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oci (1 + 2 √ 2 e = 2 2, 3), (1 − 2 2, 3) 2) √ √ √ √ 2), 530 13. + 2 = 1 4 y − 1 x2 2 9 1 Center 0, 1 2 Major Axis along x = 0 (the y-axis) Minor Axis along y = 1 2 Vertices (0, 2), (0, −1) Endpoints of Minor Axis −1, 1 2 Foci, √ √ 5 5 0, 1− 2, 1, 1 2 √ 0, 1+ 2 5 3 e = 14. Hooked on Conics = 1 (x − 2)2 5 + (y + 2)2 6 √ Center (2, −2) Major Axis along x = 2 Minor Axis along y = −2 Vertices 2, −2 + Endpoints of Minor Axis 2 − 2 + 5, −2 Foci (2, −1), (2, −3) e = 6, (2, −2 − √ √ 6 6 √ 6) √ 5, −2, 15. 17. 19. (x − 3)2 9 + x2 y2 34 25 (x − 5)2 25 + (y − 7)2 25 = 1 = 1 + 4(y − 2)2 75 = 1 21. Jamie and Jason should stand 100 − 25 √ 16. 18. 20. + y2 x2 64 39 (x − 8)2 25 (x − 8)2 64 = 1 + + (y − 2)2 4 (y + 9)2 81 = 1 = 1 7 ≈ 33.86 feet from opposite ends of the gallery. 22. The arch can be modeled by the top half of x2 9 + y2 corresponds to either x = ±2. Plugging in x = ±2 gives y = ±3 5 ≈ 6.71 feet. height, we choose y = 3 √ 81 = 1. One foot in from the base of the arch 5 and since y represents a √ 23. Distance from the sun to aphelion ≈ 1.0167 AU. Distance from the sun to perihelion ≈ 0.9833 AU. 7.5 Hyperbolas 7.5 Hyperbolas 531 In the deﬁnition of an ellipse, Deﬁnition 7.4, we �
��xed two points called foci and looked at points whose distances to the foci always added to a constant distance d. Those prone to syntactical tinkering may wonder what, if any, curve we’d generate if we replaced added with subtracted. The answer is a hyperbola. Deﬁnition 7.6. Given two distinct points F1 and F2 in the plane and a ﬁxed distance d, a hyperbola is the set of all points (x, y) in the plane such that the absolute value of the diﬀerence of each of the distances from F1 and F2 to (x, y) is d. The points F1 and F2 are called the foci of the hyperbola. (x1, y1) F2 F1 (x2, y2) In the ﬁgure above: the distance from F1 to (x1, y1) − the distance from F2 to (x1, y1) = d and the distance from F2 to (x2, y2) − the distance from F1 to (x2, y2) = d Note that the hyperbola has two parts, called branches. The center of the hyperbola is the midpoint of the line segment connecting the two foci. The transverse axis of the hyperbola is the line segment connecting two opposite ends of the hyperbola which also contains the center and foci. The vertices of a hyperbola are the points of the hyperbola which lie on the transverse axis. In addition, we will show momentarily that there are lines called asymptotes which the branches of the hyperbola approach for large x and y values. They serve as guides to the graph. In pictures, 532 Hooked on Conics Transverse Axis F1 V1 C V2 F2 A hyperbola with center C; foci F1, F2; and vertices V1, V2 and asymptotes (dashed) Before we derive the standard equation of the hyperbola, we need to discuss one further parameter, the conjugate axis of the hyperbola. The conjugate axis of a hyperbola is the line segment through the center which is perpendicular to the transverse axis and has the same length as the line segment through a vertex which connects the asymptotes. In
pictures we have V1 V2 Note that in the diagram, we can construct a rectangle using line segments with lengths equal to the lengths of the transverse and conjugate axes whose center is the center of the hyperbola and whose diagonals are contained in the asymptotes. This guide rectangle, much akin to the one we saw Section 7.4 to help us graph ellipses, will aid us in graphing hyperbolas. Suppose we wish to derive the equation of a hyperbola. For simplicity, we shall assume that the center is (0, 0), the vertices are (a, 0) and (−a, 0) and the foci are (c, 0) and (−c, 0). We label the 7.5 Hyperbolas 533 endpoints of the conjugate axis (0, b) and (0, −b). (Although b does not enter into our derivation, we will have to justify this choice as you shall see later.) As before, we assume a, b, and c are all positive numbers. Schematically we have y (0, b) (x, y) (−c, 0) (−a, 0) (a, 0) x (c, 0) (0, −b) Since (a, 0) is on the hyperbola, it must satisfy the conditions of Deﬁnition 7.6. That is, the distance from (−c, 0) to (a, 0) minus the distance from (c, 0) to (a, 0) must equal the ﬁxed distance d. Since all these points lie on the x-axis, we get distance from (−c, 0) to (a, 0) − distance from (c, 0) to (a, 0) = d (a + c) − (c − a) = d 2a = d In other words, the ﬁxed distance d from the deﬁnition of the hyperbola is actually the length of the transverse axis! (Where have we seen that type of coincidence before?) Now consider a point (x, y) on the hyperbola. Applying Deﬁnition 7.6, we get distance from (−c, 0) to (x, y) − distance from (c, 0) to (x, y) = 2a (x − c)2 + (y −
0)2 = 2a (x − c)2 + y2 = 2a (x − (−c))2 + (y − 0)2 − (x + c)2 + y2 − Using the same arsenal of Intermediate Algebra weaponry we used in deriving the standard formula of an ellipse, Equation 7.4, we arrive at the following.1 1It is a good exercise to actually work this out. 534 Hooked on Conics a2 − c2 x2 + a2y2 = a2 a2 − c2 What remains is to determine the relationship between a, b and c. To that end, we note that since a and c are both positive numbers with a < c, we get a2 < c2 so that a2 − c2 is a negative number. Hence, c2 − a2 is a positive number. For reasons which will become clear soon, we re-write the equation by solving for y2/x2 to get a2 − c2 x2 + a2y2 = a2 a2 − c2 − c2 − a2 x2 + a2y2 = −a2 c2 − a2 a2y2 = c2 − a2 x2 − a2 c2 − a2 y2 x2 = − c2 − a2 c2 − a2 a2 x2 (c2−a2) x2 → 0 so that y2 a x as \|x\| grows large. Thus y = ± b (c2−a2) a2 x2 → As x and y attain very large values, the quantity. By setting b2 = c2 − a2 we get y2 x2 → b2 a x are the asymptotes to the graph as predicted and our choice of labels for the endpoints of the conjugate axis is justiﬁed. In our equation of the hyperbola we can substitute a2 − c2 = −b2 which yields a2. This shows that y → ± b a2 − c2 x2 + a2y2 = a2 a2 − c2 −b2x2 + a2y2 = −a2b2 y2 b2 = 1 x2 a2 − The equation above is for a hyperbola whose center is the origin and which opens to the left and right. If the hyperbola were centered at a point
(h, k), we would get the following. Equation 7.6. The Standard Equation of a Horizontala Hyperbola For positive numbers a and b, the equation of a horizontal hyperbola with center (h, k) is aThat is, a hyperbola whose branches open to the left and right (x − h)2 a2 − (y − k)2 b2 = 1 If the roles of x and y were interchanged, then the hyperbola’s branches would open upwards and downwards and we would get a ‘vertical’ hyperbola. Equation 7.7. The Standard Equation of a Vertical Hyperbola For positive numbers a and b, the equation of a vertical hyperbola with center (h, k) is: (y − k)2 b2 − (x − h)2 a2 = 1 The values of a and b determine how far in the x and y directions, respectively, one counts from the center to determine the rectangle through which the asymptotes pass. In both cases, the distance 7.5 Hyperbolas 535 a2 + b2. from the center to the foci, c, as seen in the derivation, can be found by the formula c = Lastly, note that we can quickly distinguish the equation of a hyperbola from that of a circle or ellipse because the hyperbola formula involves a diﬀerence of squares where the circle and ellipse formulas both involve the sum of squares. √ Example 7.5.1. Graph the equation (x−2)2 4 − y2 transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes. 25 = 1. Find the center, the lines which contain the Solution. We ﬁrst see that this equation is given to us in the standard form of Equation 7.6. Here x − h is x − 2 so h = 2, and y − k is y so k = 0. Hence, our hyperbola is centered at (2, 0). We see that a2 = 4 so a = 2, and b2 = 25 so b = 5. This means we move 2 units to the left and right of the center and 5 units up and down from the center to arrive at points on the guide rectangle. The asymptotes pass through the center of the
hyperbola as well as the corners of the rectangle. This yields the following set up2 −1 1 2 3 4 5 6 x −1 −2 −3 −4 −5 −6 −7 Since the y2 term is being subtracted from the x2 term, we know that the branches of the hyperbola open to the left and right. This means that the transverse axis lies along the x-axis. Hence, the conjugate axis lies along the vertical line x = 2. Since the vertices of the hyperbola are where the hyperbola intersects the transverse axis, we get that the vertices are 2 units to the left and right of (2, 0) at (0, 0) and (4, 0). To ﬁnd the foci, we need c = 29. Since the foci 29, 0) lie on the transverse axis, we move (approximately (−3.39, 0)) and (2 + 29, 0) (approximately (7.39, 0)). To determine the equations of the asymptotes, recall that the asymptotes go through the center of the hyperbola, (2, 0), as well as the corners of guide rectangle, so they have slopes of ± b 2. Using the point-slope equation 29 units to the left and right of (2, 0) to arrive at (2 − √ a2 + b2 = a = ± 5 4 + 25 = √ √ √ √ √ 536 Hooked on Conics of a line, Equation 2.2, yields y − 0 = ± 5 it all together, we get 2 (x − 2), so we get y = 5 2 x − 5 and y = − 5 2 x + 5. Putting y 7 6 5 4 3 2 1 −3 −2 −1 −2 −3 −4 −5 −6 −7 Example 7.5.2. Find the equation of the hyperbola with asymptotes y = ±2x and vertices (±5, 0). Solution. Plotting the data given to us, we have y 5 −5 x 5 −5 This graph not only tells us that the branches of the hyperbola open to the left and to the right, it also tells us that the center is (0, 0). Hence, our standard form is x2 b2 = 1. Since the vertices are (
