would be: A + B = 8 A + 3B + C = 12 2A + 3C = −20 Find the partial fraction decomposition of the expression with a nonrepeating irreducible quadratic 11.18 factor. 5x2 −6x + 7 ⎞ ⎛ ⎝x2 + 1 (x−1) ⎠ Decomposing P(x) Q(x) When Q(x) Has a Repeated Irreducible Quadratic Factor Now that we can decompose a simplified rational expression with an irreducible quadratic factor, we will learn how to do partial fraction decomposition when the simplified rational expression has repeated irreducible quadratic factors. The This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1267 decomposition will consist of partial fractions with linear numerators over each irreducible quadratic factor represented in increasing powers. Decomposition of P(x) Q(x) When Q(x) Has a Repeated Irreducible Quadratic Factor The partial fraction decomposition of P(x) Q(x) P(x) is less than the degree of Q(x), is, when Q(x) has a repeated irreducible quadratic factor and the degree of P(x) ⎝ax2 + bx + c⎞ ⎛ ⎠ n = A1 x + B1 ⎝ax2 + bx + c⎞ ⎛ ⎠ + A2 x + B2 ⎝ax2 + bx + c⎞ ⎛ ⎠ 2 + A3 x + B3 ⎝ax2 + bx + c⎞ ⎛ ⎠ 3 + ⋅ ⋅ ⋅ + An x + Bn ⎝ax2 + bx + c⎞ ⎛ ⎠ n Write the denominators in increasing powers. Given a rational expression that has a repeated irreducible factor, decompose it. 1. Use variables like A, B, or C for the constant numerators over linear
factors, and linear expressions such as A1 x + B1, A2 x + B2, etc., for the numerators of each quadratic factor in the denominator written in increasing powers, such as P(x) Q(x) = A ax + b + A1 x + B1 (ax2 + bx + c) + A2 x + B2 (ax2 + bx + c)2 + ⋯ + An + Bn (ax2 + bx + c) n 2. Multiply both sides of the equation by the common denominator to eliminate fractions. 3. Expand the right side of the equation and collect like terms. 4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators. Example 11.25 Decomposing a Rational Function with a Repeated Irreducible Quadratic Factor in the Denominator Decompose the given expression that has a repeated irreducible factor in the denominator. x4 + x3 + x2 − x + 1 ⎞ x⎛ ⎝x2 + 1 ⎠ 2 Solution The factors of the denominator are x, (x2 + 1), and (x2 + 1)2. Recall that, when a factor in the denominator is a quadratic that includes at least two terms, the numerator must be of the linear form Ax + B. So, let’s begin the decomposition. x4 + x3 + x2 − x + 1 ⎞ x⎛ ⎝x2 + 1 ⎠ 2 = A x + Bx + C ⎞ ⎛ ⎝x2 + 1 ⎠ + Dx + E 2 ⎞ ⎛ ⎝x2 + 1 ⎠ We eliminate the denominators by multiplying each term by x⎛ ⎞ ⎝x2 + 1 ⎠ 2. Thus, 1268 Chapter 11 Systems of Equations and Inequalities x4 + x3 + x2 − x + 1 = A⎛ ⎞ ⎝x2 + 1 ⎠ 2 ⎞ ⎛ ⎝x2 + 1 ⎠ + (Dx + E)(x) + (Bx + C)(x) Expand the right
side. x4 + x3 + x2 − x + 1 = A(x4 + 2x2 + 1) + Bx4 + Bx2 + Cx3 + Cx + Dx2 + Ex = Ax4 + 2Ax2 + A + Bx4 + Bx2 + Cx3 + Cx + Dx2 + Ex Now we will collect like terms. x4 + x3 + x2 − x + 1 = (A + B)x4 + (C)x3 + (2A + B + D)x2 + (C + E)x + A Set up the system of equations matching corresponding coefficients on each side of the equal sign. A + B = 1 C = 1 2A + 1 A = 1 We can use substitution from this point. Substitute A = 1 into the first equation. Substitute A = 1 and B = 0 into the third equation. 1 + B = 1 B = 0 Substitute C = 1 into the fourth equation. 2(11 1 + E = −1 E = −2 Now we have solved for all of the unknowns on the right side of the equal sign. We have A = 1, B = 0, C = 1, D = −1, and E = −2. We can write the decomposition as follows: x4 + x3 + x2 − x + 1 x⎛ ⎞ ⎝x2 + x2 + 1 ⎠ − x + 2 ⎛ ⎞ ⎝x2 + 1 ⎠ 2 11.19 Find the partial fraction decomposition of the expression with a repeated irreducible quadratic factor. x3 −4x2 + 9x−5 2 ⎞ ⎛ ⎝x2 −2x + 3 ⎠ Access these online resources for additional instruction and practice with partial fractions. • Partial Fraction Decomposition (http://openstaxcollege.org/l/partdecomp) • Partial Fraction Decomposition With Repeated Linear Factors (http://openstaxcollege.org/ l/partdecomprlf) • Partial Fraction Decomposition With Linear and Quadratic Factors (http://openstaxcollege.org/l/partdecomlqu) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of
Equations and Inequalities 1269 11.4 EXERCISES Verbal 206. Can any quotient of polynomials be decomposed into at least two partial fractions? If so, explain why, and if not, give an example of such a fraction Can you explain why a partial fraction decomposition 207. is unique? (Hint: Think about it as a system of equations.) Can you explain how to verify a partial fraction 208. decomposition graphically? 209. You are unsure if you correctly decomposed the partial fraction correctly. Explain how you could doublecheck your answer. 220. 221. 222. 223. 224. 6x x2 −4 2x−3 x2 −6x + 5 4x−1 x2 − x−6 4x + 3 x2 + 8x + 15 3x−1 x2 −5x + 6 For the following exercises, find the decomposition of the partial fraction for the repeating linear factors. eventually 225. −5x−19 (x + 4)2 210. Once you have a system of equations generated by the partial fraction decomposition, can you explain another had method solve you if to 7x + 13 3x2 + 8x + 15 simplify it? For + B example, we 3x + 5 = A x + 1 to 7x + 13 = A(3x + 5) + B(x + 1). Explain how you could intelligently choose an x -value that will eliminate either A or B and solve for A and B. Algebraic For the following exercises, find the decomposition of the partial fraction for the nonrepeating linear factors. 211. 212. 213. 214. 215. 216. 217. 218. 219. 5x + 16 x2 + 10x + 24 3x−79 x2 −5x−24 −x−24 x2 −2x−24 10x + 47 x2 + 7x + 10 x 6x2 + 25x + 25 32x−11 20x2 −13x + 2 x + 1 x2 + 7x + 10 5x x2 −9 10x x2 −25 226. x (x−2)2 227. 7x + 14 (x + 3)2 228. −24x−27 (4x + 5)2 229. −24x−27 (6x−7)2 230. 5 − x (x−7)2 231. 232. 233.
234. 235. 5x + 14 2x2 + 12x + 18 5x2 + 20x + 8 2x(x + 1)2 4x2 + 55x + 25 5x(3x + 5)2 54x3 + 127x2 + 80x + 16 2x2 (3x + 2)2 x3 −5x2 + 12x + 144 ⎞ x2 ⎛ ⎝x2 + 12x + 36 ⎠ 1270 Chapter 11 Systems of Equations and Inequalities For the following exercises, find the decomposition of the partial fraction for the irreducible nonrepeating quadratic factor. x3 + 6x2 + 5x + 9 ⎞ ⎛ ⎝x2 + 1 ⎠ 2 236. 237. 238. 239. 240. 241. 242. 243. 244. 245. 246. 247. 248. 4x2 + 6x + 11 ⎞ ⎛ ⎝x2 + x + 3 (x + 2) ⎠ 4x2 + 9x + 23 ⎞ ⎛ ⎝x2 + 6x + 11 (x−1) ⎠ −2x2 + 10x + 4 ⎛ ⎞ ⎝x2 + 3x + 8 (x−1) ⎠ x2 + 3x + 1 ⎞ ⎛ ⎝x2 + 5x−2 (x + 1) ⎠ 4x2 + 17x−1 ⎛ ⎞ ⎝x2 + 6x + 1 (x + 3) ⎠ 4x2 ⎛ ⎞ ⎝x2 + 7x−5 (x + 5) ⎠ 4x2 + 5x + 3 x3 −1 −5x2 + 18x−4 x3 + 8 3x2 −7x + 33 x3 + 27 x2 + 2x + 40 x3 −125 4x2 + 4x + 12 8x3 −27 −50x2 + 5x−3 125x3 −1 −2x3 −30x2 + 36x + 216 x4 + 216x For the following exercises, find the decomposition of the partial fraction for the irreducible repeating quadratic factor. 3x3 + 2x2 + 14x +
15 ⎞ ⎛ ⎝x2 + 4 ⎠ 2 249. 250. This content is available for free at https://cnx.org/content/col11758/1.5 251. 252. 253. 254. 255. x3 − x2 + x−1 ⎞ ⎛ ⎝x2 −3 ⎠ 2 x2 + 5x + 5 (x + 2)2 x3 + 2x2 + 4x 2 ⎞ ⎛ ⎝x2 + 2x + 9 ⎠ x2 + 25 ⎞ ⎛ ⎝x2 + 3x + 25 ⎠ 2 2x3 + 11x + 7x + 70 ⎛ ⎞ ⎝2x2 + x + 14 ⎠ 2 256. 5x + 2 x⎛ ⎞ ⎝x2 + 4 ⎠ 2 257. x4 + x3 + 8x2 + 6x + 36 x⎛ ⎞ ⎝x2 + 6 ⎠ 2 258. 2x−9 ⎛ ⎝x2 − x⎞ ⎠ 2 259. 5x3 −2x + 1 2 ⎛ ⎝x2 + 2x⎞ ⎠ Extensions the following exercises, For expansion. find the partial fraction 260. 261. x2 + 4 (x + 1)3 x3 −4x2 + 5x + 4 (x−2)3 For the following exercises, perform the operation and then find the partial fraction decomposition. 7 x + 8 + 5 x−2 − x−1 x2 −6x−16 262. 263. Chapter 11 Systems of Equations and Inequalities 1271 1 x−4 − 3 x + 6 − 2x + 7 x2 + 2x−24 264. 2x x2 −16 − 1−2x x2 + 6x + 8 − x−5 x2 −4x 1272 Chapter 11 Systems of Equations and Inequalities 11.5 \| Matrices and Matrix Operations Learning Objectives In this section, you will: 11.5.1 Find the sum and difference of two matrices. 11.5.2 Find scalar multiples of a matrix. 11.5.3 Find the product of two matrices. Figure 11.29 (
credit: “SD Dirk,” Flickr) Two club soccer teams, the Wildcats and the Mud Cats, are hoping to obtain new equipment for an upcoming season. Table 11.1 shows the needs of both teams. Wildcats Mud Cats Goals Balls Jerseys Table 11.1 6 30 14 10 24 20 A goal costs $300; a ball costs $10; and a jersey costs $30. How can we find the total cost for the equipment needed for each team? In this section, we discover a method in which the data in the soccer equipment table can be displayed and used for calculating other information. Then, we will be able to calculate the cost of the equipment. Finding the Sum and Difference of Two Matrices To solve a problem like the one described for the soccer teams, we can use a matrix, which is a rectangular array of numbers. A row in a matrix is a set of numbers that are aligned horizontally. A column in a matrix is a set of numbers that are aligned This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1273 vertically. Each number is an entry, sometimes called an element, of the matrix. Matrices (plural) are enclosed in [ ] or ( ), and are usually named with capital letters. For example, three matrices named A, B, and C are shown below. A = ⎡ 1 2 ⎣ 3 4 ⎤ ⎦, B = ⎡ 1 2 7 ⎢ 0 −5 6 ⎣ 7 8 2 ⎤ ⎥, C = ⎦ ⎡ − Describing Matrices A matrix is often referred to by its size or dimensions: m × n indicating m rows and n columns. Matrix entries are defined first by row and then by column. For example, to locate the entry in matrix A identified as ai j, we look for the entry in row i, column j. In matrix A, shown below, the entry in row 2, column 3 is a23. A = a11 a12 a13 ⎡ a21 a22 a23 ⎢ ⎣ a31 a32 a33 ⎤ ⎥ ⎦ A square matrix is a matrix with dimensions n × n, meaning that it has the same number of rows as columns. The 3×3 matrix above is an example of a square matrix. A row
matrix is a matrix consisting of one row with dimensions 1 × n. [a11 a12 a13] A column matrix is a matrix consisting of one column with dimensions m × 1. a11 ⎡ a21 ⎢ ⎣ a31 ⎤ ⎥ ⎦ A matrix may be used to represent a system of equations. In these cases, the numbers represent the coefficients of the variables in the system. Matrices often make solving systems of equations easier because they are not encumbered with variables. We will investigate this idea further in the next section, but first we will look at basic matrix operations. Matrices A matrix is a rectangular array of numbers that is usually named by a capital letter: A, B, C, and so on. Each entry in a matrix is referred to as ai j, such that i represents the row and j represents the column. Matrices are often referred to by their dimensions: m × n indicating m rows and n columns. Example 11.26 Finding the Dimensions of the Given Matrix and Locating Entries Given matrix A : a. What are the dimensions of matrix A? b. What are the entries at a31 and a22? 2 ⎤ ⎥ ⎦ Solution a. The dimensions are 3 × 3 because there are three rows and three columns. 1274 Chapter 11 Systems of Equations and Inequalities b. Entry a31 is the number at row 3, column 1, which is 3. The entry a22 is the number at row 2, column 2, which is 4. Remember, the row comes first, then the column. Adding and Subtracting Matrices We use matrices to list data or to represent systems. Because the entries are numbers, we can perform operations on matrices. We add or subtract matrices by adding or subtracting corresponding entries. In order to do this, the entries must correspond. Therefore, addition and subtraction of matrices is only possible when the matrices have the same dimensions. We can add or subtract a 3 × 3 matrix and another 3 × 3 matrix, but we cannot add or subtract a 2 × 3 matrix and a 3 × 3 matrix because some entries in one matrix will not have a corresponding entry in the other matrix. Adding and Subtracting Matrices Given matrices A and B of like dimensions, addition and subtraction of A and B will produce matrix C or matrix D of the same dimension. Matrix addition is commutative.
