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yx 0 2xy 0 y A0, 24B 10 x+y=32 A16, 16B x+2y=48 A0, 0B 10 A32, 0B x FIGURE 1 The solution of this system (with vertices labeled) is sketched in Figure 1. The only values that satisfy the restrictions of the problem are the ones that correspond to points of the shaded region in Figure 1. This is called the feasible region for the problem. d Linear programming helps the telephone industry to determine the most efﬁcient way to route telephone calls. The computerized routing decisions must be made very rapidly so that callers are not kept waiting for connections. Since the database of customers and routes is huge, an extremely fast method for solving linear programming problems is essential. In 1984 the 28-year-old mathematician Narendra Karmarkar, working at Bell Labs in Murray Hill, New Jersey, discovered just such a method. His idea is so ingenious and his method so fast that the discovery caused a sensation in the mathematical world. Although mathematical discoveries rarely make the news, this one was reported in Time, on December 3, 1984. Today airlines routinely use Karmarkar’s technique to minimize costs in scheduling passengers, ﬂight personnel, fuel, baggage, and maintenance workers. Find maximum proﬁt. As x or y increases, proﬁt increases as well. Thus, it seems reasonable that the maximum proﬁt will occur at a point on one of the outside edges of the feasible region, where it is impossible to increase x or y without going outside the region. In fact, it can be shown that the maximum value occurs at a vertex. This means that we need to check the proﬁt only at the vertices. The largest value of P occurs at the point, where P $560. Thus, the manufacturer should make 16 pairs of oxfords and 16, 16 1 2 16 pairs of loafers, for a maximum daily proﬁt of $560. Vertex 0, 0 2 1 0, 24 1 2 16, 16 1 32, 0 1 2 2 P 15x 20y 0 15 15 15 0 1 2 16 1 32 1 2 2 20 20 20 2 2 24 1 16 1 0 1 2 $480 $560 $480 Maximum proﬁt The linear programming problems that we consider all follow the pattern of Example 1. Each problem involves two variables. The
problem describes restrictions, called constraints, that lead to a system of linear inequalities whose solution is called the feasible region. The function that we wish to maximize or minimize is called the objective function. This function always attains its largest and smallest values at the vertices of the feasible region. This modeling technique involves four steps, summarized in the following box. 488 Focus on Modeling GUIDELINES FOR LINEAR PROGRAMMING 1. Choose the Variables. Decide what variable quantities in the problem should be named x and y. 2. Find the Objective Function. Write an expression for the function we want to maximize or minimize. 3. Graph the Feasible Region. Express the constraints as a system of inequali- ties, and graph the solution of this system (the feasible region). 4. Find the Maximum or Minimum. Evaluate the objective function at the vertices of the feasible region to determine its maximum or minimum value. E X AM P L E 2 \| A Shipping Problem A car dealer has warehouses in Millville and Trenton and dealerships in Camden and Atlantic City. Every car that is sold at the dealerships must be delivered from one of the warehouses. On a certain day the Camden dealers sell 10 cars, and the Atlantic City dealers sell 12. The Millville warehouse has 15 cars available, and the Trenton warehouse has 10. The cost of shipping one car is $50 from Millville to Camden, $40 from Millville to Atlantic City, $60 from Trenton to Camden, and $55 from Trenton to Atlantic City. How many cars should be moved from each warehouse to each dealership to ﬁll the orders at minimum cost? ▼ SO LUTI O N Our ﬁrst step is to organize the given information. Rather than construct a table, we draw a diagram to show the ﬂow of cars from the warehouses to the dealerships (see Figure 2 below). The diagram shows the number of cars available at each warehouse or required at each dealership and the cost of shipping between these locations. Choose the variables. The arrows in Figure 2 indicate four possible routes, so the problem seems to involve four variables. But we let x number of cars to be shipped from Millville to Camden y number of cars to be shipped from Millville to Atlantic City To ﬁll the orders, we must have 10 x number of cars shipped from Trenton to Camden 12 y number of cars shipped from Trenton to Atlantic City So the only variables in the
problem are x and y. Camden Sell 10 Millville 15 cars $50 Ship x cars Ship y cars $40 Ship 10-x cars Ship 12-y cars $60 $55 Trenton 10 cars Atlantic City Sell 12 FIGURE 2 Linear Programming 489 Find the objective function. The objective of this problem is to minimize cost. From Figure 2 we see that the total cost C of shipping the cars is 55 C 50x 40y 60 10 x 12 y 1 50x 40y 600 60x 660 55y 1260 10x 15y 2 1 2 This is the objective function. Graph the feasible region. Now we derive the constraint inequalities that deﬁne the feasible region. First, the number of cars shipped on each route can’t be negative, so we have x 0 y 0 10 x 0 12 y 0 Second, the total number of cars shipped from each warehouse can’t exceed the number of cars available there, so 2 Simplifying the latter inequality, we get 1 10 x x y 15 10 12 y 2 1 y x=10 22 x y 10 (3, 12) y=12 (0, 12) x+y=12 (10, 2) (10, 5) x+y=15 x x y 12 x y 12 The inequalities 10 x 0 and 12 y 0 can be rewritten as x 10 and y 12. Thus, the feasible region is described by the constraints x y 15 x y 12 0 x 10 0 y 12 FIGURE 3 The feasible region is graphed in Figure 3. d Find minimum cost. We check the value of the objective function at each vertex of the feasible region. Vertex 0, 12 1 3, 12 1 10, 5 1 10, 2 1 2 2 2 2 C 1260 10x 15y 1260 10 1260 10 1260 10 1260 10 0 2 1 3 1 2 10 2 1 10 2 1 2 2 15 12 1 15 12 1 15 5 2 1 15 2 2 1 $1080 $1050 $1085 $1130 Minimum cost The lowest cost is incurred at the point Ó3, 12Ô. Thus, the dealer should ship 3 cars from Millville to Camden 12 cars from Millville to Atlantic City 7 cars from Trenton to Camden 0 cars from Trenton to Atlantic City ▲ In the 1940s mathematicians developed matrix methods for solving linear programming problems that involve more than two variables. These methods were ﬁrst used by the Allies in World War II to
solve supply problems similar to (but, of course, much more complicated than) Example 2. Improving such matrix methods is an active and exciting area of current mathematical research. 490 Focus on Modeling Problems 1–4 ■ Find the maximum and minimum values of the given objective function on the indicated feasible region. 1. M 200 x y 2. N x y 40 1 2 1 4 y 5 2 0 y=. P 140 x 3y x 0, y 00 2x y 10 2x 4y 28 4. Q 70x 82y x 0, y 0 x 10, y 20 x y 5 x 2y 18 c 5. Making Furniture A furniture manufacturer makes wooden tables and chairs. The producd tion process involves two basic types of labor: carpentry and ﬁnishing. A table requires 2 hours of carpentry and 1 hour of ﬁnishing, and a chair requires 3 hours of carpentry and hour of ﬁnishing. The proﬁt is $35 per table and $20 per chair. The manufacturer’s employees can supply a maximum of 108 hours of carpentry work and 20 hours of ﬁnishing work per day. How many tables and chairs should be made each day to maximize proﬁt? 1 2 6. A Housing Development A housing contractor has subdivided a farm into 100 building lots. She has designed two types of homes for these lots: colonial and ranch style. A colonial requires $30,000 of capital and produces a proﬁt of $4000 when sold. A ranch-style house requires $40,000 of capital and provides an $8000 proﬁt. If the contractor has $3.6 million of capital on hand, how many houses of each type should she build for maximum proﬁt? Will any of the lots be left vacant? 7. Hauling Fruit A trucker hauls citrus fruit from Florida to Montreal. Each crate of oranges is 4 ft3 in volume and weighs 80 lb. Each crate of grapefruit has a volume of 6 ft3 and weighs 100 lb. His truck has a maximum capacity of 300 ft3 and can carry no more than 5600 lb. Moreover, he is not permitted to carry more crates of grapefruit than crates of oranges. If his proﬁt is $2.50 on each crate of oranges and $4 on each crate of grapefruit, how many crates of each fruit should he carry
for maximum proﬁt? Linear Programming 491 8. Manufacturing Calculators A manufacturer of calculators produces two models: standard and scientiﬁc. Long-term demand for the two models mandates that the company manufacture at least 100 standard and 80 scientiﬁc calculators each day. However, because of limitations on production capacity, no more than 200 standard and 170 scientiﬁc calculators can be made daily. To satisfy a shipping contract, a total of at least 200 calculators must be shipped every day. (a) If the production cost is $5 for a standard calculator and $7 for a scientiﬁc one, how many of each model should be produced daily to minimize this cost? (b) If each standard calculator results in a $2 loss but each scientiﬁc one produces a $5 proﬁt, how many of each model should be made daily to maximize proﬁt? 9. Shipping Stereos An electronics discount chain has a sale on a certain brand of stereo. The chain has stores in Santa Monica and El Toro and warehouses in Long Beach and Pasadena. To satisfy rush orders, 15 sets must be shipped from the warehouses to the Santa Monica store, and 19 must be shipped to the El Toro store. The cost of shipping a set is $5 from Long Beach to Santa Monica, $6 from Long Beach to El Toro, $4 from Pasadena to Santa Monica, and $5.50 from Pasadena to El Toro. If the Long Beach warehouse has 24 sets and the Pasadena warehouse has 18 sets in stock, how many sets should be shipped from each warehouse to each store to ﬁll the orders at a minimum shipping cost? 10. Delivering Plywood A man owns two building supply stores, one on the east side and one on the west side of a city. Two customers order some 1 -inch plywood. Customer A needs 2 50 sheets, and customer B needs 70 sheets. The east-side store has 80 sheets, and the westside store has 45 sheets of this plywood in stock. The east-side store’s delivery costs per sheet are $0.50 to customer A and $0.60 to customer B. The west-side store’s delivery costs per sheet are $0.40 to A and $0.55 to B. How many sheets should be shipped from each store to each customer to minimize delivery costs? 11.
Packaging Nuts A confectioner sells two types of nut mixtures. The standard-mixture package contains 100 g of cashews and 200 g of peanuts and sells for $1.95. The deluxemixture package contains 150 g of cashews and 50 g of peanuts and sells for $2.25. The confectioner has 15 kg of cashews and 20 kg of peanuts available. On the basis of past sales, the confectioner needs to have at least as many standard as deluxe packages available. How many bags of each mixture should he package to maximize his revenue? 12. Feeding Lab Rabbits A biologist wishes to feed laboratory rabbits a mixture of two types of foods. Type I contains 8 g of fat, 12 g of carbohydrate, and 2 g of protein per ounce. Type II contains 12 g of fat, 12 g of carbohydrate, and 1 g of protein per ounce. Type I costs $0.20 per ounce and type II costs $0.30 per ounce. The rabbits each receive a daily minimum of 24 g of fat, 36 g of carbohydrate, and 4 g of protein, but get no more than 5 oz of food per day. How many ounces of each food type should be fed to each rabbit daily to satisfy the dietary requirements at minimum cost? 13. Investing in Bonds A woman wishes to invest $12,000 in three types of bonds: municipal bonds paying 7% interest per year, bank investment certiﬁcates paying 8%, and high-risk bonds paying 12%. For tax reasons she wants the amount invested in municipal bonds to be at least three times the amount invested in bank certiﬁcates. To keep her level of risk manageable, she will invest no more than $2000 in high-risk bonds. How much should she invest in [Hint: Let x amount in municieach type of bond to maximize her annual interest yield? pal bonds and y amount in bank certiﬁcates. Then the amount in high-risk bonds will be 12,000 x y.] 14. Annual Interest Yield Refer to Problem 13. Suppose the investor decides to increase the maximum invested in high-risk bonds to $3000 but leaves the other conditions unchanged. By how much will her maximum possible interest yield increase? 15. Business Strategy A small software company publishes computer games and educational and utility software. Their business strategy is to market a total of 36 new programs each year, at least four of
these being games. The number of utility programs published is never more than twice the number of educational programs. On average, the company makes an annual proﬁt of $5000 on each computer game, $8000 on each educational program, and $6000 on each utility program. How many of each type of software should they publish annually for maximum proﬁt? 492 Focus on Modeling 16. Feasible Region All parts of this problem refer to the following feasible region and objec- tive function. x 0 x y x 2y 12 x 0y 10 P x 4y d (a) Graph the feasible region. (b) On your graph from part (a), sketch the graphs of the linear equations obtained by setting P equal to 40, 36, 32, and 28. (c) If you continue to decrease the value of P, at which vertex of the feasible region will these lines ﬁrst touch the feasible region? (d) Verify that the maximum value of P on the feasible region occurs at the vertex you chose in part (c). 7.1 Matrices and Systems of Linear Equations 7.2 The Algebra of Matrices 7.3 Inverses of Matrices and Matrix Equations 7.4 Determinants and Cramer’s Rule © CHAPTER 7 Matrices and Determinants Fantastic action? The images and action in a video game are out of this world, allowing us to enjoy fantasies of speed, power, and dexterity that are not achievable in real life. But the real action in a video game takes place at the digital level. An image, such as that of the ninja warrior pictured here, is stored in the computer memory as a rectangular array of numbers called a matrix. Each number in the matrix determines the color and brightness of a pixel in the monitor, so the matrix “draws” the image on the monitor. To change an image—to stretch, squeeze, tilt, rotate, or otherwise distort it—the matrix of the image is changed or combined with other matrices using the rules of matrix algebra (see Focus on Modeling: Computer Graphics, pages 547–550). One of the key uses of matrices is the solution of systems of linear equations. In this chapter we learn how a system of linear equations can be expressed as a single matrix equation and then solved by using the rules of matrix algebra. 493493 493 494 CHAPTER 7 \| Matrices and Determinants 7
.1 Matrices and Systems of Linear Equations LEARNING OBJECTIVES After completing this section, you will be able to: ■ Find the augmented matrix of a linear system ■ Solve a linear system using elementary row operations ■ Solve a linear system using the row-echelon form of its matrix ■ Solve a system using the reduced row-echelon form of its matrix ■ Determine the number of solutions of a linear system from the row-echelon form of its matrix ■ Model using linear systems A matrix is simply a rectangular array of numbers. Matrices* are used to organize information into categories that correspond to the rows and columns of the matrix. For example, a scientist might organize information on a population of endangered whales as follows: Immature Juvenile Adult Male Female 12 15 52 42 18 11 This is a compact way of saying there are 12 immature males, 15 immature females, 18 adult males, and so on. B R In this section we represent a linear system by a matrix, called the augmented matrix of the system: Linear system Augmented matrix 2x y 5 x 4y 7 Equation 1 Equation The augmented matrix contains the same information as the system, but in a simpler form. The operations we learned for solving systems of equations can now be performed on the augmented matrix. ■ Matrices We begin by deﬁning the various elements that make up a matrix. *The plural of matrix is matrices. SE CTI O N 7. 1 \| Matrices and Systems of Linear Equations 495 DEFINITION OF MATRIX An m n matrix is a rectangular array of numbers with m rows and n columns. a11 a21 a31 o am1 c a12 a22 a32 o am2 c a13 a23 a33 o am3 c p a1n p a2n p a3n ∞ o p amn c ⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭ d d d d ⎫ ⎪ ⎪ ⎬ m rows ⎪ ⎪ ⎭ U E n columns We say that the matrix has dimension m n. The numbers aij are the entries of the matrix. The subscript on the entry aij indicates that it is in the ith row and the jth column. Here are some examples of matrices. Matrix Dimension rows by 3 columns 1
row by 4 columns R ■ The Augmented Matrix of a Linear System We can write a system of linear equations as a matrix, called the augmented matrix of the system, by writing only the coefﬁcients and constants that appear in the equations. Here is an example. Linear system 3x 2y z 5 x 3y z 0 x 4z 11 Augmented matrix 11 Notice that a missing variable in an equation corresponds to a 0 entry in the augmented matrix. c C S E X AM P L E 1 \| Finding the Augmented Matrix of a Linear System Write the augmented matrix of the system of equations. 6x 2y z 4 x 3z 1 7y z 5 ▼ SO LUTI O N First we write the linear system with the variables lined up in columns. c 6x 2y z 4 x 3z 1 7y z 5 c 496 CHAPTER 7 \| Matrices and Determinants The augmented matrix is the matrix whose entries are the coefﬁcients and the constants in this system ✎ Practice what you’ve learned: Do Exercise 2. S C ▲ ■ Elementary Row Operations The operations that we used in Section 6.3 to solve linear systems correspond to operations on the rows of the augmented matrix of the system. For example, adding a multiple of one equation to another corresponds to adding a multiple of one row to another. ELEMENTARY ROW OPERATIONS 1. Add a multiple of one row to another. 2. Multiply a row by a nonzero constant. 3. Interchange two rows. Note that performing any of these operations on the augmented matrix of a system does not change its solution. We use the following notation to describe the elementary row operations: Symbol kRj S Ri Ri Description Change the ith row by adding k times row j to it, and then put the result back in row i. kRi Ri 4 Rj Multiply the ith row by k. Interchange the ith and jth rows. In the next example we compare the two ways of writing systems of linear equations. E X AM P L E 2 \| Using Elementary Row Operations to Solve a Linear System Solve the system of linear equations. x y 3z 4 x 2y 2z 10 3x y 5z 14 ▼ SO LUTI O N Our goal is to eliminate the x-term from the second equation and the x- and y-terms from the third equation. For comparison, we write
both the system of equations and its augmented matrix. c 1 Add 1 3 Add 1 2 2 Equation 1 to Equation 2. Equation 1 to Equation 3. System x y 3z 4 x 2y 2z 10 3x y 5z 14 x y 3z 4 3y 5z 6 2y 4z 2 c c R2 R3 R1 3R1 R2! R3 4 10 14 3 2 2 5 Augmented matrix Multiply Equation 3 by.1 2 3 Add 1 (to eliminate y from Equation 2). Equation 3 to Equation 2 2 Interchange Equations 2 and 3. SE CTI O N 7. 1 \| Matrices and Systems of Linear Equations 497 x y 3z 4 3y 5z 6 y 2z 1 x y 3z 4 c z 3 y 2z 1 x y 3z 4 c y 2z 1 z 3 1 2R3! R2 3R3 R2! R2 4R3 Now we use back-substitution to ﬁnd that x 2, y 7, and z 3. The solution is c S 2, 7, 3 1 ✎ Practice what you’ve learned: Do Exercise 19. ▲. 2 C ■ Gaussian Elimination In general, to solve a system of linear equations using its augmented matrix, we use elementary row operations to arrive at a matrix in a certain form. This form is described in the following box. ROW-ECHELON FORM AND REDUCED ROW-ECHELON FORM OF A MATRIX A matrix is in row-echelon form if it satisﬁes the following conditions. 1. The ﬁrst nonzero number in each row (reading from left to right) is 1. This is called the leading entry. 2. The leading entry in each row is to the right of the leading entry in the row immediately above it. 3. All rows consisting entirely of zeros are at the bottom of the matrix. A matrix is in reduced row-echelon form if it is in row-echelon form and also satisﬁes the following condition. 4. Every number above and below each leading entry is a 0. In the following matrices, the ﬁrst one is not in row-echelon form. The second one is in row-echelon form, and the third one is in reduced row
-echelon form. The entries in red are the leading entries. Not in row-echelon form Row-echelon form Reduced row-echelon form. 10 Leading 1’s do not shift to the right in successive rows. T D Leading 1’s shift to the right in successive rows. T D Leading 1’s have 0’s above and below them. T Here is a systematic way to put a matrix in row-echelon form using elementary row operations: ■ Start by obtaining 1 in the top left corner. Then obtain zeros below that 1 by adding appropriate multiples of the ﬁrst row to the rows below it. 498 CHAPTER 7 \| Matrices and Determinants ■ Next, obtain a leading 1 in the next row, and then obtain zeros below that 1. ■ At each stage make sure that every leading entry is to the right of the leading entry in the row above it—rearrange the rows if necessary. ■ Continue this process until you arrive at a matrix in row-echelon form. This is how the process might work for a 3 4 matrix Once an augmented matrix is in row-echelon form, we can solve the corresponding linear system using back-substitution. This technique is called Gaussian elimination, in honor of its inventor, the German mathematician C. F. Gauss (see page 338). S C C C S S SOLVING A SYSTEM USING GAUSSIAN ELIMINATION 1. Augmented Matrix. Write the augmented matrix of the system. 2. Row-Echelon Form. Use elementary row operations to change the augmented matrix to row-echelon form. 3. Back-Substitution. Write the new system of equations that corresponds to the row-echelon form of the augmented matrix and solve by back-substitution. E X AM P L E 3 \| Solving a System Using Row-Echelon Form Solve the system of linear equations using Gaussian elimination. 4x 8y 14z 14 3x 8y 15z 11 2x 8y 12z 17 ▼ SO LUTI O N We ﬁrst write the augmented matrix of the system, and then use elementary row operations to put it in row-echelon form. c Augmented matrix R1! C R2 R3 3R1 2R1 R2! R3 1 2 R