±5, 0), we have a = 5 so a2 = 25. In order to determine b2, we recall that the slopes of the asymptotes are ± b 5 = 2, so b = 10. Hence, b2 = 100 and our ﬁnal answer is x2 a. Since a = 5 and the slope of the line y = 2x is 2, we have that b 25 − y2 a2 − y2 100 = 1. 7.5 Hyperbolas 537 As with the other conic sections, an equation whose graph is a hyperbola may not be given in either of the standard forms. To rectify that, we have the following. To Write the Equation of a Hyperbola in Standard Form 1. Group the same variables together on one side of the equation and position the constant on the other side 2. Complete the square in both variables as needed 3. Divide both sides by the constant term so that the constant on the other side of the equation becomes 1 Example 7.5.3. Consider the equation 9y2 − x2 − 6x = 10. Put this equation in to standard form and graph. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci, and the equations of the asymptotes. Solution. We need only complete the square on x: 9y2 − x2 − 6x = 10 9y2 − 1 x2 + 6x = 10 9y2 − x2 + 6x + 9 = 10 − 1(9) 9y2 − (x + 3)2 = 1 y2 (x + 3)2 1 1 9 = 1 − 9 so b = 1 Now that this equation is in the standard form of Equation 7.7, we see that x−h is x+3 so h = −3, and y − k is y so k = 0. Hence, our hyperbola is centered at (−3, 0). We ﬁnd that a2 = 1 so a = 1, and b2 = 1 3. This means that we move 1 unit to the left and right of the center and 1 3 units up and down from the center to arrive at points on the guide rectangle. Since the x2 term is being subtracted from the y2 term, we know the branches of the hyperbola open upwards and downwards. This means the transverse axis lies along
the vertical line x = −3 and the conjugate axis lies along the x-axis. Since the vertices of the hyperbola are where the hyperbola intersects and the transverse axis, we get that the vertices are 1 −3, − 1 3 3 of a unit above and below (−3, 0) at −3, 1. To ﬁnd the foci, we use 3 √ √ c = a2 + b2 = Since the foci lie on the transverse axis, we move 10 3 1 9 √ 10 3 units above and below (−3, 0) to arrive at. To determine the asymptotes, recall that the asymptotes go through the center of the hyperbola, (−3, 0), as well as the corners of guide rectangle, so they have slopes 3. Using the point-slope equation of a line, Equation 2.2, we get y = 1 of ± b 3 x + 1 and. Putting it all together, we get 10 3 −3, − + 1 = −3, and 10 3 √ 538 Hooked on Conics −6 −1 y 1 −1 x Hyperbolas can be used in so-called ‘trilateration,’ or ‘positioning’ problems. The procedure outlined in the next example is the basis of the (now virtually defunct) LOng Range Aid to Navigation (LORAN for short) system.2 Example 7.5.4. Jeﬀ is stationed 10 miles due west of Carl in an otherwise empty forest in an attempt to locate an elusive Sasquatch. At the stroke of midnight, Jeﬀ records a Sasquatch call 9 seconds earlier than Carl. If the speed of sound that night is 760 miles per hour, determine a hyperbolic path along which Sasquatch must be located. Solution. Since Jeﬀ hears Sasquatch sooner, it is closer to Jeﬀ than it is to Carl. Since the speed of sound is 760 miles per hour, we can determine how much closer Sasquatch is to Jeﬀ by multiplying 760 miles hour × 1 hour 3600 seconds × 9 seconds = 1.9 miles This means that Sasquatch is 1.9 miles closer to Jeﬀ than it is to Carl. In other words, Sasquatch must lie on a path where (the distance to Carl) − (the
distance to Jeﬀ) = 1.9 This is exactly the situation in the deﬁnition of a hyperbola, Deﬁnition 7.6. In this case, Jeﬀ and Carl are located at the foci,3 and our ﬁxed distance d is 1.9. For simplicity, we assume the hyperbola is centered at (0, 0) with its foci at (−5, 0) and (5, 0). Schematically, we have 2GPS now rules the positioning kingdom. Is there still a place for LORAN and other land-based systems? Do satellites ever malfunction? 3We usually like to be the center of attention, but being the focus of attention works equally well. 7.5 Hyperbolas 539 y 6 5 4 3 2 1 Jeﬀ −6 −5 −4 −3 −2 −1 1 2 3 4 Carl 5 6 x −1 −2 −3 −4 −5 −6 We are seeking a curve of the form x2 b2 = 1 in which the distance from the center to each focus is c = 5. As we saw in the derivation of the standard equation of the hyperbola, Equation 7.6, d = 2a, so that 2a = 1.9, or a = 0.95 and a2 = 0.9025. All that remains is to ﬁnd b2. To that end, we recall that a2 + b2 = c2 so b2 = c2 − a2 = 25 − 0.9025 = 24.0975. Since Sasquatch is closer to Jeﬀ than it is to Carl, it must be on the western (left hand) branch of a2 − y2 0.9025 − y2 x2 24.0975 = 1. In our previous example, we did not have enough information to pin down the exact location of Sasquatch. To accomplish this, we would need a third observer. Example 7.5.5. By a stroke of luck, Kai was also camping in the woods during the events of the previous example. He was located 6 miles due north of Jeﬀ and heard the Sasquatch call 18 seconds after Jeﬀ did. Use this added information to locate Sasquatch. Solution. Kai and Jeﬀ are now the foci of a second hyperbola where
the ﬁxed distance d can be determined as before 760 miles hour × 1 hour 3600 seconds × 18 seconds = 3.8 miles Since Jeﬀ was positioned at (−5, 0), we place Kai at (−5, 6). This puts the center of the new hyperbola at (−5, 3). Plotting Kai’s position and the new center gives us the diagram below on the left. The second hyperbola is vertical, so it must be of the form (y−3)2 a2 = 1. As before, the distance d is the length of the major axis, which in this case is 2b. We get 2b = 3.8 so that b = 1.9 and b2 = 3.61. With Kai 6 miles due North of Jeﬀ, we have that the distance from the center to the focus is c = 3. Since a2 + b2 = c2, we get a2 = c2 − b2 = 9 − 3.61 = 5.39. Kai heard the Sasquatch call after Jeﬀ, so Kai is farther from Sasquatch than Jeﬀ. Thus Sasquatch must lie 3.61 − (x+5)2 on the southern branch of the hyperbola (y−3)2 5.39 = 1. Looking at the western branch of the b2 − (x+5)2 540 Hooked on Conics hyperbola determined by Jeﬀ and Carl along with the southern branch of the hyperbola determined by Kai and Jeﬀ, we see that there is exactly one point in common, and this is where Sasquatch must have been when it called. Kai Jeﬀ −9−8−7−6−5−4−3−2−1 y Carl 1 2 3 4 5 6 x y Carl 1 2 3 4 5 6 x Kai Jeﬀ 6 5 4 3 2 1 −9−8−7−6−5−4−3−2 Sasquatch −2 −3 −4 −5 −6 6 5 4 3 2 1 −1 −2 −3 −4 −5 −6 To determine the coordinates of this point of intersection exactly, we would need techniques for solving systems of non-linear equations (which we won’t see until Section 8.7), so we use the calculator4 Doing so, we get Sasquatch is
approximately at (−0.9629, −0.8113). Each of the conic sections we have studied in this chapter result from graphing equations of the form Ax2 + Cy2 + Dx + Ey + F = 0 for diﬀerent choices of A, C, D, E, and5 F. While we’ve seen examples6 demonstrate how to convert an equation from this general form to one of the standard forms, we close this chapter with some advice about which standard form to choose.7 Strategies for Identifying Conic Sections Suppose the graph of equation Ax2 + Cy2 + Dx + Ey + F = 0 is a non-degenerate conic section.a If just one variable is squared, the graph is a parabola. Put the equation in the form of Equation 7.2 (if x is squared) or Equation 7.3 (if y is squared). If both variables are squared, look at the coeﬃcients of x2 and y2, A and B. If A = B, the graph is a circle. Put the equation in the form of Equation 7.1. If A = B but A and B have the same sign, the graph is an ellipse. Put the equation in the form of Equation 7.4. If A and B have the diﬀerent signs, the graph is a hyperbola. Put the equation in the form of either Equation 7.6 or Equation 7.7. aThat is, a parabola, circle, ellipse, or hyperbola – see Section 7.1. 4First solve each hyperbola for y, and choose the correct equation (branch) before proceeding. 5See Section 11.6 to see why we skip B. 6Examples 7.2.3, 7.3.4, 7.4.3, and 7.5.3, in particular. 7We formalize this in Exercise 34. 7.5 Hyperbolas 7.5.1 Exercises 541 In Exercises 1 - 8, graph the hyperbola. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes. 1. 3. 5. 7. − x2 y2 16 9 (x − 2)2 4 (x + 4)2 16 (y