It is also associative. A + B = C such that ai j + bi j = ci j A − B = D such that ai j − bi j = di j A + B = B + A (A + B) + C = A + (B + C) Example 11.27 Finding the Sum of Matrices Find the sum of A and B, given Solution Add corresponding entries. A = a b ⎡ ⎣ c d ⎤ ⎦ and ⎤ ⎦ ⎤ ⎦ Example 11.28 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1275 Adding Matrix A and Matrix B Find the sum of A and B. A = ⎡ 4 1 ⎣ 3 2 ⎤ ⎦ and B = ⎡ 5 9 ⎣ 0 7 ⎤ ⎦ Solution Add corresponding entries. Add the entry in row 1, column 1, a11, of matrix A to the entry in row 1, column 1, b11, of B. Continue the pattern until all entries have been added 10 ⎦ ⎣ 3 9 Example 11.29 Finding the Difference of Two Matrices Find the difference of A and B. A = ⎡ −2 3 ⎣ 0 1 ⎤ ⎦ and B = ⎡ 8 1 ⎣ 5 4 ⎤ ⎦ Solution We subtract the corresponding entries of each matrix10 ⎦ ⎣ −5 −3 Example 11.30 Finding the Sum and Difference of Two 3 x 3 Matrices Given A and B : a. Find the sum. b. Find the difference. A = ⎤ ⎡ 2 −10 −2 ⎥ and B = ⎢ 12 10 14 ⎦ ⎣ 2 4 −2 ⎡ ⎢ ⎣ −5 ⎤ 10 −2 6 ⎥ 0 −12 −4 ⎦ 2 −2 1276 Chapter 11 Systems of Equations and Inequalities Solution a. Add the corresponding entries5 10 − 2 2 − 10 − − 12 − 4 10 12 14 ⎦ ⎣ ⎦ − 10 + 10 − 2 − 2 ⎤ ⎡ ⎥ ⎢ 10 − 4 12 − 12 14 + 14 0
⎣ ⎦ 0 −1 0 b. Subtract the corresponding entries5 ⎤ ⎤ ⎡ 10 −2 6 2 −10 −2 ⎥ ⎥ − ⎢ 0 −12 −4 12 10 14 ⎦ ⎣ ⎦ 2 4 −2 2 −2 2 − 6 −10 − 10 −2 + 2 ⎤ ⎡ ⎥ ⎢ 12 + 12 10 + 4 14 − 0 ⎣ ⎦ −4 −20 0 ⎤ ⎡ ⎥ ⎢ 24 14 14 ⎦ ⎣ −4 4 9 11.20 Add matrix A and matrix B. A = ⎡ ⎤ 6 2 ⎢ ⎥ and B = 0 1 ⎣ ⎦ 1 −3 ⎤ ⎡ 3 −2 ⎥ ⎢ 5 1 ⎦ ⎣ 3 −4 Finding Scalar Multiples of a Matrix Besides adding and subtracting whole matrices, there are many situations in which we need to multiply a matrix by a constant called a scalar. Recall that a scalar is a real number quantity that has magnitude, but not direction. For example, time, temperature, and distance are scalar quantities. The process of scalar multiplication involves multiplying each entry in a matrix by a scalar. A scalar multiple is any entry of a matrix that results from scalar multiplication. Consider a real-world scenario in which a university needs to add to its inventory of computers, computer tables, and chairs in two of the campus labs due to increased enrollment. They estimate that 15% more equipment is needed in both labs. The school’s current inventory is displayed in Table 11.2. Lab A Lab B 15 16 16 27 34 34 Computers Computer Tables Chairs Table 11.2 Converting the data to a matrix, we have This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1277 C2013 = ⎡ 15 ⎢ 16 ⎣ 16 ⎤ ⎥ ⎦ 27 34 34 To calculate how much computer equipment will be needed, we multiply all entries in matrix C by 0.15. (0.15)C2013 = ⎡ (0.15)15 ⎢ (0.15)
16 ⎢ ⎣ (0.15)16 (0.15)27 (0.15)34 (0.15)34 ⎤ ⎥ ⎥ ⎦ = ⎡ 2.25 ⎢ 2.4 ⎣ 2.4 4.05 5.1 5.1 ⎤ ⎥ ⎦ We must round up to the next integer, so the amount of new equipment needed is Adding the two matrices as shown below, we see the new inventory amounts This means ⎡ 15 ⎢ 16 ⎣ 16 ⎤ ⎥ + ⎦ 27 34 34 ⎡ ⎡ 18 ⎢ 19 ⎣ 19 ⎤ ⎥ ⎦ 32 40 40 C2014 = ⎡ 18 ⎢ 19 ⎣ 19 ⎤ ⎥ ⎦ 32 40 40 Thus, Lab A will have 18 computers, 19 computer tables, and 19 chairs; Lab B will have 32 computers, 40 computer tables, and 40 chairs. Scalar Multiplication Scalar multiplication involves finding the product of a constant by each entry in the matrix. Given the scalar multiple cA is A = a11 ⎡ ⎣ a21 a12 a22 ⎤ ⎦ a11 a12 ⎤ ⎡ cA = c a21 a22 ⎦ ⎣ ca12 ca11 ⎤ ⎡ ca22 ca21 ⎦ ⎣ = Scalar multiplication is distributive. For the matrices A, B, and C with scalars a and b, a(A + B) = aA + aB (a + b)A = aA + bA Example 11.31 Multiplying the Matrix by a Scalar Multiply matrix A by the scalar 3. A = ⎡ 8 1 ⎣ 5 4 ⎤ ⎦ 1278 Chapter 11 Systems of Equations and Inequalities Solution Multiply each entry in A by the scalar 3. 3A = 24 ⎣ ⎦ 15 12 = = 11.21 Given matrix B, find −2B where B = ⎡ 4 1 ⎣ 3 2 ⎤ ⎦ Example 11.32 Finding the Sum of Scalar Multiples Find the sum 3A + 2B. A = ⎡ 1 −2
⎢ 0 −1 ⎣ 4 ⎤ 0 ⎥ and B = 2 ⎦ 3 −6 Solution First, find 3A, then 2B. ⎡ −1 ⎢ ⎣ 2 0 −3 0 ⎤ 1 ⎥ 2 ⎦ 1 −4 3A = = 3 ⋅ 1 3(−2(−1) 3 ⋅ 2 ⎥ ⎢ 3(−66 ⎡ ⎤ ⎢ ⎥ 0 −3 6 ⎣ ⎦ 9 −18 12 2B = = 2(−12 ⎢ ⎣ 2 ⋅ 1 ⎤ ⎥ 2(−3) 2 ⋅ 2 ⎥ 2 ⋅ 1 2(−4) ⎦ ⎤ 2 4 ⎥ 0 −6 4 ⎦ 2 −8 0 Now, add 3A + 2B. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1279 3A + 2B = = = ⎡ −2 ⎢ ⎣ ⎤ 4 2 ⎥ 0 −6 4 ⎦ 0 2 −18−8 ⎤ ⎡ 0 3 −6 ⎥ + ⎢ 6 0 −3 ⎣ ⎦ 12 9 −18 3 − 2 −6 + 4 ⎡ ⎢ 0 + 0 −3 − 6 ⎣ 12 + 0 1 −2 ⎡ 2 ⎢ 0 −9 10 ⎣ 12 11 −26 ⎤ ⎥ ⎦ Finding the Product of Two Matrices In addition to multiplying a matrix by a scalar, we can multiply two matrices. Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. If A is an m × r matrix and B is an r × n matrix, then the product matrix AB is an m × n matrix. For example, the product AB is possible because the number of columns in A is the same as the number of rows in B. If the inner dimensions do not match, the product is not defined. We multiply entries of A with entries of B according
to a specific pattern as outlined below. The process of matrix multiplication becomes clearer when working a problem with real numbers. To obtain the entries in row i of AB, we multiply the entries in row i of A by column j in B and add. For example, the product of AB given matrices A and B, where the dimensions of A are 2 × 3 and the dimensions of B are 3 × 3, will be a 2 × 3 matrix. A = a11 a12 a13 ⎡ a21 a22 a23 ⎣ ⎤ ⎦ and B = b11 b12 b13 ⎡ ⎢ b21 b22 b23 ⎢ b31 b32 b33 ⎣ ⎤ ⎥ ⎥ ⎦ Multiply and add as follows to obtain the first entry of the product matrix AB. 1. To obtain the entry in row 1, column 1 of AB, multiply the first row in A by the first column in B, and add. [a11 a12 a13] ⋅ b11 ⎡ ⎢ b21 ⎢ b31 ⎣ ⎤ ⎥ ⎥ ⎦ = a11 ⋅ b11 + a12 ⋅ b21 + a13 ⋅ b31 2. To obtain the entry in row 1, column 2 of AB, multiply the first row of A by the second column in B, and add. [a11 a12 a13] ⋅ b12 ⎡ ⎢ b22 ⎢ b32 ⎣ ⎤ ⎥ ⎥ ⎦ = a11 ⋅ b12 + a12 ⋅ b22 + a13 ⋅ b32 3. To obtain the entry in row 1, column 3 of AB, multiply the first row of A by the third column in B, and add. [a11 a12 a13] ⋅ b13 ⎡ ⎢ b23 ⎢ b33 ⎣ ⎤ ⎥ ⎥ ⎦ = a11 ⋅ b13 + a12 ⋅ b23 + a13 ⋅ b33 1280 Chapter 11 Systems of Equations and Inequalities We proceed the same way to obtain the second row of AB. In other words, row 2 of A times column 1 of B; row 2 of A times
column 2 of B; row 2 of A times column 3 of B. When complete, the product matrix will be AB = a11 ⋅ b11 + a12 ⋅ b21 + a13 ⋅ b31 ⎡ ⎢ a21 ⋅ b11 + a22 ⋅ b21 + a23 ⋅ b31 ⎣ a11 ⋅ b12 + a12 ⋅ b22 + a13 ⋅ b32 a21 ⋅ b12 + a22 ⋅ b22 + a23 ⋅ b32 a11 ⋅ b13 + a12 ⋅ b23 + a13 ⋅ b33 a21 ⋅ b13 + a22 ⋅ b23 + a23 ⋅ b33 ⎤ ⎥ ⎦ Properties of Matrix Multiplication For the matrices A, B, and C the following properties hold. • Matrix multiplication is associative: (AB)C = A(BC). • Matrix multiplication is distributive: C(A + B) = CA + CB, (A + B)C = AC + BC. Note that matrix multiplication is not commutative. Example 11.33 Multiplying Two Matrices Multiply matrix A and matrix B. A = ⎡ 1 2 ⎣ 3 4 ⎤ ⎦ and B = ⎡ 5 6 ⎣ 7 8 ⎤ ⎦ Solution First, we check the dimensions of the matrices. Matrix A has dimensions 2 × 2 and matrix B has dimensions 2 × 2. The inner dimensions are the same so we can perform the multiplication. The product will have the dimensions 2 × 2. We perform the operations outlined previously. Example 11.34 Multiplying Two Matrices Given A and B : a. Find AB. b. Find BA. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1281 A = ⎡ −1 2 3 ⎣ 4 0 5 ⎤ ⎦ and B = 5 ⎡ ⎢ −4 ⎣ 2 ⎤ −1 ⎥ 0 ⎦ 3 Solution a. As the dimensions of A are 2×3 and the dimensions of B are 3×2, these matrices can be multiplied together because
the number of columns in A matches the number of rows in B. The resulting product will be a 2×2 matrix, the number of rows in A by the number of columns in B. AB = ⎡ −4 ⎣ 2 ⎤ 5 −1 ⎥ 0 ⎦ 3 = = 4(5) + 0(−4) + 5(2) −1(5) + 2(−4) + 3(2) −1(−1) + 2(0) + 3(3) ⎡ ⎤ ⎣ ⎦ 4(−1) + 0(0) + 5(3) ⎡ −7 10 ⎣ 30 11 ⎤ ⎦ b. The dimensions of B are 3×2 and the dimensions of A are 2×3. The inner dimensions match so the product is defined and will be a 3×3 matrix. BA = = = ⎤ ⎦ ⎡ −1 ⎥ 0 ⎦ 3 ⎡ ⎢ −4 ⎣ 2 5(−1) + −1(4) 5(2) + −1(0) 5(3) + −1(5) ⎡ ⎤ ⎢ ⎥ −4(−1) + 0(4) −4(2) + 0(0) −4(3) + 0(5) ⎢ ⎥ 2(3) + 3(5) 2(2) + 3(0) ⎣ ⎦ ⎡ −9 10 ⎢ ⎣ 2(−1) + 3(4) ⎤ 10 ⎥ 4 −8 −12 ⎦ 21 4 10 Analysis Notice that the products AB and BA are not equal. AB = ⎡ −7 10 ⎣ 30 11 ⎤ ⎦ ≠ ⎤ ⎡ −9 10 10 ⎥ = BA ⎢ 4 −8 −12 ⎦ ⎣ 21 10 4 This illustrates the fact that matrix multiplication is not commutative. Is it possible for AB to be defined but not BA? Yes, consider a matrix A with dimension 3 × 4 and matrix B with dimension 4 × 2. For the product AB the inner dimensions are 4 and the product is defined, but for the product BA the inner dimensions are 2 and
3 so the product is undefined. Example 11.35 Using Matrices in Real-World Problems Let’s return to the problem presented at the opening of this section. We have Table 11.3, representing the equipment needs of two soccer teams. 1282 Chapter 11 Systems of Equations and Inequalities Wildcats Mud Cats 6 30 14 Goals Balls Jerseys Table 11.3 10 24 20 We are also given the prices of the equipment, as shown in Table 11.4. Goal $300 Ball $10 Jersey $30 Table 11.4 We will convert the data to matrices. Thus, the equipment need matrix is written as The cost matrix is written as E = ⎡ 6 ⎢ 30 ⎣ 14 ⎤ ⎥ ⎦ 10 24 20 C = [300 10 30] We perform matrix multiplication to obtain costs for the equipment. CE = [300 10 30] ⋅ ⎡ 6 10 ⎢ 30 24 ⎣ 14 20 ⎤ ⎥ ⎦ = ⎡ = ⎡ ⎣300(6) + 10(30) + 30(14) 300(10) + 10(24) + 30(20)⎤ ⎣2,520 3,840⎤ ⎦ ⎦ The total cost for equipment for the Wildcats is $2,520, and the total cost for equipment for the Mud Cats is $3,840. Given a matrix operation, evaluate using a calculator. 1. Save each matrix as a matrix variable [A], [B], [C],... 2. Enter the operation into the calculator, calling up each matrix variable as needed. 3. If the operation is defined, the calculator will present the solution matrix; if the operation is undefined, it will display an error message. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1283 Example 11.36 Using a Calculator to Perform Matrix Operations Find AB − C given A = ⎡ −15 25 ⎢ ⎣ ⎤ 32 ⎥, B = 41 −7 −28 ⎦ 10 34 −2 ⎡ 45 ⎢ −24 ⎣ ⎤ 21 −37 ⎥, and C = 52 19 ⎦ 6 −48 −31 ⎤ ⎡ −100 −89 −
98 ⎥. ⎢ 74 ⎦ ⎣ 42 −75 25 −56 −67 Solution On the matrix page of the calculator, we enter matrix A above as the matrix variable [A], matrix B above as the matrix variable [B], and matrix C above as the matrix variable [C]. On the home screen of the calculator, we type in the problem and call up each matrix variable as needed. The calculator gives us the following matrix. [A]×[B] − [C] 136 −983 − 462 ⎡ ⎤ 1, 820 1, 897 − 856 ⎢ ⎥ ⎣ ⎦ 413 −311 2, 032 Access these online resources for additional instruction and practice with matrices and matrix operations. • Dimensions of a Matrix (http://openstaxcollege.org/l/matrixdimen) • Matrix Addition and Subtraction (http://openstaxcollege.org/l/matrixaddsub) • Matrix Operations (http://openstaxcollege.org/l/matrixoper) • Matrix Multiplication (http://openstaxcollege.org/l/matrixmult) 1284 Chapter 11 Systems of Equations and Inequalities 11.5 EXERCISES Verbal Can we add any two matrices together? If so, explain 265. why; if not, explain why not and give an example of two matrices that cannot be added together. Can we multiply any column matrix by any row 266. matrix? Explain why or why not. Can both the products AB and BA be defined? If so, 267. explain how; if not, explain why. 268. Can any two matrices of the same size be multiplied? If so, explain why, and if not, explain why not and give an example of two matrices of the same size that cannot be multiplied together. Does matrix multiplication commute? That is, does 269. AB = BA? If so, prove why it does. If not, explain why it does not. Algebraic For the following exercises, use the matrices below and perform the matrix addition or subtraction. Indicate if the operation is undefined. A = ⎤ ⎡ 1 3 ⎦, B = ⎣ 0 7 ⎤ ⎡ 2 14 ⎦, C = ⎣ 22 6 ⎤ ⎡ 1 5 ⎥,
D = ⎢ 8 92 ⎦ ⎣ 12 6 ⎤ ⎡ 10 14 ⎥, E = ⎢ 7 2 ⎦ ⎣ 5 61 ⎤ ⎡ 6 12 ⎦, F = ⎣ 14 5 ⎡ ⎤ 0 9 ⎢ ⎥ 78 17 ⎣ ⎦ 15 4 270. A + B 271. C + D 272. A + C 273. B − E 274. C + F 275. D − B For the following exercises, use the matrices below to perform scalar multiplication. A = ⎤ ⎡ 4 6 ⎦, B = ⎣ 13 12 ⎡ 3 9 ⎢ 21 12 ⎣ 0 64 ⎤ ⎥, C = ⎦ ⎤ ⎡ 16 3 7 18 ⎦, D = ⎣ 90 5 3 29 ⎡ ⎤ 18 12 13 ⎢ ⎥ 8 14 6 ⎣ ⎦ 7 4 21 276. 5A 277. 3B 278. −2B 279. −4C 280. C 1 2 This content is available for free at https://cnx.org/content/col11758/1.5 281. 100D For the following exercises, use the matrices below to perform matrix multiplication. A = ⎤ ⎡ −1 5 ⎦, 8 0 12 ⎤ ⎦, C = ⎡ 4 10 ⎢ −2 6 ⎣ 9 5 ⎤ ⎥, D = ⎦ ⎤ ⎡ 12 2 −10 0 282. AB 283. BC 284. CA 285. BD 286. DC 287. CB For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. A = ⎤ ⎡ 2 −5 ⎦, B = ⎣ 7 6 ⎡ ⎤ −9 6 ⎦, C = ⎣ −4 2 ⎡ ⎤ 0 9 ⎦, D = ⎣ 7 1 ⎤ ⎡ −8 7 −5 ⎥, 6 −5 ⎦ ⎣ 9 0 1 288. A + B − C 289. 4A + 5D
290. 2C + B 291. 3D + 4E 292. C−0.5D 293. 100D−10E For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. (Hint: A2 = A ⋅ A ) A = ⎡ −10 20 ⎣ 5 25 ⎤ ⎦, B = ⎤ ⎡ 40 10 ⎦, C = ⎣ −20 30 ⎡ −1 ⎢ ⎣ ⎤ 0 ⎥ 0 −1 ⎦ 0 1 294. AB 295. BA 296. CA 297. BC Chapter 11 Systems of Equations and Inequalities 1285 298. A2 299. B2 300. C 2 301. B2 A2 302. A2 B2 303. (AB)2 304. (BA)2 For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. (Hint: A2 = A ⋅ A ) 317. BC 318. ABC Extensions For the following exercises, use the matrix below to perform the indicated operation on the given matrix ⎤ ⎥ ⎦ 319. 320. 321. 322. B2 B3 B4 B5 A = ⎤ ⎡ 1 0 ⎦, B = ⎣ 2 3 ⎤ ⎡ 4 −2 3 ⎦, C = ⎣ −1 1 −5 ⎤ ⎡ 0.5 0.1 ⎥, D = ⎢ 1 0.2 ⎦ ⎣ −0.5 0.3 ⎡ ⎢ −6 7 ⎣ 4 2 ⎤ 1 0 −1 ⎥ 5 ⎦ 1 Using the above questions, find a formula for Bn 323. Test the formula for B201 and B202, using a calculator.. 305. AB 306. BA 307. BD 308. DC 309. D2 310. A2 311. D3 312. (AB)C 313. A(BC) Technology For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. Use a calculator to verify your solution. A = −2