2! R3 5R2 R3 11 12 17 2 1 8 1 1 5 11 12 17 S 2 1 2 5 1 8 14 10 15 S 2 1 1 7 4 1 10 15 5 S 1 2 1 0 10 1 4 7 20 S S Need a 1 here. Need 0’s here. Need a 1 here. Need a 0 here. Need a 1 here. SE CTI O N 7. 1 \| Matrices and Systems of Linear Equations 499 Reduced row-echelon form N 1 10 R3 We now have an equivalent matrix in row-echelon form, and the corresponding system of equations is C S x 2y z 1 x 2y 4z 7 x 2y 4z 2 Back-substitute N ref([A]) [[1 2 -1 1 ] [0 1 2 -3] [0 0 1 -2]] FIGURE 1 We use back-substitution to solve the system Solve for y 2 y 1 1 2 x 3 So the solution of the system is ✎ Practice what you’ve learned: Do Exercise 21. 3, 1, 2 1 Solve for x. 2 1 1 2 Back-substitute z = –2 into Equation 2 Back-substitute y = 1 and z = –2 into Equation 1 ▲ Graphing calculators have a “row-echelon form” command that puts a matrix in row-echelon form. (On the TI-83 this command is ref.) For the augmented matrix in Example 3 the ref command gives the output shown in Figure 1. Notice that the rowechelon form that is obtained by the calculator differs from the one we got in Example 3. This is because the calculator used different row operations than we did. You should check that your calculator’s row-echelon form leads to the same solution as ours. ■ Gauss-Jordan Elimination If we put the augmented matrix of a linear system in reduced row-echelon form, then we don’t need to back-substitute to solve the system. To put a matrix in reduced row-echelon form, we use the following steps. ■ Use the elementary row operations to put the matrix in row-echelon form. ■ Obtain zeros above each leading entry by adding multiples of the row containing that entry to the rows above it. Begin with the last leading entry and work up. Here is
how the process works for a 3 4 matrix Using the reduced row-echelon form to solve a system is called Gauss-Jordan eliminaS tion. The process is illustrated in the next example. C S C S C E X AM P L E 4 \| Solving a System Using Reduced Row-Echelon Form Solve the system of linear equations, using Gauss-Jordan elimination. 4x 8y 4z 14 3x 8y 5z 11 2x y 12z 17 c 500 CHAPTER 7 \| Matrices and Determinants ▼ SO LUTI O N In Example 3 we used Gaussian elimination on the augmented matrix of this system to arrive at an equivalent matrix in row-echelon form. We continue using elementary row operations on the last matrix in Example 3 to arrive at an equivalent matrix in reduced row-echelon form. Need 0’s here. Need a 0 here. 1 0 0 C R2 4R3 R2 SSSSSSSO R1 R3 R1 R1 2R2 R1 SSSSSSSO We now have an equivalent matrix in reduced row-echelon form, and the corresponding system of equations is S C Since the system in reduced rowechelon form, back-subsitution is not required to get the solution x 3 y 1 z 2 Hence we immediately arrive at the solution. 2 ✎ Practice what you’ve learned: Do Exercise 23. c 1 3, 1, 2 ▲ Graphing calculators also have a command that puts a matrix in reduced row-echelon form. (On the TI-83 this command is rref.) For the augmented matrix in Example 4, the rref command gives the output shown in Figure 2. The calculator gives the same reduced row-echelon form as the one we got in Example 4. This is because every matrix has a unique reduced row-echelon form. rref([A]) [[1 0 0 -3] [0 1 0 1 ] [0 0 1 -2]] FIGURE 2 ■ Inconsistent and Dependent Systems The systems of linear equations that we considered in Examples 1–4 had exactly one solution. But as we know from Section 6.3, a linear system may have one solution, no solution, or inﬁnitely many solutions. Fortunately, the row-echelon form of a system allows us to determine which of these cases applies, as described in
the following box. First we need some terminology. A leading variable in a linear system is one that corresponds to a leading entry in the row-echelon form of the augmented matrix of the system. SE CTI O N 7. 1 \| Matrices and Systems of Linear Equations 501 THE SOLUTIONS OF A LINEAR SYSTEM IN ROW-ECHELON FORM Suppose the augmented matrix of a system of linear equations has been transformed by Gaussian elimination into row-echelon form. Then exactly one of the following is true. 1. No solution. If the row-echelon form contains a row that represents the equation 0 c, where c is not zero, then the system has no solution. A system with no solution is called inconsistent. 2. One solution. If each variable in the row-echelon form is a leading variable, then the system has exactly one solution, which we ﬁnd using back-substitution or Gauss-Jordan elimination. 3. Inﬁnitely many solutions. If the variables in the row-echelon form are not all leading variables and if the system is not inconsistent, then it has inﬁnitely many solutions. In this case the system is called dependent. We solve the system by putting the matrix in reduced row-echelon form and then expressing the leading variables in terms of the nonleading variables. The nonleading variables may take on any real numbers as their values. The matrices below, all in row-echelon form, illustrate the three cases described above. No solution One solution Inﬁnitely many solutions Last equation S says 0 = 1. Each variable is a leading variable. C S C z is not a leading variable. S E X AM P L E 5 \| A System with No Solution Solve the system. x 3y 2z 12 2x 5y 5z 14 x 2y 3z 20 ▼ SO LUTI O N We transform the system into row-echelon form. c 12 14 20 R2 2R1 R2 SSSSSSSO R3 R1 R3 2 12 1 10 8 1 R2 R3 R3 C SSSSSSSO 1 3 1 0 0 0 S 2 12 1 10 18 0 1 R3SSSO 12 S 1 10 1 0 C This last matrix is in row-echelon form, so we can stop the Gaussian elimination process. Now if we translate the
last row back into equation form, we get 0x 0y 0z 1, or 0 1, which is false. No matter what values we pick for x, y, and z, the last equation will never be a true statement. This means that the system has no solution. ✎ Practice what you’ve learned: Do Exercise 29. ▲ S C S ref([A]) [[1 -2.5 2.5 7 ] [0 1 1 -10] [0 0 0 1 ]] FIGURE 3 matrix in Example 5. You should check that this gives the same solution. Figure 3 shows the row-echelon form produced by a TI-83 calculator for the augmented 502 CHAPTER 7 \| Matrices and Determinants E X AM P L E 6 \| A System with Inﬁnitely Many Solutions Find the complete solution of the system. 3x 5y 36z 10 x 7z 5 x y 10z 4 ▼ SO LUTI O N We transform the system into reduced row-echelon form. c 3 5 10 36 1 0 5 7 1 10 4 1 R1 PRRO R3 SSSSSO 1 10 4 1 1 5 7 0 3 5 10 36 R1 R2 C R2 SSSSSSSO R3 3R1 R3 1 10 R1 R2 R1 SSSSSSSO S 1 0 0 S 1 10 4 3 1 1 0 0 0 S 2R2 R3 C R3 SSSSSSSSO The third row corresponds to the equation 0 0. This equation is always true, no matter C what values are used for x, y, and z. Since the equation adds no new information about the variables, we can drop it from the system. So the last matrix corresponds to the system S Reduced row-echelon form on the TI-83 calculator: rref([A]) [[1 0 -7 -5] [0 1 -3 1 ] [0 0 0 0 ]] x 7z 5 y 3z 1 e Equation 1 Equation 2 Leading variables Now we solve for the leading variables x and y in terms of the nonleading variable z: x 7z 5 y 3z 1 Solve for x in Equation 1 Solve for y in Equation 2 To obtain the complete solution, we let t represent any real number, and we express x, y, and z in terms of t: x 7t 5 y 3t 1 z
t We can also write the solution as the ordered triple number. 1 7t 5, 3t 1, t 2, where t is any real ✎ Practice what you’ve learned: Do Exercise 31. ▲ In Example 6, to get speciﬁc solutions, we give a speciﬁc value to t. For example, if t 1, then SE CTI O N 7. 1 \| Matrices and Systems of Linear Equations 503 Here are some other solutions of the system obtained by substituting other values for the parameter t. Parameter t Solution 7t 5, 3t 1, t 1 2 1 0 2 5 12, 2, 1 2 1 5, 1, 0 1 9, 7, 2 2 1 30, 16, 5 1 2 2 E X AM P L E 7 \| A System with Inﬁnitely Many Solutions Find the complete solution of the system. x 2y 3z 4„ 10 x 3y 3z 4„ 15 2x 2y 6z 8„ 10 ▼ SO LUTI O N We transform the system into reduced row-echelon form 10 15 10 R2 R2 R1 SSSSSSSO R3 2R1 R3 2 3 4 10 1 0 0 0 5 1 0 10 0 2 0 C 2R2 R3 R3 SSSSSSSSO 10 5 0 C 2R2 R1 R1 SSSSSSSSO This is in reduced row-echelon form. Since the last row represents the equation S S 0 0, we may discard it. So the last matrix corresponds to the system C C x 3z 4„ 0 5 y e Leading variables To obtain the complete solution, we solve for the leading variables x and y in terms of the nonleading variables z and „, and we let z and „ be any real numbers. Thus, the complete solution is x 3s 4t y 5 z s „ t where s and t are any real numbers. ✎ Practice what you’ve learned: Do Exercise 51. ▲ Note that s and t do not have to be the same real number in the solution for Example 7. We can choose arbitrary values for each if we wish to construct a speciﬁc solution to the system. For example, if we let s 1 and t 2, then we get the solution. You 2 should check that this does indeed satisfy all three
of the original equations in Example 7. 11, 5, 1, 2 1 504 CHAPTER 7 \| Matrices and Determinants Examples 6 and 7 illustrate this general fact: If a system in row-echelon form has n, then the complete solution will have m n nonm n nonzero equations in m variables 2 1 leading variables. For instance, in Example 6 we arrived at two nonzero equations in the three variables x, y, and z, which gave us 3 2 1 nonleading variable. ■ Modeling with Linear Systems Linear equations, often containing hundreds or even thousands of variables, occur frequently in the applications of algebra to the sciences and to other ﬁelds. For now, let’s consider an example that involves only three variables. E X AM P L E 8 \| Nutritional Analysis Using a System of Linear Equations A nutritionist is performing an experiment on student volunteers. He wishes to feed one of his subjects a daily diet that consists of a combination of three commercial diet foods: MiniCal, LiquiFast, and SlimQuick. For the experiment it is important that the subject consume exactly 500 mg of potassium, 75 g of protein, and 1150 units of vitamin D every day. The amounts of these nutrients in one ounce of each food are given in the table. How many ounces of each food should the subject eat every day to satisfy the nutrient requirements exactly? MiniCal LiquiFast SlimQuick Potassium (mg) Protein (g) Vitamin D (units) 50 5 90 75 10 100 10 3 50 ▼ SO LUTI O N Let x, y, and z represent the number of ounces of MiniCal, LiquiFast, and SlimQuick, respectively, that the subject should eat every day. This means that he will get 50x mg of potassium from MiniCal, 75y mg from LiquiFast, and 10z mg from SlimQuick, for a total of 50x 75y 10z mg potassium in all. Since the potassium requirement is 500 mg, we get the ﬁrst equation below. Similar reasoning for the protein and vitamin D requirements leads to the system 50x 75y 10z 500 5x 10y 3z 75 90x 100y 50z 1150 Potassium Protein Vitamin D Dividing the ﬁrst equation by 5 and the third one by 10 gives the system c 10x 15y 2z 100 5x 10y 3z 75 9x 10y 5z 115 c We
can solve this system using Gaussian elimination, or we can use a graphing calculator to ﬁnd the reduced row-echelon form of the augmented matrix of the system. Using the rref command on the TI-83, we get the output in Figure 4. From the reduced row-echelon form we see that x 5, y 2, z 10. The subject should be fed 5 oz of MiniCal, 2 oz of LiquiFast, and 10 oz of SlimQuick every day. ✎ Practice what you’ve learned: Do Exercise 55. ▲ 100 75 115 2 2 2 ✔ rref([A]) [[1 0 0 5 ] [0 1 0 2 ] [0 0 1 10]] FIGURE 4 Check Your Answer x 5, y 2, z 10: 15 2 3 10 5 10 10 10 10 10 1 1 1 c A more practical application might involve dozens of foods and nutrients rather than just three. Such problems lead to systems with large numbers of variables and equations. Computers or graphing calculators are essential for solving such large systems. 7. ▼ CONCE PTS 1. If a system of linear equations has inﬁnitely many solutions, then the system is called. If a system of linear equations has no solution, then the system is called. ✎ 2. Write the augmented matrix of the following system of equations. System x y z 1 x 2z 3 2y z 3 Augmented matrix............ 3. The matrix below is the augmented matrix of a system of linear equations in the variables x, y, and z. (It is given in reduced row-echelon form.) a) The leading variables are S (b) Is the system inconsistent or dependent? C. (c) The solution of the system is: x _____, y _____, z _____ 4. The augmented matrix of a system of linear equations is given in reduced row-echelon form. Find the solution of the system. (abc _____ y _____ z _____ S C x _____ y _____ z _____ S C x _____ y _____ z _____ ▼ SKI LLS 5–10 ■ State the dimension of the matrix. 5. 8. 1 0 c 5 2 4 11 0 3 d 7. 12 35 d c 9. 3 1 4 7 4 10. 1 0 c 0 1 d C S 11–18 ■
A matrix is given. (a) Determine whether the matrix is in row-echelon form. (b) Determine whether the matrix is in reduced row-echelon form. (c) Write the system of equations for which the given matrix is the augmented matrix. 11. 1 0 c 0 3 5 d 1 12. 1 0 c 3 3 5 d 1 SE CTI O N 7. 1 \| Matrices and Systems of Linear Equations 505 13. C 15. 1 0 0 1 0 0 17 14. C 16. 1 0 0 1 0 0 18 19–28 ■ The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. D T ✎ 19. ✎ 21. ✎ 23. x 2y z 1 y 2z 5 x y 3z 8 x y z 2 c 2x 3y 2z 4 4x y 3z 1 x 2y z 2 c x 2y z 0 2x y z 3 20. 22. 24. x y 6z 3 x y 3z 3 x 2y 4z 7 c x y z 4 x 2y 3z 17 2x y 3z 7 2y z 4 c x y z 4 3x 3y z 10 25. 27. c 9 2 22 2x2 5x2 2x2 x3 x3 2x3 x1 2x1 3x1 2x 3y z 13 c x 2y 5z 6 5x y z 49 c 26. c 28. 17 16 11 2x2 2x2 2x2 4x3 2x1 4x3 2x1 4x3 3x1 10x 10y 20z 60 15x 20y 30z 25 5x 30y 10z 45 29–38 ■ Determine whether the system of linear equations is inconsistent or dependent. If it is dependent, ﬁnd the complete solution. c c ✎ 29. ✎ 31. x y z 2 y 3z 1 2x y 5z 0 30. x y 3z 3 2x y 2z 5 y 8z 8 c 2x 3y 9z 5 x 3z 2 3x y 4z 3 c 32. x 2y 5z 3 2x 6y 11z 1 3x 16y 20z 26 33. x y 3z 3 c 4x 8y 32z 24 2x 3y 11z 4 c 35. x 4y
2z 3 2x y 5z 12 8x 5y 11z 30 37. 2x y 2z 12 c 2y z 6 x 1 2y 3z 18 3x 3 34. 36. c 2x 6y 2z 12 x 3y 2z 10 x 3y 2z 6 3r 2s 3t 10 c r s t 5 r 4s t 20 c 38. y 5z 7 3x 2y 12 3x 10z 80 c c 506 CHAPTER 7 \| Matrices and Determinants 39–54 ■ Solve the system of linear equations. 4x 3y 3z 8 2x 3y 3z 4 2x 3y 2z 3 2x y 3z 9 c x 7z 10 3x 2y z 4 c 2x 2y 3z 15 2x 4y 6z 1 10 3x 7y 2z 13 c x y 6z 8 x z 5 x 3y 14z 4 40. 42. 44. 46. 2x 3y 5z 14 4x 3y 2z 17 x 3y 5z 13 4x y 36z 24 c 2y 9z 3 2x y 6z 6 x c 3x 3y z 2 4x 3y z 4 2x 5y z 0 3x y 2z 1 c 4x 2y z 7 x 3y 2z 1 c d x 2y 4z 3„ 3 3x 4y 4z 3„ 9 x 4y 4z 3„ 0 2x 4y 4z 2„ 3 x y 2z 2„ 2 d 3y z 2„ 2 x y 3„ 2 3x z 2„ 5 3x 3y 2z 2„ 12 3x 2y 2z 2„ 10 3x 3y 2z 5„ 15 3x 3y 2z 5„ 13 x y „ 0 d 3x z 2„ 0 x 4y z 2 2y 3z „ 12 2x 2z 5„ 1 48. x y z 4„ 6 2x z 3„ 8 x y 4„ 10 3x 5y z 4„ 20 c d 52. c 54. 2x y 2z „ 5 x y 4z „ 3 3x 2y 4z 0 2x 2y 2z 2„ 10 3x 2y 2z 2„ 10 2x 2y 2
z 4„ 12 2x 2y 2z 5„ 16 39. 41. 43. 45. 47. 49. 50. ✎ 51. 53. make 100 mL of 18% solution, if she has to use four times as much of the 10% solution as the 40% solution? 57. Distance, Speed, and Time Amanda, Bryce, and Corey enter a race in which they have to run, swim, and cycle over a marked course. Their average speeds are given in the table. Corey ﬁnishes ﬁrst with a total time of 1 h 45 min. Amanda comes in second with a time of 2 h 30 min. Bryce ﬁnishes last with a time of 3 h. Find the distance (in miles) for each part of the race. Average speed (mi/h) Running Swimming Cycling Amanda Bryce Corey 10 71 2 15 4 6 3 20 15 40 58. Classroom Use A small school has 100 students who oc- cupy three classrooms: A, B, and C. After the ﬁrst period of the school day, half the students in room A move to room B, oneﬁfth of the students in room B move to room C, and one-third of the students in room C move to room A. Nevertheless, the total number of students in each room is the same for both periods. How many students occupy each room? 59. Manufacturing Furniture A furniture factory makes wooden tables, chairs, and armoires. Each piece of furniture requires three operations: cutting the wood, assembling, and ﬁnishing. Each operation requires the number of hours (h) given in the table. The workers in the factory can provide 300 hours of cutting, 400 hours of assembling, and 590 hours of ﬁnishing each work week. How many tables, chairs, and armoires should be produced so that all available labor-hours are used? Or is this impossible? Table Chair Armoire Cutting (h) Assembling (h) Finishing (h) 1 2 1 2 1 1 11 2 11 2 1 1 2 d d 60. Trafﬁc Flow A section of a city’s street network is shown in ✎ ▼ APPLICATIONS 55. Nutrition A doctor recommends that a patient take 50 mg each of niacin, riboﬂavin, and thiamin daily to alleviate a vitamin de�
��ciency. In his medicine chest at home the patient ﬁnds three brands of vitamin pills. The amounts of the relevant vitamins per pill are given in the table. How many pills of each type should he take every day to get 50 mg of each vitamin? VitaMax Vitron VitaPlus Niacin (mg) Riboﬂavin (mg) Thiamin (mg) 5 15 10 10 20 10 15 0 10 56. Mixtures A chemist has three acid solutions at various con- centrations. The ﬁrst is 10% acid, the second is 20%, and the third is 40%. How many milliliters of each should she use to the ﬁgure. The arrows indicate one-way streets, and the numbers show how many cars enter or leave this section of the city via the indicated street in a certain one-hour period. The variables x, y, z, and „ represent the number of cars that travel along the portions of First, Second, Avocado, and Birch Streets during this period. Find x, y, z, and „, assuming that none of the cars stop or park on any of the streets shown. 200 400 180 70 x z FIRST STREET „ AVOCADO STREET BIRCH STREET y SECOND STREET 200 30 20 200 ▼ DISCOVE RY • DISCUSSION • WRITI NG 61. Polynomials Determined by a Set of Points We all know that two points uniquely determine a line y ax b in the coordinate plane. Similarly, three points uniquely determine a quadratic (second-degree) polynomial y ax2 bx c four points uniquely determine a cubic (third-degree) polynomial y ax3 bx2 cx d SE CTI O N 7.2 \| The Algebra of Matrices 507 and so on. (Some exceptions to this rule are if the three points actually lie on a line, or the four points lie on a quadratic or line, and so on.) For the following set of ﬁve points, ﬁnd the line that contains the ﬁrst two points, the quadratic that contains the ﬁrst three points, the cubic that contains the ﬁrst four points, and the fourth-degree polynomial that contains all ﬁve points. 0, 0, 1, 12
, 2, 40, 3, 6, 1, 14 2 1 1 2 Graph the points and functions in the same viewing rectangle using a graphing device. 2 1 2 1 2 1 7.2 The Algebra of Matrices LEARNING OBJECTIVES After completing this section, you will be able to: ■ Determine whether two matrices are equal ■ Use addition, subtraction, and scalar multiplication of matrices ■ Multiply matrices ■ Write a linear system in matrix form Thus far, we have used matrices simply for notational convenience when solving linear systems. Matrices have many other uses in mathematics and the sciences, and for most of these applications a knowledge of matrix algebra is essential. Like numbers, matrices can be added, subtracted, multiplied, and divided. In this section we learn how to perform these algebraic operations on matrices. ■ Equality of Matrices Two matrices are equal if they have the same entries in the same positions. EQUALITY OF MATRICES The matrices A ”aij’ and B ”bij’ are equal if and only if they have the same dimension m n, and corresponding entries are equal, that is, for i 1, 2,..., m and j 1, 2,..., n. aij bij Equal matrices 24 0.5 c 22 1 e0 Unequal matrices AM P L E 1 \| Equal Matrices Find a, b, c, and d, if Since the two matrices are equal, corresponding entries must be the ▼ SO LUTI O N same. So we must have a 1, b 3, c 5, and d 2. ✎ Practice what you’ve learned: Do Exercise 5. ▲ 508 CHAPTER 7 \| Matrices and Determinants ■ Addition, Subtraction, and Scalar Multiplication of Matrices Two matrices can be added or subtracted if they have the same dimension. (Otherwise, their sum or difference is undeﬁned.) We add or subtract the matrices by adding or subtracting corresponding entries. To multiply a matrix by a number, we multiply every element of the matrix by that number. This is called the scalar product. SUM, DIFFERENCE, AND SCALAR PRODUCT OF MATRICES Let A ”aij’ and B ”bij’ be matrices of the same dimension m n, and
let c be any real number. 1. The sum A B is the m n matrix obtained by adding corresponding entries of A and B. A B aij 3 bij4 2. The difference A B is the m n matrix obtained by subtracting corresponding entries of A and B. 3 3. The scalar product cA is the m n matrix obtained by multiplying each entry A B aij bij4 of A by c. cA caij4 3 E X AM P L E 2 \| Performing Algebraic Operations on Matrices Let Carry out each indicated operation, or explain why it cannot be performed. (a) A B (b) C D (c) C A (d) 5A Julia Robinson (1919–1985) was born in St. Louis, Missouri, and grew up at Point Loma, California. Because of an illness, Robinson missed two years of school, but later, with the aid of a tutor, she completed ﬁfth, sixth, seventh, and eighth grades, all in one year. Later, at San Diego State University, reading biographies of mathematicians in E. T. Bell’s Men of Mathematics awakened in her what became a lifelong passion for mathematics. She said, “I cannot overemphasize the impor tance of such books... in the intellectual life of a student.” Robinson is famous for her work on Hilbert’s tenth problem (page 531), which asks for a general procedure for determining whether an equation has integer solutions. Her ideas led to a complete answer to the problem. Interestingly, the answer involved certain properties of the Fibonacci numbers (page 604) discovered by the then 22-year-old Russian mathematician Yuri Matijaseviˇc. As a result of her brilliant work on Hilbert’s tenth problem, Robinson was offered a professorship at the University of California, Berkeley, and became the ﬁrst woman mathematician elected to the National Academy of Sciences. She also served as president of the American Mathematical Society. SE CTI O N 7.2 \| The Algebra of Matrices 509 ▼ SO LUTI O N (a) A B (bc) C A is undeﬁned because we can’t add matrices of different dimensions. (d) 5A 5 2 3 0 5 7 1 2 10 15 0 25 5 35 2 ✎ Practice what you’ve learned:
Do Exercises 21 and 23. T D D T ▲ The properties in the box follow from the deﬁnitions of matrix addition and scalar mul- tiplication and the corresponding properties of real numbers. PROPERTIES OF ADDITION AND SCALAR MULTIPLICATION OF MATRICES Let A, B, and C be m n matrices and let c and d be scalars Associative Property of Matrix Addition Commutative Property of Matrix Addition 2 1 1 c dA cdA A cA dA cA cB 1 2 Associative Property of Scalar Multiplication Distributive Properties of Scalar Multiplication E X AM P L E 3 \| Solving a Matrix Equation Solve the matrix equation 2X A B for the unknown matrix X, where ▼ SO LUTI O N We use the properties of matrices to solve for X. 2X A B Given equation 2X B A X 1 21 B A 2 Add the matrix A to each side Multiply each side by the scalar 1 2 So Substitute the matrices A and B 510 CHAPTER 7 \| Matrices and Determinants Add matrices 1 Multiply by the scalar 2 ✎ Practice what you’ve learned: Do Exercise 15. ▲ ■ Multiplication of Matrices Multiplying two matrices is more difﬁcult to describe than other matrix operations. In later examples we will see why taking the matrix product involves a rather complex procedure, which we now describe. First, the product AB of two matrices A and B is deﬁned only when the number of columns in A is equal to the number of rows in B. This means that if we write their dimensions side by side, the two inner numbers must match: or A B 2 1 Matrices Dimensions A m n B n k Columns in A Rows in B If the dimensions of A and B match in this fashion, then the product AB is a matrix of dimension m k. Before describing the procedure for obtaining the elements of AB, we deﬁne the inner product of a row of A and a column of B. If ”a1 a2... an’ is a row of A, and if b1 b2 o bn is a column of B, then their inner product is the number a1b1 [2 1 0 4] and a2b2 5