+ 2)2 16 = 1 − − − (y + 3)2 9 (y − 4)2 1 (x − 5)2 20 = 1 = 1 = 1 2. 4. 6. 8. − x2 y2 9 16 (y − 3)2 11 (x + 1)2 9 (x − 4)2 8 = 1 − − − (x − 1)2 10 (y − 3)2 4 (y − 2)2 18 = 1 = 1 = 1 In Exercises 9 - 12, put the equation in standard form. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes. 9. 12x2 − 3y2 + 30y − 111 = 0 10. 18y2 − 5x2 + 72y + 30x − 63 = 0 11. 9x2 − 25y2 − 54x − 50y − 169 = 0 12. −6x2 + 5y2 − 24x + 40y + 26 = 0 In Exercises 13 - 18, ﬁnd the standard form of the equation of the hyperbola which has the given properties. 13. Center (3, 7), Vertex (3, 3), Focus (3, 2) 14. Vertex (0, 1), Vertex (8, 1), Focus (−3, 1) 15. Foci (0, ±8), Vertices (0, ±5). 16. Foci (±5, 0), length of the Conjugate Axis 6 17. Vertices (3, 2), (13, 2); Endpoints of the Conjugate Axis (8, 4), (8, 0) 18. Vertex (−10, 5), Asymptotes y = ± 1 2 (x − 6) + 5 In Exercises 19 - 28, ﬁnd the standard form of the equation using the guidelines on page 540 and then graph the conic section. 19. x2 − 2x − 4y − 11 = 0 20. x2 + y2 − 8x + 4y + 11 = 0 21. 9x2 + 4y2 − 36x + 24y + 36 = 0 22. 9x2 − 4y2 − 36x − 24y − 36 = 0 542 Hooked on Conics 23. y2 + 8y − 4x
+ 16 = 0 24. 4x2 + y2 − 8x + 4 = 0 25. 4x2 + 9y2 − 8x + 54y + 49 = 0 26. x2 + y2 − 6x + 4y + 14 = 0 27. 2x2 + 4y2 + 12x − 8y + 25 = 0 28. 4x2 − 5y2 − 40x − 20y + 160 = 0 29. The graph of a vertical or horizontal hyperbola clearly fails the Vertical Line Test, Theorem 1.1, so the equation of a vertical of horizontal hyperbola does not deﬁne y as a function of x.8 However, much like with circles, horizontal parabolas and ellipses, we can split a hyperbola into pieces, each of which would indeed represent y as a function of x. With the help of your classmates, use your calculator to graph the hyperbolas given in Exercises 1 - 8 above. How many pieces do you need for a vertical hyperbola? How many for a horizontal hyperbola? 30. The location of an earthquake’s epicenter − the point on the surface of the Earth directly above where the earthquake actually occurred − can be determined by a process similar to how we located Sasquatch in Example 7.5.5. (As we said back in Exercise 75 in Section 6.1, earthquakes are complicated events and it is not our intent to provide a complete discussion of the science involved in them. Instead, we refer the interested reader to a course in Geology or the U.S. Geological Survey’s Earthquake Hazards Program found here.) Our technique works only for relatively small distances because we need to assume that the Earth is ﬂat in order to use hyperbolas in the plane.9 The P-waves (“P” stands for Primary) of an earthquake in Sasquatchia travel at 6 kilometers per second.10 Station A records the waves ﬁrst. Then Station B, which is 100 kilometers due north of Station A, records the waves 2 seconds later. Station C, which is 150 kilometers due west of Station A records the waves 3 seconds after that (a total of 5 seconds after Station A). Where is the epicenter? 31. The notion of eccentricity introduced for ellipses in Deﬁnition 7.5 in Section 7.4 is the same for hyper
bolas in that we can deﬁne the eccentricity e of a hyperbola as e = distance from the center to a focus distance from the center to a vertex (a) With the help of your classmates, explain why e > 1 for any hyperbola. (b) Find the equation of the hyperbola with vertices (±3, 0) and eccentricity e = 2. (c) With the help of your classmates, ﬁnd the eccentricity of each of the hyperbolas in Exercises 1 - 8. What role does eccentricity play in the shape of the graphs? 32. On page 510 in Section 7.3, we discussed paraboloids of revolution when studying the design of satellite dishes and parabolic mirrors. In much the same way, ‘natural draft’ cooling towers are often shaped as hyperboloids of revolution. Each vertical cross section of these towers 8We will see later in the text that the graphs of certain rotated hyperbolas pass the Vertical Line Test. 9Back in the Exercises in Section 1.1 you were asked to research people who believe the world is ﬂat. What did you discover? 10Depending on the composition of the crust at a speciﬁc location, P-waves can travel between 5 kps and 8 kps. 7.5 Hyperbolas 543 is a hyperbola. Suppose the a natural draft cooling tower has the cross section below. Suppose the tower is 450 feet wide at the base, 275 feet wide at the top, and 220 feet at its narrowest point (which occurs 330 feet above the ground.) Determine the height of the tower to the nearest foot. 275 ft 220 ft 450 ft 330 ft 33. With the help of your classmates, research the Cassegrain Telescope. It uses the reﬂective property of the hyperbola as well as that of the parabola to make an ingenious telescope. 34. With the help of your classmates show that if Ax2 + Cy2 + Dx + Ey + F = 0 determines a non-degenerate conic11 then AC < 0 means that the graph is a hyperbola AC = 0 means that the graph is a parabola AC > 0 means that the graph is an ellipse or circle NOTE: This result will be generalized in Theorem 11.11 in Section 11.6.1. 11Recall that this means its graph is either
a circle, parabola, ellipse or hyperbola. 544 7.5.2 Answers Hooked on Conics 1. 2. − = 1 y2 x2 9 16 Center (0, 0) Transverse axis on y = 0 Conjugate axis on x = 0 Vertices (4, 0), (−4, 0) Foci (5, 0), (−5, 0) Asymptotes y = ± 3 4 x − = 1 y2 x2 16 9 Center (0, 0) Transverse axis on x = 0 Conjugate axis on y = 0 Vertices (0, 3), (0, −3) Foci (0, 5), (0, −5) Asymptotes y = ± 3 4 x 3. (x − 2)2 4 − (y + 3)2 9 = 1 Center (2, −3) Transverse axis on y = −3 Conjugate axis on x = 2 Vertices (0, −3), (4, −3) √ 13, −3), (2 − Foci (2 + Asymptotes y = ± 3 √ 13, −3) 2 (x − 26 −5 −4 −3 −2 −1 −1 −2 1 2 3 4 5 6 x −3 −4 −5 −6 y 6 5 4 3 2 1 −6 −5 −4 −3 −2 −1 −1 1 2 3 4 5 6 x −2 −3 −4 −5 −6 y 4 3 2 1 −3 −2 −1 −2 −3 −4 −5 −6 −7 −8 −9 −10 7.5 Hyperbolas 545 4. (y − 3)2 11 − (x − 1)2 10 = 1 Center (1, 3) Transverse axis on x = 1 Conjugate axis on y = 3 Vertices (1, 3 + Foci (1, 3 + Asymptotes y = ± √ √ 11), (1, 3 − √ 21), (1, 3 − 21) √ 110 10 (x − 1) + 3 √ 11) 5. (x + 4)2 16 − (y − 4)2 1 = 1 Center (−4, 4) Transverse axis on y = 4 Conjugate axis on x = −4 Vertices (−8, 4), (0, 4) Foci (−
4 + Asymptotes y = ± 1 17, 4), (−4 − √ √ 17, 4) 4 (x + 4) + 4 6. (x + 1)2 9 − (y − 3)2 4 = 1 Center (−1, 3) Transverse axis on y = 3 Conjugate axis on x = −1 Vertices (2, 3), (−4, 3) Foci −1 + Asymptotes y = ± 2 13, 3, −1 − √ 3 (x + 1) + 3 √ 13, 3 7. (y + 2)2 16 − (x − 5)2 20 = 1 Center (5, −2) Transverse axis on x = 5 Conjugate axis on y = −2 Vertices (5, 2), (5, −6) Foci (5, 4), (5, −8) Asymptotes y = ± 2 √ 5 5 (x − 55 −4 −3 −2 −1 −1 −2 −11−10 −9 −8 −7 −6 −5 −4 −3 −2 −7 −6 −5 −4 −3 −2 − 10 11 x 4 3 2 1 −1 −1 −2 −3 −4 −5 −6 −7 −8 546 Hooked on Conics 8. (x − 4)2 8 − (y − 2)2 18 = 1 Center (4, 2) Transverse axis on y = 2 Conjugate axis on x = 4 Vertices 4 + 2 Foci 4 + Asymptotes y = ± 3 26, 2, 4 − √ √ 2, 2, 4 − 2 2, 2 √ √ 26, 2 2 (x − 4) + 2 9. − = 1 (y − 5)2 12 x2 3 Center (0, 5) Transverse axis on y = 5 Conjugate axis on x = 0 Vertices ( 3, 5) 15, 5), (− Foci ( 15, 5) Asymptotes y = ±2x + 5 3, 5), (− √ √ √ √ 10 x −2−1 −1 −2 10. (y + 2)2 5 − (x − 3)2 18 = 1 Center (3, −2) Transverse axis on x = 3 Conjugate axis on y = −2 Vertices (3,
−2 + Foci (3, −2 + Asymptotes y = ± √ √ 5), (3, −2 − √ 23), (3, −2 − 23) √ 10 6 (x − 3) − 2 √ 5) 11. (x − 3)2 25 − (y + 1)2 9 = 1 12. Center (3, −1) Transverse axis on y = −1 Conjugate axis on x = 3 Vertices (8, −1), (−2, −1) Foci 3 + 34, −1, 3 − Asymptotes y = ± 3 √ √ 34, −1 5 (x − 3) − 1 − = 1 (y + 4)2 6 (x + 2)2 5 Center (−2, −4) Transverse axis on x = −2 Conjugate axis on y = −4 Vertices −2, −4 + √ Foci −2, −4 + Asymptotes y = ± √ 6, −2, −4 − √ 11, −2, −4 − √ 30 5 (x + 2) − 4 √ 6 11 13. (y − 7)2 16 − (x − 3)2 9 = 1 14. (x − 4)2 16 − (y − 1)2 33 = 1 15. y2 25 − x2 39 = 1 16. x2 16 − y2 9 = 1 17. (x − 8)2 25 − (y − 2)2 4 = 1 18. (x − 6)2 256 − (y − 5)2 64 = 1 7.5 Hyperbolas 547 19. (x − 1)2 = 4(y + 3) 20. (x − 4)2 + (y + 2)2 = 9 y y −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 21. (x − 2)2 4 + (y + 3)1 −2 −3 −4 −5 −6 1 4 7 x 1 −2 −5 22. (x − 2)2 4 − (y + 3)3 −2 −1 −1 −3 −4 −5 −6 −7 −8 −9 −10 (x − 1)2 1 y2 4 + = 0 The graph is the point (1, 0) only. 23. (
y + 4)2 = 4x y −1 1 2 3 4 x 24. −1 −2 −3 −4 −5 −6 −7 −8 548 Hooked on Conics 25. (x − 1)2 9 + (y + 3)2 4 = 1 y −2 −1 1 2 3 4 x 26. (x − 3)2 + (y + 2)2 = −1 There is no graph. −1 −2 −3 −4 −5 27. (x + 3)2 2 + There is no graph. (y − 1)2 1 = − 3 4 28. (y + 2)2 16 − (x − 5)2 20 = 10 11 x 4 3 2 1 −1 −1 −2 −3 −4 −5 −6 −7 −8 30. By placing Station A at (0, −50) and Station B at (0, 50), the two second time diﬀerence yields the hyperbola y2 2464 = 1 with foci A and B and center (0, 0). Placing Station C at (−150, −50) and using foci A and C gives us a center of (−75, −50) and the hyperbola 225 − (y+50)2 (x+75)2 5400 = 1. The point of intersection of these two hyperbolas which is closer to A 36 − x2 than B and closer to A than C is (−57.8444, −9.21336) so that is the epicenter. 31. (b) x2 9 − y2 27 = 1. 32. The tower may be modeled (approximately)12 by x2 34203 = 1. To ﬁnd the height, we plug in x = 137.5 which yields y ≈ 191 or y ≈ 469. Since the top of the tower is above the narrowest point, we get the tower is approximately 469 feet tall. 12100 − (y−330)2 12The exact value underneath (y − 330)2 is 52707600 1541 in case you need more precision. Chapter 8 Systems of Equations and Matrices 8.1 Systems of Linear Equations: Gaussian Elimination Up until now, when we concerned ourselves with solving diﬀerent types of equations there was only one equation to solve at a time. Given an equation f (x) = g(x),