0 ⎡ ⎢ ⎣ 0.5 4 ⎤ 9 ⎥, B = 1 8 −3 ⎦ 5 ⎡ 0.5 3 0 ⎢ −4 1 6 ⎣ 8 7 2 ⎤ ⎥, ⎤ ⎥ ⎦ 314. AB 315. BA 316. CA 1286 Chapter 11 Systems of Equations and Inequalities 11.6 \| Solving Systems with Gaussian Elimination Learning Objectives In this section, you will: 11.6.1 Write the augmented matrix of a system of equations. 11.6.2 Write the system of equations from an augmented matrix. 11.6.3 Perform row operations on a matrix. 11.6.4 Solve a system of linear equations using matrices. Figure 11.30 German mathematician Carl Friedrich Gauss (1777–1855). Carl Friedrich Gauss lived during the late 18th century and early 19th century, but he is still considered one of the most prolific mathematicians in history. His contributions to the science of mathematics and physics span fields such as algebra, number theory, analysis, differential geometry, astronomy, and optics, among others. His discoveries regarding matrix theory changed the way mathematicians have worked for the last two centuries. We first encountered Gaussian elimination in Systems of Linear Equations: Two Variables. In this section, we will revisit this technique for solving systems, this time using matrices. Writing the Augmented Matrix of a System of Equations A matrix can serve as a device for representing and solving a system of equations. To express a system in matrix form, we extract the coefficients of the variables and the constants, and these become the entries of the matrix. We use a vertical line to separate the coefficient entries from the constants, essentially replacing the equal signs. When a system is written in this form, we call it an augmented matrix. For example, consider the following 2 × 2 system of equations. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1287 We can write this system as an augmented matrix: 3x + 4y = 7 4x−2y = 5 We can also write a matrix containing just the coefficients. This is called the coefficient matrix. ⎡ 4 3 ⎣ 4 −2 ⎤ ⎦ 7 5 \| A three-by-three
system of equations such as has a coefficient matrix and is represented by the augmented matrix ⎤ ⎡ 3 4 ⎦ ⎣ 4 −2 3x − 2x−3z = 2 ⎡ ⎤ 3 −1 −3 2 Notice that the matrix is written so that the variables line up in their own columns: x-terms go in the first column, y-terms in the second column, and z-terms in the third column. It is very important that each equation is written in standard form ax + by + cz = d so that the variables line up. When there is a missing variable term in an equation, the coefficient is 0. ⎡ 3 −1 −1 ⎢ 1 0 1 ⎣ 0 −3 2 ⎤ ⎥ ⎦ 0 5 2 \| Given a system of equations, write an augmented matrix. 1. Write the coefficients of the x-terms as the numbers down the first column. 2. Write the coefficients of the y-terms as the numbers down the second column. 3. If there are z-terms, write the coefficients as the numbers down the third column. 4. Draw a vertical line and write the constants to the right of the line. Example 11.37 Writing the Augmented Matrix for a System of Equations Write the augmented matrix for the given system of equations. x + 2y − z = 3 2x − y + 2z = 6 x − 3y + 3z = 4 Solution The augmented matrix displays the coefficients of the variables, and an additional column for the constants. ⎡ 1 ⎢ 2 −1 ⎣ 1 −3 2 − 1288 Chapter 11 Systems of Equations and Inequalities 11.22 Write the augmented matrix of the given system of equations. 4x−3y = 11 3x + 2y = 4 Writing a System of Equations from an Augmented Matrix We can use augmented matrices to help us solve systems of equations because they simplify operations when the systems are not encumbered by the variables. However, it is important to understand how to move back and forth between formats in order to make finding solutions smoother and more intuitive. Here, we will use the information in an augmented matrix to write the system of equations in standard form. Example 11.38 Writing a System of Equations from an Augmented Matrix Form Find the system of equations from the augmented matrix. ⎡ ⎢ ⎣ −3
1 −3 −5 2 −5 −4 4 5 ⎤ ⎥ ⎦ \| −2 5 6 Solution When the columns represent the variables x, y, and z, ⎡ ⎢ ⎣ −3 1 −3 −5 2 −5 −4 4 5 \| −2 5 6 ⎤ ⎥ → ⎦ x − 3y − 5z = − 2 2x − 5y − 4z = 5 −3x + 5y + 4z = 6 11.23 Write the system of equations from the augmented matrix. 1 −1 1 ⎡ ⎢ 2 −9 \| Performing Row Operations on a Matrix Now that we can write systems of equations in augmented matrix form, we will examine the various row operations that can be performed on a matrix, such as addition, multiplication by a constant, and interchanging rows. Performing row operations on a matrix is the method we use for solving a system of equations. In order to solve the system of equations, we want to convert the matrix to row-echelon form, in which there are ones down the main diagonal from the upper left corner to the lower right corner, and zeros in every position below the main diagonal as shown. Row-echelon form ⎤ ⎥ ⎦ We use row operations corresponding to equation operations to obtain a new matrix that is row-equivalent in a simpler form. Here are the guidelines to obtaining row-echelon form. 1. In any nonzero row, the first nonzero number is a 1. It is called a leading 1. 2. Any all-zero rows are placed at the bottom on the matrix. 3. Any leading 1 is below and to the right of a previous leading 1. 4. Any column containing a leading 1 has zeros in all other positions in the column. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1289 To solve a system of equations we can perform the following row operations to convert the coefficient matrix to row-echelon form and do back-substitution to find the solution. 1. Interchange rows. (Notation: Ri ↔ R j ) 2. Multiply a row by a constant. (Notation: cRi ) 3. Add the product of a row multiplied by a constant to another row.
(Notation: Ri + cR j) Each of the row operations corresponds to the operations we have already learned to solve systems of equations in three variables. With these operations, there are some key moves that will quickly achieve the goal of writing a matrix in rowechelon form. To obtain a matrix in row-echelon form for finding solutions, we use Gaussian elimination, a method that uses row operations to obtain a 1 as the first entry so that row 1 can be used to convert the remaining rows. Gaussian Elimination The Gaussian elimination method refers to a strategy used to obtain the row-echelon form of a matrix. The goal is to write matrix A with the number 1 as the entry down the main diagonal and have all zeros below. A = a11 a12 a13 ⎡ a21 a22 a23 ⎢ ⎣ a31 a32 a33 ⎤ ⎥ ⎦ After Gaussian elimination A = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ → 1 b12 b13 ⎡ ⎢ 1 b23 0 ⎢ ⎣ 1 0 0 ⎤ ⎥ ⎥ ⎦ The first step of the Gaussian strategy includes obtaining a 1 as the first entry, so that row 1 may be used to alter the rows below. Given an augmented matrix, perform row operations to achieve row-echelon form. 1. The first equation should have a leading coefficient of 1. Interchange rows or multiply by a constant, if necessary. 2. Use row operations to obtain zeros down the first column below the first entry of 1. 3. Use row operations to obtain a 1 in row 2, column 2. 4. Use row operations to obtain zeros down column 2, below the entry of 1. 5. Use row operations to obtain a 1 in row 3, column 3. 6. Continue this process for all rows until there is a 1 in every entry down the main diagonal and there are only zeros below. 7. If any rows contain all zeros, place them at the bottom. Example 11.39
Solving a 2×2 System by Gaussian Elimination Solve the given system by Gaussian elimination. 2x + 3y = 6 x − y = 1 2 Solution First, we write this as an augmented matrix. 1290 Chapter 11 Systems of Equations and Inequalities We want a 1 in row 1, column 1. This can be accomplished by interchanging row 1 and row 2. ⎡ 3 ⎢2 1 −1 ⎣ ⎤ ⎥ ⎦ \| 6 1 2 We now have a 1 as the first entry in row 1, column 1. Now let’s obtain a 0 in row 2, column 1. This can be accomplished by multiplying row 1 by −2, and then adding the result to row 2. R1 ↔ R2 → ⎡ ⎢1 −2R1 + R2 = R2 → We only have one more step, to multiply row 2 by 1 5. ⎤ ⎥ ⎦ ⎡ ⎢1 − R2 = R2 → 1 5 ⎡ ⎢1 −1 ⎣ 0 Use back-substitution. The second row of the matrix represents y = 1. Back-substitute y = 1 into the first equation. x − (1) = 1 2 x = 3 2 The solution is the point ⎛ ⎝ ⎞, 1 ⎠. 3 2 11.24 Solve the given system by Gaussian elimination. 4x + 3y = 11 x−3y = −1 Example 11.40 Using Gaussian Elimination to Solve a System of Equations Use Gaussian elimination to solve the given 2 × 2 system of equations. Solution Write the system as an augmented matrix. 2x + y = 1 4x + 2y = This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1291 Obtain a 1 in row 1, column 1. This can be accomplished by multiplying the first row by 1 2. Next, we want a 0 in row 2, column 1. Multiply row 1 by −4 and add row 1 to row 2. 1 2 R1 = R1 → ⎡ ⎢ \| The second row represents the equation 0 = 4. Therefore, the system is inconsistent and has no solution. −4R
1 + R2 = R2 → ⎡ ⎢ \| Example 11.41 Solving a Dependent System Solve the system of equations. 3x + 4y = 12 6x + 8y = 24 Solution Perform row operations on the augmented matrix to try and achieve row-echelon form \| 12 24 − 1 2 R2 + R1 = R1 → ⎡ 6 ⎣ 0 R1 ↔ R2 → 8 ⎤ ⎦ 0 \| 24 0 ⎡ 0 ⎣ 6 0 24 ⎤ ⎦ 0 8 \| The matrix ends up with all zeros in the last row: 0y = 0. Thus, there are an infinite number of solutions and the system is classified as dependent. To find the generic solution, return to one of the original equations and solve for y. 3x + 4y = 12 4y = 12−3x y = 3 − 3 4 x So the solution to this system is ⎛ ⎝x, 3 − 3 4 x⎞ ⎠. Example 11.42 1292 Chapter 11 Systems of Equations and Inequalities Performing Row Operations on a 3×3 Augmented Matrix to Obtain Row-Echelon Form Perform row operations on the given matrix to obtain row-echelon form. ⎡ ⎢ ⎣ −3 1 −3 4 2 − Solution The first row already has a 1 in row 1, column 1. The next step is to multiply row 1 by −2 and add it to row 2. Then replace row 2 with the result. −2R1 + R2 = R2 → Next, obtain a zero in row 3, column 1. 3R1 + R3 = R3 → Next, obtain a zero in row 3, column 2. 6R2 + R3 = R3 → The last step is to obtain a 1 in row 3, column 3. ⎡ 1 −3 ⎢ 0 ⎣ −3 4 1 −2 ⎡ 1 −3 ⎢ 0 ⎣ 0 −6 ⎡ 1 −3 ⎢ 0 ⎣ 0 4 1 − 15 4 1 −2 3 4 \| 3 16 \| 3 4 \| 3 \| ⎤ 3 ⎥ −6 ⎥ 21 ⎦ 2 0 15 ⎤ ⎥ ⎦ 1 2 R3 = R3 → �
�� ⎢1 −3 0 ⎢ 0 ⎣ 4 1 −2 1 0 11.25 Write the system of equations in row-echelon form. x − 2y + 3z = 9 − x + 3y = − 4 2x − 5y + 5z = 17 Solving a System of Linear Equations Using Matrices We have seen how to write a system of equations with an augmented matrix, and then how to use row operations and backsubstitution to obtain row-echelon form. Now, we will take row-echelon form a step farther to solve a 3 by 3 system of linear equations. The general idea is to eliminate all but one variable using row operations and then back-substitute to solve for the other variables. Example 11.43 Solving a System of Linear Equations Using Matrices Solve the system of linear equations using matrices. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1293 x − y + z = 8 2x + 3y − z = −2 3x − 2y − 9z = 9 Solution First, we write the augmented matrix. Next, we perform row operations to obtain row-echelon form. ⎡ 1 1 −1 ⎢ 3 −1 2 ⎣ 3 −2 −9 ⎤ 8 ⎥ −2 ⎦ 9 \| −2R1 + R2 = R2 → 1 5 −3 ⎡ 1 −1 ⎢ 0 ⎣ 3 −2 −9 \| 8 −18 9 ⎤ ⎥ ⎦ −3R1 + R3 = R3 → The easiest way to obtain a 1 in row 2 of column 1 is to interchange R2 and R3. Interchange R2 and R3 → ⎡ 1 −12 −15 ⎦ −3 −18 5 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 8 −18 −15 ⎡ ⎢1 −3 1 −12\| 8 ⎤ ⎥ ⎥ ⎥ −15 ⎥ ⎥ ⎦ 1 1 1\| Then −5R2 + R3 = R3 → ⎡ 1 −1 ⎢ 0 ⎣ 0 1 1 −12 0
57 \| ⎤ 8 ⎥ −15 ⎦ 57 − 1 57 R3 = R3 → ⎡ ⎢1 −12 The last matrix represents the equivalent system. x − y + z = 8 y − 12z = −15 z = 1 Using back-substitution, we obtain the solution as (4, −3, 1). Example 11.44 Solving a Dependent System of Linear Equations Using Matrices Solve the following system of linear equations using matrices. −x−2y + z = −1 2x + 3y = 2 y−2z = 0 1294 Chapter 11 Systems of Equations and Inequalities Solution Write the augmented matrix. First, multiply row 1 by −1 to get a 1 in row 1, column 1. Then, perform row operations to obtain row-echelon form. ⎡ 1 −1 −2 ⎢ 0 3 ⎣ 1 −2 2 0 ⎤ ⎥ ⎦ \| −1 2 0 −R1 → R2 ↔ R3 → ⎡ 2R1 + R3 = R3 → R2 + R3 = R3 → ⎤ ⎥ ⎦ ⎤ ⎥ ⎦ 2 −1 1 3 2 0 1 −2 0 2 −1 1 −2 0 0 3 2 ⎡ 2 −1 1 ⎢ 1 −2 0 ⎣ 0 −1 ⎡ 1 ⎢ 0 ⎣ 0 2 −1 1 − ⎤ ⎥ ⎦ The last matrix represents the following system. x + 2y − z = 1 y − 2z = 0 0 = 0 We see by the identity 0 = 0 that this is a dependent system with an infinite number of solutions. We then find the generic solution. By solving the second equation for y and substituting it into the first equation we can solve for z in terms of x. x + 2y − z = 1 y = 2z x + 2(2z) − z = 1 x + 3z = 1 z = 1 − x 3 Now we substitute the expression for z into the second equation to solve for y in terms of x. y − 2z = ⎝ The generic solution is ⎛ ⎝x, 2−2x 3, 1 − x 3 ⎞ ⎠. − 2x 3 This content is
available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1295 11.26 Solve the system using matrices. x + 4y − z = 4 2x + 5y + 8z = 15 x + 3y−3z = 1 Can any system of linear equations be solved by Gaussian elimination? Yes, a system of linear equations of any size can be solved by Gaussian elimination. Given a system of equations, solve with matrices using a calculator. 1. Save the augmented matrix as a matrix variable [A], [B], [C], …. 2. Use the ref( function in the calculator, calling up each matrix variable as needed. Example 11.45 Solving Systems of Equations with Matrices Using a Calculator Solve the system of equations. 5x + 3y + 9z = −1 −2x + 3y − z = −2 −x−4y + 5z = 1 Solution Write the augmented matrix for the system of equations. On the matrix page of the calculator, enter the augmented matrix above as the matrix variable [A]. ⎡ 5 ⎢ −2 ⎣ −1 −4 3 9 3 −1 5 ⎤ 5 ⎥ −2 ⎦ −1 \| [A] = ⎡ 5 ⎢ −2 ⎣ −1 −4 ⎤ 9 −1 3 ⎥ 3 −1 −2 ⎦ 1 5 Use the ref( function in the calculator, calling up the matrix variable [A]. ref([A]) Evaluate 13 − 4 ⎢ 7 21 ⎢ 1 − 24 0 0 ⎣ 187 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ → + 13 21 z = − 24 187 Using back-substitution, the solution is ⎛ ⎝ 61 187, − 92 187, − 24 187 ⎞ ⎠. 1296 Chapter 11 Systems of Equations and Inequalities Example 11.46 Applying 2 × 2 Matrices to Finance Carolyn invests a total of $12,000 in two municipal bonds, one paying 10.5% interest and the other paying 12% interest. The annual interest earned on the two investments last year was $1,335. How much was invested at each rate? Solution We have a system of two equations in two