4 3 1 2... anbn. For example, taking the inner product of D T gives 2 # 5 T We now deﬁne the product AB of two matrices MATRIX MULTIPLICATION If A ”aij’ is an m n matrix and B ”bij’ an n k matrix, then their product is the m k matrix C ”cij’ where cij is the inner product of the ith row of A and the jth column of B. We write the product as C AB This deﬁnition of matrix product says that each entry in the matrix AB is obtained from a row of A and a column of B as follows: The entry cij in the ith row and jth column of the matrix AB is obtained by multiplying the entries in the ith row of A with the corresponding entries in the jth column of B and adding the results. SE CTI O N 7.2 \| The Algebra of Matrices 511 ith row of A jth column of B Entry in ith row and jth column of AB cij E X AM P L E 4 \| Multiplying Matrices C S C S C S Let A 1 1 c 3 0 d and B 1 0 c 5 4 2 7 d Calculate, if possible, the products AB and BA. ▼ SO LUTI O N AB is deﬁned and has dimension 2 3. We can therefore write Since A has dimension 2 2 and B has dimension 2 3, the product AB where the question marks must be ﬁlled in using the rule deﬁning the product of two matrices. If we deﬁne C AB ”cij’, then the entry c11 is the inner product of the ﬁrst row of A and the ﬁrst column of B Similarly, we calculate the remaining entries of the product as follows. Entry c12 c Inner product of Value 1 # 5 3 # 4 17 1 # 2 3 # 7 23 Product matrix 1 17 c c 1 17 23 17 23 # 5 0 # 4 5 c 1 17 1 5 23 d # 2 0 # 7 2 c 1 23 17 1 5 2 d d d d AB 1 17 23 1 5 2 d c c13 c c21 c c22 c c23 c 1 1 1 1 1 1 1 1 Thus, we have The product BA is
not deﬁned, however, because the dimensions of B and A are 2 3 and 2 2 The inner two numbers are not the same, so the rows and columns won’t match up when we try to calculate the product. ✎ Practice what you’ve learned: Do Exercise 25. ▲ Inner numbers match, so product is deﬁned. 2 2 2 3 Outer numbers give dimension of product: 2 3. Not equal, so product not deﬁned. 2 3 2 2 512 CHAPTER 7 \| Matrices and Determinants [A]*[B] [[ -1 17 23] [1 -5 -2]] Graphing calculators and computers are capable of performing matrix algebra. For instance, if we enter the matrices in Example 4 into the matrix variables [A]and [B] on a TI-83 calculator, then the calculator ﬁnds their product as shown in Figure 1. FIGURE 1 ■ Properties of Matrix Multiplication Although matrix multiplication is not commutative, it does obey the Associative and Distributive Properties. PROPERTIES OF MATRIX MULTIPLICATION Let A, B, and C be matrices for which the following products are deﬁned. Then 1 2 BC A 1 B C B C C AB 2 AB AC A BA CA 2 2 A 1 1 Associative Property Distributive Property The next example shows that even when both AB and BA are deﬁned, they aren’t necessarily equal. This result proves that matrix multiplication is not commutative. E X AM P L E 5 \| Matrix Multiplication Is Not Commutative Let A 5 3 c 7 0 d and B 2 1 9 1 d c Calculate the products AB and BA. ▼ SO LUTI O N and BA are deﬁned, and each product is also a 2 2 matrix. Since both matrices A and B have dimension 2 2, both products AB AB BA 68 48 c 7 63 d 1 2 This shows that, in general, AB BA. In fact, in this example AB and BA don’t even have an entry in common. ✎ Practice what you’ve learned: Do Exercise 27. ▲ 2 1 2 1 ■ Applications of Matrix Multiplication We now consider some applied examples that give some indication of why mathematicians chose to deﬁne the matrix product in such an apparently
bizarre fashion. Example 6 shows how our deﬁnition of matrix product allows us to express a system of linear equations as a single matrix equation. SE CTI O N 7.2 \| The Algebra of Matrices 513 E X AM P L E 6 \| Writing a Linear System as a Matrix Equation Show that the following matrix equation is equivalent to the system of equations in Example 2 of Section 7.1. Matrix equations like this one are described in more detail on page 524 10 14 ▼ SO LUTI O N we get If we perform matrix multiplication on the left side of the equation, C S C S C S x y 3z x 2y 2z 3x y 5z 4 10 14 Because two matrices are equal only if their corresponding entries are equal, we equate entries to get S C C S x y 3z 4 x 2y 2z 10 3x y 5z 14 This is exactly the system of equations in Example 2 of Section 7.1. c ✎ Practice what you’ve learned: Do Exercise 39. ▲ E X AM P L E 7 \| Representing Demographic Data by Matrices In a certain city the proportions of voters in each age group who are registered as Democrats, Republicans, or Independents are given by the following matrix. Age 18–30 31–50 Over 50 Democrat Republican Independent 0.30 0.50 0.20 0.60 0.35 0.05 0.50 0.25 0.25 A The next matrix gives the distribution, by age and sex, of the voting population of this city. C S Male Female Age 18–30 31–50 Over 50 5,000 10,000 12,000 6,000 12,000 15,000 B C S For this problem, let’s make the (highly unrealistic) assumption that within each age group, political preference is not related to gender. That is, the percentage of Democrat males in the 18–30 group, for example, is the same as the percentage of Democrat females in this group. (a) Calculate the product AB. (b) How many males are registered as Democrats in this city? (c) How many females are registered as Republicans? 514 CHAPTER 7 \| Matrices and Determinants ▼ SO LUTI O N (a) AB 0.30 0.50 0.20 0.60 0.35 0.05 0.50 0.25 0.25 5
,000 10,000 12,000 6,000 12,000 15,000 13,500 9,000 4,500 16,500 10,950 5,550 Olga Taussky-Todd (1906–1995) was instrumental in developing applications of matrix theory. Described as “in love with anything matrices can do,” she successfully applied matrices to aerodynamics, a ﬁeld used in the design of airplanes and rockets. Taussky-Todd was also famous for her work in number theory, which deals with prime numbers and divisibility. Although number theory has often been called the least applicable branch of mathematics, it is now used in signiﬁcant ways throughout the computer industry. Taussky-Todd studied mathe- matics at a time when young women rarely aspired to be mathematicians. She said, “When I entered university I had no idea what it meant to study mathematics.” One of the most respected mathematicians of her day, she was for many years a professor of mathematics at Caltech in Pasadena. 0 1 2 3 4 5 6 7 FIGURE b) When we take the inner product of a row in A with a column in B, we are adding the number of people in each age group who belong to the category in question. For example, the entry c21 of AB is obtained by taking the inner product of the Republican row in A with the Male column in B. This number is therefore the total number of male Republicans in this city. We can label the rows and columns of AB as follows. C S the 9000 2 S S C 1 Democrat Republican Independent Male Female 13,500 9,000 4,500 16,500 10,950 5,550 AB Thus, 13,500 males are registered as Democrats in this city. (c) There are 10,950 females registered as Republicans. ✎ Practice what you’ve learned: Do Exercise 45. C S ▲ In Example 7 the entries in each column of A add up to 1. (Can you see why this has to be true, given what the matrix describes?) A matrix with this property is called stochastic. Stochastic matrices are used extensively in statistics, where they arise frequently in situations like the one described here. ■ Computer Graphics One important use of matrices is in the digital representation of images. A digital camera or a scanner converts an image into a matrix by dividing the image into a rectangular array
of elements called pixels. Each pixel is assigned a value that represents the color, brightness, or some other feature of that location. For example, in a 256-level gray-scale image each pixel is assigned a value between 0 and 255, where 0 represents white, 255 represents black, and the numbers in between represent increasing gradations of gray. The gradations of a much simpler 8-level gray scale are shown in Figure 2. We use this 8-level gray scale to illustrate the process. To digitize the black and white image in Figure 3(a), we place a grid over the picture as shown in Figure 3(b). Each cell in the grid is compared to the gray scale, and then assigned a) Original image (b) 10 x 10 grid (c) Matrix representation (d) Digital image FIGURE 3 SE CTI O N 7.2 \| The Algebra of Matrices 515 a value between 0 and 7 depending on which gray square in the scale most closely matches the “darkness” of the cell. (If the cell is not uniformly gray, an average value is assigned.) The values are stored in the matrix shown in Figure 3(c). The digital image corresponding to this matrix is shown in Figure 3(d). Obviously the grid that we have used is far too coarse to provide good image resolution. In practice, currently available high-resolution digital cameras use matrices with dimension as large as 2048 2048. Once the image is stored as a matrix, it can be manipulated by using matrix operations. For example, to darken the image, we add a constant to each entry in the matrix; to lighten the image, we subtract a constant. To increase the contrast, we darken the darker areas and lighten the lighter areas, so we could add 1 to each entry that is 4, 5, or 6 and subtract 1 from each entry that is 1, 2, or 3. (Note that we cannot darken an entry of 7 or lighten a 0.) Applying this process to the matrix in Figure 3(c) produces the new matrix in Figure 4(a). This generates the high-contrast image shown in Figure 4(b). FIGURE 4 (a) Matrix modified to increase contrast (b) High contrast image Other ways of representing and manipulating images using matrices are discussed in Focus on Modeling on pages 547–550. 7. ▼ CONCE PTS 1. We can add (or subtract) two matrices only if
they have the same. 2. (a) We can multiply two matrices only if the number of in the ﬁrst matrix is the same as the number of in the second matrix. (b) If A is a 3 3 matrix and B is a 4 3 matrix, which of the following matrix multiplications are possible? (i) AB (ii) BA (iii) AA (iv) BB 3. Which of the following operations can we perform for a matrix A of any dimension? (i) A A (ii) 2A (iii) A # A 4. Fill in the missing entries in the product matrix ▼ SKI LLS 5–6 ■ Determine whether the matrices A and B are equal. ✎ 5. A 1 4 2 c ln 1 3 d B 0.25 14 c 0 6 2 d 7–14 ■ Perform the matrix operation, or if it is impossible, explain why. 7. 3 2 1 4 1 0 1 11 10 12 516 CHAPTER 7 \| Matrices and Determinants 13 14 15–20 ■ Solve the matrix equation for the unknown matrix X, or explain why no solution exists 20 10 20 30 0 10 3X B C S C D X C 5 1 2A B 3X 2 ✎ 15. 17. 19. 2X 51 2 C S 16. 18. 20. 21–34 ■ The matrices A, B, C, D, E, F, G and H are deﬁned as follows 10 Carry out the indicated algebraic operation, or explain why it cannot be performed. B C C B 21. (a) 22. (a) (b) (b) S ✎ ✎ 23. (a) 24. (a) 5A 3B 2C (b) (b) B F 2C 6B C 5A 2H D ✎ ✎ 25. (a) AD 26. (a) DH 27. (a) AH 28. (a) BC 29. (a) GF 30. (a) 31. (a) B2 A2 32. (a) 33. (a) 34. (a) DA B 2 1 ABE DB DC (b) DA (b) HD (b) HA (b) BF (b) GE (b) (b) (b) (b) (b) F2 A3 D AB 2 1 AHE BF FE 35–38 ■
Solve for x and y. 35. x 4 c 2y 6 d 2 2 2x 6y d c 36 37 38 39–42 ■ Write the system of equations as a matrix equation (see Example 6). ✎ 39. 2x 5y 7 3x 2y 4 e 40. 6x y z 12 2x z 7 y 2z 4 41. 42. 2x2 3x1 x1 3x2 x3 x4 x3 x4 x3 0 5 4 c x y z 2 c 4x 2y z 2 x y 5z 2 x y z 2 43. Let Determine which of the following products are deﬁned, and calculate the ones that are: ABC BCA ACB CAB BAC CBA 44. (a) Prove that if A and B are 2 2 matrices, then A B 2 A2 AB BA B2 1 2 2 2 (b) If A and B are matrices, is it necessarily true that A B 1 2? 2 A2 2AB B2 ✎ ▼ APPLICATIONS 45. Fast-Food Sales A small fast-food chain with restaurants in Santa Monica, Long Beach, and Anaheim sells only hamburgers, hot dogs, and milk shakes. On a certain day, sales were distributed according to the following matrix. Number of items sold Santa Monica Long Beach Anaheim Hamburgers Hot dogs Milk shakes 4000 400 700 1000 300 500 3500 200 9000 A The price of each item is given by the following matrix. C S Hamburger Hot dog Milk shake ”$0.90 $0.80 $1.10’ B (a) Calculate the product BA. (b) Interpret the entries in the product matrix BA. 46. Car-Manufacturing Proﬁts A specialty-car manufacturer has plants in Auburn, Biloxi, and Chattanooga. Three models are produced, with daily production given in the following matrix. Cars produced each day Model K Model R Model W number of pounds of each product sold by each sibling on Saturday and Sunday. SE CTI ON 7.2 \| The Algebra of Matrices 517 Auburn Biloxi Chattanooga 12 4 8 10 4 9 0 20 12 A C Because of a wage increase, February proﬁts are lower than January proﬁts. The proﬁt per car is tabulated by model in the following matrix. S January February Model K Model R Model W $1000 $2000
$1500 $500 $1200 $1000 B (a) Calculate AB. (b) Assuming that all cars produced were sold, what was the C S daily proﬁt in January from the Biloxi plant? (c) What was the total daily proﬁt (from all three plants) in February? Saturday Melons Squash Tomatoes Amy Beth Chad 120 40 60 50 25 30 60 30 20 A C Melons Sunday S Squash Tomatoes Amy Beth Chad 100 35 60 60 20 25 30 20 30 B The matrix C gives the price per pound (in dollars) for each type of produce that they sell. S C Price per pound Melons Squash Tomatoes 0.10 0.50 1.00 C Perform each of the following matrix operations, and interpret the entries in each result. (a) AC S C (c) A B (d) ÓA BÔC (b) BC 49. Digital Images A four-level gray scale is shown below. 0 1 2 3 47. Canning Tomato Products Jaeger Foods produces tomato sauce and tomato paste, canned in small, medium, large, and giant sized cans. The matrix A gives the size (in ounces) of each container. (a) Use the gray scale to ﬁnd a 6 6 matrix that digitally rep- resents the image in the ﬁgure. Small Medium Giant 6 10 14 28 Large A 4 Ounces 3 The matrix B tabulates one day’s production of tomato sauce and tomato paste. Cans of Cans of sauce Small Medium Large Giant 2000 3000 2500 1000 paste 2500 1500 1000 500 B (a) Calculate the product of AB. (b) Interpret the entries in the product matrix AB. D T 48. Produce Sales A farmer’s three children, Amy, Beth, and Chad, run three roadside produce stands during the summer months. One weekend they all sell watermelons, yellow squash, and tomatoes. The matrices A and B tabulate the (b) Find a matrix that represents a darker version of the image in the ﬁgure. (c ) The negative of an image is obtained by reversing light and dark, as in the negative of a photograph. Find the matrix that represents the negative of the image in the ﬁgure. How do you change the elements of the matrix to create the negative? (d) Increase the contrast of the image by changing each 1 to a
0 and each 2 to a 3 in the matrix you found in part (b). Draw the image represented by the resulting matrix. Does this clarify the image? 518 CHAPTER 7 \| Matrices and Determinants (e) Draw the image represented by the matrix I. Can you rec- ognize what this is? If you don’t, try increasing the contrast. 52. Powers of a Matrix Let A 1 1 c 1 1 d. Calculate A2 ▼ DISCOVE RY • DISCUSSION • WRITI NG 50. When Are Both Products Deﬁned? What must be true about the dimensions of the matrices A and B if both products AB and BA are deﬁned? 51. Powers of a Matrix Let A 1 0 c 1 1 d Calculate A2, A3, A4,... until you detect a pattern. Write a general formula for An. A3, A4,... until you detect a pattern. Write a general formula for An. 53. Square Roots of Matrices A square root of a matrix B is a matrix A with the property that A2 B. (This is the same deﬁnition as for a square root of a number.) Find as many square roots as you can of each matrix: 4 0 c b d d [Hint: If, write the equations that a, b, c, and d would have to satisfy if A is the square root of the given matrix.] DISCOVERY PR OJECT WILL THE SPECIES SURVIVE? To study how species survive, scientists model their populations by observing the different stages in their life. Scientists consider, for example, the stage at which the animal is fertile, the proportion of the population that reproduces, and the proportion of the young that survive each year. For a certain species there are three stages: immature, juvenile, and adult. An animal is considered immature for the ﬁrst year of its life, juvenile for the second year, and an adult from then on. Conservation biologists have collected the following ﬁeld data for this species: Immature 0 0.1 0 A Juvenile Adult 0.4 0 0.8 0 0 0.3 Immature Juvenile Adult X0 600 400 3500 Immature Juvenile Adult The entries in the matrix A indicate the proportion of the population that survives to the next year. For example, the ﬁr
st column describes what happens to the immature population: None remain immature, 10% survive to become juveniles, and of course none become adults. The second column describes what happens to the juvenile population: None become immature or remain juvenile, and 30% survive to adulthood. The third column describes the adult population: The number of their new offspring is 40% of the adult population, no adults become juveniles, and 80% survive to live another year. The entries in the population matrix X0 indicate the current population (year 0) of immature, juvenile, and adult animals. AX0, X2 Let X1 1. Explain why X1 gives the population in year 1, X2 the population in year 2, and so on. 2. Find the population matrix for years 1, 2, 3, and 4. (Round fractional entries to AX2, and so on. AX1, X3 the nearest whole number.) Do you see any trend? A2X0, X3 3. Show that X2 4. Find the population after 50 years—that is, ﬁnd X50. (Use your results in Problem 3 and a graphing calculator.) Does it appear that the species will survive? A3X0, and so on. 5. Suppose the environment has improved so that the proportion of immatures that become juveniles each year increases to 0.3 from 0.1, the proportion of juveniles that become adults increases to 0.7 from 0.3, and the proportion of adults that survives to the next year increases to 0.95. Find the population after 50 years with the new matrix A. Does it appear that the species will survive under these new conditions? 6. The matrix A in the example is called a transition matrix. Such matrices occur in many applications of matrix algebra. The following transition matrix T predicts the calculus grades of a class of students who must take a four-semester sequence of calculus courses. The ﬁrst column of the matrix, for instance, indicates that of those students who get an A in one course, 70% will get an A in the following course, 15% will get a B, and 10% will get a C. (Students who receive D or F are not permitted to go on to the next course and so are not included in the matrix.) The entries in the matrix Y0 give the number of incoming students who got A, B, and C, respectively, in their ﬁnal high school mathematics course
. T Y2, and Y4 T Y3. Calculate and interpret the Let Y1 TY0, Y2 entries of Y1, Y2, Y3, and Y4. T Y1, Y3 A 0.70 0.15 0.10 B C 0.25 0.50 0.15 0.05 0.25 0.45 A B C T Y0 140 320 400 A B C C S C S 519 520 CHAPTER 7 \| Matrices and Determinants 7.3 Inverses of Matrices and Matrix Equations LEARNING OBJECTIVES After completing this section, you will be able to: ■ Determine when two matrices are inverses of each other ■ Find the inverse of a 2 2 matrix ■ Find the inverse of an n n matrix ■ Solve a linear system by writing it as a matrix equation ■ Model using matrix equations In the preceding section we saw that when the dimensions are appropriate, matrices can be added, subtracted, and multiplied. In this section we investigate division of matrices. With this operation we can solve equations that involve matrices. ■ The Inverse of a Matrix First, we deﬁne identity matrices, which play the same role for matrix multiplication as the number 1 does for ordinary multiplication of numbers; that is, 1 a a 1 a for all numbers a. In the following deﬁnition the term main diagonal refers to the entries of a square matrix whose row and column numbers are the same. These entries stretch diagonally down the matrix, from top left to bottom right. The identity matrix In is the n n matrix for which each main diagonal entry is a 1 and for which all other entries are 0. Thus, the 2 2, 3 3, and 4 4 identity matrices are I2 1 0 0 1 I3 1 0 0 0 1 0 0 0 1 I4 Identity matrices behave like the number 1 in the sense that D In B B C A A In and B R S 0 0 0 1 T whenever these products are deﬁned. E X AM P L E 1 \| Identity Matrices The following matrix products show how multiplying a matrix by an identity matrix of the appropriate dimension leaves the matrix unchanged 12 12 SE CTI O N 7. 3 \| Inverses of Matrices and Matrix Equations 521 If A and B are n n matrices, and if AB BA In, then we say that B is the inverse of
A, and we write B A1. The concept of the inverse of a matrix is analogous to that of the reciprocal of a real number. INVERSE OF A MATRIX Let A be a square n n matrix. If there exists an n n matrix A1 with the property that then we say that A1 is the inverse of A. AA1 A1A In E X AM P L E 2 \| Verifying That a Matrix Is an Inverse Verify that B is the inverse of A, where A 2 5 c 1 3 d and B 3 1 2 d 5 c ▼ SO LUTI O N We perform the matrix multiplications to show that AB I and BA I ✎ Practice what you’ve learned: Do Exercise 3. ▲ ■ Finding the Inverse of a 2 2 Matrix The following rule provides a simple way for ﬁnding the inverse of a 2 2 matrix, when it exists. For larger matrices there is a more general procedure for ﬁnding inverses, which we consider later in this section. INVERSE OF A 2 2 MATRIX If A a c c b d d then A1 1 ad bc d b a d c. c If ad bc 0, then A has no inverse. E X AM P L E 3 \| Finding the Inverse of a 2 2 Matrix Let A 4 2 c 5 3 d Find A1 and verify that AA1 A1A I2. 522 CHAPTER 7 \| Matrices and Determinants ▼ SO LUTI O N Using the rule for the inverse of a 2 2 matrix, we get A1 To verify that this is indeed the inverse of A, we calculate AA1 and A1A AA1 A1A 4 2 ✎ Practice what you’ve learned: Do Exercise 9 ▲ The quantity ad bc that appears in the rule for calculating the inverse of a 2 2 matrix is called the determinant of the matrix. If the determinant is 0, then the matrix does not have an inverse (since we cannot divide by 0). ■ Finding the Inverse of an n n Matrix For 3 3 and larger square matrices the following technique provides the most efﬁcient way to calculate their inverses. If A is an n n matrix, we ﬁrst construct the n 2n matrix that has the entries of A on the left and of the identity matrix In on the right: a11 a21 o an