we could check our solutions geometrically by ﬁnding where the graphs of y = f (x) and y = g(x) intersect. The x-coordinates of these intersection points correspond to the solutions to the equation f (x) = g(x), and the ycoordinates were largely ignored. If we modify the problem and ask for the intersection points of the graphs of y = f (x) and y = g(x), where both the solution to x and y are of interest, we have what is known as a system of equations, usually written as y = f (x) y = g(x) The ‘curly bracket’ notation means we are to ﬁnd all pairs of points (x, y) which satisfy both equations. We begin our study of systems of equations by reviewing some basic notions from Intermediate Algebra. Deﬁnition 8.1. A linear equation in two variables is an equation of the form a1x + a2y = c where a1, a2 and c are real numbers and at least one of a1 and a2 is nonzero. For reasons which will become clear later in the section, we are using subscripts in Deﬁnition 8.1 to indicate diﬀerent, but ﬁxed, real numbers and those subscripts have no mathematical meaning beyond that. For example, 3x − y 2 = 0.1 is a linear equation in two variables with a1 = 3, a2 = − 1 2 and c = 0.1. We can also consider x = 5 to be a linear equation in two variables1 by identifying a1 = 1, a2 = 0, and c = 5. If a1 and a2 are both 0, then depending on c, we get either an equation which is always true, called an identity, or an equation which is never true, called a If c = 0, then we’d have contradiction. (If c = 0, then we get 0 = 0, which is always true. 0 = 0, which is never true.) Even though identities and contradictions have a large role to play 1Critics may argue that x = 5 is clearly an equation in one variable. It can also be considered an equation in 117 variables with the coeﬃcients of 116 variables set to 0. As with many conventions in Mathematics, the context will clarify the situation. 550 Systems
of Equations and Matrices in the upcoming sections, we do not consider them linear equations. The key to identifying linear equations is to note that the variables involved are to the ﬁrst power and that the coeﬃcients of the variables are numbers. Some examples of equations which are non-linear are x2 + y = 1, xy = 5 and e2x + ln(y) = 1. We leave it to the reader to explain why these do not satisfy Deﬁnition 8.1. From what we know from Sections 1.2 and 2.1, the graphs of linear equations are lines. If we couple two or more linear equations together, in eﬀect to ﬁnd the points of intersection of two or more lines, we obtain a system of linear equations in two variables. Our ﬁrst example reviews some of the basic techniques ﬁrst learned in Intermediate Algebra. Example 8.1.1. Solve the following systems of equations. Check your answer algebraically and graphically. 2x − y = 1 y = 3 1. 3x + 4y = −2 −3x − y = 5 2. 2x − 4y = 6 3x − 6y = 9 4. 6x + 3y = 9 4x + 2y = 12 5. 3. 6. Solution. x 3 − 4y 9 + y 5 = 7 3 = 1 2x 2x + y = −2 1. Our ﬁrst system is nearly solved for us. The second equation tells us that y = 3. To ﬁnd the corresponding value of x, we substitute this value for y into the the ﬁrst equation to obtain 2x − 3 = 1, so that x = 2. Our solution to the system is (2, 3). To check this algebraically, we substitute x = 2 and y = 3 into each equation and see that they are satisﬁed. We see 2(2) − 3 = 1, and 3 = 3, as required. To check our answer graphically, we graph the lines 2x − y = 1 and y = 3 and verify that they intersect at (2, 3). 2. To solve the second system, we use the addition method to eliminate the variable x. We take the two equations as given and ‘add equals to equals’ to obtain 3x
+ 4y = −2 + (−3x − y = 5) 3 3y = This gives us y = 1. We now substitute y = 1 into either of the two equations, say −3x−y = 5, to get −3x − 1 = 5 so that x = −2. Our solution is (−2, 1). Substituting x = −2 and y = 1 into the ﬁrst equation gives 3(−2) + 4(1) = −2, which is true, and, likewise, when we check (−2, 1) in the second equation, we get −3(−2) − 1 = 5, which is also true. Geometrically, the lines 3x + 4y = −2 and −3x − y = 5 intersect at (−2, 1). 8.1 Systems of Linear Equations: Gaussian Elimination 551 y 4 2 1 (2, 3) −1 1 2 3 4 x 2x − y = 1 y = 3 y 2 1 (−2, 1) −4 −3 −2 −1 x −1 −2 3x + 4y = −2 −3x − y = 5 3. The equations in the third system are more approachable if we clear denominators. We multiply both sides of the ﬁrst equation by 15 and both sides of the second equation by 18 to obtain the kinder, gentler system 5x − 12y = 21 4x + 6y = 9 Adding these two equations directly fails to eliminate either of the variables, but we note that if we multiply the ﬁrst equation by 4 and the second by −5, we will be in a position to eliminate the x term 20x − 48y = 84 + (−20x − 30y = −45) 39 −78y = From this we get y = − 1 2. We can temporarily avoid too much unpleasantness by choosing to substitute y = − 1 2 into one of the equivalent equations we found by clearing denominators, say into 5x − 12y = 21. We get 5x + 6 = 21 which gives x = 3. Our answer is 3, − 1. 2 At this point, we have no choice − in order to check an answer algebraically, we must see if the answer satisﬁes both of the original equations, so we substitute x = 3 and y = − 1 2 into both x 2.
We leave it to the reader to verify that the solution is correct. Graphing both of the lines involved with considerable care yields an intersection point of 3, − 1 2 5 and 2x 3 − 4y. An eerie calm settles over us as we cautiously approach our fourth system. Do its friendly integer coeﬃcients belie something more sinister? We note that if we multiply both sides of the ﬁrst equation by 3 and both sides of the second equation by −2, we are ready to eliminate the x 552 Systems of Equations and Matrices 6x − 12y = 18 + (−6x + 12y = −18) 0 0 = 2 x − 3 2. For each value of x, the formula y = 1 We eliminated not only the x, but the y as well and we are left with the identity 0 = 0. This means that these two diﬀerent linear equations are, in fact, equivalent. In other words, if an ordered pair (x, y) satisﬁes the equation 2x − 4y = 6, it automatically satisﬁes the equation 3x − 6y = 9. One way to describe the solution set to this system is to use the roster method2 and write {(x, y) \| 2x − 4y = 6}. While this is correct (and corresponds exactly to what’s happening graphically, as we shall see shortly), we take this opportunity to introduce the notion of a parametric solution to a system. Our ﬁrst step is to solve 2x − 4y = 6 for one of the variables, say y = 1 2 x − 3 2 determines the corresponding y-value of a solution. Since we have no restriction on x, it is called a free variable. We let x = t, a so-called ‘parameter’, and get y = 1 2. Our set of solutions can then be described as t, 1 \| − ∞ < t < ∞.3 For speciﬁc values of t, we can generate solutions. For example, t = 0 gives us the solution 0, − 3 ; t = 117 2 gives us (117, 57), and while we can readily check each of these particular solutions satisfy both equations, the question is how do we check our general answer algebraically? Same as always. We claim that for any real number t, the pair t, 1
satisﬁes both equations. = 6. Simplifying, Substituting x = t and we get 2t − 2t + 6 = 6, which is always true. Similarly, when we make these substitutions in the equation 3x − 6y = 9, we get 3t − 6 1 = 9 which reduces to 3t − 3t + 9 = 9, so it 2 t − 3 2 checks out, too. Geometrically, 2x − 4y = 6 and 3x − 6y = 9 are the same line, which means that they intersect at every point on their graphs. The reader is encouraged to think about how our parametric solution says exactly that. 2 t − 3 2 into 2x − 4y = 6 gives 2t −, − 1 2 1 −1 −1 −2 −3 −4 x 3 − 4y 2x y 2 1 −1 1 2 3 4 x 2x − 4y = 6 3x − 6y = 9 (Same line.) 2See Section 1.2 for a review of this. 3Note that we could have just as easily chosen to solve 2x − 4y = 6 for x to obtain x = 2y + 3. Letting y be the parameter t, we have that for any value of t, x = 2t + 3, which gives {(2t + 3, t) \| − ∞ < t < ∞}. There is no one correct way to parameterize the solution set, which is why it is always best to check your answer. 8.1 Systems of Linear Equations: Gaussian Elimination 553 5. Multiplying both sides of the ﬁrst equation by 2 and the both sides of the second equation by −3, we set the stage to eliminate x 12x + 6y = 18 + (−12x − 6y = −36) 0 = −18 As in the previous example, both x and y dropped out of the equation, but we are left with an irrevocable contradiction, 0 = −18. This tells us that it is impossible to ﬁnd a pair (x, y) which satisﬁes both equations; in other words, the system has no solution. Graphically, the lines 6x + 3y = 9 and 4x + 2y = 12 are distinct and parallel, so they do not intersect. 6. We can begin to solve our last system by adding