variables. Let x = the amount invested at 10.5% interest, and y = the amount invested at 12% interest. x + y = 12,000 0.105x + 0.12y = 1,335 As a matrix, we have Multiply row 1 by −0.105 and add the result to row 2. ⎡ 1 1 ⎣ 0.105 0.12 12,000 1,335 ⎤ ⎦ \| Then, ⎡ 1 1 ⎣ 0 0.015 12,000 75 ⎤ ⎦ \| 0.015y = 75 y = 5,000 So 12,000−5,000 = 7,000. Thus, $5,000 was invested at 12% interest and $7,000 at 10.5% interest. Example 11.47 Applying 3 × 3 Matrices to Finance Ava invests a total of $10,000 in three accounts, one paying 5% interest, another paying 8% interest, and the third paying 9% interest. The annual interest earned on the three investments last year was $770. The amount invested at 9% was twice the amount invested at 5%. How much was invested at each rate? Solution We have a system of three equations in three variables. Let x be the amount invested at 5% interest, let y be the amount invested at 8% interest, and let z be the amount invested at 9% interest. Thus, x + y + z = 10, 000 0.05x + 0.08y + 0.09z = 770 2x − z = 0 As a matrix, we have This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1297 Now, we perform Gaussian elimination to achieve row-echelon form. 1 1 ⎡ 1 ⎢ 0.05 0.08 0.09 ⎣ 0 −1 2 ⎤ 10, 000 ⎥ 770 ⎦ 0 \| −0.05R1 + R2 = R2 → −2R1 + R3 = R3 → 1 0.03 R2 = R2 → 2R2 + R3 = R3 → 1 0.04 1 0.04 ⎡ ⎣ ⎤ ⎥ 270 ⎦ 0 ⎤ 10,000 ⎥ 270 ⎦
−20,000 1 0.03 0 1 0.03 −2 1 −1 \| 10,000 −3 \| 0 −2 −3\| 3\| 10,000 ⎤ ⎥ ⎥ 9,000 ⎥ −2,000 ⎦ ⎤ 10,000 ⎥ 9,000 ⎥ −20,000 ⎦ The third row tells us − 1 3 z = −2,000; thus z = 6,000. The second row tells us y + 4 3 z = 9,000. Substituting z = 6,000, we get (6,000) = 9,000 y + 4 3 y + 8,000 = 9,000 y = 1,000 The first row tells us x + y + z = 10, 000. Substituting y = 1, 000 and z = 6, 000, we get x + 1, 000 + 6, 000 = 10,000 x = 3,000 The answer is $3,000 invested at 5% interest, $1,000 invested at 8%, and $6,000 invested at 9% interest. 11.27 A small shoe company took out a loan of $1,500,000 to expand their inventory. Part of the money was borrowed at 7%, part was borrowed at 8%, and part was borrowed at 10%. The amount borrowed at 10% was four times the amount borrowed at 7%, and the annual interest on all three loans was $130,500. Use matrices to find the amount borrowed at each rate. Access these online resources for additional instruction and practice with solving systems of linear equations using Gaussian elimination. • Solve a System of Two Equations Using an Augmented Matrix (http://openstaxcollege.org/l/system2augmat) • Solve a System of Three Equations Using an Augmented Matrix (http://openstaxcollege.org/l/system3augmat) • Augmented Matrices on the Calculator (http://openstaxcollege.org/l/augmatcalc) 1298 Chapter 11 Systems of Equations and Inequalities For the following exercises, solve the system by Gaussian elimination. 11.6 EXERCISES Verbal 324. Can any system of linear equations be written as an augmented matrix? Explain why or why not. Explain how to write that augmented matrix. 325. Can any matrix be written as a system of linear equations? Explain why or why
not. Explain how to write that system of equations. 336. 337. 338. 326. Is there only one correct method of using row operations on a matrix? Try to explain two different row operations possible to solve the augmented matrix ⎡ 3 9 ⎣ 1 −2 ⎤ ⎦. 6 \| 0 Can a matrix whose entry is 0 on the diagonal be 327. solved? Explain why or why not. What would you do to remedy the situation? Can a matrix that has 0 entries for an entire row have 328. one solution? Explain why or why not. Algebraic For the following exercises, write the augmented matrix for the linear system. 329. 8x−37y = 8 2x + 12y = 3 330. 16y = 4 9x − y = 2 331. 332. 333. 3x + 2y + 10z = 3 −6x + 2y + 5z = 13 4x + z = 18 x + 5y + 8z = 19 12x + 3y = 4 3x + 4y + 9z = −7 6x + 12y + 16z = 4 19x−5y + 3z = −9 x + 2y = −8 For the following exercises, write the linear system from the augmented matrix. 334. ⎡ −2 ⎣ 5 6 −18 335. ⎡ 4 3 ⎣ 10 17 ⎤ ⎦ 5 26 \| \| 10 439 ⎤ ⎦ This content is available for free at https://cnx.org/content/col11758/1.5 3 ⎡ 2 0 ⎢ −1 −9 4 ⎣ 5 7 8 ⎡ 8 29 1 ⎢ −2 ⎢ 0 1 58 ⎣ 8 7 −3 ⎤ 3 ⎥ −1 ⎦ 8 ⎤ ⎥ ⎦ 43 38 10 \| \| \| 12 ⎤ ⎥ 2 ⎦ −5 339. 340. 341. 342. 343. 3441 ⎣ 2 4 −5 ⎤ −3 ⎦ 6 ⎡ −2 0 ⎣ 0 2 ⎤ 1 ⎦ −1 2x − 3y = − 9 5x + 4y = 58 345. 6x + 2y = −4 3x + 4y = −17 346. 2x + 3y = 12 4
x + y = 14 347. −4x−3y = −2 3x−5y = −13 348. −5x + 8y = 3 10x + 6y = 5 349. 3x + 4y = 12 −6x−8y = −24 350. −60x + 45y = 12 20x−15y = −4 351. 11x + 10y = 43 15x + 20y = 65 Chapter 11 Systems of Equations and Inequalities 1299 352. 2x − y = 2 3x + 2y = 17 y − z = −3 ⎡ −0.1 0.3 −0.1 ⎢ 0.1 −0.4 0.2 ⎣ 0.7 0.6 0.1 ⎤ 0.2 ⎥ 0.8 ⎦ −0.8 Extensions For the following exercises, use Gaussian elimination to solve the system. −1.06x−2.25y = 5.51 −5.03x−1.08y = 5.40 1 y = 3 ⎤ ⎥ ⎦ 31 45 87 ⎤ 50 ⎥ 20 ⎦ −90 ⎤ ⎥ ⎦ 2x + 3y − 2z = 3 4x + 2y − z = 9 4x − 8y + 2z = −6 x + y − 4z = −4 5x − 3y − 2z = 0 2x + 6y + 7z = 30 2x + 3y + 2z = 1 −4x − 6y − 4z = −2 10x + 15y + 10z = 5 x + 2y − z = 1 −x − 2y + 2z = −2 3x + 6y − 3z = 5 x + 2y − z = 1 −x−2y + 2z = −2 3x + 6y−3z = 3 353. 354. 355. 356. 357. 358. 359. 360. 361. 362. 363. 364. 365. 366. 367. 368. 369. x + y + z = 100 x + 2z = 125 −y + 2z = 25 = − 53 14 z = 3 z = 23 15 − 10 z = − 29 6 z = 431 210 z = − 49 45 370. x−1 7 + y−2 8 + z−
+ 2y + z−3 3 = 5 371. x−1 4 − y + 1 4 + 3z = −−2 2 − z = 4 = 1 372. 373. x−3 4 x + 5 2 + − y−1 3 y + 5 2 + + 2z = −−3 10 x + 5 4 x−1 4 + y + 3 2 y−1 8 y + 4 2 − −2z = 3 + z = 3 2 + 3z = 3 2 Chapter 11 Systems of Equations and Inequalities ice cream, find out the percentage of ice cream sales each individual ice cream made last year. A bag of mixed nuts contains cashews, pistachios, 383. and almonds. There are 1,000 total nuts in the bag, and there are 100 less almonds than pistachios. The cashews weigh 3 g, pistachios weigh 4 g, and almonds weigh 5 g. If the bag weighs 3.7 kg, find out how many of each type of nut is in the bag. A bag of mixed nuts contains cashews, pistachios, 384. and almonds. Originally there were 900 nuts in the bag. 30% of the almonds, 20% of the cashews, and 10% of the pistachios were eaten, and now there are 770 nuts left in the bag. Originally, there were 100 more cashews than almonds. Figure out how many of each type of nut was in the bag to begin with. 1300 374. x−3 4 x + 5 2 + − y−1 3 y + 5 2 + 2z = − Real-World Applications For the following exercises, set up the augmented matrix that describes the situation, and solve for the desired solution. Every day, a cupcake store sells 5,000 cupcakes in 375. chocolate and vanilla flavors. If the chocolate flavor is 3 times as popular as the vanilla flavor, how many of each cupcake sell per day? At a competing cupcake store, $4,520 worth of 376. cupcakes are sold daily. The chocolate cupcakes cost $2.25 and the red velvet cupcakes cost $1.75. If the total number of cupcakes sold per day is 2,200, how many of each flavor are sold each day? You invested $10,000 into two accounts: one that has 377. simple 3% interest, the other with 2.5% interest. If your total interest payment after one year was $283.
50, how much was in each account after the year passed? You invested $2,300 into account 1, and $2,700 into 378. account 2. If the total amount of interest after one year is $254, and account 2 has 1.5 times the interest rate of account 1, what are the interest rates? Assume simple interest rates. Bikes’R’Us manufactures bikes, which sell for $250. 379. It costs the manufacturer $180 per bike, plus a startup fee of $3,500. After how many bikes sold will the manufacturer break even? A major appliance store is considering purchasing 380. vacuums from a small manufacturer. The store would be able to purchase the vacuums for $86 each, with a delivery fee of $9,200, regardless of how many vacuums are sold. If the store needs to start seeing a profit after 230 units are sold, how much should they charge for the vacuums? The three most popular ice cream flavors are 381. chocolate, strawberry, and vanilla, comprising 83% of the flavors sold at an ice cream shop. If vanilla sells 1% more than twice strawberry, and chocolate sells 11% more than vanilla, how much of the total ice cream consumption are the vanilla, chocolate, and strawberry flavors? At an ice cream shop, three flavors are increasing in 382. demand. Last year, banana, pumpkin, and rocky road ice cream made up 12% of total ice cream sales. This year, the same three ice creams made up 16.9% of ice cream sales. The rocky road sales doubled, the banana sales increased by 50%, and the pumpkin sales increased by 20%. If the rocky road ice cream had one less percent of sales than the banana This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1301 11.7 \| Solving Systems with Inverses Learning Objectives In this section, you will: 11.7.1 Find the inverse of a matrix. 11.7.2 Solve a system of linear equations using an inverse matrix. Nancy plans to invest $10,500 into two different bonds to spread out her risk. The first bond has an annual return of 10%, and the second bond has an annual return of 6%. In order to receive an 8.5% return from the two bonds, how much should Nancy invest in each bond
? What is the best method to solve this problem? There are several ways we can solve this problem. As we have seen in previous sections, systems of equations and matrices are useful in solving real-world problems involving finance. After studying this section, we will have the tools to solve the bond problem using the inverse of a matrix. Finding the Inverse of a Matrix We know that the multiplicative inverse of a real number a is a−1, and aa−1 = a−1 a = ⎛ ⎝ 1 a ⎞ ⎠a = 1. For example, 1 2 and ⎛ ⎝ 2−1 = 1 2 ⎞ ⎠2 = 1. The multiplicative inverse of a matrix is similar in concept, except that the product of matrix A and its inverse A−1 equals the identity matrix. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. We identify identity matrices by In where n represents the dimension of the matrix. Equation 11.1 and Equation 11.2 are the identity matrices for a 2×2 matrix and a 3×3 matrix, respectively. I2 = ⎡ 1 ⎣ 0 ⎤ ⎦ 0 1 I3 = ⎡ 11.1) (11.2) The identity matrix acts as a 1 in matrix algebra. For example, AI = IA = A. A matrix that has a multiplicative inverse has the properties AA−1 = I A−1 A = I A matrix that has a multiplicative inverse is called an invertible matrix. Only a square matrix may have a multiplicative inverse, as the reversibility, AA−1 = A−1 A = I, is a requirement. Not all square matrices have an inverse, but if A is invertible, then A−1 is unique. We will look at two methods for finding the inverse of a 2×2 matrix and a third method that can be used on both 2×2 and 3×3 matrices. The Identity Matrix and Multiplicative Inverse The identity matrix, In, is a square matrix containing ones down the main diagonal and zeros everywhere else. ⎤ ⎦ I2 = ⎡ I3 = ⎤ ⎥ ⎦ If A is an n × n matrix and B is an n × n matrix such that AB = BA = In, then B = A
−1, the multiplicative inverse of a matrix A. 1302 Chapter 11 Systems of Equations and Inequalities Example 11.48 Showing That the Identity Matrix Acts as a 1 Given matrix A, show that AI = IA = A2 5 Solution Use matrix multiplication to show that the product of A and the identity is equal to the product of the identity and A. AI = AI = ⎤ ⎡ 1 ⎣ ⎦ 0 ⎡ 4 3 ⎣ −2 5 ⎡ ⎤ 4 3 ⎦ ⎣ − ⋅ (−2) ⎣ 0 ⋅ 3 + 1 ⋅ (−22 5 ⎡ 4 3 ⎣ −2 5 ⎤ ⎦ ⎤ ⎦ Given two matrices, show that one is the multiplicative inverse of the other. 1. Given matrix A of order n × n and matrix B of order n × n multiply AB. 2. If AB = I, then find the product BA. If BA = I, then B = A−1 and A = B−1. Example 11.49 Showing That Matrix A Is the Multiplicative Inverse of Matrix B Show that the given matrices are multiplicative inverses of each other. A = 1 ⎤ ⎡ 5 ⎦, B = ⎣ −2 −9 ⎤ ⎡ −9 −5 ⎦ ⎣ 1 2 Solution Multiply AB and BA. If both products equal the identity, then the two matrices are inverses of each other. AB = = = 1 ⎤ ⎦ ⎡ ⎤ −9 −2 −9 2 ⎤ ⎡ 1(−5) + 5(1) 1(−9) + 5(2) ⎦ ⎣ −2(−9)−9(2) −2(−5)−9(19 −5 ⎦ · ⎣ 1 ⎤ ⎡ −9(5)−5(−9) −9(1)−5(−2) ⎦ ⎣ 2(1) + 1(−2) 2(−5) + 1(−92 −9 0 1 ⎤ ⎦ 1 2 BA = = = A and B are inverses of each other. This content is available for
free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1303 11.28 Show that the following two matrices are inverses of each other. A = 1 ⎤ ⎡ 4 ⎦, B = ⎣ −1 −3 ⎤ ⎡ −3 −4 ⎦ ⎣ 1 1 Finding the Multiplicative Inverse Using Matrix Multiplication We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using matrix multiplication. Example 11.50 Finding the Multiplicative Inverse Using Matrix Multiplication Use matrix multiplication to find the inverse of the given matrix. A = ⎤ ⎡ 1 −2 ⎦ ⎣ 2 −3 Solution For this method, we multiply A by a matrix containing unknown constants and set it equal to the identity. ⎤ ⎡ 1 −2 ⎦ ⎣ 2 − ⎤ ⎦ Find the product of the two matrices on the left side of the equal sign. ⎤ ⎡ 1 −2 ⎦ ⎣ 2 −3 a b ⎡ ⎣ c d ⎤ ⎦ = 1a−2c 1b−2d ⎡ ⎣ 2a−3c 2b−3d ⎤ ⎦ Next, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0. 1a−2c = 1 R1 2a−3c = 0 R2 Using row operations, multiply and add as follows: (−2)R1 + R2 → R2. Add the equations, and solve for c. 1a − 2c = 1 0 + 1c = − 2 c = − 2 Back-substitute to solve for a. a−2(−23 Write another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in
row 2, column 2 equal to the corresponding entry of the identity. Using row operations, multiply and add as follows: (−2)R1 + R2 = R2. Add the two equations and solve for d. 1b−2d = 0 R1 2b−3d = 1 R2 1304 Chapter 11 Systems of Equations and Inequalities Once more, back-substitute and solve for b. 1b−2d = 0 0 + 1d = 1 d = 1 b−2(1) = 0 b−2 = 0 b = 2 ⎤ ⎡ −3 2 ⎦ ⎣ −2 1 A−1 = Finding the Multiplicative Inverse by Augmenting with the Identity Another way to find the multiplicative inverse is by augmenting with the identity. When matrix A is transformed into I, the augmented matrix I transforms into A−1. For example, given augment A with the identity Perform row operations with the goal of turning A into the identity. Switch row 1 and row 2. 2. Multiply row 2 by −2 and add to row 1. 3. Multiply row 1 by −2 and add to row 2. 4. Add row 2 to row 1. 5. Multiply row 2 by −1. The matrix we have found is A−11 \| −2 ⎤ 1 ⎦ 5 −2 ⎡ 0 1 ⎣ 0 −1 \| ⎤ 3 −1 ⎦ 5 −1 ⎦ 2 −5 A−1 = ⎡ ⎣ −5 ⎤ 3 −1 ⎦ 2 Finding the Multiplicative Inverse of 2×2 Matrices Using a Formula When we need to find the multiplicative inverse of a 2 × 2 matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1305 If A is a 2×2 matrix, such as the multiplicative inverse of A is given by the formula −1 = 1 ad − bc ⎡ ⎣ −c ⎤ d −b ⎦ a (11.3) where ad − bc ≠ 0. If ad − bc = 0, then A has no inverse. Example 11.51 Using the Formula to