1 a12 a22 o an2 p a1n p a2n ∞ o p ann \| \| \| \| We then use the elementary row operations on this new large matrix to change the left side into the identity matrix. (This means that we are changing the large matrix to reduced row-echelon form.) The right side is transformed automatically into A1. (We omit the proof of this fact.) Arthur Cayley (1821–1895) was an English mathematician who was instrumental in developing the theory of matrices. He was the ﬁrst to use a single symbol such as A to represent a matrix, thereby introducing the idea that a matrix is a single entity rather than just a collection of numbers. Cayley practiced law until the age of 42, but his primary interest from adolescence was mathematics, and he published almost 200 articles on the subject in his spare time. In 1863 he accepted a professorship in mathematics at Cambridge, where he taught until his death. Cayley’s work on matrices was of purely theoretical interest in his day, but in the 20th century many of his results found application in physics, the social sciences, business, and other ﬁelds. One of the most common uses of matrices today is in computers, where matrices are employed for data storage, error correction, image manipulation, and many other purposes. These applications have made matrix algebra more useful than ever. E X AM P L E 4 \| Finding the Inverse of a 3 3 Matrix Let A be the matrix A 1 2 4 2 3 6 15 6 3 A1. (a) Find (b) Verify that AA1 A1A I3. C S ▼ SO LUTI O N (a) We begin with the 3 6 matrix whose left half is A and whose right half is the identity matrix. 1 2 4 2 3 6 15 SE CTI O N 7. 3 \| Inverses of Matrices and Matrix Equations 523 We then transform the left half of this new matrix into the identity matrix by performing the following sequence of elementary row operations on the entire new matrix: R2 2R1 R2 SSSSSSSO R3 3R1 R3 1 R3SSSO 3 R1 2R2 R1 SSSSSSSO R2 2R3 R2 SSSSSSSO We have now transformed the left half of this matrix into an identity matrix. (This means that we have put the entire
matrix in reduced row-echelon form.) Note that to do this in as systematic a fashion as possible, we ﬁrst changed the elements below the main diagonal to zeros, just as we would if we were using Gaussian elimination. We then changed each main diagonal element to a 1 by multiplying by the appropriate constant(s). Finally, we completed the process by changing the remaining entries on the left side to zeros. S The right half is now A1. A1 b) We calculate AA1 and A1A and verify that both products give the identity matrix I3. C S AA1 1 2 4 2 3 6 15 A1A 15 6 3 ✎ Practice what you’ve learned: Do Exercise 17A]-1 Frac [[ -3 2 0 ] [ -4 1 -2/3] [1 0 1/3 ]] Graphing calculators are also able to calculate matrix inverses. On the TI-82 and TI-83 calculators, matrices are stored in memory using names such as [A], [B], [C],.... To ﬁnd the inverse of [A], we key in FIGURE 1 [A] 1 X ENTER For the matrix of Example 4 this results in the output shown in Figure 1 (where we have also used the Frac command to display the output in fraction form rather than in decimal form). The next example shows that not every square matrix has an inverse. 524 CHAPTER 7 \| Matrices and Determinants E X AM P L E 5 \| A Matrix That Does Not Have an Inverse Find the inverse of the matrix ▼ SO LUTI O N We proceed as follows PRRO R2 R1 SSSSSO R2 2R1 R2 SSSSSSSO R3 R1 R3 S C 1 7 2 0 7 21 R2SSSO R3 R2 R3 SSSSSSSO R1 2R2 R1 and 3, 1 1 3, 3 1 position of this matrix to a 1 without At this point we would like to change the 0 in the changing the zeros in the positions. But there is no way to accomplish this, because no matter what multiple of rows 1 and/or 2 we add to row 3, we can’t change the third zero in row 3 without changing the ﬁrst or second zero as well. Thus, we cannot change the left half to the identity
matrix, so the original matrix doesn’t have an inverse. ✎ Practice what you’ve learned: Do Exercise 19. C 3, 2 1 ▲ S 2 2 2 ERR:SINGULAR MAT 1:Quit 2:Goto If we encounter a row of zeros on the left when trying to ﬁnd an inverse, as in Example 5, then the original matrix does not have an inverse. If we try to calculate the inverse of the matrix from Example 5 on a TI-83 calculator, we get the error message shown in Figure 2. (A matrix that has no inverse is called singular.) FIGURE 2 ■ Matrix Equations We saw in Example 6 in Section 7.2 that a system of linear equations can be written as a single matrix equation. For example, the system x 2y 4z 7 2x 3y 6z 5 3x 6y 15z 0 is equivalent to the matrix equation c 1 2 4 2 3 6 15 Solving the matrix equation AX B is very similar to solving the simple real-number equation 3x 12 which we do by multiplying each side by the reciprocal (or inverse) of 3: 1 3x 3 2 1 x 4 12 1 1 3 2 SE CTI O N 7. 3 \| Inverses of Matrices and Matrix Equations 525 If we let A 1 2 4 2 3 6 15 then this matrix equation can be written as S AX B C The matrix A is called the coefﬁcient matrix. C S C S We solve this matrix equation by multiplying each side by the inverse of A (provided that this inverse exists): AX B A1 A1B AX 2 1 X A1B A1A 2 1 I3X A1B X A1B Multiply on left by A 1 Associative Property Property of inverses Property of identity matrix In Example 4 we showed that A1 So from X A1B we have 11 23 Thus, x 11, y 23, z 7 is the solution of the original system. We have proved that the matrix equation AX B can be solved by the following method. SOLVING A MATRIX EQUATION If A is a square n n matrix that has an inverse A1 and if X is a variable matrix and B a known matrix, both with n rows, then the solution of the matrix equation is given by AX B X A1B E X AM P L E 6 \| Solving a System Using a Matrix Inverse
(a) Write the system of equations as a matrix equation. (b) Solve the system by solving the matrix equation. 2x 5y 15 3x 6y 36 b 526 CHAPTER 7 \| Matrices and Determinants ▼ SO LUTI O N (a) We write the system as a matrix equation of the form AX B: 2 5 3 6 x y 15 36 b) Using the rule for ﬁnding the inverse of a 2 2 matrix, we get A1 Multiplying each side of the matrix equation by this inverse matrix, we get 15 36 d 30 9 d c X A 1 B So x 30 and y 9. ✎ Practice what you’ve learned: Do Exercise 25. ▲ ■ Modeling with Matrix Equations Suppose we need to solve several systems of equations with the same coefﬁcient matrix. Then converting the systems to matrix equations provides an efﬁcient way to obtain the solutions, because we need to ﬁnd the inverse of the coefﬁcient matrix only once. This procedure is particularly convenient if we use a graphing calculator to perform the matrix operations, as in the next example. E X AM P L E 7 \| Modeling Nutritional Requirements Using Matrix Equations A pet-store owner feeds his hamsters and gerbils different mixtures of three types of rodent food: KayDee Food, Pet Pellets, and Rodent Chow. He wishes to feed his animals the correct amount of each brand to satisfy their daily requirements for protein, fat, and carbohydrates exactly. Suppose that hamsters require 340 mg of protein, 280 mg of fat, and 440 mg of carbohydrates, and gerbils need 480 mg of protein, 360 mg of fat, and 680 mg of carbohydrates each day. The amount of each nutrient (in mg) in one gram of each brand is given in the following table. How many grams of each food should the storekeeper feed his hamsters and gerbils daily to satisfy their nutrient requirements? KayDee Food Pet Pellets Rodent Chow Protein (mg) Fat (mg) Carbohydrates (mg) 10 10 5 0 20 10 20 10 30 SE CTI O N 7. 3 \| Inverses of Matrices and Matrix Equations 527 ▼ SO LUTI O N We let x 1, x 2, and x 3 be the respective amounts (in grams) of KayDee Food, Pet Pellets,
and Rodent Chow that the hamsters should eat and y 1, y 2, and y 3 be the corresponding amounts for the gerbils. Then we want to solve the matrix equations 10 10 5 10 10 5 C C 0 20 10 0 20 10 20 10 30 20 10 30 S C x1 x2 x3 y1 y2 y3 S C 340 280 440 S 480 360 680 B C S C C S 340 280 440 S 480 360 680 Let A 10 10 5 0 20 10 20 10 30 Then we can write these matrix equations as S S C C AX B AY C C Hamster equation S Gerbil equation Hamster equation Gerbil equation X x1 x2 x3 Y y1 y2 y3 C S C S We want to solve for X and Y, so we multiply both sides of each equation by A1, the inverse of the coefﬁcient matrix. We could ﬁnd A1 by hand, but it is more convenient to use a graphing calculator as shown in Figure 3. [A]-1[B] [[10] [3 ] [12]] [A]-1[C] [[8 ] [4 ] [20]] FIGURE 3 (a) (b) MATHEMATICS IN THE MODERN WORLD Mathematical Ecology © In the 1970s humpback whales became a center of controversy. Environmentalists believed that whaling threatened the whales with imminent extinction; whalers saw their livelihood threatened by any attempt to stop whaling. Are whales really threatened to extinction by whaling? What level of whaling is safe to guarantee survival of the whales? These questions motivated mathematicians to study population patterns of whales and other species more closely. As early as the 1920s Lotka and Volterra had founded the ﬁeld of mathematical biology by creating predator-prey models. Their models, which draw on a branch of mathematics called differential equations, take into account the rates at which predator eats prey and the rates of growth of each population. Note that as predator eats prey, the prey population decreases; this means less food supply for the predators, so their population begins to decrease; with fewer predators the prey population begins to increase, and so on. Normally, a state of equilibrium develops, and the two populations alternate between a minimum and a maximum. Notice that if the predators eat the prey too fast, they will be left without food and will thus ensure their own extinction. Since Lotka and Volter
ra’s time, more detailed mathematical models of animal populations have been developed. For many species the population is divided into several stages: immature, juvenile, adult, and so on. The proportion of each stage that survives or reproduces in a given time period is entered into a matrix (called a transition matrix); matrix multiplication is then used to predict the population in succeeding time periods. (See the Discovery Project, page 519.) As you can see, the power of mathematics to model and predict is an invaluable tool in the ongoing debate over the environment. 528 CHAPTER 7 \| Matrices and Determinants So X A1B 10 3 12 Y A1C 8 4 20 Thus, each hamster should be fed 10 g of KayDee Food, 3 g of Pet Pellets, and 12 g of Rodent Chow, and each gerbil should be fed 8 g of KayDee Food, 4 g of Pet Pellets, and 20 g of Rodent Chow daily. ✎ Practice what you’ve learned: Do Exercise 47. ▲ C S S C 7. ▼ CONCE PTS 1. (a) The matrix I is called an c (b) If A is a 2 2 matrix, then –8 ■ Find the inverse of the matrix and verify that A1A AA1 I2 and B1B BB1 I3. matrix. and 7. A 7 3 c 4 2 d 8c) If A and B are 2 2 matrices with AB I, then B is the 9–24 ■ Find the inverse of the matrix if it exists. C S of A. 2. (a) Write the following system as a matrix equation AX B. System Matrix equation # A X B 5x 3y 4 3x 2y 3 (b) The inverse of A is A1. R B d c B R B R (c) The solution of the matrix equation is X A1 B X A1 B. x y d c (d) The solution of the system is ▼ SKI LLS 3–6 ■ Calculate the products AB and BA to verify that B is the inverse of A. ✎ 3. A 4. A 5 12 1 S B 9 10 8 C 12 11 14. 11. 13. 15. 3 5 3 d 2 2 5 5 13.4 1.2 0.6 d 0.3 c ✎ 17. ✎ 19. C C 21 10 23 10. 12. 14
. 16 18. C 20. C 22 24 25–32 ■ Solve the system of equations by converting to a matrix equation and using the inverse of the coefﬁcient matrix, as in Example 6. Use the inverses from Exercises 9–12, 17, 18, 21, and 23. ✎ 25. 5x 3y 4 3x 2y 0 27. 2x 5y 2 5x 13y 20 b 26. 28. 3x 4y 10 7x 9y 20 7x 4y 0 8x 5y 100 b b b SE CTI O N 7. 3 \| Inverses of Matrices and Matrix Equations 529 29. 31. 2x 4y z 7 x y z 0 x 4y 2 1x2y 2z 12 c 3x 1y 3z 2 1x 2y 3z 08 30. 32. 5x 7y 4z 1 3x 1y 3z 1 6x 7y 5z 1 x 2y z 3„ 0 c 11y z 1„ 1 1y z 1„ 2 x 2y z 2„ 3 of food available, and each type contains the following amounts of these nutrients per ounce: Type A Type B Type C Folic acid (mg) Choline (mg) Inositol (mg 35. 34. 33. 3x 4y 2z 2 2x 3y 5z 5 5x 2y 2z 33–38 ■ Use a calculator that can perform matrix operations to solve the system, as in Example 7. x 1y 2z 03 2x 1y 5z 11 2x 3y 1z 12 12x 1 2 y 7z 21 c 11x 2y 3z 43 13x y 4z 29 x 2y 3„ 10 c 2z 18 x 2y 2z 2„ 15 2„ 13 2x 3y 1x 1y 1z 1„ 15 1x 1y 1z 1„ 15 1x 2y 3z 4„ 26 1x 2y 3z 4„ 12 38. 37. 36. c d d 39–40 ■ Solve the matrix equation by multiplying each side by the appropriate inverse matrix. 39. 40 12 0 § 41–42 ■ Find the inverse of the matrix. S C 41 42 abcd 0 D 1 T 2 43–46 ■ Find the inverse of the matrix. For what value(s) of x, if
any, does the matrix have no inverse? 43. 2 x c x x2 d 45. ex 1 ex e2x 0 0 0 0 2 44. ex e2x e3x e2x d c x 46 ▼ APPLICATIONS 47. Nutrition A nutritionist is studying the effects of the ✎ nutrients folic acid, choline, and inositol. He has three types (a) Find the inverse of the matrix 3 4 3 1 2 2 3 4 4 and use it to solve the remaining parts of this problem. (b) How many ounces of each food should the nutritionist feed his laboratory rats if he wants their daily diet to contain 10 mg of folic acid, 14 mg of choline, and 13 mg of inositol? C S (c) How much of each food is needed to supply 9 mg of folic acid, 12 mg of choline, and 10 mg of inositol? (d) Will any combination of these foods supply 2 mg of folic acid, 4 mg of choline, and 11 mg of inositol? 48. Nutrition Refer to Exercise 47. Suppose food type C has been improperly labeled, and it actually contains 4 mg of folic acid, 6 mg of choline, and 5 mg of inositol per ounce. Would it still be possible to use matrix inversion to solve parts (b), (c), and (d) of Exercise 47? Why or why not? 49. Sales Commissions An encyclopedia saleswoman works for a company that offers three different grades of bindings for its encyclopedias: standard, deluxe, and leather. For each set that she sells, she earns a commission based on the set’s binding grade. One week she sells one standard, one deluxe, and two leather sets and makes $675 in commission. The next week she sells two standard, one deluxe, and one leather set for a $600 commission. The third week she sells one standard, two deluxe, and one leather set, earning $625 in commission. (a) Let x, y, and z represent the commission she earns on standard, deluxe, and leather sets, respectively. Translate the given information into a system of equations in x, y, and z. (b) Express the system of equations you found in part (a) as a matrix equation of the form AX B. (c) Find the inverse of the coefﬁcient matrix A and use
it to solve the matrix equation in part (b). How much commission does the saleswoman earn on a set of encyclopedias in each grade of binding? ▼ DISCOVE RY • DISCUSSION • WRITI NG 50. No Zero-Product Property for Matrices We have used the Zero-Product Property to solve algebraic equations. Matrices do not have this property. Let O represent the 2 2 zero matrix: O 0 0 c 0 0 d Find 2 2 matrices A O and B O such that AB O. Can you ﬁnd a matrix A O such that A2 O? 530 CHAPTER 7 \| Matrices and Determinants 7.4 Determinants and Cramer’s Rule LEARNING OBJECTIVES After completing this section, you will be able to: ■ Find the determinant of a 2 2 matrix ■ Find the determinant of an n n matrix ■ Use the Invertibility Criterion ■ Use row and column transformations in ﬁnding the determinant of a matrix ■ Use Cramer’s Rule to solve a linear system ■ Use determinants to ﬁnd the area of a triangle in the coordinate plane If a matrix is square (that is, if it has the same number of rows as columns), then we can assign to it a number called its determinant. Determinants can be used to solve systems of linear equations, as we will see later in this section. They are also useful in determining whether a matrix has an inverse. ■ Determinant of a 2 2 Matrix We denote the determinant of a square matrix A by the symbol det We ﬁrst deﬁne det The following box gives the deﬁnition of a 2 2 determinant. A 2 1 for the simplest cases. If A ”a’ is a 1 1 matrix, then det A 2 1 or A. a. A 2 1 A 2 1 We will use both notations, det and A, for the determinant of A. Although the symbol A looks like the absolute value symbol, it will be clear from the context which meaning is intended. DETERMINANT OF A 2 2 MATRIX The determinant of the 2 2 matrix A a c c b d d is det ` ad bc E X AM P L E 1 \| Determinant of a 2 2 Matrix To evaluate a 2 2 determinant, we take the product of the diagonal from top left
to bottom right and subtract the product from top right to bottom left, as indicated by the arrows. Evaluate A for A 6 3 3 d 2. c ▼ SO LUTI O N 6 3 —— 18 6 1 2 24 ✎ Practice what you’ve learned: Do Exercise 5. ▲ ■ Determinant of an n n Matrix To deﬁne the concept of determinant for an arbitrary n n matrix, we need the following terminology. I S B R O C © David Hilbert (1862–1943) was born in Königsberg, Germany, and became a professor at Göttingen University. He is considered by many to be the greatest mathematician of the 20th century. At the International Congress of Mathematicians held in Paris in 1900, Hilbert set the direction of mathematics for the about-todawn 20th century by posing 23 problems that he believed to be of crucial importance. He said that “these are problems whose solutions we expect from the future.” Most of Hilbert’s problems have now been solved (see Julia Robinson, page 508, and Alan Turing, page 160), and their solutions have led to important new areas of mathematical research. Yet as we enter the new millennium, some of his problems remain unsolved. In his work, Hilbert emphasized structure, logic, and the foundations of mathematics. Part of his genius lay in his ability to see the most general possible statement of a problem. For instance, Euler proved that every whole number is the sum of four squares; Hilbert proved a similar statement for all powers of positive integers. SE CTI O N 7. 4 \| Determinants and Cramer’s Rule 531 Let A be an n n matrix. 1. The minor Mij of the element aij is the determinant of the matrix obtained by deleting the ith row and jth column of A. 2. The cofactor Aij of the element aij is For example, if A is the matrix Aij 1 1 2 ijMij 2 3 1 4 0 2 2 5 6 § £ then the minor M12 is the determinant of the matrix obtained by deleting the ﬁrst row and second column from A. Thus, M12 So the cofactor A12 3 1 1 2 12M12 3 8. Similarly, M33 So A33 1 33M33 4. 3 1 2 3 Note that the cofactor of aij is simply the minor of
aij multiplied by either 1 or 1, is even or odd. Thus, in a 3 3 matrix we obtain the cofactor depending on whether of any element by preﬁxing its minor with the sign obtained from the following checkerboard pattern. i j § £ We are now ready to deﬁne the determinant of any square matrix. THE DETERMINANT OF A SQUARE MATRIX If A is an n n matrix, then the determinant of A is obtained by multiplying each element of the ﬁrst row by its cofactor, and then adding the results. In symbols. det A 1 2 \| A \| a11 a12 a21 a22 o an1 an2 o p a1n p a2n ∞ o p ann 4 4 a11 A11 a12 A12 # # # a1n A1n 532 CHAPTER 7 \| Matrices and Determinants E X AM P L E 2 \| Determinant of a 3 3 Matrix Evaluate the determinant of the matrix. ▼ SO LUTI det 16 24 4 2 3 1 44 ✎ Practice what you’ve learned: Do Exercise 19. ▲ In our deﬁnition of the determinant we used the cofactors of elements in the ﬁrst row only. This is called expanding the determinant by the ﬁrst row. In fact, we can expand the determinant by any row or column in the same way, and obtain the same result in each case (although we won’t prove this). The next example illustrates this principle. E X AM P L E 3 \| Expanding a Determinant About a Row and a Column Let A be the matrix of Example 2. Evaluate the determinant of A by expanding (a) by the second row (b) by the third column Verify that each expansion gives the same value. ▼ SO LUTI O N (a) Expanding by the second row, we get det 20 64 44 b) Expanding by the third column gives Graphing calculators are capable of computing determinants. Here is the output when the TI-83 is used to calculate the determinant in Example 3: det A] [[2 3 -1] [0 2 4 ] [ - 2 5 6 ]] det([A]) -44 64 24 44 SE CTI O N 7. 4 \| Determinants and Cramer�
�s Rule 533 In both cases we obtain the same value for the determinant as when we expanded by the ﬁrst row in Example 2. ✎ Practice what you’ve learned: Do Exercise 31. ▲ The following criterion allows us to determine whether a square matrix has an inverse without actually calculating the inverse. This is one of the most important uses of the determinant in matrix algebra, and it is the reason for the name determinant. INVERTIBILITY CRITERION If A is a square matrix, then A has an inverse if and only if det 0. A 2 1 We will not prove this fact, but from the formula for the inverse of a 2 2 matrix (page 521) you can see why it is true in the 2 2 case. E X AM P L E 4 \| Using the Determinant to Show That a Matrix Is Not Invertible Show that the matrix A has no inverse ▼ SO LUTI O N We begin by calculating the determinant of A. Since all but one of the elements of the second row is zero, we expand the determinant by the second row. If we do this, we see from the following equation that only the cofactor A24 will have to be calculated. D T det # A21 4 0 # A22 4 0 # A23 3 # A24 3A24 Expand this by column Since the determinant of A is zero, A cannot have an inverse, by the Invertibility Criterion. ✎ Practice what you’ve learned: Do Exercise 23. ▲ 534 CHAPTER 7 \| Matrices and Determinants ■ Row and Column Transformations The preceding example shows that if we expand a determinant about a row or column that contains many zeros, our work is reduced considerably because we don’t have to evaluate the cofactors of the elements that are zero. The following principle often simpliﬁes the process of ﬁnding a determinant by introducing zeros into it without changing its value. ROW AND COLUMN TRANSFORMATIONS OF A DETERMINANT If A is a square matrix and if the matrix B is obtained from A by adding a multiple of one row to another or a multiple of one column to another, then det detÓBÔ. A 2 1 E X AM P L E 5 \| Using Row and Column Transformations to Calculate a Determinant Find the determinant of the matrix