the ﬁrst two equations x − y = 0 + (x + y = 2) 2x = 2 which gives x = 1. Substituting this into the ﬁrst equation gives 1 − y = 0 so that y = 1. We seem to have determined a solution to our system, (1, 1). While this checks in the ﬁrst two equations, when we substitute x = 1 and y = 1 into the third equation, we get −2(1)+(1) = −2 which simpliﬁes to the contradiction −1 = −2. Graphing the lines x−y = 0, x + y = 2, and −2x + y = −2, we see that the ﬁrst two lines do, in fact, intersect at (1, 1), however, all three lines never intersect at the same point simultaneously, which is what is required if a solution to the system is to be found. y 6 5 4 3 2 1 −1 −2 −3 1 2 x 6x + 3y = 9 4x + 2y = 12 y 1 −2x + y = −2 A few remarks about Example 8.1.1 are in order. It is clear that some systems of equations have solutions, and some do not. Those which have solutions are called consistent, those with no solution are called inconsistent. We also distinguish the two diﬀerent types of behavior among 554 Systems of Equations and Matrices consistent systems. Those which admit free variables are called dependent; those with no free variables are called independent.4 Using this new vocabulary, we classify numbers 1, 2 and 3 in Example 8.1.1 as consistent independent systems, number 4 is consistent dependent, and numbers 5 and 6 are inconsistent.5 The system in 6 above is called overdetermined, since we have more equations than variables.6 Not surprisingly, a system with more variables than equations is called underdetermined. While the system in number 6 above is overdetermined and inconsistent, there exist overdetermined consistent systems (both dependent and independent) and we leave it to the reader to think about what is happening algebraically and geometrically in these cases. Likewise, there are both consistent and inconsistent underdetermined systems,7 but a consistent underdetermined system of linear equations is necessarily dependent.8 In order to move this section beyond a review of Intermediate Algebra, we now deﬁ
ne what is meant by a linear equation in n variables. Deﬁnition 8.2. A linear equation in n variables, x1, x2,..., xn, is an equation of the form a1x1 + a2x2 +... + anxn = c where a1, a2,... an and c are real numbers and at least one of a1, a2,..., an is nonzero. Instead of using more familiar variables like x, y, and even z and/or w in Deﬁnition 8.2, we use subscripts to distinguish the diﬀerent variables. We have no idea how many variables may be involved, so we use numbers to distinguish them instead of letters. (There is an endless supply of distinct numbers.) As an example, the linear equation 3x1 −x2 = 4 represents the same relationship between the variables x1 and x2 as the equation 3x − y = 4 does between the variables x and y. In addition, just as we cannot combine the terms in the expression 3x − y, we cannot combine the terms in the expression 3x1 − x2. Coupling more than one linear equation in n variables results in a system of linear equations in n variables. When solving these systems, it becomes increasingly important to keep track of what operations are performed to which equations and to develop a strategy based on the kind of manipulations we’ve already employed. To this end, we ﬁrst remind ourselves of the maneuvers which can be applied to a system of linear equations that result in an equivalent system.9 4In the case of systems of linear equations, regardless of the number of equations or variables, consistent independent systems have exactly one solution. The reader is encouraged to think about why this is the case for linear equations in two variables. Hint: think geometrically. 5The adjectives ‘dependent’ and ‘independent’ apply only to consistent systems – they describe the type of solu- tions. Is there a free variable (dependent) or not (independent)? 6If we think if each variable being an unknown quantity, then ostensibly, to recover two unknown quantities, we need two pieces of information - i.e., two equations. Having more than two equations suggests we have more information than necessary to determine the values of the unknowns. While this is not necessarily the case,
it does explain the choice of terminology ‘overdetermined’. 7We need more than two variables to give an example of the latter. 8Again, experience with systems with more variables helps to see this here, as does a solid course in Linear Algebra. 9That is, a system with the same solution set. 8.1 Systems of Linear Equations: Gaussian Elimination 555 Theorem 8.1. Given a system of equations, the following moves will result in an equivalent system of equations. Interchange the position of any two equations. Replace an equation with a nonzero multiple of itself.a Replace an equation with itself plus a nonzero multiple of another equation. aThat is, an equation which results from multiplying both sides of the equation by the same nonzero number. We have seen plenty of instances of the second and third moves in Theorem 8.1 when we solved the systems in Example 8.1.1. The ﬁrst move, while it obviously admits an equivalent system, seems silly. Our perception will change as we consider more equations and more variables in this, and later sections. Consider the system of equations    1 3 7 2 3, 7 2 and z = −1 into the ﬁrst equation to get x − 1 2 (−1) = 4 to obtain y = 7 Clearly z = −1, and we substitute this into the second equation y − 1 2. + 1 Finally, we substitute y = 7 2 (−1) = 1, so that x = 8 3. The reader can verify that these values of x, y and z satisfy all three original equations. It is tempting for us to write the solution to this system by extending the usual (x, y) 2, −1. The question quickly becomes what does notation to (x, y, z) and list our solution as 8 3, 7 an ‘ordered triple’ like 8 2, −1 represent? Just as ordered pairs are used to locate points on the two-dimensional plane, ordered triples can be used to locate points in space.10 Moreover, just as equations involving the variables x and y describe graphs of one-dimensional lines and curves in the two-dimensional plane, equations involving variables x, y, and z describe objects called surfaces in three-dimensional space. Each of the equations in the above system can be visualized as a plane situated in three-space. Ge
ometrically, the system is trying to ﬁnd the intersection, or common point, of all three planes. If you imagine three sheets of notebook paper each representing a portion of these planes, you will start to see the complexities involved in how three such planes can intersect. Below is a sketch of the three planes. It turns out that any two of these planes intersect in a line,11 so our intersection point is where all three of these lines meet. 10You were asked to think about this in Exercise 40 in Section 1.1. 11In fact, these lines are described by the parametric solutions to the systems formed by taking any two of these equations by themselves. 556 Systems of Equations and Matrices Since the geometry for equations involving more than two variables is complicated, we will focus our eﬀorts on the algebra. Returning to the system    1 we note the reason it was so easy to solve is that the third equation is solved for z, the second equation involves only y and z, and since the coeﬃcient of y is 1, it makes it easy to solve for y using our known value for z. Lastly, the coeﬃcient of x in the ﬁrst equation is 1 making it easy to substitute the known values of y and z and then solve for x. We formalize this pattern below for the most general systems of linear equations. Again, we use subscripted variables to describe the general case. The variable with the smallest subscript in a given equation is typically called the leading variable of that equation. Deﬁnition 8.3. A system of linear equations with variables x1, x2,... xn is said to be in triangular form provided all of the following conditions hold: 1. The subscripts of the variables in each equation are always increasing from left to right. 2. The leading variable in each equation has coeﬃcient 1. 3. The subscript on the leading variable in a given equation is greater than the subscript on the leading variable in the equation above it. 4. Any equation without variablesa cannot be placed above an equation with variables. anecessarily an identity or contradiction 8.1 Systems of Linear Equations: Gaussian Elimination 557 In our previous system, if we make the obvious choices x = x1, y = x2, and z = x3, we