Find the Multiplicative Inverse of Matrix A Use the formula to find the multiplicative inverse of A = ⎤ ⎡ 1 −2 ⎦ ⎣ 2 −3 Solution Using the formula, we have ⎡ −3 2 ⎣ −2 1 ⎤ ⎦ A−1 = = = 1 (1)(−3) − (−2)(2) ⎤ ⎡ −3 2 ⎦ ⎣ −2 1 1 −3 + 4 ⎤ ⎡ −3 2 ⎦ ⎣ −2 1 Analysis We can check that our formula works by using one of the other methods to calculate the inverse. Let’s augment A with the identity. Perform row operations with the goal of turning A into the identity. ⎡ 1 −2 ⎣ 2 −3 ⎤ ⎦ \| 1 0 0 1 1. Multiply row 1 by −2 and add to row 2. 2. Multiply row 1 by 2 and add to row 1. So, we have verified our original solution2 ⎣ 0 1 ⎤ ⎦ −2 1 \| 1 0 \| −3 2 ⎤ ⎦ −2 1 A−1 = ⎡ −3 2 ⎣ −2 1 ⎤ ⎦ Use the formula to find the inverse of matrix A. Verify your answer by augmenting with the identity 11.29 matrix. A = ⎤ ⎡ 1 −1 ⎦ ⎣ 2 3 1306 Chapter 11 Systems of Equations and Inequalities Example 11.52 Finding the Inverse of the Matrix, If It Exists Find the inverse, if it exists, of the given matrix. A = ⎡ 3 6 ⎣ 1 2 ⎤ ⎦ Solution We will use the method of augmenting with the identity. 1. Switch row 1 and row 2. Multiply row 1 by −3 and add it to row 23 1 ⎤ ⎦ 3. There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse. Finding the Multiplicative Inverse of 3×3 Matrices Unfortunately, we do not have a formula similar to the one for a 2×2 matrix to find the inverse of a 3×3 matrix. Instead, we will augment the original matrix with the identity matrix and use row operations
to obtain the inverse. Given a 3×3 matrix augment A with the identity matrix \| ⎤ ⎥ ⎦ \| To begin, we write the augmented matrix with the identity on the right and A on the left. Performing elementary row operations so that the identity matrix appears on the left, we will obtain the inverse matrix on the right. We will find the inverse of this matrix in the next example. Given a 3 × 3 matrix, find the inverse 1. Write the original matrix augmented with the identity matrix on the right. 2. Use elementary row operations so that the identity appears on the left. 3. What is obtained on the right is the inverse of the original matrix. 4. Use matrix multiplication to show that AA−1 = I and A−1 A = I. Example 11.53 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1307 Finding the Inverse of a 3 × 3 Matrix Given the 3 × 3 matrix A, find the inverse ⎤ ⎥ ⎦ Solution Augment A with the identity matrix, and then begin row operations until the identity matrix replaces A. The matrix on the right will be the inverse of A. Interchange R2 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ and R1 → ⎡ R2 + R1 = R1 → −R2 + R3 = R3 → ⎡ R3 ↔ R2 → ⎤ ⎥ ⎦ \| 1 1 0 ⎥ 1 0 0 ⎦ −1 1 0 −1 0 1 1 0 0 −1 −2 −3 −1 −1 −2R1 + R3 = R3 → −3R2 + R3 = R3 → Thus, A−1 = B = 0 −1 1 ⎡ ⎢ −1 0 1 ⎣ 6 −2 −3 ⎤ ⎥ ⎦ Analysis To prove that B = A−1, let’s multiply the two matrices together to see if the product equals the identity, if AA−1
= I and A−1 A = I. AA−1 = = = ⎤ ⎥ ⎦ ⎡ −1 ⎢ −2 −(−1) + 3(−1) + 1(6) 2(1) + 3(0) + 1(−2) 2(0) + 3(1) + 1(−3) ⎤ ⎡ ⎥ ⎢ 3(−1) + 3(−1) + 1(6) 3(1) + 3(0) + 1(−2) 3(0) + 3(1) + 1(−3) ⎥ ⎢ 2(−1) + 4(−1) + 1(6) 2(1) + 4(0) + 1(−2) 2(0) + 4(1) + 1(−3 ⎤ ⎥ ⎦ 1308 Chapter 11 Systems of Equations and Inequalities A−2 −3 2 4 1 −1(2) + 1(3) + 0(2) −1(2) + 0(3) + 1(2) ⎡ −1 ⎢ −1 ⎣ −1(1) + 1(1) + 0(1) −1(3) + 1(3) + 0(4) ⎤ ⎡ ⎥ ⎢ −1(1) + 0(1) + 1(1) −1(3) + 0(3) + 1(4) ⎥ ⎢ 6(2) + −2(3) + −3(2) 6(3) + −2(3) + −3(4) 6(1) + −2(1) + −3(1 ⎤ ⎥ ⎦ 11.30 Find the inverse of the 3×3 matrix. A = 2 −17 11 ⎡ ⎤ ⎢ ⎥ −1 11 −7 ⎣ ⎦ 0 3 −2 Solving a System of Linear Equations Using the Inverse of a Matrix Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: X is the matrix representing the variables of the system, and B is the matrix representing the constants. Using matrix multiplication, we may define a system of
equations with the same number of equations as variables as To solve a system of linear equations using an inverse matrix, let A be the coefficient matrix, let X be the variable matrix, and let B be the constant matrix. Thus, we want to solve a system AX = B. For example, look at the following system of equations. AX = B From this system, the coefficient matrix is The variable matrix is And the constant matrix is Then AX = B looks like a1 x + b1 y = c1 a2 x + b2 y = c2 A = a1 b1 ⎡ ⎣ a2 b2 ⎤ ⎦ X = x ⎡ y ⎣ ⎤ ⎦ B = c1 ⎡ c2 ⎣ ⎤ ⎦ a1 b1 ⎡ ⎣ a2 b2 ⎤ ⎦ x ⎡ y ⎣ ⎤ ⎦ = c1 ⎡ c2 ⎣ ⎤ ⎦ Recall the discussion earlier in this section regarding multiplying a real number by its inverse, (2−1) 2 = ⎞ ⎠ 2 = 1. To solve a single linear equation ax = b for x, we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of a. Thus, 1 2 ⎛ ⎝ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1309 ⎛ ⎝ ⎞ ⎠b ax = b ⎛ ⎞ 1 1 ⎠ax = a a ⎝ (a−1 )ax = (a−1)b [(a−1)a]x = (a−1)b 1x = (a−1)b x = (a−1)b The only difference between a solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same—to isolate the variable. We will investigate this idea in detail, but it is helpful to begin with a 2 × 2 system and then move on to a 3 × 3 system. Solving a System of Equations Using the Inverse of a Matrix Given a system of
equations, write the coefficient matrix A, the variable matrix X, and the constant matrix B. Then Multiply both sides by the inverse of A to obtain the solution. AX = B ⎛ ⎝A−1⎞ ⎠AX = ⎡ ⎠A⎤ ⎛ ⎝A−1⎞ ⎣ ⎦X = IX = X = ⎛ ⎝A−1⎞ ⎠B ⎛ ⎝A−1⎞ ⎠B ⎛ ⎝A−1⎞ ⎠B ⎛ ⎝A−1⎞ ⎠B If the coefficient matrix does not have an inverse, does that mean the system has no solution? No, if the coefficient matrix is not invertible, the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions. Example 11.54 Solving a 2 × 2 System Using the Inverse of a Matrix Solve the given system of equations using the inverse of a matrix. 3x + 8y = 5 4x + 11y = 7 Solution Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix. Then A = ⎡ 3 8 ⎣ 4 11 ⎤ ⎦, X = x ⎡ y ⎣ ⎤ ⎦, 11 ⎤ ⎦ First, we need to calculate A−1. Using the formula to calculate the inverse of a 2 by 2 matrix, we have: 1310 Chapter 11 Systems of Equations and Inequalities ⎤ ⎦ d −b ⎡ ⎣ −c a ⎤ ⎡ 11 −8 ⎦ ⎣ −4 3 A−1 = = 1 ad − bc 1 3(11)−8(4) ⎤ ⎡ 11 −8 ⎦ ⎣ −4 3 = 1 1 So, Now we are ready to solve. Multiply both sides of the equation by A−1. A−1 = ⎤ ⎡ 11 −8 ⎦ ⎣ −4 3 ⎤ ⎡ 11 −8 ⎦ ⎣ 3 −4 The solution is (−1, 1). ⎤ ⎦ ⎤ ⎦ x
⎡ y ⎣ ⎡ 5 ⎣ 7 (A−1)AX = (A−1)B ⎤ ⎡ ⎡ 11 −4 3 4 11 ⎡ ⎤ 11(5) + (−8)4(5) + 3(7) ⎡ ⎤ − ⎣ Can we solve for X by finding the product BA−1? No, recall that matrix multiplication is not commutative, so A−1 B ≠ BA−1. Consider our steps for solving the matrix equation. ⎛ ⎝A−1⎞ ⎠AX = ⎡ ⎠A⎤ ⎛ ⎝A−1⎞ ⎣ ⎦X = IX = ⎛ ⎝A−1⎞ ⎠B ⎛ ⎝A−1⎞ ⎠B ⎛ ⎝A−1⎞ ⎠B ⎛ ⎝A−1⎞ ⎠B X = Notice in the first step we multiplied both sides of the equation by A−1, but the A−1 was to the left of A on the left side and to the left of B on the right side. Because matrix multiplication is not commutative, order matters. Example 11.55 Solving a 3 × 3 System Using the Inverse of a Matrix Solve the following system using the inverse of a matrix. 5x + 15y + 56z = 35 −4x−11y−41z = −26 −x−3y−11z = −7 Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1311 Write the equation AX = B. ⎤ ⎡ 56 15 5 ⎥ ⎢ −4 −11 −41 ⎦ ⎣ −1 −3 −11 x ⎡ y ⎢ ⎣ z ⎤ ⎥ = ⎦ ⎤ ⎡ 35 ⎥ ⎢ −26 ⎦ ⎣ −7 First, we will find the inverse of A by augmenting with the identity. Multiply row 1 by 1 5. Multiply row 1 by 4 and add to row
2. Add row 1 to row 3. Multiply row 2 by −3 and add to row 1. Multiply row 3 by 5. Multiply row 3 by 1 5 and add to row 1 19 ⎢ 19 ⎢ 5 ⎣ 0 0 1 5 ⎡ 56 15 ⎢ −4 −11 −41 ⎣ −1 −3 −11 56 5 ⎡ ⎢ ⎢ −4 −11 −41 ⎣ −1 −3 −11 1 −3 −11 56 5 19 56 ⎢ 5 ⎢ 0 1 19 ⎢ − 11 5 4 5 1 5 − 11 19 ⎢ 5 ⎣ 0 0 1 −2 −3 1 4 5 1 1 0 0 5 ⎤ ⎥ ⎥ ⎦ 1312 Chapter 11 Systems of Equations and Inequalities Multiply row 3 by − 19 5 and add to row 2. So−1 = 1 ⎤ −2 −3 ⎥ −3 1 −19 ⎦ 0 1 5 \| 1 ⎤ ⎡ −2 −3 ⎥ ⎢ −3 1 −19 ⎦ ⎣ 0 1 5 Multiply both sides of the equation by A−1. We want A−1 AX = A−1 B : ⎡ −2 −3 ⎢ −3 ⎣ 1 ⎤ 1 ⎥ 1 −19 ⎦ 5 0 5 ⎤ ⎡ 56 15 ⎥ ⎢ −4 −11 −41 ⎦ ⎣ −1 −3 −11 2 −3 ⎢ −3 ⎣ 1 ⎤ 1 ⎥ 1 −19 ⎦ 5 0 ⎤ ⎡ 35 ⎥ ⎢ −26 ⎦ ⎣ −7 Thus, The solution is (1, 2, 0). A−1 B = ⎡ ⎤ −70 + 78−7 ⎢ ⎥ = −105−26 + 133 ⎣ ⎦ 35 + 0−35 ⎤ ⎥ ⎦ ⎡ 1 ⎢ 2 ⎣ 0 11.31 Solve the system using the inverse of the coefficient matrix. 2x − 17y + 11z = 0 − x + 11y − 7z = 8 3y −
2z = −2 Given a system of equations, solve with matrix inverses using a calculator. 1. Save the coefficient matrix and the constant matrix as matrix variables [A] and [B]. 2. Enter the multiplication into the calculator, calling up each matrix variable as needed. 3. If the coefficient matrix is invertible, the calculator will present the solution matrix; if the coefficient matrix is not invertible, the calculator will present an error message. Example 11.56 Using a Calculator to Solve a System of Equations with Matrix Inverses Solve the system of equations with matrix inverses using a calculator 2x + 3y + z = 32 3x + 3y + z = −27 2x + 4y + z = −2 Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1313 On the matrix page of the calculator, enter the coefficient matrix as the matrix variable [A], and enter the constant matrix as the matrix variable [B]. [A ⎤ ⎥, [B] = ⎦ ⎤ ⎡ 32 ⎥ ⎢ −27 ⎦ ⎣ −2 On the home screen of the calculator, type in the multiplication to solve for X, calling up each matrix variable as needed. Evaluate the expression. [A]−1×[B] ⎤ ⎡ −59 ⎥ ⎢ −34 ⎦ ⎣ 252 Access these online resources for additional instruction and practice with solving systems with inverses. • The Identity Matrix (http://openstaxcollege.org/l/identmatrix) • Determining Inverse Matrices (http://openstaxcollege.org/l/inversematrix) • Using a Matrix Equation to Solve a System of Equations (http://openstaxcollege.org/l/ matrixsystem) 1314 Chapter 11 Systems of Equations and Inequalities 11.7 EXERCISES Verbal 385. In a previous section, we showed that matrix multiplication is not commutative, that is, AB ≠ BA in most cases. Can you explain why matrix multiplication is commutative for matrix inverses, that is, A−1 A = AA−1? Does every 2×2 matrix have an inverse? Explain 386. why or why
not. Explain what condition is necessary for an inverse to exist. Can you explain whether a 2×2 matrix with an entire 387. row of zeros can have an inverse? Can a matrix with an entire column of zeros have an 388. inverse? Explain why or why not. Can a matrix with zeros on the diagonal have an 389. inverse? If so, find an example. If not, prove why not. For simplicity, assume a 2×2 matrix. Algebraic In the following exercises, show that matrix A is the inverse of matrix B. 390. 391. 392. 393. 394. 395. 396. A = ⎡ 1 0 ⎣ −1 1 ⎤ ⎦, ⎤ ⎦, ⎤ ⎦, B = ⎡ −0 ⎢ 1 ⎣ 5 1 7 − 4 35 ⎤ ⎥ ⎥ ⎦ A = ⎡ ⎤ ⎢−2 1 ⎥, B = 2 ⎣ ⎦ 3 −1 ⎤ ⎡ −2 −1 ⎦ ⎣ −6 −4 A = ⎡ ⎤ 1 0 1 ⎥, B = 1 ⎢ 0 1 − ⎤ ⎥, B = 1 4 ⎦ ⎤ ⎡ 0 −2 6 ⎥ ⎢ 17 −3 −5 ⎦ ⎣ 4 −12 12 ⎤ ⎥, B = 1 36 ⎦ ⎤ ⎡ −6 84 −6 ⎥ ⎢ 7 −26 1 ⎦ ⎣ −1 −22 5 For the following exercises, find the multiplicative inverse of each matrix, if it exists. This content is available for free at https://cnx.org/content/col11758/1.5 397. 398. 399. 400. 401. 402. 403. 404. 405. 406. 407. 408. 409. 410. ⎡ ⎤ 3 −2 ⎣ ⎦ 1 9 ⎡ −2 2 ⎣ 3 1 ⎤ ⎦ ⎡ −4 −3 ⎣ ⎦ −.5 1.5 ⎦ ⎣ 1 −0.5 ⎡ 1 0 6 ⎢ −1 1 �
�� ⎢ 1 −3 4 ⎦ ⎣ −2 −4 −5 ⎤ ⎡ 1 9 −2 7 3 ⎤ ⎡ 1 −2 ⎥ ⎢ −4 8 −12 ⎦ ⎣ ⎤ ⎥ ⎦ For the following exercises, solve the system using the inverse of a 2 × 2 matrix. 411. Chapter 11 Systems of Equations and Inequalities 1315 5x − 6y = − 61 4x + 3y = − 2 412. 8x + 4y = −100 3x−4y = 1 413. 3x−2y = 6 −x + 5y = −2 414. 5x−4y = −5 4x + y = 2.3 415. −3x−4y = 9 12x + 4y = −6 416. 417. 418. −2x + 3y = 3 10 − x + 5y = 10 For the following exercises, solve a system using the inverse of a 3×3 matrix. 3x−2y + 5z = 21 5x + 4y = 37 x−2y−5z = 5 4x + 4y + 4z = 40 2x − 3y + 4z = −12 − x + 3y + 4z = 9 6x − 5y − z = 31 − x + 2y + z = −6 3x + 3y + 2z = 13 6x−5y + 2z = −4 2x + 5y − z = 12 2x + 5y + z = 12 4x−2y + 3z = −12 2x + 2y−9z = 33 6y−4z = 1 419. 420. 421. 422. 423. 424. y + 4z = −41 2 z = −101 x − 1 1 10 5 x−20y + 2 1 5 5 3 10 x + 4y − 3 10 z = 23 425. z = 31 100 40 426. 0.1x + 0.2y + 0.3z = −1.4 0.1x−0.2y + 0.3z = 0.6 0.4y + 0.9z = −2 Technology For the following exercises, use a calculator to solve the system of equations with matrix inverses. 427. 2x − y = −3 −x +
2y = 2.3 428. 429. 430 + 11 2 5 y = − 43 20 y = 31 4 12.3x−2y−2.5z = 2 36.9x + 7y−7.5z = −7 8y−5z = −10 0.5x−3y + 6z = −0.8 0.7x−2y = −0.06 0.5x + 4y + 5z = 0 Extensions For the following exercises, find the inverse of the given matrix. 431. 432. 4331 ⎢ 0 ⎢ 0 ⎣ 1 −2 ⎢ 1 0 0 2 ⎢ 4 −2 3 1 ⎣ 1 1 0 −5 ⎤ ⎥ ⎥ ⎦ Chapter 11 Systems of Equations and Inequalities Three roommates shared a package of 12 ice cream 443. bars, but no one remembers who ate how many. If Tom ate twice as many ice cream bars as Joe, and Albert ate three less than Tom, how many ice cream bars did each roommate eat? A farmer constructed a chicken coop out of chicken 444. wire, wood, and plywood. The chicken wire cost $2 per square foot, the wood $10 per square foot, and the plywood $5 per square foot. The farmer spent a total of $51, and the total amount of materials used was 14 ft2. He used 3 ft2 more chicken wire than plywood. How much of each material in did the farmer use? Jay has lemon, orange, and pomegranate trees in his 445. backyard. An orange weighs 8 oz, a lemon 5 oz, and a pomegranate 11 oz. Jay picked 142 pieces of fruit weighing a total of 70 lb, 10 oz. He picked 15.5 times more oranges than pomegranates. How many of each fruit did Jay pick? 1316 434. 435 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ Real-World Applications For the following exercises, write a system of equations that represents the situation. Then, solve the system using the inverse of a matrix. 2,400 tickets were sold for a basketball game. If the 436. prices for floor 1 and floor 2 were different, and the total amount of money brought in is $64,000, how much was the price of each ticket? In the previous