A. Does it have an inverse? A 8 3 24 2 2 1 5 3 6 2 4 11 1 12 1 7 ▼ SO LUTI O N If we add 3 times row 1 to row 3, we change all but one element of row 3 to zeros 11 4 0 0 7 1 2 This new matrix has the same determinant as A, and if we expand its determinant by the third row, we get T D det 11 2 1 Now, adding 2 times column 3 to column 1 in this determinant gives us 3 det A 1 2 4 0 25 0 3 2 4 5 11 2 1 Expand this by column 1 4 4 3 1 1 25 2 3 2 4 2 1 25 600 Since the determinant of A is not zero, A does have an inverse. ✎ Practice what you’ve learned: Do Exercise 27 Emmy Noether (1882–1935) was one of the foremost mathematicians of the early 20th century. Her groundbreaking work in abstract algebra provided much of the foundation for this ﬁeld, and her work in invariant theory was essential in the development of Einstein’s theory of general relativity. Although women weren’t allowed to study at German universities at that time, she audited courses unofﬁcially and went on to receive a doctorate at Erlangen summa cum laude, despite the opposition of the academic senate, which declared that women students would “overthrow all academic order.” She subsequently taught mathematics at Göttingen, Moscow, and Frankfurt. In 1933 she left Germany to escape Nazi persecution, accepting a position at Bryn Mawr College in suburban Philadelphia. She lectured there and at the Institute for Advanced Study in Princeton, New Jersey, until her untimely death in 1935. SE CTI O N 7. 4 \| Determinants and Cramer’s Rule 535 ■ Cramer’s Rule The solutions of linear equations can sometimes be expressed by using determinants. To illustrate, let’s solve the following pair of linear equations for the variable x. ax by r cx dy s e To eliminate the variable y, we multiply the ﬁrst equation by d and the second by b, and subtract. adx bdy rd bcx bdy bs adx bcx rd bs Factoring the left-hand side, we get 1 we can now solve this equation for x: ad bc 2 x
rd bs. Assuming that ad bc Z 0, Similarly, we ﬁnd x rd bs ad bc y as cr ad bc The numerator and denominator of the fractions for x and y are determinants of 2 2 matrices. So we can express the solution of the system using determinants as follows. CRAMER’S RULE FOR SYSTEMS IN TWO VARIABLES ax by r cx dy The linear system has the solution x provided that a b c d 0. 2 2 Using the notation D a c c b d d Dx r s c b d d Dy a c c r s d Coefﬁcient matrix Replace ﬁrst column of D by r and s. Replace second column of D by r and s. 536 CHAPTER 7 \| Matrices and Determinants we can write the solution of the system as x 0 Dx 0 D 0 0 and y 0 Dy 0 D 0 0 E X AM P L E 6 \| Using Cramer’s Rule to Solve a System with Two Variables Use Cramer’s Rule to solve the system. 2x 6y 1 x 8y 2 e ▼ SO LUTI O N For this system we have 10 0 0 0 The solution is Dx 0 2 Dy 20 1 1 2 1 5 x 0 y 0 Dx 0 D 0 Dy 0 D 0 0 0 20 10 2 5 10 1 2 ✎ Practice what you’ve learned: Do Exercise 33. ▲ Cramer’s Rule can be extended to apply to any system of n linear equations in n variables in which the determinant of the coefﬁcient matrix is not zero. As we saw in the preceding section, any such system can be written in matrix form as a11 a12 a21 a22 o an1 an2 o ≥ p a1n p a2n ∞ o p ann x1 x2 o xn ¥ ≥ ¥ ≥ b1 b2 o bn ¥ By analogy with our derivation of Cramer’s Rule in the case of two equations in two unknowns, we let D be the coefﬁcient matrix in this system, and Dxi be the matrix obtained by replacing the ith column of D by the numbers b1, b2,..., bn that appear to the right of the equal sign. The solution of the system is then given by the following
rule. SE CTI O N 7. 4 \| Determinants and Cramer’s Rule 537 CRAMER’S RULE If a system of n linear equations in the n variables x 1, x 2,..., x n is equivalent to the matrix equation DX B, and if D 0, then its solutions are 0 x1 Dx1 0 D 0 0 x2 0 Dx2 0 D 0 0 p xn 0 Dxn 0 D 0 0 where Dxi is the matrix obtained by replacing the ith column of D by the n 1 matrix B. E X AM P L E 7 \| Using Cramer’s Rule to Solve a System with Three Variables Use Cramer’s Rule to solve the system. 2x 3y 4z 1 x 6z 0 3x 2y 5 ▼ SO LUTI O N First, we evaluate the determinants that appear in Cramer’s Rule. Note that D is the coefﬁcient matrix and that Dx, Dy, and Dz are obtained by replacing the ﬁrst, second, and third columns of D by the constant terms. c D 0 0 3 Dy 38 3 22 Dx 0 Dz 0 0 0 Now we use Cramer’s Rule to get the solution: 3 3 78 13 Dx 0 D 0 Dy 0 D 0 Dz 0 D 0 78 38 39 19 22 38 11 19 13 38 13 38 0 ✎ Practice what you’ve learned: Do Exercise 39. ▲ Solving the system in Example 7 using Gaussian elimination would involve matrices whose elements are fractions with fairly large denominators. Thus, in cases like Examples 6 and 7, Cramer’s Rule gives us an efﬁcient way to solve systems of linear equations. But in systems with more than three equations, evaluating the various determinants that are involved is usually a long and tedious task (unless you are using a graphing calculator). Moreover, the rule doesn’t apply if D 0 or if D is not a square matrix. So Cramer’s Rule is a useful alternative to Gaussian elimination, but only in some situations. 538 CHAPTER 7 \| Matrices and Determinants ■ Areas of Triangles Using Determinants Determinants provide a simple way to calculate the area of a triangle in the coordinate plane. AREA OF A TRIANGLE If a triangle in
the coordinate plane has vertices then its area is a1, b12, 1 1 a2, b22, and a3, b32 1, area 1 2 a1 a2 a3 b1 b2 b3 1 1 1 3 3 y 0 (a‹, b‹) (a⁄, b⁄) x (a¤, b¤) where the sign is chosen to make the area positive. You are asked to prove this formula in Exercise 63. E X AM P L E 8 \| Area of a Triangle Find the area of the triangle shown in Figure 1. y 6 4 2 0 1 3 x FIGURE 1 We can calculate the determinant by hand or by using a graphing calculator. ▼ SO LUTI O N The vertices are ceding box, we get 1, 2 1, 2 3, 6 1, and 2 1, 4 1. Using the formula in the pre2 [A] [[ -1 4 1] [3 6 1] [1 2 1]] det([A]) -12 area 12 1 2 To make the area positive, we choose the negative sign in the formula. Thus, the area of the triangle is 3 3 ✎ Practice what you’ve learned: Do Exercise 55. ▲ area 1 2 1 12 2 6 7. ▼ CONCE PTS A 1. True or false? det 1 2 A 2. True or false? det 2 1 A 3. True or false? If det 2 1 is deﬁned only for a square matrix A. is a number, not a matrix. 0, then A is not invertible. 4. Fill in the blanks with appropriate numbers to calculate the determinant. Where there is “”, choose the appropriate sign or. 2 1 2 3 1 4 (ab ▼ SKI LLS 5–12 ■ Find the determinant of the matrix, if it exists. 6. c c ✎ 7. 9. 11.2 1.4 1.0 d 0.5 8. 10. 12. c c c 13–18 ■ Evaluate the minor and cofactor using the matrix A 13. M11, A11 16. M13, A13 14. 17. M33, A33 C M23, A23 15. S 18. M12, A12 M32, A32 19–26 ■ Find the determinant of the matrix. Determine whether the matrix
has an inverse, but don’t calculate the inverse. ✎ 19 21. 20 30 0 S 0 10 20 10 0 40 S ✎ 23 20. 1 2 2 3 5 3 C 22. 2 3 2 4 2 2 1 2 24 SE CTI O N 7. 4 \| Determinants and Cramer’s Rule 539 25 26 27–30 ■ Evaluate the determinant, using row or column operations whenever possible to simplify your work 12 27. 28. T 29 31. Let 10 1 30a) Evaluate det B 1 (b) Evaluate det B 1 (c) Do your results in parts (a) and (b) agree? by expanding by the second row. by expanding by the third column. 2 2 C S 32. Consider the system x 2y 6z 5 3x 6y 5z 8 2x 6y 9z 7 (a) Verify that x 1, y 0, z 1 is a solution of the system. (b) Find the determinant of the coefﬁcient matrix. (c) Without solving the system, determine whether there are c any other solutions. (d) Can Cramer’s Rule be used to solve this system? Why or why not? 33–48 ■ Use Cramer’s Rule to solve the system. ✎ 33. 35. 37. ✎ 39. 41. e • • 8 e 2x y 9 x 2y x 6y 3 3x 2y 1 e 0.4x 1.2y 0.4 1.2x 1.6y 3.2 x 2y 2z 10 3x 2y 2z 11 x 2y 2z 10 2x1 x1 3x2 x2 2x2 5x3 x3 x3 1 2 8 34. 36. 38. 40. 42. e e e • • 6x 12y 33 4x 17y 20 10x 17y 21 20x 31y 39 1 2 5x 03y z 06 0 4y 6z 22 7x 10y 13 2a 2b 2c 02 a 2b 2c 09 3a 5b 2c 22 540 CHAPTER 7 \| Matrices and Determinants 43. 45. • • 47. µ 1 10 10 11 2x 3y 5z 04 2x 7y 5z 10 4x 7y 5z 00 2x y 2z „ 0 2x y 2z „ 0
2x y 2z „ 0 2x y 2z „ 1 44. 46. • • 48. µ 2x y 5 3z 19 5x 4y 7z 17 2x 5y 5z 4 x 5y 0z 8 3x 5y 5z the heights of three points on the bridge, as shown in the ﬁgure. He wishes to ﬁnd an equation of the form y ax2 bx c to model the shape of the arch. (a) Use the surveyed points to set up a system of linear equa- tions for the unknown coefﬁcients a, b, and c. (b) Solve the system using Cramer’s Rule. y (ft) 49–50 ■ Evaluate the determinants 49. a 0 0 0 0 5 51–54 ■ Solve for x. 51. x 0 0 12 x 1 0 13 23 x 2 50. 0 52 53. 3 1 x2 x 0 1 0 x 0 1 0 54. 3 3 3 55–58 ■ Sketch the triangle with the given vertices and use a determinant to ﬁnd its area. 3 ✎ 55. 57. 1 1 0, 0 6, 2, 1 2 3, 8, 1 2 2 56. 1, 0 3, 2 2 1, 3 2, 9, 1 2, 1 2 5, 6 2 58. 2, 5 1 7, 2, 1 2, 1 2 3, 4 2 59. Show that 1 1 1 x y z x2 y2 z2 ▼ APPLICATIONS 60. Buying Fruit A roadside fruit stand sells apples at 75¢ a pound, peaches at 90¢ a pound, and pears at 60¢ a pound. Muriel buys 18 pounds of fruit at a total cost of $13.80. Her peaches and pears together cost $1.80 more than her apples. (a) Set up a linear system for the number of pounds of apples, peaches, and pears that she bought. (b) Solve the system using Cramer’s Rule. 61. The Arch of a Bridge The opening of a railway bridge over a roadway is in the shape of a parabola. A surveyor measures 3 33 ft 4 25 ft 40 ft 10 15 40 62. A Triangular Plot of Land An outdoors club is purchasing land to set up a conservation area. The last remaining piece they need to
buy is the triangular plot shown in the ﬁgure. Use the determinant formula for the area of a triangle to ﬁnd the area of the plot. 6000 4000 2000 ) 2000 4000 6000 E-W baseline (ft) ▼ DISCOVE RY • DISCUSSION • WRITI NG 63. Determinant Formula for the Area of a Triangle The ﬁgure on the next page shows a triangle in the plane with vera1, b12 a3, b32 tices., 1 1 (a) Find the coordinates of the vertices of the surrounding a2, b22 1, and rectangle, and ﬁnd its area. (b) Find the area of the red triangle by subtracting the areas of the three blue triangles from the area of the rectangle. (c) Use your answer to part (b) to show that the area of the red triangle is given by area 1 2 a1 a2 a3 b1 b2 b3 1 1 1 3 3 (a‹, b‹) y 0 (a⁄, b⁄) (a¤, b¤) x 64. Collinear Points and Determinants (a) If three points lie on a line, what is the area of the “triangle” that they determine? Use the answer to this question, together with the determinant formula for the a1, b12 area of a triangle, to explain why the points, 1 a2, b22 are collinear if and only if 1, and a3, b32 1 a1 a2 a3 b1 b2 b3 1 1 1 0 (b) Use a determinant to check whether each set of points is 3 collinear. Graph them to verify your answer. i(i) (ii) 6, 4 2, 10, 1 2 1 5, 10, 2, 6 1 6, 13 2 1 15 CHAPTER 7 \| Review 541 65. Determinant Form for the Equation of a Line (a) Use the result of Exercise 64(a) to show that the equation of the line containing the points x1, y12 1 x2, y22 and 1 is x x1 x2 y y1 y2 1 1 1 0 (b) Use the result of part (a) to ﬁnd an equation for the line 3 20, 50 1 containing the points
10, 25 1 and. 2 2 3 66. Matrices with Determinant Zero Use the deﬁnition of determinant and the elementary row and column operations to explain why matrices of the following types have determinant 0. (a) A matrix with a row or column consisting entirely of zeros (b) A matrix with two rows the same or two columns the same (c) A matrix in which one row is a multiple of another row, or one column is a multiple of another column 67. Solving Linear Systems Suppose you have to solve a linear system with ﬁve equations and ﬁve variables without the assistance of a calculator or computer. Which method would you prefer: Cramer’s Rule or Gaussian elimination? Write a short paragraph explaining the reasons for your answer. CHAPTER 7 \| REVIEW ▼ P R O P E RTI LAS Matrices (p. 495) A matrix A of dimension with m rows and n columns: m n is a rectangular array of numbers A a11 a21 o am1 a12 a22 o am2 p a1n p a2n ∞ p amn o D Augmented Matrix of a System (p. 495) The augmented matrix of a system of linear equations is the matrix consisting of the coefﬁcients and the constant terms. For example, for the two-variable system T the augmented matrix is a11x a12 x b1 a21x a22 x b2 a11 a21 c a12 a22 b1 b2 d Elementary Row Operations (p. 496) To solve a system of linear equations using the augmented matrix of the system, the following operations can be used to transform the rows of the matrix: 1. Add a nonzero multiple of one row to another. 2. Multiply a row by a nonzero constant. 3. Interchange two rows. Row-Echelon Form of a Matrix (p. 497) A matrix is in row-echelon form if its entries satisfy the following conditions: 1. The ﬁrst nonzero entry in each row (the leading entry) is the number 1. 2. The leading entry of each row is to the right of the leading entry in the row above it. 3. All rows consisting entirely of zeros are at the bottom of the matrix. If the matrix also satisﬁes the following condition, it is in reduced row
-echelon form: 4. If a column contains a leading entry, then every other entry in that column is a 0. 542 CHAPTER 7 \| Matrices and Determinants Number of Solutions of a Linear System (p. 501) If the augmented matrix of a system of linear equations has been reduced to row-echelon form using elementary row operations, then the system has: 1. No solution if the row-echelon form contains a row that 0 1. In this case the system is represents the equation inconsistent. 2. One solution if each variable in the row-echelon form is a leading variable. 3. Inﬁnitely many solutions if the system is not inconsistent but not every variable is a leading variable. In this case the system is dependent. Operations on Matrices (p. 508) m n If A and B are matrices and c is a scalar (real number), then: 1. The sum A B is the m n matrix that is obtained by adding corresponding entries of A and B. 2. The difference A B is the m n matrix that is obtained by subtracting corresponding entries of A and B. 3. The scalar product cA is the m n matrix that is obtained by multiplying each entry of A by c. m n Multiplication of Matrices (p. 510) n k If A is an matrix (so the number of columns of A is the same as the number of rows of B), then the matrix product AB is the matrix whose ij-entry is the inner product of the ith row of A and the jth column of B. matrix and B is an m k Properties of Matrix Operations (pp. 509, 512) If A, B, and C are matrices of compatible dimensions then the following properties hold: 1. Commutativity of addition: A B B A 2. Associativity: A B 1 C A C A 1 BC 2 AB 1 2 1 2 B C 2 3. Distributivity: B C A 1 B C AB AC 2 A BA CA 1 2 (Note that matrix multiplication is not commutative.) Identity Matrix (p. 520) The identity matrix entries are all 1 and whose other entries are all 0: n n is the In matrix whose main diagonal In If A is an m n matrix, then D T AIn A and Im A A Inverse of a Matrix (p. 521) If A is an A1 with the following
properties: n n matrix, then the inverse of A is the n n matrix A1A In and AA1 In To ﬁnd the inverse of a matrix, we use a procedure involving elementary row operations (explained on page 496). (Note that some square matrices do not have an inverse.) Inverse of a 2 2 Matrix (p. 521) For for ﬁnding the inverse: 2 2 matrices the following special rule provides a shortcut A a c b d 1 A1 1 ad bc d b c a Writing a Linear System as a Matrix Equation (p. 524) B linear equations in variables can be written as a A system of single matrix equation R R n n B AX B A is the n n where matrix of coefﬁcients, n 1 B of the variables, and example, the linear system of two equations in two variables is the matrix of the constants. For n 1 is the X matrix can be expressed as a11x a12 x b1 a21x a22 x b2 a11 a21 a12 a22 x y b1 b2 Solving Matrix Equations (p. 525) R B B n 1 matrix, X is an If A is an invertible and B is an B n n constant matrix, then the matrix equation n 1 R R variable matrix, has the unique solution AX B X A1B Determinant of a 2 2 Matrix (p. 530) The determinant of the matrix A a c b d is the number det A 1 2 ƒ A ƒ ad bc B R A Minors and Cofactors (p. 531) aij 0 If of the entry is the determinant of the matrix obtained by deleting the ith row and the jth column of A. matrix, then the minor n n is an Mij 0 aij The cofactor Aij of the entry aij is 1 Aij ijMij 1 (Thus, the minor and the cofactor of each entry either are the same or are negatives of each other.) 2 CHAPTER 7 \| Review 543 Cramer’s Rule (pp. 535–537) If a system of n linear equations in the n variables equivalent to the matrix equation solutions of the system are DX B and if 0 x1, x2, p, xn 0 D, then the is 0 0 x1 Dx1 0 D 0 0 x2 0 D
x2 0 D 0 0 p xn 0 Dxn 0 D 0 0 Dxi where column by the constant matrix B. is the matrix that is obtained from D by replacing its ith Area of a Triangle Using Determinants (p. 538) a1, b12 If a triangle in the coordinate plane has vertices, and, then the area of the triangle is given by 1 a3, b32 1 a2, b2 2 1, area 1 2 a1 a2 a3 b1 b2 b3 1 1 1 where the sign is chosen to make the area positive. 3 3 Determinant of an n A n Matrix (p. 531) To ﬁnd the determinant of the n n matrix A a11 a21 o an1 a12 a22 o an2 p a1n p a2n ∞ p ann o D we choose a row or column to expand, and then we calculate the number that is obtained by multiplying each element of that row or column by its cofactor and then adding the resulting products. For example, if we choose to expand about the ﬁrst row, we get p a1n A1n a11A11 a12 A12 det A A T 1 2 0 0 Invertibility Criterion (p. 533) A square matrix has an inverse if and only if its determinant is not 0. Row and Column Transformations (p. 534) If we add a nonzero multiple of one row to another row in a square matrix or a nonzero multiple of one column to another column, then the determinant of the matrix is unchanged. ▼ CO N C E P T S U M MARY Section 7.1 ■ Find the augmented matrix of a linear system ■ Solve a linear system using elementary row operations ■ Solve a linear system using the row-echelon form of its matrix ■ Solve a system using the reduced row-echelon form of its matrix ■ Determine the number of solutions of a linear system from the row-echelon form of its matrix ■ Model using linear systems Section 7.2 ■ Determine whether two matrices are equal ■ Use addition, subtraction, and scalar multiplication of matrices ■ Multiply matrices ■ Write a linear system in matrix form Section 7.3 ■ Determine when two matrices are inverses of each other ■ Find the inverse of a 2 2 matrix ■ Find the inverse
of an n n matrix ■ Solve a linear system by writing it as a matrix equation ■ Model using matrix equations Section 7.4 ■ Find the determinant of a 2 2 matrix ■ Find the determinant of an n n matrix ■ Use the Invertibility Criterion ■ Use row and column transformations in computing the determinant of a matrix ■ Use Cramer’s Rule to solve a linear system ■ Use determinants to ﬁnd the area of a triangle in the coordinate plane Review Exercises 1–20 7–20 7–12 13–20 7–20 63–64 Review Exercises 21–22 23–26 27–34 51–54 Review Exercises 35–36 43–45 46–50 51–54 55–56 Review Exercises 43–45 46–50 43–50 47–48, 50 57–60 61–62 544 CHAPTER 7 \| Matrices and Determinants ▼ E X E RC I S E S 1–6 ■ A matrix is given. (a) State the dimension of the matrix. (b) Is the matrix in row-echelon form? (c) Is the matrix in reduced row-echelon form? (d) Write the system of equations for which the given matrix is the augmented matrix. 1. 1 0 2 5 3 1 3. B C 5. 4. B 1 0 1 2 0 6–12 ■ Use Gaussian elimination to ﬁnd the complete solution of the system, or show that no solution exists. T S D C 7. 9. x 2y 2z 6 1 x y 2x y 3z 7 x 2y 3z 2 c 2x y z 2 2x 7y 11z 9 11. x y z „ 0 c x y 4z „ 1 x 2y 4„ 7 2x 2y 3z 4„ 3 8. 10. 12. x y z 2 x y 3z 6 2y 3z 5 x y z 2 c x y 3z 6 3x y 5z 10 y x c 3z 1 4„ 5 2y z „ 0 2x y 5z 4„ 4 d 13–20 ■ Use Gauss-Jordan elimination to ﬁnd the complete solution of the system, or show that no solution exists. d 13. x y 3z 2 2x y z 2 3x 4z 4 15. x y z „ 0 c
3x y z „ 2 17. 19. b x y z 0 3x 2y z 6 x 4y 3z „ 2 2x 2x 4y 4z 2„ 6 14. 16. 18. 20. x y 1 x y 2z 3 x 3y 2z 1 c x y 3 2x y 6 x 2y 9 x 2y 3z 2 c 2x y 5z 1 4x 3y z 6 c x y 2z 3„ 0 y z „ 1 3x 2y 7z 10„ 2 21–22 ■ Determine whether the matrices A and B are equal. c d 21 22. A C c 225 0 1 S 21 d B 5 log 1 c e0 1 2 d 23–34 ■ Let Carry out the indicated operation, or explain why it cannot be performed. A B C D 2C 3D 23. 25. 24. 26. 5B 2C 29. BC 32. FC 27. GA 30. CB 28. AG 31. BF 33. 1 C D E 2 34. F 1 2C D 2 35–36 ■ Verify that the matrices A and B are inverses of each other by calculating the products AB and BA. 35. A 36 37–42 ■ Solve the matrix equation for the unknown matrix, X, or show that no solution exists, where 3X 2B B A R 2 B R 37. 38. 39. 40. 1 21 2 X A 3B 2 1 2X C 5A 41. AX C 42. AX B 43–50 ■ Find the determinant and, if possible, the inverse of the matrix. 43. 1 2 4 9 45. 4 12 R 6 2 B 47 44. 2 2 1 3 46. B 48 49 50 51–54 ■ Express the system of linear equations as a matrix equation. Then solve the matrix equation by multiplying each side by the inverse of the coefﬁcient matrix. T T D 51. 12x 5y 10 5x 2y 17 53. b 2x 2y 5z 1 3 x 2y 2z 1 4 x 2y 3z 1 6 52. 6x 5y 1 8x 7y 1 54. b 2x y 3z 5 x y 6z 0 3x y 6z 5 55. Magda and Ivan grow tomatoes, onions, and zucchini in their c backyard and sell them at a roadside stand on Saturdays and
Sundays. They price tomatoes at $1.50 per pound, onions at $1.00 per pound, and zucchini at 50 cents per pound. The following table shows the number of pounds of each type of produce that they sold during the last weekend in July. c Tomatoes Onions Zucchini Saturday Sunday 25 14 16 12 30 16 (a) Let A 25 14 c 16 12 30 16 d and B 1.50 1.00 0.50 Compare these matrices to the data given in the problem, and describe what their entries represent. C S (b) Only one of the products AB or BA is deﬁned. Calculate the product that is deﬁned, and describe what its entries represent. 56. An ATM at a bank in Qualicum Beach, British Columbia, dispenses $20 and $50 bills. Brodie withdraws $600 from this machine and receives a total of 18 bills. Let x be the number of $20 bills and y the number of $50 bills that he receives. (a) Find a system of two linear equations in x and y that express the information given in the problem. (b) Write your linear system as a matrix equation of the form AX B. A1, and use it to solve your matrix equation in (c) Find part (b). How many bills of each type did Brodie receive? CHAPTER 7 \| Review 545 57–60 ■ Solve the system using Cramer’s Rule. 57. 2x 7y 13 6x 16y 30 59. b 2x y 5z 0 x 7y 9 5x 4y 3z 9 58. 12x 11y 140 7x 19y 20 60. b 3x 4y 5z 10 4z 20 x 2x y 5z 30 c 61–62 ■ Use the determinant formula for the area of a triangle to ﬁnd the area of the triangle in the ﬁgure. c 61. 62 63–64 ■ Use any of the methods you have learned in this chapter to solve the problem. 63. Clarisse invests $60,000 in money-market accounts at three different banks. Bank A pays 2% interest per year, bank B pays 2.5%, and bank C pays 3%. She decides to invest twice as much in bank B as in the other two banks. After one year, Clarisse has earned $1575 in interest. How much did she invest in each bank