see that the system is in triangular form.12 An example of a more complicated system in triangular form is    x1 − 4x3 + x4 − x6 = 6 x2 + 2x3 = 1 x4 + 3x5 − x6 = 8 x5 + 9x6 = 10 Our goal henceforth will be to transform a given system of linear equations into triangular form using the moves in Theorem 8.1. Example 8.1.2. Use Theorem 8.1 to put the following systems into triangular form and then solve the system if possible. Classify each system as consistent independent, consistent dependent, or inconsistent.    1. 3x − y + z = 3 2x − 4y + 3z = 16. 2x + 3y − z = 1 10x − z = 2 4x − 9y + 2z = 5    3. 3x1 + x2 + x4 = 6 2x1 + x2 − x3 = 4 x2 − 3x3 − 2x4 = 0 Solution. 1. For deﬁnitiveness, we label the topmost equation in the system E1, the equation beneath that E2, and so forth. We now attempt to put the system in triangular form using an algorithm known as Gaussian Elimination. What this means is that, starting with x, we transform the system so that conditions 2 and 3 in Deﬁnition 8.3 are satisﬁed. Then we move on to the next variable, in this case y, and repeat. Since the variables in all of the equations have a consistent ordering from left to right, our ﬁrst move is to get an x in E1’s spot with a coeﬃcient of 1. While there are many ways to do this, the easiest is to apply the ﬁrst move listed in Theorem 8.1 and interchange E1 and E3.    3x − y + z = 3 (E1) (E2) 2x − 4y + 3z = 16 x − y + z = 5 (E3) Switch E1 and E3 −−−−−−−−−−−→ �
��   x − y + z = 5 (E1) (E2) 2x − 4y + 3z = 16 3x − y + z = 3 (E3) To satisfy Deﬁnition 8.3, we need to eliminate the x’s from E2 and E3. We accomplish this by replacing each of them with a sum of themselves and a multiple of E1. To eliminate the x from E2, we need to multiply E1 by −2 then add; to eliminate the x from E3, we need to multiply E1 by −3 then add. Applying the third move listed in Theorem 8.1 twice, we get    x − y + z = 5 (E1) (E2) 2x − 4y + 3z = 16 3x − y + z = 3 (E3) Replace E2 with −2E1 + E2 −−−−−−−−−−−−−−−−−−→ Replace E3 with −3E1 + E3    (E1) x − y + z = (E2) −2y + z = (E3) 5 6 2y − 2z = −12 12If letters are used instead of subscripted variables, Deﬁnition 8.3 can be suitably modiﬁed using alphabetical order of the variables instead of numerical order on the subscripts of the variables. 558 Systems of Equations and Matrices Now we enforce the conditions stated in Deﬁnition 8.3 for the variable y. To that end we need to get the coeﬃcient of y in E2 equal to 1. We apply the second move listed in Theorem 8.1 and replace E2 with itself times − 1 2.    (E1) x − y + z = (E2) −2y + z = (E3) 5 6 2y − 2z = −12 Replace E2 with − 1 2 E2 −−−−−−−−−−−−−−→    (E1) x − y + z = (E2) (E3) 5 y − 1 2 z = −3 2y − 2
z = −12 To eliminate the y in E3, we add −2E2 to it.    (E1) x − y + z = (E2) (E3) 5 y − 1 2 z = −3 2y − 2z = −12 Replace E3 with −2E2 + E3 −−−−−−−−−−−−−−−−−−→    (E1) x − y + z = (E2) (E3) 5 2 z = −3 −z = −6 y − 1 Finally, we apply the second move from Theorem 8.1 one last time and multiply E3 by −1 to satisfy the conditions of Deﬁnition 8.3 for the variable z.    (E1) x − y + z = (E2) (E3) 5 2 z = −3 −z = −6 y − 1 Replace E3 with −1E3 −−−−−−−−−−−−−−→    (E1) x − y + z = (E2) (E3) 5 2 z = −3 6 z = y − 1 Now we proceed to substitute. Plugging in z = 6 into E2 gives y − 3 = −3 so that y = 0. With y = 0 and z = 6, E1 becomes x − 0 + 6 = 5, or x = −1. Our solution is (−1, 0, 6). We leave it to the reader to check that substituting the respective values for x, y, and z into the original system results in three identities. Since we have found a solution, the system is consistent; since there are no free variables, it is independent. 2. Proceeding as we did in 1, our ﬁrst step is to get an equation with x in the E1 position with 1 as its coeﬃcient. Since there is no easy ﬁx, we multiply E1 by 1 2.    2x + 3y − z = 1 (E1) 10x − z = 2 (E2) (E3) 4x − 9y + 2z = 5 Replace E1 with 1 2 E1 −
−−−−−−−−−−−−→    E1) 2 10x − z = 2 (E2) (E3) 4x − 9y + 2z = 5 Now it’s time to take care of the x’s in E2 and E3E1) 2 10x − z = 2 (E2) (E3) 4x − 9y + 2z = 5 Replace E2 with −10E1 + E2 −−−−−−−−−−−−−−−−−−→ Replace E3 with −4E1 + E3    2 y − 1 (E1) x + 3 (E2) (E3) 1 2 −15y + 4z = −3 −15y + 4z = 3 2 z = 8.1 Systems of Linear Equations: Gaussian Elimination 559 Our next step is to get the coeﬃcient of y in E2 equal to 1. To that end, we have    (E1) x + 3 (E215y + 4z = −3 −15y + 4z = 3 (E3) Replace E2 with − 1 15 E2 −−−−−−−−−−−−−−−→    Finally, we rid E3 of y. (E1) x + 3 (E2 15 z = 1 5 −15y + 4z = 3 2 (E3)    (E1) x + 3 (E2 15 z = 1 5 −15y + 4z = 3 2 (E3) Replace E3 with 15E2 + E3 −−−−−−−−−−−−−−−−−→    (E1) x − y + z = (E2) y − 1 5 2 z = −3 6 0 = (E3) The last equation, 0 = 6, is a contradiction so the system has no solution. According to Theorem 8.1, since this system has no solutions, neither does the original, thus we have
an inconsistent system. 3. For our last system, we begin by multiplying E1 by 1 3 to get a coeﬃcient of 1 on x1.    3x1 + x2 + x4 = 6 (E1) 2x1 + x2 − x3 = 4 (E2) (E3) x2 − 3x3 − 2x4 = 0 Replace E1 with 1 3 E1 −−−−−−−−−−−−−→    3 x2 + 1 (E1) x1 + 1 3 x4 = 2 2x1 + x2 − x3 = 4 (E2) (E3) x2 − 3x3 − 2x4 = 0 Next we eliminate x1 from E2    3 x2 + 1 (E1) x1 + 1 3 x4 = 2 2x1 + x2 − x3 = 4 (E2) (E3) x2 − 3x3 − 2x4 = 0 Replace E2 −−−−−−−−−−→ with −2E1 + E2    3 x2 + 1 (E1) x1 + 1 3 x4 = 2 3 x2 − x3 − 2 1 3 x4 = 0 (E2) (E3) x2 − 3x3 − 2x4 = 0 We switch E2 and E3 to get a coeﬃcient of 1 for x2.    (E1) x1 + 1 3 x2 + 1 3 x4 = 2 3 x2 − x3 − 2 1 3 x4 = 0 (E2) (E3) x2 − 3x3 − 2x4 = 0 Switch E2 and E3 −−−−−−−−−−−→    (E1) x1 + 1 3 x2 + 1 3 x4 = 2 (E2) x2 − 3x3 − 2x4 = 0 3 x2 − x3 − 2 3 x4 = 0 (E3) 1 Finally, we eliminate x2
in E3. 560 Systems of Equations and Matrices    (E1) x1 + 1 3 x2 + 1 3 x4 = 2 (E2) x2 − 3x3 − 2x4 = 0 3 x2 − x3 − 2 3 x4 = 0 (E3) 1 Replace E3 −−−−−−−−−−→ with − 1 3 E2 + E3    (E1) x1 + 1 3 x2 + 1 3 x4 = 2 (E2) x2 − 3x3 − 2x4 = 0 0 = 0 (E3) Equation E3 reduces to 0 = 0,which is always true. Since we have no equations with x3 or x4 as leading variables, they are both free, which means we have a consistent dependent system. We parametrize the solution set by letting x3 = s and x4 = t and obtain from E2 that x2 = 3s + 2t. Substituting this and x4 = t into E1, we have x1 + 1 3 t = 2 which gives x1 = 2 − s − t. Our solution is the set {(2 − s − t, 2s + 3t, s, t) \| − ∞ < s, t < ∞}.13 We leave it to the reader to verify that the substitutions x1 = 2 − s − t, x2 = 3s + 2t, x3 = s and x4 = t satisfy the equations in the original system. 3 (3s + 2t) + 1 Like all algorithms, Gaussian Elimination has the advantage of always producing what we need, but it can also be ineﬃcient at times. For example, when solving 2 above, it is clear after we eliminated the x’s in the second step to get the system    (E1) x + 3 (E215y + 4z = −3 −15y + 4z = 3 (E3) that equations E2 and E3 when taken together form a contradiction since we have identical left hand sides and diﬀerent right hand sides. The algorithm takes two more steps to reach this contradiction. We also note that
substitution in Gaussian Elimination is delayed until all the elimination is done, thus it gets called back-substitution. This may also be ineﬃcient in many cases. Rest assured, the technique of substitution as you may have learned it in Intermediate Algebra will once again take center stage in Section 8.7. Lastly, we note that the system in 3 above is underdetermined, and as it is consistent, we have free variables in our answer. We close this section with a standard ‘mixture’ type application of systems of linear equations. Example 8.1.3. Lucas needs to create a 500 milliliters (mL) of a 40% acid solution. He has stock solutions of 30% and 90% acid as well as all of the distilled water he wants. Set-up and solve a system of linear equations which determines all of the possible combinations of the stock solutions and water which would produce the required solution. Solution. We are after three unknowns, the amount (in mL) of the 30% stock solution (which we’ll call x), the amount (in mL) of the 90% stock solution (which we’ll call y) and the amount (in mL) of water (which we’ll call w). We now need to determine some relationships between these variables. Our goal is to produce 500 milliliters of a 40% acid solution. This product has two deﬁning characteristics. First, it must be 500 mL; second, it must be 40% acid. We take each 13Here, any choice of s and t will determine a solution which is a point in 4-dimensional space. Yeah, we have trouble visualizing that, too. 8.1 Systems of Linear Equations: Gaussian Elimination 561 of these qualities in turn. First, the total volume of 500 mL must be the sum of the contributed volumes of the two stock solutions and the water. That is amount of 30% stock solution + amount of 90% stock solution + amount of water = 500 mL Using our deﬁned variables, this reduces to x + y + w = 500. Next, we need to make sure the ﬁnal solution is 40% acid. Since water contains no acid, the acid will come from the stock solutions only. We ﬁnd 40% of 500 mL to be 200 mL which means the ﬁnal solution must contain 200 mL of acid. We