exercise, if you were told there were 437. 400 more tickets sold for floor 2 than floor 1, how much was the price of each ticket? A food drive collected two different types of canned 438. goods, green beans and kidney beans. The total number of collected cans was 350 and the total weight of all donated food was 348 lb, 12 oz. If the green bean cans weigh 2 oz less than the kidney bean cans, how many of each can was donated? Students were asked to bring their favorite fruit to 439. class. 95% of the fruits consisted of banana, apple, and oranges. If oranges were twice as popular as bananas, and apples were 5% less popular than bananas, what are the percentages of each individual fruit? A sorority held a bake sale to raise money and sold 440. brownies and chocolate chip cookies. They priced the brownies at $1 and the chocolate chip cookies at $0.75. They raised $700 and sold 850 items. How many brownies and how many cookies were sold? A clothing store needs to order new inventory. It has 441. three different types of hats for sale: straw hats, beanies, and cowboy hats. The straw hat is priced at $13.99, the beanie at $7.99, and the cowboy hat at $14.49. If 100 hats were sold this past quarter, $1,119 was taken in by sales, and the amount of beanies sold was 10 more than cowboy hats, how many of each should the clothing store order to replace those already sold? Anna, Ashley, and Andrea weigh a combined 370 lb. 442. If Andrea weighs 20 lb more than Ashley, and Anna weighs 1.5 times as much as Ashley, how much does each girl weigh? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1317 11.8 \| Solving Systems with Cramer's Rule Learning Objectives In this section, you will: 11.8.1 Evaluate 2 × 2 determinants. 11.8.2 Use Cramer’s Rule to solve a system of equations in two variables. 11.8.3 Evaluate 3 × 3 determinants. 11.8.4 Use Cramer’s Rule to solve a system of three equations in three variables. 11.8.5 Know the properties of determinants. We have learned how to
solve systems of equations in two variables and three variables, and by multiple methods: substitution, addition, Gaussian elimination, using the inverse of a matrix, and graphing. Some of these methods are easier to apply than others and are more appropriate in certain situations. In this section, we will study two more strategies for solving systems of equations. Evaluating the Determinant of a 2×2 Matrix A determinant is a real number that can be very useful in mathematics because it has multiple applications, such as calculating area, volume, and other quantities. Here, we will use determinants to reveal whether a matrix is invertible by using the entries of a square matrix to determine whether there is a solution to the system of equations. Perhaps one of the more interesting applications, however, is their use in cryptography. Secure signals or messages are sometimes sent encoded in a matrix. The data can only be decrypted with an invertible matrix and the determinant. For our purposes, we focus on the determinant as an indication of the invertibility of the matrix. Calculating the determinant of a matrix involves following the specific patterns that are outlined in this section. Find the Determinant of a 2 × 2 Matrix The determinant of a 2 × 2 matrix, given is defined as A = a b ⎡ ⎣ c d ⎤ ⎦ Notice the change in notation. There are several ways to indicate the determinant, including det(A) and replacing the brackets in a matrix with straight lines, \|A\|. Example 11.57 Finding the Determinant of a 2 × 2 Matrix Find the determinant of the given matrix. A = ⎡ 5 2 ⎣ −6 3 ⎤ ⎦ Solution 1318 Chapter 11 Systems of Equations and Inequalities det(A) = \| 5 2 −6 3\| = 5(3) − (−6)(2) = 27 Using Cramer’s Rule to Solve a System of Two Equations in Two Variables We will now introduce a final method for solving systems of equations that uses determinants. Known as Cramer’s Rule, this technique dates back to the middle of the 18th century and is named for its innovator, the Swiss mathematician Gabriel Cramer (1704-1752), who introduced it in 1750 in Introduction à l'Analyse des lignes Courbes algébriques. Cramer’s Rule is a viable
and efficient method for finding solutions to systems with an arbitrary number of unknowns, provided that we have the same number of equations as unknowns. Cramer’s Rule will give us the unique solution to a system of equations, if it exists. However, if the system has no solution or an infinite number of solutions, this will be indicated by a determinant of zero. To find out if the system is inconsistent or dependent, another method, such as elimination, will have to be used. To understand Cramer’s Rule, let’s look closely at how we solve systems of linear equations using basic row operations. Consider a system of two equations in two variables. a1 x + b1 y = c1 (1) a2 x + b2 y = c2 (2) We eliminate one variable using row operations and solve for the other. Say that we wish to solve for x. If equation (2) is multiplied by the opposite of the coefficient of y in equation (1), equation (1) is multiplied by the coefficient of y in equation (2), and we add the two equations, the variable y will be eliminated. Now, solve for x. b2 a1 x + b2 b1 y = b2 c1 −b1 a2 x − b1 b2 y = − b1 c2 Multiply R2 by − b1 ________________________________________________________ Multiply R1 by b2 b2 a1 x − b1 a2 x = b2 c1 − b1 c2 b2 a1 x − b1 a2 x = b2 c1 − b1 c2 x(b2 a1 − b1 a2) = b2 c1 − b1 c2 x = b2 c1 − b1 c2 b2 a1 − b1 a2 = c1 b1 ⎤ ⎡ ⎦ ⎣ c2 b2 a1 b1 ⎤ ⎡ ⎦ ⎣ a2 b2 Similarly, to solve for y, we will eliminate x. a2 a1 x + a2 b1 y = a2 c1 −a1 a2 x − a1 b2 y = − a1 c2 Multiply R2 by − a1 ________________________________________________________ a2 b1 y − a1 b2 y = a2 c1 − a1 c2 Multiply R1 by a2 Solving
for y gives (11.4) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1319 a2 b1 y − a1 b2 y = a2 c1 − a1 c2 y(a2 b1 − a1 b2) = a2 c1 − a1 c2 y = a2 c1 − a1 c2 a2 b1 − a1 b2 = a1 c2 − a2 c1 a1 b2 − a2 b1 a1 c1 a2 c2\| = \| \|a1 b1 a2 b2\| Notice that the denominator for both x and y is the determinant of the coefficient matrix. We can use these formulas to solve for x and y, but Cramer’s Rule also introduces new notation: • D : determinant of the coefficient matrix • D x : determinant of the numerator in the solution of x • D y : determinant of the numerator in the solution of The key to Cramer’s Rule is replacing the variable column of interest with the constant column and calculating the determinants. We can then express x and y as a quotient of two determinants. Cramer’s Rule for 2×2 Systems Cramer’s Rule is a method that uses determinants to solve systems of equations that have the same number of equations as variables. Consider a system of two linear equations in two variables. The solution using Cramer’s Rule is given as a1 x + b1 y = c1 a2 x + b2 y = c2 x =, D ≠ 0; y = D x D = \|c1 b1 c2 b2\| \|a1 b1 a2 b2\| D y D = \| a1 c1 a2 c2\| \|a1 b1 a2 b2\|, D ≠ 0. (11.5) If we are solving for x, the x column is replaced with the constant column. If we are solving for y, the y column is replaced with the constant column. Example 11.58 Using Cramer’s Rule to Solve a 2 × 2 System Solve the following 2 × 2 system using Cramer’s Rule. 12x + 3y = 15 2x − 3y = 13 Solution 1320 Chapter 11 Systems of Equ
ations and Inequalities Solve for x. Solve for y. x = D x 3 D = \|15 13 −3\| \|12 2 −3\| 3 = −45 − 39 −36 − 6 = −84 −42 = 2 y = The solution is (2, −3). D y D = \|12 15 2 13\| \|12 2 −3\| 3 = 156 − 30 −36 − 6 = − 126 42 = −3 11.32 Use Cramer’s Rule to solve the 2 × 2 system of equations. x + 2y = −11 −2x + y = −13 Evaluating the Determinant of a 3 × 3 Matrix Finding the determinant of a 2×2 matrix is straightforward, but finding the determinant of a 3×3 matrix is more complicated. One method is to augment the 3×3 matrix with a repetition of the first two columns, giving a 3×5 matrix. Then we calculate the sum of the products of entries down each of the three diagonals (upper left to lower right), and subtract the products of entries up each of the three diagonals (lower left to upper right). This is more easily understood with a visual and an example. Find the determinant of the 3×3 matrix. 1. Augment A with the first two columns. A = a1 b1 c1 ⎡ ⎢ a2 b2 c2 ⎢ a3 b3 c3 ⎣ ⎤ ⎥ ⎥ ⎦ 2. From upper left to lower right: Multiply the entries down the first diagonal. Add the result to the product of entries down the second diagonal. Add this result to the product of the entries down the third diagonal. a1 b1 c1 a2 b2 c2 a3 b3 c3 a1 a2 a3 \| det(A) =\| b1 b2 b3\| 3. From lower left to upper right: Subtract the product of entries up the first diagonal. From this result subtract the product of entries up the second diagonal. From this result, subtract the product of entries up the third diagonal. The algebra is as follows: \|A\| = a1 b2 c3 + b1 c2 a3 + c1 a2 b3 − a3 b2 c1 − b3 c2 a1 − c3 a2 b1 This content is available for free at https://
cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1321 Example 11.59 Finding the Determinant of a 3 × 3 Matrix Find the determinant of the 3 × 3 matrix given A = ⎡ 0 2 1 ⎢ 3 −1 1 ⎣ 4 0 1 ⎤ ⎥ ⎦ Solution Augment the matrix with the first two columns and then follow the formula. Thus, \|A\| =\|0 2 1 3 −1 0\| = 0(−1)(1) + 2(1)(4) + 1(3)(0) − 4(−1)(1) − 0(1)(0) − 1(3)(2 11.33 Find the determinant of the 3 × 3 matrix. 1 1 1 det(A) =\|1 −3 7 1 −2 3\| Can we use the same method to find the determinant of a larger matrix? No, this method only works for 2 × 2 and 3 × 3 matrices. For larger matrices it is best to use a graphing utility or computer software. Using Cramer’s Rule to Solve a System of Three Equations in Three Variables Now that we can find the determinant of a 3 × 3 matrix, we can apply Cramer’s Rule to solve a system of three equations in three variables. Cramer’s Rule is straightforward, following a pattern consistent with Cramer’s Rule for 2 × 2 matrices. As the order of the matrix increases to 3 × 3, however, there are many more calculations required. When we calculate the determinant to be zero, Cramer’s Rule gives no indication as to whether the system has no solution or an infinite number of solutions. To find out, we have to perform elimination on the system. Consider a 3 × 3 system of equations. where = Dz D, D ≠ 0 1322 Chapter 11 Systems of Equations and Inequalities If we are writing the determinant D x, we replace the x column with the constant column. If we are writing the determinant D y, we replace the y column with the constant column. If we are writing the determinant Dz, we replace the z column with the constant column. Always check the answer. Example 11.60 Solving a 3 × 3 System Using Cramer’s Rule Find the solution to the given 3 × 3
system using Cramer’s Rule. x + y − z = 6 3x − 2y + z = −5 x + 3y − 2z = 14 Solution Use Cramer’s Rule. 3 −2 1 D =\|1 1 −1 1 3 −2\|, D x =\| 6 1 −1 −5 −2 1 14 3 −2\|, D y =\|1 6 −1 1 14 −2\|, Dz =\|1 1 1 3 14\| 6 3 −2 −5 3 −5 1 Then3 −3 D y D = −9 −3 Dz D = 6 −3 = 1 = 3 = − 2 The solution is (1, 3, −2). 11.34 Use Cramer’s Rule to solve the 3 × 3 matrix. x − 3y + 7z = 13 x + y + z = 1 x − 2y + 3z = 4 (11.6) Example 11.61 Using Cramer’s Rule to Solve an Inconsistent System Solve the system of equations using Cramer’s Rule. 3x − 2y = 4 (1) 6x − 4y = 0 (2) Solution We begin by finding the determinants D, D x, and D y. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1323 We know that a determinant of zero means that either the system has no solution or it has an infinite number of solutions. To see which one, we use the process of elimination. Our goal is to eliminate one of the variables. D = \|3 −2 6 −4\| = 3(−4) − 6(−2) = 0 1. Multiply equation (1) by −2. 2. Add the result to equation (2). −6x + 4y = −8 6x − 4y = 0 _______________ 0 = −8 (11.7) We obtain the equation 0 = −8, which is false. Therefore, the system has no solution. Graphing the system reveals two parallel lines. See Figure 11.31. Figure 11.31 Example 11.62 Use Cramer’s Rule to Solve a Dependent System Solve the system with an infinite number of solutions. x − 2y + 3z = 0 (1) 3x +
y − 2z = 0 (2) 2x − 4y + 6z = 0 (3) Solution Let’s find the determinant first. Set up a matrix augmented by the first two columns. 1324 Chapter 11 Systems of Equations and Inequalities Then, 3 1 −2 6 3 2 −4 \|1 −2 \| 1 −2 1 3 2 −4\| 1(1)(6) + (−2)(−2)(2) + 3(3)(−4) − 2(1)(3) − (−4)(−2)(1) − 6(3)(−2) = 0 As the determinant equals zero, there is either no solution or an infinite number of solutions. We have to perform elimination to find out. 1. Multiply equation (1) by −2 and add the result to equation (3): −2x + 4y − 6x = 0 2x − 4y + 6z = 0 0 = 0 2. Obtaining an answer of 0 = 0, a statement that is always true, means that the system has an infinite number of solutions. Graphing the system, we can see that two of the planes are the same and they both intersect the third plane on a line. See Figure 11.32. Figure 11.32 Understanding Properties of Determinants There are many properties of determinants. Listed here are some properties that may be helpful in calculating the determinant of a matrix. Properties of Determinants 1. If the matrix is in upper triangular form, the determinant equals the product of entries down the main diagonal. 2. When two rows are interchanged, the determinant changes sign. 3. 4. If either two rows or two columns are identical, the determinant equals zero. If a matrix contains either a row of zeros or a column of zeros, the determinant equals zero. 5. The determinant of an inverse matrix A−1 is the reciprocal of the determinant of the matrix A. 6. If any row or column is multiplied by a constant, the determinant is multiplied by the same factor. Example 11.63 Illustrating Properties of Determinants Illustrate each of the properties of determinants. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1325 Solution Property 1 states that if the matrix is in upper triangular form, the determinant is
the product of the entries down the main diagonal1 ⎤ ⎥ ⎦ Augment A with the first two columns. A = Then ⎡ 1 det(A) = 1(2)(−1) + 2(1)(0) + 3(0)(0) − 0(2)(3) − 0(1)(1) + 1(0)(2) = −2 Property 2 states that interchanging rows changes the sign. Given A = ⎤ ⎡ −1 5 ⎦, det(A) = (−1)(−3) − (4)(5) = 3 − 20 = −17 ⎣ 4 −3 B = ⎡ ⎤ 4 −3 ⎦, det(B) = (4)(5) − (−1)(−3) = 20 − 3 = 17 ⎣ −1 5 Property 3 states that if two rows or two columns are identical, the determinant equals zero1 \| det(A) = 1(2)(2) + 2(2)(−1) + 2(2)(2) + 1(2)(2) − 2(2)(1) − 2(2)(2 Property 4 states that if a row or column equals zero, the determinant equals zero. Thus, A = ⎡ 1 2 ⎣ 0 0 ⎤ ⎦, det(A) = 1(0) − 2(0) = 0 Property 5 states that the determinant of an inverse matrix A−1 is the reciprocal of the determinant A. Thus, A = ⎡ 1 2 ⎣ 3 4 ⎤ ⎦, det(A) = 1(4) − 3(2) = −2 A−1 = ⎡ −2 1 ⎢ 3 − 1 ⎣ 2 2 ⎤ ⎥, det ⎦ ⎝A−1⎞ ⎛ ⎛ ⎝− ⎞ ⎠(1) = − 1 2 Property 6 states that if any row or column of a matrix is multiplied by a constant, the determinant is multiplied by the same factor. Thus, A = ⎡ 1 2 ⎣ 3 4 ⎤ ⎦, det(A) = 1(4) − 2(3) = −2 B = ⎤ ⎡ 2(1) 2