? 64. A commercial ﬁsherman ﬁshes for haddock, sea bass, and red snapper. He is paid $1.25 a pound for haddock, $0.75 a pound for sea bass, and $2.00 a pound for red snapper. Yesterday he caught 560 lb of ﬁsh worth $575. The haddock and red snapper together are worth $320. How many pounds of each ﬁsh did he catch? ■ CHAPTER 7 \| TEST 1–4 ■ Determine whether the matrix is in reduced row-echelon form, row-echelon form, or neither 10. 3. 4–6 ■ Use Gaussian elimination to ﬁnd the complete solution of the system, or show that no solution exists. 5. x y 2z 0 2x 4y 5z 5 2y 3z 5 6. S C 2x 3y z 3 x 2y 2z 1 4x y 5z 4 7. Use Gauss-Jordan elimination to ﬁnd the complete solution of the system. c c x 3y z 0 3x 4y 2z 1 1 x 2y 8–15 ■ Let Carry out the indicated operation, or explain why it cannot be performed. 8. 12. A B A1 B R 9. AB B1 13. 1 16. (a) Write a matrix equation equivalent to the following system. 2 14. det B C S C BA 3B 10. S 11. CBA 15. det C 1 2 4x 3y 10 3x 2y 30 (b) Find the inverse of the coefﬁcient matrix, and use it to solve the system. 17. Only one of the following matrices has an inverse. Find the determinant of each matrix, and b use the determinants to identify the one that has an inverse. Then ﬁnd the inverse 18. Solve using Cramer’s Rule: C C S 2x 2y 5z 14 3x 2y 5z 0 4x 2y 3z 2 S 19. A shopper buys a mixture of nuts; the almonds cost $4.75 a pound, and the walnuts cost $3.45 c a pound. Her total purchase weighs 3 lb and costs $11.91. Use Cramer’s Rule to determine how much of each nut she bought. 546 COMPUTER GRAPH
ICS Matrix algebra is the basic tool used in computer graphics to manipulate images on a computer screen. We will see how matrix multiplication can be used to “move” a point in the plane to a prescribed location. Combining such moves enables us to stretch, compress, rotate, and otherwise transform a ﬁgure, as we see in the images below. Image Compressed Rotated Sheared Moving Points in the Plane Let’s represent the point x, y 1 2 in the plane by a 2 1 matrix: 4 x, y 1 2 x y d c For example, the point 3, 2 1 2 in the ﬁgure is represented by the matrix P 3 2 d c y 1 0 (3, 2) 1 x Multiplying by a 2 2 matrix moves the point in the plane. For example, if T 1 0 0 1 d c then multiplying P by T, we get TP 3, _2) 2 1 3, 2 has been moved to the point We see that the point. In general, multiplication by this matrix T reﬂects points in the x-axis. If every point in an image is multiplied by this matrix, then the entire image will be ﬂipped upside down about the x-axis. Matrix multiplication “transforms” a point to a new point in the plane. For this reason a matrix used in this way is called a transformation. 2 1 3, 2 Table 1 gives some standard transformations and their effects on the gray square in the ﬁrst quadrant. 547 548 Focus on Modeling TABLE 1 Transformation matrix Effect T 0 1 0 1 d c Reﬂection in x-axis c c 0 T 0 1 d Expansion (or contraction) in the x-direction T 1 c 0 Shear in x-direction +1 x Moving Images in the Plane Simple line drawings such as the house in Figure 1 consist of a collection of vertex points and connecting line segments. The house in Figure 1 can be represented in a computer by the 2 11 data matrix The columns of D represent the vertex points of the image. To draw the house, we connect successive points (columns) in D by line segments. Now we can transform the whole house by multiplying D by an appropriate transformation matrix. For example, if we apply the shear transformation matrix. T TD 1 0 c 2 0 c 0..5 3 4.5 5 5..5 1 d 3 2 2 2,
we get the following 3 0 d 2 0 3 0 d To draw the image represented by TD, we start with the point 2 0 d c, connect it by a line segment to the point, then follow that by a line segment to c resulting tilted house is shown in Figure 2. 0 0 d 1.5 3 d c, and so on. The A convenient way to draw an image corresponding to a given data matrix is to use a graphing calculator. The TI-83 program in the margin at the top of the next page converts y 1 0 1 x FIGURE 1 y 1 0 1 x FIGURE 2 PROGRAM:IMAGE :For(N,1,10) :Line([A](1,N), [A](2,N),[A](1,N+1), [A](2,N+1)) :End Computer Graphics 549 a data matrix stored in [A] into the corresponding image, as shown in Figure 3. (To use this program for a data matrix with m columns, store the matrix in [A] and change the “10” in the For command to m 1.) 6 _1 _1 (a) House with data matrix D FIGURE 3 6 _1 (b) 7 _1 7 Tilted house with data matrix TD Problems 1. The gray square in Table 1 has the following vertices Apply each of the three transformations given in Table 1 to these vertices and sketch the result to verify that each transformation has the indicated effect. Use c 2 in the expansion matrix and c 1 in the shear matrix. 2. Verify that multiplication by the given matrix has the indicated effect when applied to the gray square in the table. Use c 3 in the expansion matrix and c 1 in the shear matrix. T1 1 0 0 1 d c T2 1 0 c d 0 c T3 1 0 c 1 d c Reﬂection in y-axis Expansion (or contraction) in y-direction Shear in y-direction 3. Let T 1 0 c 1.5 1 d. (a) What effect does T have on the gray square in the Table 1? (b) Find T 1. (c) What effect does T 1 have on the gray square? (d) What happens to the square if we ﬁrst apply T, then T 1? 4. (a) Let T (b) Let. What effect does T have on the gray square in Table 1?.
What effect does S have on the gray square in Table 1? (c) Apply S to the vertices of the square, and then apply T to the result. What is the effect of the combined transformation? (d) Find the product matrix W TS. (e) Apply the transformation W to the square. Compare to your ﬁnal result in part (c). What do you notice? 550 Focus on Modeling 5. The ﬁgure shows three outline versions of the letter F. The second one is obtained from the ﬁrst by shrinking horizontally by a factor of 0.75, and the third is obtained from the ﬁrst by shearing horizontally by a factor of 0.25. (a) Find a data matrix D for the ﬁrst letter F. (b) Find the transformation matrix T that transforms the ﬁrst F into the second. Calculate TD, and verify that this is a data matrix for the second F. (c) Find the transformation matrix S that transforms the ﬁrst F into the third. Calcu- late SD, and verify that this is a data matrix for the third F. Here is a data matrix for a line drawinga) Draw the image represented by D. (b) Let T 1 1 0 1 d c. Calculate the matrix product TD, and draw the image repre- sented by this product. What is the effect of the transformation T? (c) Express T as a product of a shear matrix and a reﬂection matrix. (See Problem 2.) 8.1 Parabolas 8.2 Ellipses 8.3 Hyperbolas 8.4 Shifted Conics CHAPTER 8 Conic Sections Farming sunlight? The sun is incredibly hot, with temperatures reaching into the millions of degrees; but fortunately, the amount of sunlight that reaches us is just right for keeping our planet at a comfortable temperature. Yet we can use the little sunlight we receive to unlock some of the sun’s immense power. The key to doing this is a simple geometric shape called a parabola. A mirror in the shape of a parabola can concentrate sunlight to a single point (see page 558). Many such parabolic mirrors, assembled into “solar farms,” can concentrate sunlight from a large area. The concentrated sunlight can heat water or other liquids to thousands of degrees, driving steam turbines for generating electricity.
Two such solar power plants have been built in the Mojave Desert in California, each producing more than 10 megawatts of electric power. One of these is shown in the photo above. In this chapter we study parabolas as well as the other conic sections. We will see that all the conic sections have impressive real-world applications. 551551 551 552 CHAPTER 8 \| Conic Sections 8.1 Parabolas LEARNING OBJECTIVES After completing this section, you will be able to: ■ Find geometric properties of a parabola from its equation ■ Find the equation of a parabola from some of its geometric properties Conic sections are the curves we get when we make a straight cut in a cone, as shown in the ﬁgure. For example, if a cone is cut horizontally, the cross section is a circle. So a circle is a conic section. Other ways of cutting a cone produce parabolas, ellipses, and hyperbolas. Circle Ellipse Parabola Hyperbola Our goal in this chapter is to ﬁnd equations whose graphs are the conic sections. We already know from Section 2.2 that the graph of the equation is a circle. We will ﬁnd equations for each of the other conic sections by analyzing their geometric properties. 2 y 2 r x 2 ■ Geometric Deﬁnition of a Parabola We saw in Section 4.1 that the graph of the equation y ax 2 bx c is a U-shaped curve called a parabola that opens either upward or downward, depending on whether the sign of a is positive or negative. In this section we study parabolas from a geometric rather than an algebraic point of view. We begin with the geometric deﬁnition of a parabola and show how this leads to the algebraic formula that we are already familiar with. GEOMETRIC DEFINITION OF A PARABOLA A parabola is the set of points in the plane that are equidistant from a ﬁxed point F (called the focus) and a ﬁxed line l (called the directrix). SECTION 8.1 \| Parabolas 553 This deﬁnition is illustrated in Figure 1. The vertex V of the parabola lies halfway between the focus and the directrix, and the axis of symmetry is the line that runs through the
focus perpendicular to the directrix. parabola axis focus F V FIGURE 1 vertex directrix l In this section we restrict our attention to parabolas that are situated with the vertex at the origin and that have a vertical or horizontal axis of symmetry. (Parabolas in more general positions will be considered in Section 8.4.) If the focus of such a parabola is the, then the axis of symmetry must be vertical, and the directrix has the equapoint tion y p. Figure 2 illustrates the case p 0. 0, p F 1 2 y P(x, y) F(0, p) p p 0 y x y=_p FIGURE 2 If P x, y is any point on the parabola, then the distance from P to the focus F (using 2 the Distance Formula) is 1 2x 2 y p 2 2 1 The distance from P to the directrix is y p y p 0 By the deﬁnition of a parabola these two distances must be equal: 2 0 1 0 0 1 2 2x y p y p x 2 1 2 2py 2py p 2 2 0 2 y x 2 2py 2py 2 4py x y p 2 2 Square both sides 2 Expand Simplify p 0, then the parabola opens upward, but if, it opens downward. When x is reIf placed by x, the equation remains unchanged, so the graph is symmetric about the y-axis. p 0 ■ Equations and Graphs of Parabolas The following box summarizes what we have just proved about the equation and features of a parabola with a vertical axis. 554 CHAPTER 8 \| Conic Sections PARABOLA WITH VERTICAL AXIS The graph of the equation is a parabola with the following properties. 2 4py x vertex FOCUS directrix 0, 0 V 1 2 1 0, p F 2 y p The parabola opens upward if p 0 or downward if p 0. F(0, p) y 0 x y=_p y 0 y=_p x F(0, p) ≈=4py with p>0 ≈=4py with p<0 E X AM P L E 1 \| Finding the Equation of a Parabola y 3 0 _3 ≈=8y F(0, 2) 3 x y=_2 _3 FIGURE 3 Find the equation of the par
abola with vertex graph. V 1 0, 0 2 and focus F 0, 2 1 2, and sketch its ▼ SO LUTI O N y 2 Since the focus is ). Thus, the equation of the parabola is 0, 2 F 1 2, we conclude that p 2 (so the directrix is x 2 4 2 1 2 8y y 2 x 2 = 4py with p = 2 x Since p 2 0, the parabola opens upward. See Figure 3. ✎ Practice what you’ve learned: Do Exercises 29 and 41. ▲ MATHEMATICS IN THE MODERN WORLD Looking Inside Your Head © How would you like to look inside your head? The idea isn’t particularly appealing to most of us, but doctors often need to do just that. If they can look without invasive surgery, all the better. An X-ray doesn’t really give a look inside, it simply gives a “graph” of the density of tissue the X-rays must pass through. So an X-ray is a “ﬂattened” view in one direction. Suppose you get an X-ray view from many different directions. Can these “graphs” be used to reconstruct the three-dimensional inside view? This is a purely mathematical problem and was solved by mathematicians a long time ago. However, reconstructing the inside view requires thousands of tedious computations. Today, mathematics and highspeed computers make it possible to “look inside” by a process called computer-aided tomography (or CAT scan). Mathematicians continue to search for better ways of using mathematics to reconstruct images. One of the latest techniques, called magnetic resonance imaging (MRI), combines molecular biology and mathematics for a clear “look inside.” SECTION 8.1 \| Parabolas 555 E X AM P L E 2 \| Finding the Focus and Directrix of a Parabola from Its Equation Find the focus and directrix of the parabola y x 2, and sketch the graph. x p 2 y. 1 4 ▼ SO LUTI O N To ﬁnd the focus and directrix, we put the given equation in the standard 4p 1, Comparing this to the general equation form F A so. The graph of the parabola, together with the focus and the directrix, is shown in Figure 4(a). We can also draw the graph
using a graphing calculator as shown in Figure 4(b)., and the directrix is. Thus, the focus is we see that 2 4py, y 1 4 0, 1 4B x y 1 _2 _2 F!0, _ @1 4 y= 1 4 2 x y=_≈ _2 FIGURE 4 (a) 2 1 _4 (b) ✎ Practice what you’ve learned: Do Exercise 11. ▲ Reﬂecting the graph in Figure 2 about the diagonal line y x has the effect of interchanging the roles of x and y. This results in a parabola with horizontal axis. By the same method as before, we can prove the following properties. PARABOLA WITH HORIZONTAL AXIS The graph of the equation 2 4px y is a parabola with the following properties. vertex FOCUS directrix 0, 0 V 1 2 1 p, 0 F 2 x p The parabola opens to the right if p 0 or to the left if p 0. x=_p y y F( p, 0) 0 x F( p, 0) 0 x ¥=4px with p>0 ¥=4px with p<0 x=_p 556 CHAPTER 8 \| Conic Sections E X AM P L E 3 \| A Parabola with Horizontal Axis A parabola has the equation (a) Find the focus and directrix of the parabola, and sketch the graph. (b) Use a graphing calculator to draw the graph. 6x y 2 0. ▼ SO LUTI O N (a) To ﬁnd the focus and directrix, we put the given equation in the standard form Comparing this to the general equation F A. Thus, the focus is 2 6x. 4p 6, y. Since p 0, the p 3 so 2 parabola opens to the left. The graph of the parabola, together with the focus and the directrix, is shown in Figure 5(a) below. 2 4px, and the directrix is we see that x 3 2 3 2, 0 y B, (b) To draw the graph using a graphing calculator, we need to solve for y. 6x y 2 0 2 6x y Subtract 6x y ; 16x Take square roots To obtain the graph of the parabola, we graph both functions
y 16x and y 16x as shown in Figure 5(b). y y = –6x 6x+¥=0 1 1 0 3 2_F!, 0@ _6 x 6 _6 2 x= 3 2 (a) FIGURE 5 ✎ Practice what you’ve learned: Do Exercise 13. y = – –6x (b) ▲ 2 4px, y The equation does not deﬁne y as a function of x (see page 220). So to use a graphing calculator to graph a parabola with horizontal axis, we must ﬁrst solve for y. This y 14px leads to two functions,. We need to graph both functions to get the complete graph of the parabola. For example, in Figure 5(b) we had to graph both y 16x to graph the parabola y 16x y 14px 2 6x. and and y We can use the coordinates of the focus to estimate the “width” of a parabola when sketching its graph. The line segment that runs through the focus perpendicular to the axis, with endpoints on the parabola, is called the latus rectum, and its length is the focal diameter of the parabola. From Figure 6 we can see that the distance from an endpoint Q. Thus, the distance from Q to the focus must of the latus rectum to the directrix is be. In the next example we use the focal diameter to determine the “width” of a parabola when graphing it. 0 as well (by the deﬁnition of a parabola), so the focal diameter is 2p 4p 2p 0 0 0 0 0 y p p Q 2p 0 F( p, 0) x latus rectum x=_p FIGURE 6 E X AM P L E 4 \| The Focal Diameter of a Parabola SECTION 8.1 \| Parabolas 557 y 1 Find the focus, directrix, and focal diameter of the parabola ▼ SO LUTI O N We ﬁrst put the equation in the form x 2 4py. y 1 2 x 2 x 2 2y Multiply each side by 2 2 x 2, and sketch its graph., so the focus is From this equation we see that 4p 2, so the focal diameter is 2. Solving for p
gives p 1. Since the focal diameter is 2, the 2 latus rectum extends 1 unit to the left and 1 unit to the right of the focus. The graph is sketched in Figure 7. ✎ Practice what you’ve learned: Do Exercise 15. and the directrix is y 0, 1 2B ▲ 1 2 A In the next example we graph a family of parabolas, to show how changing the distance between the focus and the vertex affects the “width” of a parabola. E X AM P L E 5 \| A Family of Parabolas 1 y= x™ 2 y 2 1 F!0, @ 2 1!_1, @ 2 1 1 1!1, @ 2 x y=_ 1 2 FIGURE 7 (a) Find equations for the parabolas with vertex at the origin and foci F41 (b) Draw the graphs of the parabolas in part (a). What do you conclude? 0, 1 8B 0, 1 2B, F3A, F2A, and F1A 0, 4 0, 1 B. 2 ▼ SO LUTI O N (a) Since the foci are on the positive y-axis, the parabolas open upward and have equa- tions of the form x 2 4py. This leads to the following equations. Focus p Equation x2 4py Form of the equation for graphing calculator F1A F2A F31 F41 0, 1 8B 0, 1 2B 0, 1 2 0 x2 1 2 y x 2 2y x 2 4y x 2 16y y 2x 2 y 0.5x 2 y 0.25x 2 y 0.0625x 2 Archimedes (287–212 b.c.) was the greatest mathematician of the ancient world. He was born in Syracuse, a Greek colony on Sicily, a generation after Euclid (see page 69). One of his many discoveries is the Law of the Lever (see page 84). He famously said, “Give me a place to stand and a fulcrum for my lever, and I can lift the earth.” Renowned as a mechanical genius for his many engineering inventions, he designed pulleys for lifting heavy ships and the spiral screw for transporting water to higher levels. He is said to have used parabolic mirrors to concentrate the rays of the sun to set ﬁre to
Roman ships attacking Syracuse. King Hieron II of Syracuse once suspected a goldsmith of keeping part of the gold intended for the king’s crown and replacing it with an equal amount of silver. The king asked Archimedes for advice. While in deep thought at a public bath, Archimedes discovered the solution to the king’s problem when he noticed that his body’s volume was the same as the volume of water it displaced from the tub. Using this insight he was able to measure the volume of each crown, and so determine which was the denser, all-gold crown. As the story is told, he ran home naked, shouting “Eureka, eureka!” (“I have found it, I have found it!”) This incident attests to his enormous powers of concentration. In spite of his engineering prowess, Archimedes was most proud of his mathematical discoveries. These include the formulas for the volume of a sphere, S 4pr2 a sphere parabolas and other conics. B and a careful analysis of the properties of and the surface area of V 4 3 pr3 A B A 558 CHAPTER 8 \| Conic Sections (b) The graphs are drawn in Figure 8. We see that the closer the focus to the vertex, the narrower the parabola. 5 5 5 5 _5 5 _5 5 _5 5 _5 _0.5 y=2≈ _0.5 y=0.5≈ _0.5 y=0.25≈ _0.5 y=0.0625≈ FIGURE 8 A family of parabolas ✎ Practice what you’ve learned: Do Exercise 51. 5 ▲ ■ Applications Parabolas have an important property that makes them useful as reﬂectors for lamps and telescopes. Light from a source placed at the focus of a surface with parabolic cross section will be reﬂected in such a way that it travels parallel to the axis of the parabola (see Figure 9). Thus, a parabolic mirror reﬂects the light into a beam of parallel rays. Conversely, light approaching the reﬂector in rays parallel to its axis of symmetry is concentrated to the focus. This reﬂection property, which can be proved by using calculus, is used in the construction of reﬂecting telescopes. F FIGURE 9 Parabolic
reﬂector E X AM P L E 6 \| Finding the Focal Point of a Searchlight Reﬂector A searchlight has a parabolic reﬂector that forms a “bowl,” which is 12 in. wide from rim to rim and 8 in. deep, as shown in Figure 10. If the ﬁlament of the light bulb is located at the focus, how far from the vertex of the reﬂector is it? 12 in. 8 in. FIGURE 10 A parabolic reﬂector ▼ SO LUTI O N We introduce a coordinate system and place a parabolic cross section of the reﬂector so that its vertex is at the origin and its axis is vertical (see Figure 11). Then y (6, 8) 8 6 x 1 1 8 0 12 _6 FIGURE 11 SECTION 8.1 \| Parabolas 559 the equation of this parabola has the form x 2 4py. From Figure 11 we see that the point 6, 8 lies on the parabola. We use this to ﬁnd p. 1 2 The point (6, 8) satisﬁes the equation x2 = 4py 2 62 4p 8 1 36 32p p 9 8 9, so the distance between the vertex and the focus is 8 0, 9 8B F A The focus is 1 8 in the ﬁlament is positioned at the focus, it is located ✎ Practice what you’ve learned: Do Exercise 53. 1 1. Because. from the vertex of the reﬂector. 1 8 in ▲. ▼ CONCE PTS 1. A parabola is the set of all points in the plane that are equidistant from a ﬁxed point called the and a ﬁxed line called the of the parabola. x2 4py 2. The graph of the equation is a parabola with focus F, 1 x2 12y y and directrix y 2 is a parabola with focus F 1. So the graph of, 2 and directrix. 3. The graph of the equation y2 4px is a parabola with focus F, 1 y2 12x x and directrix x 2 is a parabola with focus F 1. So the graph of, 2 and directrix. II IV I III. Label the focus, direct
rix, and vertex on the graphs given for V y VI the parabolas in Exercises 2 and 3. (a) x2 12y (b) y2 12x ✎ ✎ ✎ ▼ SKI LLS 5–10 ■ Match the equation with the graphs labeled I–VI. Give reasons for your answers. 5. y2 2x 7. x 2 6y 9. y 2 8x 0 y2 6. 8. 2x 2 y 10. 12y x 2 0 1 4 x 1 10 x 11–22 ■ Find the focus, directrix, and focal diameter of the parabola, and sketch its graph. 11. x 2 9y 13. y 2 4x 15. y 5x 2 17. x 8y 2 19. x 2 6y 0 21. 5x 3y 2 0 12. x 2 y 14. y 2 3x 16. y 2x 2 x 1 18. 20. x 7y 2 0 22. 8x 2 12y 0 2 y2 23–28 ■ Use a graphing device to graph the parabola. 23. x 2 16y y2 1 3 x 24. x 2 8y 26. 8y 2 x 25. 560 CHAPTER 8 \| Conic Sections 27. 4x y 2 0 28. x 2y 2 0 49. 50. 29–40 ■ Find an equation for the parabola that has its vertex at the origin and satisﬁes the given condition(s). ✎ 1 F F 29. Focus 0, 2 2 8, 0 31. Focus 2 33. Directrix x 2 35. Directrix y 10 1 30. Focus 0, 1 2B F A F 1 5, 0 32. Focus 34. Directrix y 6 x 36. Directrix 2 1 8 y 0 focus x focus shaded region has area 8 y 0 slope= 1 2 2 x 37. Focus on the positive x-axis, 2 units away from the directrix 38. Directrix has y-intercept 6 39. Opens upward with focus 5 units from the vertex ✎ 51. (a) Find equations for the family of parabolas with vertex at the origin and with directrixes y 1 2, y 1, y 4, and y 8. (b) Draw the graphs. What do you conclude? 40. Focal diameter 8 and focus on the negative y-axis 52. (a) Find equations for