have amount of acid in 30% stock solution + amount of acid 90% stock solution = 200 mL The amount of acid in x mL of 30% stock is 0.30x and the amount of acid in y mL of 90% solution is 0.90y. We have 0.30x + 0.90y = 200. Converting to fractions,14 our system of equations becomes We ﬁrst eliminate the x from the second equation x + y + w = 500 10 y = 200 10 x + 9 3 (E1) x + y + w = 500 10 x + 9 10 y = 200 (E2) 3 Replace E2 with − 3 10 E1 + E2 −−−−−−−−−−−−−−−−−−→ (E1) x + y + w = 500 5 y − 3 10 w = 50 (E2) 3 Next, we get a coeﬃcient of 1 on the leading variable in E2 (E1) x + y + w = 500 5 y − 3 10 w = 50 (E2) 3 Replace E2 with 5 3 E2 −−−−−−−−−−−−−→ (E1) x + y + w = 500 y − 1 2 w = 250 (E2) 3 3 3, 1 2 t + 250 2 t + 250 2 t + 250 3. Substituting into E1 gives x + 1 + t = 500 so that x = − 3 2 t + 1250 Notice that we have no equation to determine w, and as such, w is free. We set w = t and from E2 get y = 1 2 t + 1250 3. This system is consistent, dependent and its solution set is {− 3 3, t \| − ∞ < t < ∞}. While this answer checks algebraically, we have neglected to take into account that x, y and w, being amounts of acid and water, need to be nonnegative. That is, x ≥ 0, y ≥ 0 and w ≥ 0. The 3 ≥ 0 or t ≥ − 500 constraint x ≥ 0 gives us − 3 3 ≥ 0, or t ≤ 2500 3. The condition z ≥ 0 yields t ≥ 0, and we see that when we take the set theoretic intersection of 3, t \| 0 ≤ t ≤ 2500 9. Our ﬁnal answer is {− 3 these intervals, we get
0 ≤ t ≤ 2500 9 }. Of what practical use is our answer? Suppose there is only 100 mL of the 90% solution remaining and it is due to expire. Can we use all of it to make our required solution? We would have y = 100 so that 1 3. This means the amount of 30% solution required is x = − 3 3 mL. The reader is invited to check that mixing these three amounts of our constituent solutions produces the required 40% acid mix. 3 = 100, and we get t = 100 + 1250 3 = − 3 3 mL, and for the water, w = t = 100 9. From y ≥ 0, we get 1 2 t + 250 2 t + 1250 3 = 1100 2 t + 1250 2 t + 1250 2 t + 250 2 t + 250 3, 1 100 3 2 14We do this only because we believe students can use all of the practice with fractions they can get! 562 Systems of Equations and Matrices 8.1.1 Exercises (Review Exercises) In Exercises 1 - 8, take a trip down memory lane and solve the given system using substitution and/or elimination. Classify each system as consistent independent, consistent dependent, or inconsistent. Check your answers both algebraically and graphically. 1. 3. 5. 7. x + 2y = 5 x = 6 x+2y 4 3x−y 2 = −5 = 1 1 2 x − 1 2y − 3x = 3 y = −1 6 2 x = − 15 3y −. 4. 6. 8. 2y − 3x = 1 y = − + 4y = 6 3 y = 1 12 − 20 3 y = − 7 3 3 y = 10 − 10 In Exercises 9 - 26, put each system of linear equations into triangular form and solve the system if possible. Classify each system as consistent independent, consistent dependent, or inconsistent. −5x + y = 17 x + y = 5 9. 11. 13. 15. 17. 19.    4x − y + z = 5 2y + 6z = 30 17 y − 3z = 0          3x − 2y + z = −5 x + 3y − z = 12 0 x + y + 2z
= x − y + z = −4 −3x + 2y + 4z = −5 x − 5y + 2z = −18 2x − y + z = 1 2x + 2y − z = 1 3x + 6y + 4z = 9                   10. 12. 14. 16. 18. 20. x + y + z = 3 2x − y + z = 0 −3x + 5y + 7z = 7 4x − y + z = 5 2y + 6z = 30 x + z = 6 x − 2y + 3z = 7 −3x + y + 2z = −5 3 2x + 2y + z = 2x − y + z = −1 1 4 4x + 3y + 5z = 5y + 3z = 2x − 4y + z = −7 x − 2y + 2z = −2 3 −x + 4y − 2z = x − 3y − 4z = 3 3x + 4y − z = 13 2x − 19y − 19z = 2 8.1 Systems of Linear Equations: Gaussian Elimination 563 21. 23. 25.          x + y + z = 4 2x − 4y − z = −1 x − y = 2 2x − 3y + z = −1 4x − 4y + 4z = −13 6x − 5y + 7z = −25 x1 − x3 = −2 2x2 − x4 = 0 x1 − 2x2 + x3 = 0 −x3 + x4 = 1 22. 24. 26.          x − y + z = 8 3x + 3y − 9z = −6 7x − 2
y + 5z = 39 2x1 + x2 − 12x3 − x4 = 16 −x1 + x2 + 12x3 − 4x4 = −5 3x1 + 2x2 − 16x3 − 3x4 = 25 x1 + 2x2 − 5x4 = 11 x1 − x2 − 5x3 + 3x4 = −1 x1 + x2 + 5x3 − 3x4 = 0 x2 + 5x3 − 3x4 = 1 x1 − 2x2 − 10x3 + 6x4 = −1 27. Find two other forms of the parametric solution to Exercise 11 above by reorganizing the equations so that x or y can be the free variable. 28. A local buﬀet charges $7.50 per person for the basic buﬀet and $9.25 for the deluxe buﬀet (which includes crab legs.) If 27 diners went out to eat and the total bill was $227.00 before taxes, how many chose the basic buﬀet and how many chose the deluxe buﬀet? 29. At The Old Home Fill’er Up and Keep on a-Truckin’ Cafe, Mavis mixes two diﬀerent types of coﬀee beans to produce a house blend. The ﬁrst type costs $3 per pound and the second costs $8 per pound. How much of each type does Mavis use to make 50 pounds of a blend which costs $6 per pound? 30. Skippy has a total of $10,000 to split between two investments. One account oﬀers 3% simple interest, and the other account oﬀers 8% simple interest. For tax reasons, he can only earn $500 in interest the entire year. How much money should Skippy invest in each account to earn $500 in interest for the year? 31. A 10% salt solution is to be mixed with pure water to produce 75 gallons of a 3% salt solution. How much of each are needed? 32. At The Crispy Critter’s Head Shop and Patchouli Emporium along with their dried up weeds, sunﬂower seeds and astrological postcards they sell an herbal tea blend. By weight, Type I herbal tea is 30% peppermint, 40% rose hips and
30% chamomile, Type II has percents 40%, 20% and 40%, respectively, and Type III has percents 35%, 30% and 35%, respectively. How much of each Type of tea is needed to make 2 pounds of a new blend of tea that is equal parts peppermint, rose hips and chamomile? 33. Discuss with your classmates how you would approach Exercise 32 above if they needed to use up a pound of Type I tea to make room on the shelf for a new canister. 34. If you were to try to make 100 mL of a 60% acid solution using stock solutions at 20% and 40%, respectively, what would the triangular form of the resulting system look like? Explain. 564 Systems of Equations and Matrices 8.1.2 Answers 1. Consistent independent Solution 6, − 1 2 3. Consistent independent Solution − 16 7, − 62 7 2. Consistent independent Solution − 7 3, −3 4. Consistent independent Solution 49 12, − 25 18 5. Consistent dependent Solution t, 3 for all real numbers t 2 t + 3 7. Inconsistent No solution 6. Consistent dependent Solution (6 − 4t, t) for all real numbers t 8. Inconsistent No solution Because triangular form is not unique, we give only one possible answer to that part of the question. Yours may be diﬀerent and still be correct.          10. 11. 12 + 3z = 15 + 3z = 15 0 = 1 13. x + y + z = −17 y − 3z = 0    14. x − 2y + 3z = y − 11 7 5 z = − 16 5 1 z = Consistent independent Solution (−2, 7) Consistent independent Solution (1, 2, 0) Consistent dependent Solution (−t + 5, −3t + 15, t) for all real numbers t Inconsistent No solution Consistent dependent Solution (−4t − 17, 3t, t) for all real numbers t Consistent independent Solution (2, −1, 1) 8.1 Systems of Linear Equations: Gaussian Elimination 565 15. 16. 17. 18
. 19. 20. 21. 22. 23. 24. 25.                                  0 x + y + 2z = 4 y − 7z = 17 z = −2 x − 2y + 2z = − − 3y − 4z = 3 13 z = 4 13 0 = 0 y + 11 − 2z = − = − 11 2 0 0 = Consistent independent Solution (1, 3, −2) Inconsistent no solution Consistent independent Solution (1, 3, −2) Consistent independent Solution −3, 1 2, 1 Consistent independent Solution 1 3, 2 3, 1 Consistent dependent Solution 19 13, − 11 for all real numbers t 13 t + 51 13 t + 4 13, t Inconsistent no solution Consistent independent Solution (4, −3, 1) Consistent dependent Solution −2t − 35 for all real numbers t 4, −t − 11 2, t x1 + 2 3 x3 − x4 = 25 3 x2 − 16 3 x2 + 4x3 − 3x4 = 2 0 = 0 0 = 0 x1 − x3 = −2 x2 − 1 2 x4 = 0 x3 − 1 2 x4 = x4 = 1 4 Consistent dependent Solution (8s − t + 7, −4s + 3t + 2, s, t) for all real numbers s and t Consistent independent Solution (1, 2, 3, 4) 566 Systems of Equations and Matrices 26.    x1 − x2 − 5x3 + 3x4 = −1 1 x2 + 5x3 − 3x4 = 2 1 0
= 0 0 = Inconsistent No solution 27. If x is the free variable then the solution is (t, 3t, −t + 5) and if y is the free variable then the solution is 1 3 t, t, − 1 3 t + 5. 28. 13 chose the basic buﬀet and 14 chose the deluxe buﬀet. 29. Mavis needs 20 pounds of $3 per pound coﬀee and 30 pounds of $8 per pound coﬀee. 30. Skippy needs to invest $6000 in the 3% account and $4000 in the 8% account. 31. 22.5 gallons of the 10% solution and 52.5 gallons of pure water. 32. 4 3 − 1 2 t pounds of Type I, 2 3 − 1 2 t pounds of Type II and t pounds of Type III where 0 ≤ t ≤ 4 3. 8.2 Systems of Linear Equations: Augmented Matrices 567 8.2 Systems of Linear Equations: Augmented Matrices In Section 8.1 we introduced Gaussian Elimination as a means of transforming a system of linear equations into triangular form with the ultimate goal of producing an equivalent system of linear equations which is easier to solve. If we take a step back and study the process, we see that all of our moves are determined entirely by the coeﬃcients of the variables involved, and not the variables themselves. Much the same thing happened when we studied long division in Section 3.2. Just as we developed synthetic division to streamline that process, in this section, we introduce a similar bookkeeping device to help us solve systems of linear equations. To that end, we deﬁne a matrix as a rectangular array of real numbers. We typically enclose matrices with square brackets, ‘[ ’ and ‘ ]’, and we size matrices by the number of rows and columns they have. For example, the size (sometimes called the dimension) of 3 0 −1 10 2 −5 is 2 × 3 because it has 2 rows and 3 columns. The individual numbers in a matrix are called its entries and are usually labeled with double subscripts: the ﬁrst tells which row the element is in and the second tells which column it is in. The rows are numbered from top to bottom and the columns are numbered from left to right. Matrices themselves are usually denoted by uppercase letters (