(2) ⎦, det(B) = 2(4) − 3(4) = −4 ⎣ 3 4 1326 Chapter 11 Systems of Equations and Inequalities Example 11.64 Using Cramer’s Rule and Determinant Properties to Solve a System Find the solution to the given 3 × 3 system. 2x + 4y + 4z = 2 (1) 3x + 7y + 7z = −5 (2) x + 2y + 2z = 4 (3) Solution Using Cramer’s Rule, we have Notice that the second and third columns are identical. According to Property 3, the determinant will be zero, so there is either no solution or an infinite number of solutions. We have to perform elimination to find out. 3 7 7 D =\|2 4 4 1 2 2\| 1. Multiply equation (3) by –2 and add the result to equation (1). −2x − 4y − 4x = − 8 2x + 4y + 4z = 2 0 = − 6 Obtaining a statement that is a contradiction means that the system has no solution. Access these online resources for additional instruction and practice with Cramer’s Rule. • Solve a System of Two Equations Using Cramer's Rule (http://openstaxcollege.org/l/ system2cramer) • Solve a Systems of Three Equations using Cramer's Rule (http://openstaxcollege.org/l/ system3cramer) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1327 11.8 EXERCISES Verbal Explain why we can always evaluate the determinant 446. of a square matrix. 447. Examining Cramer’s Rule, explain why there is no unique solution to the system when the determinant of your matrix is 0. For simplicity, use a 2 × 2 matrix. Explain what it means in terms of an inverse for a 448. matrix to have a 0 determinant. The determinant of 2 × 2 matrix A is 3. 449. If you switch the rows and multiply the first row by 6 and the second row by 2, explain how to find the determinant and provide the answer. Algebraic For the following exercises, find the determinant. 450. 451. 452
. 453. 454. 455. 456. 457. 458. 459. 460. 461. 0 2 20 −1 \|1 2 3 4\| \|−1 3 −4\| \| 2 −5 6\| \|−8 4 −1 5\| \|1 3 −4\| \|10 0 −10\| \|10 0.2 5 0.1\| \|6 −3 4\| \|−2 3.1 4, 000\| \|−1.1 7.2 −0.5\| \|−1 0 0 0 −3\| 0 1 0.6 −3 0 0 8 463. 462. 464. 465. 0 3 2 5 −5 −4 0 2 0 1 0 3 −4 1 4 1 2 −8 −4 −3 1 \|−1 4 0 0 −3\| \|1 0 1 1 0 0\| \| 2 −3 1 6 1\| \|−2 2 −8 −3\| \| 6 −1 9 −1\| \|5 3 −6 −3\| \|1.1 4.1 −0.4 2.5\| \| \| 81 0 0 1 −1 1 3 1.1 −9.3 1 4 1 7 1 −4 0 0 2 469. 466. 467. 468. 2 −1.6 3.1 3 −8 0 2\| For the following exercises, solve the system of linear equations using Cramer’s Rule. 470. 2x − 3y = −1 4x + 5y = 9 471. 5x − 4y = 2 −4x + 7y = 6 472. 6x − 3y = 2 −8x + 9y = −1 473. 2x + 6y = 12 5x − 2y = 13 474. 1328 Chapter 11 Systems of Equations and Inequalities 4x + 3y = 23 2x − y = −1 475. 10x − 6y = 2 −5x + 8y = −1 476. 4x − 3y = −3 2x + 6y = −4 477. 4x − 5y = 7 −3x + 9y = 0 478. 4x + 10y = 180 −3x − 5y = −105 479. 8x − 2y = −3 −4x + 6y = 4 For the following exercises, solve the system of linear equations using Cramer’s
Rule. x + 2y − 4z = − 1 7x + 3y + 5z = 26 −2x − 6y + 7z = − 6 −5x + 2y − 4z = − 47 4x − 3y − z = − 94 3x − 3y + 2z = 94 4x + 5y − z = −7 −2x − 9y + 2z = 8 5y + 7z = 21 4x − 3y + 4z = 10 5x − 2z = − 2 3x + 2y − 5z = − 9 4x − 2y + 3z = 6 − 6x + y = − 2 2x + 7y + 8z = 24 5x + 2y − z = 1 −7x − 8y + 3z = 1.5 6x − 12y + z = 7 13x − 17y + 16z = 73 −11x + 15y + 17z = 61 46x + 10y − 30z = − 18 480. 481. 482. 483. 484. 485. 486. 487. This content is available for free at https://cnx.org/content/col11758/1.5 −4x − 3y − 8z = − 7 2x − 9y + 5z = 0.5 5x − 6y − 5z = − 2 488. 489. 4x − 6y + 8z = 10 −2x + 3y − 4z = − 5 x + y + z = 1 4x − 6y + 8z = 10 −2x + 3y − 4z = − 5 12x + 18y − 24z = − 30 Technology For the following exercises, use the determinant function on a graphing utility. 490. 491. 492. 4939 3 0 1 2 1 3 0 −2 −1 1 0 2 4 3\| \| 1 −2\| \| \| 7 8 9 0 100 2 2,000 0 4 5 2\| Real-World Applications For the following exercises, create a system of linear equations to describe the behavior. Then, calculate the determinant. Will there be a unique solution? If so, find the unique solution. Two numbers add up to 56. One number is 20 less 494. than the other. Two numbers add up to 104. If you add two times the 495. first number plus two times
the second number, your total is 208 Three numbers add up to 106. The first number is 3 496. less than the second number. The third number is 4 more than the first number. 497. Chapter 11 Systems of Equations and Inequalities 1329 Three numbers add to 216. The sum of the first two numbers is 112. The third number is 8 less than the first two numbers combined. For the following exercises, create a system of linear equations to describe the behavior. Then, solve the system for all solutions using Cramer’s Rule. You invest $10,000 into two accounts, which receive 498. 8% interest and 5% interest. At the end of a year, you had $10,710 in your combined accounts. How much was invested in each account? You invest $80,000 into two accounts, $22,000 in one 499. account, and $58,000 in the other account. At the end of one year, assuming simple interest, you have earned $2,470 in interest. The second account receives half a percent less than twice the interest on the first account. What are the interest rates for your accounts? A movie theater needs to know how many adult 500. tickets and children tickets were sold out of the 1,200 total tickets. If children’s tickets are $5.95, adult tickets are $11.15, and the total amount of revenue was $12,756, how many children’s tickets and adult tickets were sold? A concert venue sells single tickets for $40 each and 501. couple’s tickets for $65. If the total revenue was $18,090 and the 321 tickets were sold, how many single tickets and how many couple’s tickets were sold? You decide to paint your kitchen green. You create the 502. color of paint by mixing yellow and blue paints. You cannot remember how many gallons of each color went into your mix, but you know there were 10 gal total. Additionally, you kept your receipt, and know the total amount spent was $29.50. If each gallon of yellow costs $2.59, and each gallon of blue costs $3.19, how many gallons of each color go into your green mix? You sold two types of scarves at a farmers’ market 503. and would like to know which one was more popular. The total number of scarves sold was 56, the yellow scarf cost $10, and the purple
scarf cost $11. If you had total revenue of $583, how many yellow scarves and how many purple scarves were sold? Your garden produced two types of tomatoes, one 504. green and one red. The red weigh 10 oz, and the green weigh 4 oz. You have 30 tomatoes, and a total weight of 13 lb, 14 oz. How many of each type of tomato do you have? At a market, the three most popular vegetables make 505. up 53% of vegetable sales. Corn has 4% higher sales than broccoli, which has 5% more sales than onions. What percentage does each vegetable have in the market share? At the same market, the three most popular fruits 506. make up 37% of the total fruit sold. Strawberries sell twice as much as oranges, and kiwis sell one more percentage point than oranges. For each fruit, find the percentage of total fruit sold. Three bands performed at a concert venue. The first 507. band charged $15 per ticket, the second band charged $45 per ticket, and the final band charged $22 per ticket. There were 510 tickets sold, for a total of $12,700. If the first band had 40 more audience members than the second band, how many tickets were sold for each band? A movie theatre sold tickets to three movies. The 508. tickets to the first movie were $5, the tickets to the second movie were $11, and the third movie was $12. 100 tickets were sold to the first movie. The total number of tickets sold was 642, for a total revenue of $6,774. How many tickets for each movie were sold? Men aged 20–29, 30–39, and 40–49 made up 78% of 509. the population at a prison last year. This year, the same age groups made up 82.08% of the population. The 20–29 age group increased by 20%, the 30–39 age group increased by 2%, and the 40–49 age group decreased to 3 4 of their previous population. Originally, the 30–39 age group had 2% more prisoners than the 20–29 age group. Determine the prison population percentage for each age group last year. 510. At a women’s prison down the road, the total number of inmates aged 20–49 totaled 5,525. This year, the 20–29 age group increased by 10%, the 30–39 age group decreased by 20%, and
the 40–49 age group doubled. There are now 6,040 prisoners. Originally, there were 500 more in the 30–39 age group than the 20–29 age group. Determine the prison population for each age group last year. For the following exercises, use this scenario: A healthconscious company decides to make a trail mix out of almonds, chocolate-covered cashews. The nutritional information for these items is shown in Table 11.5. cranberries, dried and Fat (g) 6 Almonds (10) Cranberries (10) 0.02 Protein (g) Carbohydrates (g) 2 0 3 8 7 3.5 5.5 Cashews (10) Table 11.5 511. 1330 Chapter 11 Systems of Equations and Inequalities For the special “low-carb”trail mix, there are 1,000 pieces of mix. The total number of carbohydrates is 425 g, and the total amount of fat is 570.2 g. If there are 200 more pieces of cashews than cranberries, how many of each item is in the trail mix? For the “hiking” mix, there are 1,000 pieces in the 512. mix, containing 390.8 g of fat, and 165 g of protein. If there is the same amount of almonds as cashews, how many of each item is in the trail mix? For the “energy-booster” mix, there are 1,000 pieces 513. in the mix, containing 145 g of protein and 625 g of carbohydrates. If the number of almonds and cashews summed together amount of equivalent cranberries, how many of each item is in the trail mix? to the is This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1331 CHAPTER 11 REVIEW KEY TERMS addition method an algebraic technique used to solve systems of linear equations in which the equations are added in a way that eliminates one variable, allowing the resulting equation to be solved for the remaining variable; substitution is then used to solve for the first variable augmented matrix brackets a coefficient matrix adjoined with the constant column separated by a vertical line within the matrix break-even point the point at which a cost function intersects a revenue function; where profit is zero coefficient matrix a matrix that contains only the coefficients from a system of equations column a set of numbers aligned vertically in a matrix consistent
system a system for which there is a single solution to all equations in the system and it is an independent system, or if there are an infinite number of solutions and it is a dependent system cost function costs the function used to calculate the costs of doing business; it usually has two parts, fixed costs and variable Cramer’s Rule a method for solving systems of equations that have the same number of equations as variables using determinants dependent system a system of linear equations in which the two equations represent the same line; there are an infinite number of solutions to a dependent system determinant a number calculated using the entries of a square matrix that determines such information as whether there is a solution to a system of equations entry an element, coefficient, or constant in a matrix feasible region the solution to a system of nonlinear inequalities that is the region of the graph where the shaded regions of each inequality intersect Gaussian elimination using elementary row operations to obtain a matrix in row-echelon form identity matrix algebra a square matrix containing ones down the main diagonal and zeros everywhere else; it acts as a 1 in matrix inconsistent system a system of linear equations with no common solution because they represent parallel lines, which have no point or line in common independent system a system of linear equations with exactly one solution pair (x, y) main diagonal entries from the upper left corner diagonally to the lower right corner of a square matrix matrix a rectangular array of numbers multiplicative inverse of a matrix a matrix that, when multiplied by the original, equals the identity matrix nonlinear inequality an inequality containing a nonlinear expression partial fraction decomposition the process of returning a simplified rational expression to its original form, a sum or difference of simpler rational expressions partial fractions the individual fractions that make up the sum or difference of a rational expression before combining them into a simplified rational expression profit function the profit function is written as P(x) = R(x) − C(x), revenue minus cost revenue function the function that is used to calculate revenue, simply written as R = xp, where x = quantity and p = price row a set of numbers aligned horizontally in a matrix 1332 Chapter 11 Systems of Equations and Inequalities row operations adding one row to another row, multiplying a row by a constant, interchanging rows, and so on, with the goal of achieving row-echelon form row-echelon form after performing row operations, the matrix form that contains ones down the main diagonal and zeros at every space below the diagonal row-equ
ivalent two matrices A and B are row-equivalent if one can be obtained from the other by performing basic row operations scalar multiple an entry of a matrix that has been multiplied by a scalar solution set the set of all ordered pairs or triples that satisfy all equations in a system of equations substitution method an algebraic technique used to solve systems of linear equations in which one of the two equations is solved for one variable and then substituted into the second equation to solve for the second variable system of linear equations a set of two or more equations in two or more variables that must be considered simultaneously. system of nonlinear equations a system of equations containing at least one equation that is of degree larger than one system of nonlinear inequalities inequality that is not linear KEY EQUATIONS Identity matrix for a 2×2 matrix Identity matrix for a 3×3 matrix a system of two or more inequalities in two or more variables containing at least one I2 = ⎡ 1 0 ⎣ 0 1 ⎤ ⎦ I3 = ⎡ ⎤ ⎥ ⎦ Multiplicative inverse of a 2×2 matrix A−1 = 1 ad − bc d −b ⎡ ⎣ −c a ⎤ ⎦, where ad − bc ≠ 0 KEY CONCEPTS 11.1 Systems of Linear Equations: Two Variables • A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously. • The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. See Example 11.1. • Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution. • One method of solving a system of linear equations in two variables is by graphing. In this method, we graph the equations on the same set of axes. See Example 11.2. • Another method of solving a system of linear equations is by substitution. In this method, we solve for one variable in one equation and substitute the result into the second equation. See Example 11.3. • A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables. See Example 11.4. • It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two equations together. See Example