the family of parabolas with vertex at 41–50 ■ Find an equation of the parabola whose graph is shown. ✎ 41. focus y 0 2 42. directrix x x=_2 44. 43. 45. y directrix 0 x focus x=4 y 46. 3 2 3 2 focus _3 the origin, focus on the positive y-axis, and with focal diameters 1, 2, 4, and 8. (b) Draw the graphs. What do you conclude? ▼ APPLICATIONS 53. Parabolic Reﬂector A lamp with a parabolic reﬂector is shown in the ﬁgure. The bulb is placed at the focus and the focal diameter is 12 cm. (a) Find an equation of the parabola. (b) Find the diameter d C, D of the opening, 20 cm from the vertex. 1 2 O F A 6 cm 6 cm B 20 cm C D focus 5 x 54. Satellite Dish A reﬂector for a satellite dish is parabolic in cross section, with the receiver at the focus F. The reﬂector is 1 ft deep and 20 ft wide from rim to rim (see the ﬁgure). How far is the receiver from the vertex of the parabolic reﬂector? 47. 48. y 0 square has area 16 x y 0 x (4, _2) directrix F? 20 ft 1 ft 55. Suspension Bridge In a suspension bridge the shape of the suspension cables is parabolic. The bridge shown in the ﬁgure has towers that are 600 m apart, and the lowest point of the suspension cables is 150 m below the top of the towers. Find the equation of the parabolic part of the cables, placing the origin of the coordinate system at the vertex. [Note: This equation is used to ﬁnd the length of cable needed in the construction of the bridge.] 600 m 150 m SECTION 8.1 \| Parabolas 561 ▼ DISCOVE RY • DISCUSSION • WRITI NG 57. Parabolas in the Real World Several examples of the uses of parabolas are given in the text. Find other situations in real life in which parabolas occur. Consult a scientiﬁc encyclopedia in the reference section of your library, or search the Internet. 58. Light Cone from a Flashlight A ﬂash
light is held to form a lighted area on the ground, as shown in the ﬁgure. Is it possible to angle the ﬂashlight in such a way that the boundary of the lighted area is a parabola? Explain your answer. 56. Reﬂecting Telescope The Hale telescope at the Mount Palomar Observatory has a 200-in. mirror, as shown. The mirror is constructed in a parabolic shape that collects light from the stars and focuses it at the prime focus, that is, the focus of the parabola. The mirror is 3.79 in. deep at its center. Find the focal length of this parabolic mirror, that is, the distance from the vertex to the focus. Prime focus 3.79 in. 200 in. DISCOVERY PR OJECT ROLLING DOWN A RAMP Galileo was the ﬁrst to show that the motion of a falling ball can be modeled by a parabola (see page 293). To accomplish this, he rolled balls down a ramp and used his acute sense of timing to study the motion of the ball. In this project you will perform a similar experiment, using modern equipment. You will need the following: ■ A wide board, about 6–8 ft long, to be used as a ramp ■ A large ball, the size of a soccer or volleyball ■ A calculator-based motion detector, such as the Texas Instruments CBR system (Calculator Based Ranger) Use books to prop up the ramp at an angle of about 15°, clamp the motion detector to the top of the ramp, and connect it to the calculator, as shown in the ﬁgure. (Read the manual carefully to make sure you have set up the calculator and motion detector correctly.) 1. Mark a spot on the ramp about 2 ft from the top. Place the ball on the mark and release it to see how many seconds it takes to roll to the bottom. Adjust the motion detector to record the ball’s position every 0.05 s over this length of time. Now let the ball roll down the ramp again, this time with the motion detector running. The calculator should record at least 50 data points that indicate the distance between the ball and the motion detector every 0.05 s. 2. Make a scatter plot of your data, plotting time on the x-axis and distance on the y-axis. Does the plot look like half of a parabola? 3
. Use the quadratic regression command on your calculator (called QuadReg on the TI-83) to ﬁnd the parabola equation y ax 2 bx c that best ﬁts the data. Graph this equation on your scatter plot to see how well it ﬁts. Do the data points really form part of a parabola? 4. Repeat the experiment with the ramp inclined at a shallower angle. How does reducing the angle of the ramp affect the shape of the parabola? 5. Try rolling the ball up the ramp from the bottom to the spot you marked in Step 1, so that it rolls up and then back down again. If you perform the experiment this way instead of just letting the ball roll down, how does your graph in Step 3 change? 562 SECTION 8.2 \| Ellipses 563 8.2 Ellipses LEARNING OBJECTIVES After completing this section, you will be able to: ■ Find geometric properties of an ellipse from its equation ■ Find the equation of an ellipse from some of its geometric properties ■ Geometric Deﬁnition of an Ellipse An ellipse is an oval curve that looks like an elongated circle. More precisely, we have the following deﬁnition. GEOMETRIC DEFINITION OF AN ELLIPSE An ellipse is the set of all points in the plane the sum of whose distances from two ﬁxed points F1 and F2 is a constant. (See Figure 1.) These two ﬁxed points are the foci (plural of focus) of the ellipse. The geometric deﬁnition suggests a simple method for drawing an ellipse. Place a sheet of paper on a drawing board, and insert thumbtacks at the two points that are to be the foci of the ellipse. Attach the ends of a string to the tacks, as shown in Figure 2(a). With the point of a pencil, hold the string taut. Then carefully move the pencil around the foci, keeping the string taut at all times. The pencil will trace out an ellipse, because the sum of the distances from the point of the pencil to the foci will always equal the length of the string, which is constant. If the string is only slightly longer than the distance between the foci, then the ellipse that
is traced out will be elongated in shape, as in Figure 2(a), but if the foci are close together relative to the length of the string, the ellipse will be almost circular, as shown in Figure 2(b). P F⁄ F¤ FIGURE 1 y P(x, y) F⁄(_c, 0) 0 F¤(c, 0) x (a) FIGURE 2 (b) To obtain the simplest equation for an ellipse, we place the foci on the x-axis at c, 0 For later convenience we let the sum of the distances from a point on the ellipse to the so that the origin is halfway between them (see Figure 3). and c, 0 2 F11 F21 foci be 2a. Then if 2 x, y P 1 2 is any point on the ellipse, we have d P, F12 1 d P, F2 2 1 2a FIGURE 2a So from the Distance Formula 564 CHAPTER 8 \| Conic Sections or x c 2 1 2 2 y 2 2a 2 x c 1 2 2 y 2 Squaring each side and expanding, we get 2 2cx c 2 y 2 4a 2 4a2 2cx c 2 y x 2 2 which simpliﬁes to Dividing each side by 4 and squaring again, we get 4a2 1 x c 2 2 y 2 4a 2 4cx a 2x 2 2a x c 2 a 3 1 2cx 2y 2y 2 cx a 1 4 4 2a 2c x 2 1 2 2cx c 2 2x Since the sum of the distances from P to the foci must be larger than the distance between the foci, we have that 2a 2c, or a c. Thus, 2 c a, and we can divide each side to get of the preceding equation by with b 0 1 2 2 c For convenience let preceding equation then becomes 2 a b 2 1. Since 2 a b 2, it follows that b a. The 2 2 2 x a 2 y b 1 with a b 2 This is the equation of the ellipse. To graph it, we need to know the x- and y-intercepts. Setting y 0, we get 2 2 x a 1 2 x 2 a, or x a. Thus, the ellipse crosses the x-axis at so,
as in Figure 4. These points are called the vertices of the ellipse, and the segment that joins them is called the major axis. Its length is 2a. a, 0 and a, 0 1 2 1 2 (_a, 0) y b (0, b) a 0 (_c, 0) c (c, 0) (0, _b) (a, 0) x FIGURE 4 x2 a2 y2 b2 1 with a b Similarly, if we set x 0, we get y b, so the ellipse crosses the y-axis at and 0, b. The segment that joins these points is called the minor axis, and it has length 2b. 1 Note that 2a 2b, so the major axis is longer than the minor axis. The origin is the center of the ellipse. 0, b 2 2 1 If the foci of the ellipse are placed on the y-axis at 1 rather than on the x-axis, then the roles of x and y are reversed in the preceding discussion, and we get a vertical ellipse. 2 0, c The orbits of the planets are ellipses, with the sun at one focus. ■ Equations and Graphs of Ellipses The following box summarizes what we have just proved about the equation and features of an ellipse centered at the origin. SECTION 8.2 \| Ellipses 565 ELLIPSE WITH CENTER AT THE ORIGIN The graph of each of the following equations is an ellipse with center at the origin and having the given properties. In the standard equation for an ellipse, a2 is the larger denominator and b2 is the smaller. To ﬁnd c2, we subtract: larger denominator minus smaller denominator. equation vertices major axis minor axis foci graph, 0 1 Horizontal, length 2a Vertical, length 2b c, a 1 Vertical, length 2a Horizontal, length 2b 0¤(0, c) b F⁄(_c, 0) F¤(c, 0) _a 0 a x _b 0 b x _b _a F⁄(0, _c) E X AM P L E 1 \| Sketching an Ellipse An ellipse has the equation 2 x 9 2 y 4 1 (a) Find the foci, vertices, and the lengths of the major and minor axes,
and sketch the graph. (b) Draw the graph using a graphing calculator. ▼ SO LUTI O N (a) Since the denominator of x 2 is larger, the ellipse has a horizontal major axis. This 2 b Thus, a 3, b 2, and 2 9 4 5. 2 a 2 4 and, so b c 2 9 a gives c 15. foci vertices length of major axis 1 15, 0 3, 0 2 2 1 6 length of minor axis 4 The graph is shown in Figure 5(a) on the next page. (b) To draw the graph using a graphing calculator, we need to solve for y. Note that the equation of an ellipse does not deﬁne y as a function of x (see page 220). That’s why we need to graph two functions to graph an ellipse Subtract x 2 9 566 CHAPTER 8 \| Conic Sections Multiply by 4 Take square roots To obtain the graph of the ellipse, we graph both functions y 221 x 2/9 and y 221 x 2/9 as shown in Figure 5(b). y 3 0 (a) F⁄!_œ∑5, 0@ F¤!œ∑5, 0@ 1 – x2/9 y = 2 œ∑∑∑∑∑ 3.1 _4.7 4 x 4.7 _3.1 (b) 1 – x2/9 y = –2 œ∑∑∑∑∑ FIGURE 5 2 x 9 y 4 2 1 ✎ Practice what you’ve learned: Do Exercise 9. ▲ E X AM P L E 2 \| Finding the Foci of an Ellipse Find the foci of the ellipse 16x 2 9y 2 144, and sketch its graph. ▼ SO LUTI O N First we put the equation in standard form. Dividing by 144, we get y 16 Since 16 9, this is an ellipse with its foci on the y-axis, and with a 4 and b 3. We have 1 x 9 2 2 2 16 9 7 2 b 2 a c c 17. The graph is shown in Figure 6(a). Thus, the foci are 0, 17 1 2 We can also draw the graph using a graphing calculator as shown in Figure 6
(b). y 5 0 F¤!0, œ∑7@ 4 x _9 F⁄!0, _œ∑7@ 1 – x2/9 y = 4 œ∑∑∑∑∑ 9 5 5 _5 1 – x2/9 y = –4 œ∑∑∑∑∑ (a) (b) ✎ Practice what you’ve learned: Do Exercise 11. ▲ FIGURE 6 2 9y 16x 2 144 SECTION 8.2 \| Ellipses 567 E X AM P L E 3 \| Finding the Equation of an Ellipse The vertices of an ellipse are sketch the graph. 1 4, 0 2, and the foci are 2, 0 1 2. Find its equation, and ▼ SO LUTI O N Since the vertices are so c 2. To write the equation, we need to ﬁnd b. Since, we have a 4. The foci are we have 2 a 2 b 4, 0 2, c 1 2 2, 0 1, 2 y 4 F⁄(_2, 0) 0 F¤(2, 0) x 5 Thus, the equation of the ellipse is 22 42 b 2 2 16 4 12 b 2 x 16 2 y 12 1 FIGURE 7 x2 16 y2 12 1 The graph is shown in Figure 7. ✎ Practice what you’ve learned: Do Exercises 25 and 33. ▲ ■ Eccentricity of an Ellipse We saw earlier in this section (Figure 2) that if 2a is only slightly greater than 2c, the ellipse is long and thin, whereas if 2a is much greater than 2c, the ellipse is almost circular. We measure the deviation of an ellipse from being circular by the ratio of a and c. DEFINITION OF ECCENTRICITY 2 2 x a For the ellipse is the number 2 y b 1 or 2 x2 b2 y2 a2 1 with a b 0 1, the eccentricity e 2 e c a where c 2a 2 b. The eccentricity of every ellipse satisﬁes 0 e 1. 2 Thus, if e is close to 1, then c is almost equal to a, and the ellipse is elongated in shape, but if e is
close to 0, then the ellipse is close to a circle in shape. The eccentricity is a measure of how “stretched” the ellipse is. In Figure 8 we show a number of ellipses to demonstrate the effect of varying the ec- centricity e. e=0.1 e=0.5 e=0.68 e=0.86 FIGURE 8 Ellipses with various eccentricities 568 CHAPTER 8 \| Conic Sections E X AM P L E 4 \| Finding the Equation of an Ellipse from Its Eccentricity and Foci Find the equation of the ellipse with foci graph. 1 0, 8 2 and eccentricity e 4 5, and sketch its ▼ SO LUTI O N We are given e 4 5 and c 8. Thus y 10 F⁄(0, 8) 8 a 4 5 4a 40 a 10 Eccentricity e c a Cross-multiply To ﬁnd b, we use the fact that c 2 a 2 b 2. 2 82 102 b 2 102 82 36 b b 6 _6 0 6 x Thus, the equation of the ellipse is _10 F¤(0, _8) FIGURE 9 y2 x2 100 36 1 Eccentricities of the Orbits of the Planets The orbits of the planets are ellipses with the sun at one focus. For most planets these ellipses have very small eccentricity, so they are nearly circular. However, Mercury and Pluto, the innermost and outermost known planets, have visibly elliptical orbits. Planet Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto Eccentricity 0.206 0.007 0.017 0.093 0.048 0.056 0.046 0.010 0.248 2 x 36 2 y 100 1 Because the foci are on the y-axis, the ellipse is oriented vertically. To sketch the ellipse, we ﬁnd the intercepts: The x-intercepts are 6, and the y-intercepts are 10. The graph is sketched in Figure 9. ✎ Practice what you’ve learned: Do Exercise 43. ▲ Gravitational attraction causes the planets to move in elliptical orbits around the sun with the sun at one focus. This remarkable property was ﬁrst observed by Johannes Kepler and was later deduced by Isaac Newton from his inverse square
law of gravity, using calculus. The orbits of the planets have different eccentricities, but most are nearly circular (see the margin). Ellipses, like parabolas, have an interesting reﬂection property that leads to a number of practical applications. If a light source is placed at one focus of a reﬂecting surface with elliptical cross sections, then all the light will be reﬂected off the surface to the other focus, as shown in Figure 10. This principle, which works for sound waves as well as for light, is used in lithotripsy, a treatment for kidney stones. The patient is placed in a tub of water with elliptical cross sections in such a way that the kidney stone is accurately located at one focus. High-intensity sound waves generated at the other focus are reﬂected to the stone and destroy it with minimal damage to surrounding tissue. The patient is spared the trauma of surgery and recovers within days instead of weeks. The reﬂection property of ellipses is also used in the construction of whispering galleries. Sound coming from one focus bounces off the walls and ceiling of an elliptical room and passes through the other focus. In these rooms even quiet whispers spoken at one focus can be heard clearly at the other. Famous whispering galleries include the National Statuary Hall of the U.S. Capitol in Washington, D.C. (see page 595), and the Mormon Tabernacle in Salt Lake City, Utah. F⁄ F¤ FIGURE 10 SECTION 8.2 \| Ellipses 569 8. ▼ CONCE PTS 1. An ellipse is the set of all points in the plane for which the of the distances from two ﬁxed points F1 and F2 is constant. The points ellipse. F1 and F2 are called the of the III y 1 0 1 x IV y 1 0 2 x 2. The graph of the equation x2 a2 ellipse with vertices 1 2. So the graph of where c, 1 y2 b2 and 1 with a b 0 is an, x2 52 and foci 2 y2 42 and foci 1 1 c, 0 1, 2 is an el-, 2 lipse with vertices 1, and 1 2, 2 and 1,. 2 3. The graph of the equation is an ellipse with vertices 0, c where c, 2 1 is
an ellipse with vertices 1 1 y2 a2 x2 b2, 1 with a b 0 and 1 2,. So the graph of, and 1 2, and foci y2 52 and foci 1 2 x2 42 2, 1 and 1 2,. 2 4. Label the vertices and foci on the graphs given for the ellipses in Exercises 2 and 3. (a) x2 52 y2 42 1 (b) x2 42 y2 52 ▼ SKI LLS 5–8 ■ Match the equation with the graphs labeled I–IV. Give reasons for your answers. 5. 7. I 2 1 2 y x 4 16 4x2 y2 4 6. 2 x 2 y 9 1 8. 16x2 25y2 400 y 1 0 1 x II y 1 0 1 x 9–22 ■ Find the vertices, foci, and eccentricity of the ellipse. Determine the lengths of the major and minor axes, and sketch the graph. ✎ ✎ 9. 2 x 25 2 y 9 1 11. 9x2 4y2 36 13. x2 4y2 16 15. 2x2 y2 3 17. x2 4y2 1 19. 21 y2 1 2x2 10. 2 x 16 2 y 25 1 12. 4x2 25y2 100 14. 4x2 y2 16 16. 5x2 6y2 30 18. 9x2 4y2 1 20. x2 4 2y2 22. 20x2 4y2 5 23–28 ■ Find an equation for the ellipse whose graph is shown. 23. 24 ✎ 25. 27. 5 x 0 2 x 26. F(0, 2) 2 x y 4 0 F(0, 3) x 28. y (8, 6) (_1, 2) x16 0 2 x 570 CHAPTER 8 \| Conic Sections 29–32 ■ Use a graphing device to graph the ellipse. (b) What do the members of this family of ellipses have in 2 1 2 y x 20 25 6x2 y2 36 29. 31. 30. 32. 2 x 2 1 y 12 x2 2y2 8 33–44 ■ Find an equation for the ellipse that satisﬁes the given conditions. ✎ 33. Foci 34. Foci 4, 0 0,
3 1 1 2 2, vertices, vertices 5, 0 0, 5 1 1 2 2 35. Length of major axis 4, length of minor axis 2, foci on y-axis 36. Length of major axis 6, length of minor axis 4, foci on x-axis 37. Foci 38. Foci 0, 2 5, 0 1 1 2 2, length of minor axis 6, length of major axis 12 39. Endpoints of major axis 1 40. Endpoints of minor axis, distance between foci 6 2, distance between foci 8 10, 0 0, 3 2 1 41. Length of major axis 10, foci on x-axis, ellipse passes through common? How do they differ? 50. If k 0, the following equation represents an ellipse: 2 x k 2 y 4 k 1 Show that all the ellipses represented by this equation have the same foci, no matter what the value of k. ▼ APPLICATIONS 51. Perihelion and Aphelion The planets move around the sun in elliptical orbits with the sun at one focus. The point in the orbit at which the planet is closest to the sun is called perihelion, and the point at which it is farthest is called aphelion. These points are the vertices of the orbit. The earth’s distance from the sun is 147,000,000 km at perihelion and 153,000,000 km at aphelion. Find an equation for the earth’s orbit. (Place the origin at the center of the orbit with the sun on the x-axis.) the point 1 15, 2 2 42. Eccentricity, foci 1 9 ✎ 43. Eccentricity 0.8, foci 0, 2 1 2 1.5, 0 1 2 44. Eccentricity 13/2, foci on y-axis, length of major axis 4 aphelion perihelion 45–47 ■ Find the intersection points of the pair of ellipses. Sketch the graphs of each pair of equations on the same coordinate axes and label the points of intersection. 4x 4x 2 y 2 9y 2 4 2 36 45. e 46. µ 2 x 16 2 x 9 2 y 9 2 y 16 1 1 100x 47. 2 100 2 25y 2 y 9 1 2 x 48. The ancillary circle of an ellipse is the circle
with radius equal c to half the length of the minor axis and center the same as the ellipse (see the ﬁgure). The ancillary circle is thus the largest circle that can ﬁt within an ellipse. (a) Find an equation for the ancillary circle of the ellipse x2 4y2 16. (b) For the ellipse and ancillary circle of part (a), show that if is a point on the ancillary circle, then s, t 1 on the ellipse. 2 2s, t 1 2 is a point ellipse ancillary circle 49. (a) Use a graphing device to sketch the top half (the portion in the ﬁrst and second quadrants) of the family of ellipses x2 ky2 100 for k 4, 10, 25, and 50. 52. The Orbit of Pluto With an eccentricity of 0.25, Pluto’s orbit is the most eccentric in the solar system. The length of the minor axis of its orbit is approximately 10,000,000,000 km. Find the distance between Pluto and the sun at perihelion and at aphelion. (See Exercise 51.) 53. Lunar Orbit For an object in an elliptical orbit around the moon, the points in the orbit that are closest to and farthest from the center of the moon are called perilune and apolune, respectively. These are the vertices of the orbit. The center of the moon is at one focus of the orbit. The Apollo 11 spacecraft was placed in a lunar orbit with perilune at 68 mi and apolune at 195 mi above the surface of the moon. Assuming that the moon is a sphere of radius 1075 mi, ﬁnd an equation for the orbit of Apollo 11. (Place the coordinate axes so that the origin is at the center of the orbit and the foci are located on the x-axis.) apolune 195 mi moon perilune 68 mi 54. Plywood Ellipse A carpenter wishes to construct an elliptical 58. How Wide Is an Ellipse at Its Foci? A latus rectum for an SECTION 8.2 \| Ellipses 571 table top from a sheet of plywood, 4 ft by 8 ft. He will trace out the ellipse using the “thumbtack and string” method illustrated in Figures 2 and
3. What length of string should he use, and how far apart should the tacks be located, if the ellipse is to be the largest possible that can be cut out of the plywood sheet? ellipse is a line segment perpendicular to the major axis at a focus, with endpoints on the ellipse, as shown. Show that the length of a latus rectum is 2b 2/a for the ellipse y2 b2 1 with a b x2 a2 y b _a _b foci latus rectum a x 55. Sunburst Window A “sunburst” window above a doorway is constructed in the shape of the top half of an ellipse, as shown in the ﬁgure. The window is 20 in. tall at its highest point and 80 in. wide at the bottom. Find the height of the window 25 in. from the center of the base. 59. Is It an Ellipse? A piece of paper is wrapped around a cylindrical bottle, and then a compass is used to draw a circle on the paper, as shown in the ﬁgure. When the paper is laid ﬂat, is the shape drawn on the paper an ellipse? (You don’t need to prove your answer, but you might want to do the experiment and see what you get.) 20 in. h 25 in. 80 in. ▼ DISCOVE RY • DISCUSSION • WRITI NG 56. Drawing an Ellipse on a Blackboard Try drawing an ellipse as accurately as possible on a blackboard. How would a piece of string and two friends help this process? 57. Light Cone from a Flashlight A ﬂashlight shines on a wall, as shown in the ﬁgure. What is the shape of the boundary of the lighted area? Explain your answer. 572 CHAPTER 8 \| Conic Sections 8.3 Hyperbolas LEARNING OBJECTIVES After completing this section, you will be able to: ■ Find geometric properties of a hyperbola from its equation ■ Find the equation of a hyperbola from some of its geometric properties ■ Geometric Deﬁnition of a Hyperbola Although ellipses and hyperbolas have completely different shapes, their deﬁnitions and equations are similar. Instead of using the sum of distances from two �
�xed foci, as in the case of an ellipse, we use the difference to deﬁne a hyperbola. GEOMETRIC DEFINITION OF A HYPERBOLA A hyperbola is the set of all points in the plane, the difference of whose distances from two ﬁxed points F1 and F2 is a constant. (See Figure 1.) These two ﬁxed points are the foci of the hyperbola. y 0 F⁄(_c, 0) P(x, y) As in the case of the ellipse, we get the simplest equation for the hyperbola by placing lies on the 2 must equal some posi-, as shown in Figure 1. By deﬁnition, if d P, F22 c, 0 the foci on the x-axis at 1 P, F12 d d or hyperbola, then either 1 tive constant, which we call 2a. Thus, we have P, F12 P, F22 d x, y P 2 1 1 1 1 F¤(c, 0) x or x c 2 1 2 2 y d P, F12 1 2 2 d P, F22 1 2 y 2a 2 2a x c 1 2 Proceeding as we did in the case of the ellipse (Section 8.2), we simplify this to FIGURE 1 P is on the hyperbola if P, F22 0 P, F12 2a d d. 1 1 0 From triangle PF1F2 in Figure 1 we see that 2a 2c, or a c. Thus, last displayed equation to get 2 0 2y 2 a 2 2 a c 2 1 P, F12 d 0 b so we can set 1 2 P, F22 0 2 a 2. d 1 2 c 2c. It follows that We then simplify the 2 2 x a 2 y b 1 2 This is the equation of the hyperbola. If we replace x by x or y by y in this equation, it remains unchanged, so the hyperbola is symmetric about both the x- and y-axes and about the origin. The x-intercepts are a, and the points are the vertices of the hyperbola. There is no y-intercept, because setting x 0 in the equation of the hyperbola leads to which has no real solution.