11.5, Example 11.6, and Example 11.7. • Either method of solving a system of equations results in a false statement for inconsistent systems because they are made up of parallel lines that never intersect. See Example 11.8. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1333 • The solution to a system of dependent equations will always be true because both equations describe the same line. See Example 11.9. • Systems of equations can be used to solve real-world problems that involve more than one variable, such as those relating to revenue, cost, and profit. See Example 11.10 and Example 11.11. 11.2 Systems of Linear Equations: Three Variables • A solution set is an ordered triple ⎧ ⎨(x, y, z)⎫ ⎭ ⎩ ⎬ that represents the intersection of three planes in space. See Example 11.12. • A system of three equations in three variables can be solved by using a series of steps that forces a variable to be eliminated. The steps include interchanging the order of equations, multiplying both sides of an equation by a nonzero constant, and adding a nonzero multiple of one equation to another equation. See Example 11.13. • Systems of three equations in three variables are useful for solving many different types of real-world problems. See Example 11.14. • A system of equations in three variables is inconsistent if no solution exists. After performing elimination operations, the result is a contradiction. See Example 11.15. • Systems of equations in three variables that are inconsistent could result from three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location. • A system of equations in three variables is dependent if it has an infinite number of solutions. After performing elimination operations, the result is an identity. See Example 11.16. • Systems of equations in three variables that are dependent could result from three identical planes, three planes intersecting at a line, or two identical planes that intersect the third on a line. 11.3 Systems of Nonlinear Equations and Inequalities: Two Variables • There are three possible types of solutions to a system of equations representing a line and a parabola: (1) no solution, the line does not intersect the parabola;
(2) one solution, the line is tangent to the parabola; and (3) two solutions, the line intersects the parabola in two points. See Example 11.17. • There are three possible types of solutions to a system of equations representing a circle and a line: (1) no solution, the line does not intersect the circle; (2) one solution, the line is tangent to the parabola; (3) two solutions, the line intersects the circle in two points. See Example 11.18. • There are five possible types of solutions to the system of nonlinear equations representing an ellipse and a circle: (1) no solution, the circle and the ellipse do not intersect; (2) one solution, the circle and the ellipse are tangent to each other; (3) two solutions, the circle and the ellipse intersect in two points; (4) three solutions, the circle and ellipse intersect in three places; (5) four solutions, the circle and the ellipse intersect in four points. See Example 11.19. • An inequality is graphed in much the same way as an equation, except for > or <, we draw a dashed line and shade the region containing the solution set. See Example 11.20. • Inequalities are solved the same way as equalities, but solutions to systems of inequalities must satisfy both inequalities. See Example 11.21. 11.4 Partial Fractions • Decompose P(x) Q(x) by writing the partial fractions as A a1 x + b1 + B a2 x + b2. Solve by clearing the fractions, expanding the right side, collecting like terms, and setting corresponding coefficients equal to each other, then setting up and solving a system of equations. See Example 11.22. • The decomposition of P(x) Q(x) with repeated linear factors must account for the factors of the denominator in increasing powers. See Example 11.23. 1334 Chapter 11 Systems of Equations and Inequalities • The decomposition of P(x) Q(x) with a nonrepeated irreducible quadratic factor needs a linear numerator over the quadratic factor, as in A x + Bx + C ⎝ax2 + bx + c⎞ ⎛ ⎠. See Example 11.24. • In the
decomposition of P(x) Q(x), where Q(x) has a repeated irreducible quadratic factor, when the irreducible quadratic factors are repeated, powers of the denominator factors must be represented in increasing powers as Ax + B ⎝ax2 + bx + c⎞ ⎛ ⎠ + A2 x + B2 ⎝ax2 + bx + c⎞ ⎛ ⎠ 2 + ⋯ + An x + Bn ⎝ax2 + bx + c⎞ ⎛ ⎠ n. See Example 11.25. 11.5 Matrices and Matrix Operations • A matrix is a rectangular array of numbers. Entries are arranged in rows and columns. • The dimensions of a matrix refer to the number of rows and the number of columns. A 3×2 matrix has three rows and two columns. See Example 11.26. • We add and subtract matrices of equal dimensions by adding and subtracting corresponding entries of each matrix. See Example 11.27, Example 11.28, Example 11.29, and Example 11.30. • Scalar multiplication involves multiplying each entry in a matrix by a constant. See Example 11.31. • Scalar multiplication is often required before addition or subtraction can occur. See Example 11.32. • Multiplying matrices is possible when inner dimensions are the same—the number of columns in the first matrix must match the number of rows in the second. • The product of two matrices, A and B, is obtained by multiplying each entry in row 1 of A by each entry in column 1 of B; then multiply each entry of row 1 of A by each entry in columns 2 of B, and so on. See Example 11.33 and Example 11.34. • Many real-world problems can often be solved using matrices. See Example 11.35. • We can use a calculator to perform matrix operations after saving each matrix as a matrix variable. See Example 11.36. 11.6 Solving Systems with Gaussian Elimination • An augmented matrix is one that contains the coefficients and constants of a system of equations. See Example 11.37. • A matrix augmented with the constant column can be represented as the original system of equations. See Example 11.38. • Row operations include multiplying a row by a constant, adding one row to another row, and inter
changing rows. • We can use Gaussian elimination to solve a system of equations. See Example 11.39, Example 11.40, and Example 11.41. • Row operations are performed on matrices to obtain row-echelon form. See Example 11.42. • To solve a system of equations, write it in augmented matrix form. Perform row operations to obtain row-echelon form. Back-substitute to find the solutions. See Example 11.43 and Example 11.44. • A calculator can be used to solve systems of equations using matrices. See Example 11.45. • Many real-world problems can be solved using augmented matrices. See Example 11.46 and Example 11.47. 11.7 Solving Systems with Inverses • An identity matrix has the property AI = IA = A. See Example 11.48. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1335 • An invertible matrix has the property AA−1 = A−1 A = I. See Example 11.49. • Use matrix multiplication and the identity to find the inverse of a 2×2 matrix. See Example 11.50. • The multiplicative inverse can be found using a formula. See Example 11.51. • Another method of finding the inverse is by augmenting with the identity. See Example 11.52. • We can augment a 3×3 matrix with the identity on the right and use row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse. See Example 11.53. • Write the system of equations as AX = B, and multiply both sides by the inverse of A : A−1 AX = A−1 B. See Example 11.54 and Example 11.55. • We can also use a calculator to solve a system of equations with matrix inverses. See Example 11.56. 11.8 Solving Systems with Cramer's Rule a b • The determinant for ⎡ ⎣ c d ⎤ is ad − bc. See Example 11.57. ⎦ • Cramer’s Rule replaces a variable column with the constant column. Solutions are. See Example 11.58. • To find the determinant of a 3×3 matrix, augment with the first two columns. Add the three diagonal entries (upper left
to lower right) and subtract the three diagonal entries (lower left to upper right). See Example 11.59. • To solve a system of three equations in three variables using Cramer’s Rule, replace a variable column with the constant column for each desired solution = Dz D. See Example 11.60. • Cramer’s Rule is also useful for finding the solution of a system of equations with no solution or infinite solutions. See Example 11.61 and Example 11.62. • Certain properties of determinants are useful for solving problems. For example: ◦ If the matrix is in upper triangular form, the determinant equals the product of entries down the main diagonal. ◦ When two rows are interchanged, the determinant changes sign. ◦ ◦ If either two rows or two columns are identical, the determinant equals zero. If a matrix contains either a row of zeros or a column of zeros, the determinant equals zero. ◦ The determinant of an inverse matrix A−1 is the reciprocal of the determinant of the matrix A. ◦ If any row or column is multiplied by a constant, the determinant is multiplied by the same factor. See Example 11.63 and Example 11.64. CHAPTER 11 REVIEW EXERCISES Systems of Linear Equations: Two Variables For the following exercises, determine whether the ordered pair is a solution to the system of equations. 514. 3x − y = 4 x + 4y = − 3 and ( − 1, 1) 515. 6x − 2y = 24 −3x + 3y = 18 and (9, 15) For the following exercises, use substitution to solve the system of equations. 516. 10x + 5y = −5 3x − 2y = −12 517 = 43 70 y = − 2 3 1336 Chapter 11 Systems of Equations and Inequalities 518. 5x + 6y = 14 4x + 8y = 8 For the following exercises, use addition to solve the system of equations. 519. 3x + 2y = −7 2x + 4y = 6 520. 3x + 4y = 2 9x + 12y = 3 521. 8x + 4y = 2 6x − 5y = 0.7 For the following exercises, write a system of equations to solve each problem. Solve the system of equations. 522. has A factory C(x
) = 150x + 15,000 and revenue R(x) = 200x. What is the break-even point? cost of a a production function 523. A performer charges C(x) = 50x + 10,000, where x is the total number of attendees at a show. The venue charges $75 per ticket. After how many people buy tickets does the venue break even, and what is the value of the total tickets sold at that point? Systems of Linear Equations: Three Variables 529. 530. 531. 3x + 4z = −11 x − 2y = 5 4y − z = −10 2x − 3y + z = 0 2x + 4y − 3z = 0 6x − 2y − z = 0 6x − 4y − 2z = 2 3x + 2y − 5z = 4 6y − 7z = 5 For the following exercises, write a system of equations to solve each problem. Solve the system of equations. 532. Three odd numbers sum up to 61. The smaller is onethird the larger and the middle number is 16 less than the larger. What are the three numbers? 533. A local theatre sells out for their show. They sell all 500 tickets for a total purse of $8,070.00. The tickets were priced at $15 for students, $12 for children, and $18 for adults. If the band sold three times as many adult tickets as children’s tickets, how many of each type was sold? Systems of Nonlinear Equations and Inequalities: Two Variables For the following exercises, solve the system of nonlinear equations. For the following exercises, solve the system of three equations using substitution or addition. 534. y = x2 − 7 y = 5x − 13 524. 525. 526. 527. 528. 0.5x − 0.5y = 10 − 0.2y + 0.2x = 4 0.1x + 0.1z = 2 5x + 3y − z = 5 3x − 2y + 4z = 13 4x + 3y + 5z = 22 x + y + z = 1 2x + 2y + 2z = 1 3x + 3y = 2 2x − 3y + z = −1 x + y + z = −4 4x + 2y − 3z = 33 3x +
2y − z = −10 x − y + 2z = 7 −x + 3y + z = −2 This content is available for free at https://cnx.org/content/col11758/1.5 535. y = x2 − 4 y = 5x + 10 536. x2 + y2 = 16 y = x − 8 537. x2 + y2 = 25 y = x2 + 5 538. x2 + y2 = 4 y − x2 = 3 For the following exercises, graph the inequality. 539. y > x2 − 1 Chapter 11 Systems of Equations and Inequalities 1337 540. 1 4 x2 + y2 < 4 the following exercises, graph the system of For inequalities. 541. x2 + y2 + 2x < 3 y > − x2 − 3 542. x2 − 2x + y2 − 4x < 4 y < − x + 4 543. x2 + y2 < 1 y2 < x Partial Fractions the following exercises, decompose into partial For fractions. 544. −2x + 6 x2 + 3x + 2 545. 10x + 2 4x2 + 4x + 1 546. 7x + 20 x2 + 10x + 25 547. x − 18 x2 − 12x + 36 548. −x2 + 36x + 70 x3 − 125 549. −5x2 + 6x − 2 x3 + 27 553. 10D − 6E 554. B + C 555. AB 556. BA 557. BC 558. CB 559. DE 560. ED 561. EC 562. CE 563. A3 Solving Systems with Gaussian Elimination For the following exercises, write the system of linear equations from the augmented matrix. Indicate whether there will be a unique solution. 564. 565. ⎡ 1 0 −2 ⎣ 0 0 0 ⎤ 7 ⎥ −5 ⎦ 0 ⎤ −9 ⎥ 4 ⎦ 3 \| \| For the following exercises, write the augmented matrix from the system of linear equations. 550. x3 − 4x2 + 3x + 11 (x2 − 2)2 551. 4x4 − 2x3 + 22x2 − 6x + 48 x(x2 + 4
)2 Matrices and Matrix Operations the following exercises, perform the requested For operations on the given matrices. A = ⎡ 4 −2 ⎣ 3 1 ⎤ ⎦, B = ⎡ 6 ⎣ 11 −2 ⎤ 7 −3 ⎦, C = 4 ⎡ ⎤ 7 6 ⎢ ⎥, D = 11 −2 ⎣ ⎦ 14 0 ⎡ 1 −4 ⎢ 10 ⎣ 2 9 5 −7 5 8 ⎤ ⎥, E = ⎦ ⎡ ⎤ 7 −14 3 ⎢ ⎥ 2 −1 3 ⎣ ⎦ 1 9 0 552. −4A 566. 567. 568. −2x + 2y + z = 7 2x − 8y + 5z = 0 19x − 10y + 22z = 3 4x + 2y − 3z = 14 −12x + 3y + z = 100 9x − 6y + 2z = 31 x + 3z = 12 −x + 4y = 0 y + 2z = − 7 1338 Chapter 11 Systems of Equations and Inequalities For the following exercises, solve the system of linear equations using Gaussian elimination. 569. 3x − 4y = − 7 −6x + 8y = 14 570. 3x − 4y = 1 −6x + 8y = 6 571. −1.1x − 2.3y = 6.2 −5.2x − 4.1y = 4.3 572. 573. 2x + 3y + 2z = 1 −4x − 6y − 4z = − 2 10x + 15y + 10z = 0 −x + 2y − 4z = 8 3y + 8z = − 4 −7x + y + 2z = 1 Solving Systems with Inverses For the following exercises, find the inverse of the matrix. 574. ⎡ −0.2 ⎣ 1.4 1.2 −0.4 ⎤ ⎦ 575. 576. 577 ⎡ 12 9 −6 ⎢ 2 −1 3 ⎣ −4 − ⎤ ⎥ ⎦ For the following exercises, computing the inverse of the matrix. find the

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