Furthermore, the equation of the hyperbola implies that 2 b a, 0 and y a, 0 2/a thus, 2 a 2 1; and hence x a or x a. This means that the hyperbola so consists of two parts, called its branches. The segment joining the two vertices on the separate branches is the transverse axis of the hyperbola, and the origin is called its center. 2, SECTION 8.3 \| Hyperbolas 573 If we place the foci of the hyperbola on the y-axis rather than on the x-axis, then this has the effect of reversing the roles of x and y in the derivation of the equation of the hyperbola. This leads to a hyperbola with a vertical transverse axis. ■ Equations and Graphs of Hyperbolas The main properties of hyperbolas are listed in the following box. HYPERBOLA WITH CENTER AT THE ORIGIN The graph of each of the following equations is a hyperbola with center at the origin and having the given properties. equation VERTICES TRANSVERSE AXIS ASYMPTOTES FOCI GRAPH 1 2 1 a 0, 0 2 1 Horizontal, length 2a y b a x, c=_ x y= x b a F¤(c, 0) a x y b _b F⁄(_c, 0) _a 1 1 a 0, a 2 1 Vertical, length 2a y a b x 0=_ x a b _b y a _a F⁄(0, c) y= x a b b x F¤(0, _c) Asymptotes of rational functions are discussed in Section 4.5. The asymptotes mentioned in this box are lines that the hyperbola approaches for large values of x and y. To ﬁnd the asymptotes in the ﬁrst case in the box, we solve the equation for y to get y b a 2x 2 a 2 b a x As x gets large, a 2/x 2 gets closer to zero. In other words, as x q, we have a 2/x 2 0. So for large x the value of y can be approximated as. This shows that these lines are asymptotes of the hyperbola. 1 a x y b/a B x 2 1 2 2 Asymptotes are an
essential aid for graphing a hyperbola; they help us to determine its shape. A convenient way to ﬁnd the asymptotes, for a hyperbola with horizontal transverse axis, is to ﬁrst plot the points. Then sketch horizontal, 2 and vertical segments through these points to construct a rectangle, as shown in Figure 2(a) on the next page. We call this rectangle the central box of the hyperbola. The slopes of the diagonals of the central box are b/a, so by extending them, we obtain the asymptotes y, as sketched in part (b) of the ﬁgure. Finally, we plot the vertices and use the a, 0 0, b, and 0, b a, 0 b/ 574 CHAPTER 8 \| Conic Sections asymptotes as a guide in sketching the hyperbola shown in part (c). (A similar procedure applies to graphing a hyperbola that has a vertical transverse axis.) y b y b y b _a 0 a x _a a x _a a x FIGURE 2 Steps in graphing the hyperbola _b _b _b a) Central box (b) Asymptotes and vertices (c) Hyperbola HOW TO SKETCH A HYPERBOLA 1. Sketch the Central Box. This is the rectangle centered at the origin, with sides parallel to the axes, that crosses one axis at a, the other at b. 2. Sketch the Asymptotes. These are the lines obtained by extending the dia- gonals of the central box. 3. Plot the Vertices. These are the two x-intercepts or the two y-intercepts. 4. Sketch the Hyperbola. Start at a vertex, and sketch a branch of the hyperbola, approaching the asymptotes. Sketch the other branch in the same way. E X AM P L E 1 \| A Hyperbola with Horizontal Transverse Axis A hyperbola has the equation 9x 2 16y 2 144 (a) Find the vertices, foci, and asymptotes, and sketch the graph. (b) Draw the graph using a graphing calculator. ▼ SO LUTI O N (a) First we divide both sides of the equation by 144 to put it into standard form: 2 x 16 2 y 9 1
Because the x 2-term is positive, the hyperbola has a horizontal transverse axis; its vertices and foci are on the x-axis. Since a 2 16 and b 2 9, we get a 4, b 3, and c 116 9 5. Thus, we have VERTICES FOCI asymptotes 2 4, 0 1 5, 0 1 2 y 3 4 x After sketching the central box and asymptotes, we complete the sketch of the hyperbola as in Figure 3(a). Note that the equation of a hyperbola does not deﬁne y as a function of x (see page 220). That’s why we need to graph two functions to graph a hyperbola. SECTION 8.3 \| Hyperbolas 575 (b) To draw the graph using a graphing calculator, we need to solve for y. 9x 2 16y 2 144 16y 2 9x 2 144 Subtract 9x 2 2 9 y 2 x 16 a 1 b y 3 2 x 16 B 1 Divide by 16 and factor 9 Take square roots To obtain the graph of the hyperbola, we graph the functions y 32 x 2/16 1 2 1 and y 32 x 2/16 1 2 1 as shown in Figure 3(b). y = 3 4 x (5, 0) 4 x _10 y = – 3 4 x (_5, 0) _4 y 3 _3 (a) (x2/16) – 1 y = 3 œ 6 10 _ _6 y = –3œ (x2/16) – 1 (b) ✎ Practice what you’ve learned: Do Exercise 9. ▲ FIGURE 3 9x 2 16y 2 144 E X AM P L E 2 \| A Hyperbola with Vertical Transverse Axis Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. 2 9y 2 9 0 x ▼ SO LUTI O N We begin by writing the equation in the standard form for a hyperbola. 2 9y 2 9 x y 2 2 x 9 1 Divide by 9 Because the y 2-term is positive, the hyperbola has a vertical transverse axis; its foci and vertices are on the y-axis. Since a 2 1 and b 2 9, we get a 1, b 3, and c 11 9 110.
Thus, we have VERTICES FOCI asymptotes 0, 1 1 0, 1 y 2 110 2 1 3 x We sketch the central box and asymptotes, then complete the graph, as shown in Figure 4(a). 576 CHAPTER 8 \| Conic Sections We can also draw the graph using a graphing calculator, as shown in Figure 4(b). y 1 F⁄Ó0, œ∑10Ô y = œ 1 + x2/9 2 3 x _5 5 FIGURE 4 x2 9y2 9 0 F¤Ó0, _œ∑10Ô (a) _2 y = – œ 1 + x2/9 (b) ✎ Practice what you’ve learned: Do Exercise 17. ▲ E X AM P L E 3 \| Finding the Equation of a Hyperbola from Its Vertices and Foci Find the equation of the hyperbola with vertices graph. 1 3, 0 2 and foci 4, 0 1 2. Sketch the ▼ SO LUTI O N verse axis. Its equation is of the form Since the vertices are on the x-axis, the hyperbola has a horizontal trans- 2 x 32 2 y b 1 2 We have a 3 and c 4. To ﬁnd b, we use the relation a 2 b 2 c 2 : 32 b 2 42 2 42 32 7 b b 17 Thus, the equation of the hyperbola is 2 x 9 2 y 7 1 Paths of Comets The path of a comet is an ellipse, a parabola, or a hyperbola with the sun at a focus. This fact can be proved using calculus and Newton’s laws of motion.* If the path is a parabola or a hyperbola, the comet will never return. If the path is an ellipse, it can be determined precisely when and where the comet can be seen again. Halley’s comet has an elliptical path and returns every 75 years; it was last seen in 1987. The brightest comet of the 20th century was comet Hale-Bopp, seen in 1997. Its orbit is a very eccentric ellipse; it is expected to return to the inner solar system around the year 4377. *James Stewart, Calculus, 6th ed. (Belmont, CA: Brooks/Cole, 2008), pp
. 884–885. y F⁄ F¤ 1 x FIGURE 6 y2 4 x2 1 The graph is shown in Figure 5. SECTION 8.3 \| Hyperbolas 577 œ∑7 y 3 0 _3 _ œ∑7 _3 3 x 2 FIGURE 5 x 9 y 7 2 1 ✎ Practice what you’ve learned: Do Exercises 21 and 31. ▲ E X AM P L E 4 \| Finding the Equation of a Hyperbola from Its Vertices and Asymptotes Find the equation and the foci of the hyperbola with vertices y 2x. Sketch the graph. 1 0, 2 2 and asymptotes ▼ SO LUTI O N Since the vertices are on the y-axis, the hyperbola has a vertical transverse axis with a 2. From the asymptote equation we see that a/b 2. Since a 2, we get 2/b 2, and so b 1. Thus, the equation of the hyperbola is x 2 1 2 y 4 2 b 0, 15 c. The graph is shown in Figure 6. To ﬁnd the foci, we calculate are ✎ Practice what you’ve learned: Do Exercises 25 and 35. so 2 1 2 12 5, 2 2 2 a c 15. Thus, the foci ▲ Like parabolas and ellipses, hyperbolas have an interesting reﬂection property. Light aimed at one focus of a hyperbolic mirror is reﬂected toward the other focus, as shown in Figure 7. This property is used in the construction of Cassegrain-type telescopes. A hyperbolic mirror is placed in the telescope tube so that light reﬂected from the primary parabolic reﬂector is aimed at one focus of the hyperbolic mirror. The light is then refocused at a more accessible point below the primary reﬂector (Figure 8). F⁄ Hyperbolic reflector F¤ F⁄ FIGURE 7 Reﬂection property of hyperbolas Parabolic reﬂector F¤ FIGURE 8 Cassegrain-type telescope 578 CHAPTER 8 \| Conic Sections P, A d The LORAN (LOng RAnge Navigation) system was used until the early 1990s; it has now been supers
eded by the GPS system (see page 462). In the LORAN system, hyperbolas are used onboard a ship to determine its location. In Figure 9 radio stations at A and B transmit signals simultaneously for reception by the ship at P. The onboard computer converts the time difference in reception of these signals into a distance difference d. From the deﬁnition of a hyperbola this locates the ship on one branch of a hyperbola with foci at A and B (sketched in black in the ﬁgure). The same procedure is carried out with two other radio stations at C and D, and this locates the ship on a second hyperbola (shown in red in the ﬁgure). (In practice, only three stations are needed because one station can be used as a focus for both hyperbolas.) The coordinates of the intersection point of these two hyperbolas, which can be calculated precisely by the computer, give the location of P. P FIGURE 9 LORAN system for ﬁnding the location of a ship 8. ▼ CONCE PTS 1. A hyperbola is the set of all points in the plane for which the y2 42 x2 32 1 is a hyperbola with vertices of the distances from two ﬁxed points F1 and F2 is, 1 and foci 1 2, and 1 2,, and 2 1. 2 constant. The points hyperbola. F1 and F2 are called the of the 4. Label the vertices, foci, and asymptotes on the graphs given for the hyperbolas in Exercises 2 and 3. 2. The graph of the equation x2 a2 is a hyperbola with vertices 1, where c 2 foci 1 x2 42 c, 0 y2 32 1 is a hyperbola with vertices y2 b2, 1 with a 0, b 0 (a) x2 42 y2 32 1 and, 2 1. So the graph of and 2 (b) y2 42 x2 32 1 y, and and foci 1 2, 3. The graph of the equation and 1 x2 b2 2 y2 a2 is a hyperbola with vertices foci 1 0, c 2, where c, 1 1 with a 0, b 0 and, 2 1 2. So the graph of and ▼ SKI LLS
5–8 ■ Match the equation with the graphs labeled I–IV. Give reasons for your answers. ✎ 25. y 26. y=_ x1 2 y= x1 2 6. y 2 2 x 9 1 _5 x 5 SECTION 8.3 \| Hyperbolas 579 y 3 0 y=3x 1 x y=_3x 5. 7. I III 2 y 2 1 x 4 16y2 x2 144 2 x y 1 y 1 8. II IV 9x2 25y2 225 –20 ■ Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. ✎ ✎ 9. 2 x 4 2 y 16 1 11. 13. 15. 17. 19. ✎ 2 1 2 x y 25 x2 y2 1 25y2 9x2 225 x2 4y2 8 0 4y2 x2 1 2 y 9 2 x 16 1 2 y 2 1 x 2 9x2 4y2 36 x2 y2 4 0 x2 2y2 3 9x2 16y2 1 10. 12. 14. 16. 18. 20. 21–26 ■ Find the equation for the hyperbola whose graph is shown. ✎ 21. F⁄(_4, 0) y 1 0 F¤(4, 0) 1 x 23. y 4 0 _4 2 x (3, _5) 22. 24. y 12 0 F⁄(0, 13) x F¤(0, _13) _12 y 2 (4, 4) 2œ∑3 x 27–30 ■ Use a graphing device to graph the hyperbola. 3y2 4x2 24 28. 27. x2 2y2 8 2 2 y 2 x 6 1 29. 30. 2 x 100 2 y 64 1 31–42 ■ Find an equation for the hyperbola that satisﬁes the given conditions. ✎ 31. Foci 32. Foci 33. Foci 34. Foci 5, 0 2 0, 10 0, 2 6, 0 2 1 1 1 1, vertices 1, vertices 2, vertices 3, 0 2 0, 8 1 0, 1 2, 0 2 1 2, vertices 2 1, 0 0, 6 2 1 y 5x, asymptotes y 1 3 x, asymptotes 2 2
, asymptotes y 1 2 x 35. Vertices 1 36. Vertices 1 0, 8 2 0, 6 37. Foci 1 38. Vertices 1 40. Foci 39. Asymptotes 3, 0 5, 0 0, 1 41. Foci 42. Foci 1 1 2 2 1 2 5, 9 1 2 5, 3 2, hyperbola passes through 2 y x, hyperbola passes through 1, hyperbola passes through 1, length of transverse axis 6 4, 1 2, length of transverse axis 1 43. (a) Show that the asymptotes of the hyperbola x2 y2 5 are perpendicular to each other. (b) Find an equation for the hyperbola with foci with asymptotes perpendicular to each other. 1 c, 0 and 2 44. The hyperbolas 2 2 x a 2 y 2 b 1 and x 2 a 2 2 y 2 b 1 are said to be conjugate to each other. (a) Show that the hyperbolas 2 4y 2 16 0 and 4y 2 x 2 16 0 x are conjugate to each other, and sketch their graphs on the same coordinate axes. (b) What do the hyperbolas of part (a) have in common? (c) Show that any pair of conjugate hyperbolas have the relationship you discovered in part (b). 45. In the derivation of the equation of the hyperbola at the beginning of this section, we said that the equation 2a 580 CHAPTER 8 \| Conic Sections simpliﬁes to 2 a 2 c 1 2 2 a x 2y 2 a 2 a c 2 2 1 2 before it neared the solar system is at a right angle to the path it continues on after leaving the solar system. y Supply the steps needed to show this. 46. (a) For the hyperbola 2 x 9 2 y 16 1 x 2 10ª mi 50. Ripples in Pool Two stones are dropped simultaneously in a calm pool of water. The crests of the resulting waves form equally spaced concentric circles, as shown in the ﬁgures. The waves interact with each other to create certain interference patterns. (a) Explain why the red dots lie on an ellipse. (b) Explain why the blue dots lie on a hyperbola. ▼ DISCOVE RY • DISCUSS
ION • WRITI NG 51. Hyperbolas in the Real World Several examples of the uses of hyperbolas are given in the text. Find other situations in real life in which hyperbolas occur. Consult a scientiﬁc encyclopedia in the reference section of your library, or search the Internet. 52. Light from a Lamp The light from a lamp forms a lighted area on a wall, as shown in the ﬁgure. Why is the boundary of this lighted area a hyperbola? How can one hold a ﬂashlight so that its beam forms a hyperbola on the ground? determine the values of a, b, and c, and ﬁnd the coordinates of the foci F1 and F2. 5, 16 P 3 2 1 P, F22. (b) Show that the point (c) Find d and (d) Verify that the difference between d lies on this hyperbola. P, F12 d and d 1 1 P, F12 1 P, F22 1 is 2a. 47. Hyperbolas are called confocal if they have the same foci. (a) Show that the hyperbolas 2 y k 2 x 16 k 1 with 0 k 16 are confocal. (b) Use a graphing device to draw the top branches of the family of hyperbolas in part (a) for k 1, 4, 8, and 12. How does the shape of the graph change as k increases? ▼ APPLICATIONS 48. Navigation In the ﬁgure, the LORAN stations at A and B are 500 mi apart, and the ship at P receives station A’s signal 2640 microseconds (ms) before it receives the signal from station B. (a) Assuming that radio signals travel at 980 ft/ms, ﬁnd d P, A 1 2 d P, B. 1 2 (b) Find an equation for the branch of the hyperbola indicated in red in the ﬁgure. (Use miles as the unit of distance.) (c) If A is due north of B and if P is due east of A, how far is P from A? y (mi) A 250 P 0 B _250 x (mi) 49. Comet Trajectories Some comets, such as Halley’s comet, are a permanent part of the solar system, traveling in elliptical orbits around
the sun. Others pass through the solar system only once, following a hyperbolic path with the sun at a focus. The ﬁgure shows the path of such a comet. Find an equation for the path, assuming that the closest the comet comes to the sun is 2 109 mi and that the path the comet was taking SECT IO N 8.4 \| Shifted Conics 581 8.4 Shifted Conics LEARNING OBJECTIVES After completing this section, you will be able to: ■ Find geometric properties of a shifted conic from its equation ■ Find the equation of a shifted conic from some of its geometric properties In the preceding sections we studied parabolas with vertices at the origin and ellipses and hyperbolas with centers at the origin. We restricted ourselves to these cases because these equations have the simplest form. In this section we consider conics whose vertices and centers are not necessarily at the origin, and we determine how this affects their equations. ■ Shifting Graphs of Equations In Section 3.5 we studied transformations of functions that have the effect of shifting their graphs. In general, for any equation in x and y, if we replace x by x h or by x h, the graph of the new equation is simply the old graph shifted horizontally; if y is replaced by y k or by y k, the graph is shifted vertically. The following box gives the details. SHIFTING GRAPHS OF EQUATIONS If h and k are positive real numbers, then replacing x by x h or by x h and replacing y by y k or by y k has the following effect(s) on the graph of any equation in x and y. Replacement 1. x replaced by x h 2. x replaced by x h 3. y replaced by y k 4. y replaced by y k How the graph is shifted Right h units Left h units Upward k units Downward k units ■ Shifted Ellipses Let’s apply horizontal and vertical shifting to the ellipse with equation 2 2 x a 2 y b 1 2 whose graph is shown in Figure 1 on the next page. If we shift it so that its center is at the point h, k 1 2 instead of at the origin, then its equation becomes 582 CHAPTER 8 \| Conic Sections FIGURE 1 Shifted ellipse y (x-h)™ a™ + (y-k)™ b™ =1 x™ a™ y ™

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