A, B, C, etc.) while their entries are usually denoted by the corresponding letter. So, for instance, if we have A = 3 0 −1 10 2 −5 then a11 = 3, a12 = 0, a13 = −1, a21 = 2, a22 = −5, and a23 = 10. We shall explore matrices as mathematical objects with their own algebra in Section 8.3 and introduce them here solely as a bookkeeping device. Consider the system of linear equations from number 2 in Example 8.1.2    2x + 3y − z = 1 (E1) 10x − z = 2 (E2) (E3) 4x − 9y + 2z = 5 We encode this system into a matrix by assigning each equation to a corresponding row. Within that row, each variable and the constant gets its own column, and to separate the variables on the left hand side of the equation from the constants on the right hand side, we use a vertical bar, \|. Note that in E2, since y is not present, we record its coeﬃcient as 0. The matrix associated with this system is (E1) → (E2) → (E3) →   x 2 10 4 −9 y c z 3 −1 1 0 −1 2 2 5   568 Systems of Equations and Matrices This matrix is called an augmented matrix because the column containing the constants is appended to the matrix containing the coeﬃcients.1 To solve this system, we can use the same kind operations on the rows of the matrix that we performed on the equations of the system. More speciﬁcally, we have the following analog of Theorem 8.1 below. Theorem 8.2. Row Operations: Given an augmented matrix for a system of linear equations, the following row operations produce an augmented matrix which corresponds to an equivalent system of linear equations. Interchange any two rows. Replace a row with a nonzero multiple of itself.a Replace a row with itself plus a nonzero multiple of another row.b aThat is, the row obtained by multiplying each entry in the row by the same nonzero number. bWhere we add entries in corresponding columns. As a demonstration of the moves in Theorem 8.2, we revisit some of the steps that were used
in solving the systems of linear equations in Example 8.1.2 of Section 8.1. The reader is encouraged to perform the indicated operations on the rows of the augmented matrix to see that the machinations are identical to what is done to the coeﬃcients of the variables in the equations. We ﬁrst see a demonstration of switching two rows using the ﬁrst step of part 1 in Example 8.1.2.    3x − y + z = 3 (E1) (E2) 2x − 4y + 3z = 16 x − y + z = 5 (E3) Switch E1 and E3 −−−−−−−−−−−→   3 −1 2 −4 1 −1   1 3 3 16 5 1 Switch R1 and R3 −−−−−−−−−−−→      x − y + z = 5 (E1) (E2) 2x − 4y + 3z = 16 3x − y + z = 3 (E3) 1 −1 2 −4 3 −1   5 1 3 16 3 1 Next, we have a demonstration of replacing a row with a nonzero multiple of itself using the ﬁrst step of part 3 in Example 8.1.2.    3x1 + x2 + x4 = 6 (E1) 2x1 + x2 − x3 = 4 (E2) (E3) x2 − 3x3 − 2x4 = 0 Replace E1 with 1 3 E1 −−−−−−−−−−−−−→   3 2 0 1 6 1 0 1 −1 0 4 1 −3 −2 0   Replace R1 with 1 3 R1 −−−−−−−−−−−−−→      3 x2 + 1 (E1) x1 + 1 3 x4 = 2 2x1 + x2 − x3 = 4 (E2) (E3) x2 − 3x3 − 2x4 = 1 0 4 1 −3
−2 0   Finally, we have an example of replacing a row with itself plus a multiple of another row using the second step from part 2 in Example 8.1.2. 1We shall study the coeﬃcient and constant matrices separately in Section 8.3. 8.2 Systems of Linear Equations: Augmented Matrices 569    E1) 2 10x − z = 2 (E2) (E3) 4x − 9y + 2z = 5 Replace E2 with −10E1 + E2 −−−−−−−−−−−−−−−−−−→ Replace E3 with −4E1 + E3   3 1 10 4 −9 2 − 1 2 0 −1 2   1 2 2 5 Replace R2 with −10R1 + R2 −−−−−−−−−−−−−−−−−−→ Replace R3 with −4R1 + R3      2 y − 1 (E1) x + 3 (E2) (E3) 1 2 −15y + 4z = −3 −15y + 4z = 15 0 −15   1 2 2 4 −3 3 4 The matrix equivalent of ‘triangular form’ is row echelon form. The reader is encouraged to refer to Deﬁnition 8.3 for comparison. Note that the analog of ‘leading variable’ of an equation is ‘leading entry’ of a row. Speciﬁcally, the ﬁrst nonzero entry (if it exists) in a row is called the leading entry of that row. Deﬁnition 8.4. A matrix is said to be in row echelon form provided all of the following conditions hold: 1. The ﬁrst nonzero entry in each row is 1. 2. The leading 1 of a given row must be to the right of the leading 1 of the row above it. 3. Any row of all zeros cannot be placed above a row with nonzero entries. To solve a system of a linear equations using an augmented matrix, we encode the system into an augmented matrix and apply Gaussian Elimination
to the rows to get the matrix into row-echelon form. We then decode the matrix and back substitute. The next example illustrates this nicely. Example 8.2.1. Use an augmented matrix to transform the following system of linear equations into triangular form. Solve the system.    3x − y + z = 8 x + 2y − z = 4 2x + 3y − 4z = 10 Solution. We ﬁrst encode the system into an augmented matrix.    3x − y + z = 8 x + 2y − z = 4 2x + 3y − 4z = 10 Encode into the matrix −−−−−−−−−−−−−−→   3 −1 1 2 1 8 2 −1 4 3 −4 10   Thinking back to Gaussian Elimination at an equations level, our ﬁrst order of business is to get x in E1 with a coeﬃcient of 1. At the matrix level, this means getting a leading 1 in R1. This is in accordance with the ﬁrst criteria in Deﬁnition 8.4. To that end, we interchange R1 and R2.   3 −1 1 2 8 1 2 −1 4 3 −4 10   Switch R1 and R2 −−−−−−−−−−−→   1 3 −1 2 2 −1 4 8 1 3 −4 10   570 Systems of Equations and Matrices Our next step is to eliminate the x’s from E2 and E3. From a matrix standpoint, this means we need 0’s below the leading 1 in R1. This guarantees the leading 1 in R2 will be to the right of the leading 1 in R1 in accordance with the second requirement of Deﬁnition 8.4.   1 3 −1 2 2 −1 4 1 8 3 −4 10   Replace R2 with −3R1 + R2 −−−−−−−−−−−−−−−−−→ Replace R3 with −2R1 + R3   2
−1 1 0 −7 0 −1 −2 4 4 −4 2   Now we repeat the above process for the variable y which means we need to get the leading entry in R2 to be 1.   2 −1 1 0 −7 0 −1 −2 4 4 −4 2   Replace R2 with − 1 7 R2 −−−−−−−−−−−−−−→   2 −1 1 1 − 4 0 7 0 −1 −2   4 4 7 2 To guarantee the leading 1 in R3 is to the right of the leading 1 in R2, we get a 0 in the second column of R3.   2 −1 1 1 − 4 0 7 0 −1 −2   4 4 7 2 Replace R3 with R2 + R3 −−−−−−−−−−−−−−−−→    1 0 0 2 −1 1 − 4 7 0 − 18 7    4 4 7 18 7 Finally, we get the leading entry in R3 to be 1.    1 0 0 2 −1 1 − 4 7 0 − 18 7    4 4 7 18 7 Replace R3 with − 7 18 R3 −−−−−−−−−−−−−−−→   1 0 0 2 −1 0   Decoding from the matrix gives a system in triangular form    1 0 0 2 −1 0 We get z = −1 ﬁnal answer of (3, 0, −1). We leave it to the reader to check. Decode from the matrix −−−−−−−−−−−−−−→ 7 (−1) + 4     x + 2y − 1 7 = 0 and x = −2y + z + 4 = −2(0) + (−1) + 4 = 3 for a As part of Gaussian Elimination, we used row operations to obtain 0’s beneath each leading 1 to put the matrix into row eche
lon form. If we also require that 0’s are the only numbers above a leading 1, we have what is known as the reduced row echelon form of the matrix. Deﬁnition 8.5. A matrix is said to be in reduced row echelon form provided both of the following conditions hold: 1. The matrix is in row echelon form. 2. The leading 1s are the only nonzero entry in their respective columns. 8.2 Systems of Linear Equations: Augmented Matrices 571 Of what signiﬁcance is the reduced row echelon form of a matrix? To illustrate, let’s take the row echelon form from Example 8.2.1 and perform the necessary steps to put into reduced row echelon form. We start by using the leading 1 in R3 to zero out the numbers in the rows above it.   1 0 0 2 −1 0   Replace R1 with R3 + R1 −−−−−−−−−−−−−−−−−→ Replace R2 with 4 7 R3 + R2   1   Finally, we take care of the 2 in R1 above the leading 1 in R21   Replace R1 with −2R2 + R1 −−−−−−−−−−−−−−−−−→   1   To our surprise and delight, when we decode this matrix, we obtain the solution instantly without having to deal with any back-substitution at all1   Decode from the matrix −−−−−−−−−−−−−−→    3 x = 0 y = z = −1 Note that in the previous discussion, we could have started with R2 and used it to get a zero above its leading 1 and then done the same for the leading 1 in R3. By starting with R3, however, we get more zeros ﬁrst, and the more zeros there are, the faster the remaining calculations will be.2 It is also worth noting that while a matrix has several3 row echelon forms, it has only one reduced row echelon form. The process by which we
have put a matrix into reduced row echelon form is called Gauss-Jordan Elimination. Example 8.2.2. Solve the following system using an augmented matrix. Use Gauss-Jordan Elimination to put the augmented matrix into reduced row echelon form.    x2 − 3x1 + x4 = 2 2x1 + 4x3 = 5 4x2 − x4 = 3 Solution. We ﬁrst encode the system into a matrix. (Pay attention to the subscripts!)    x2 − 3x1 + x4 = 2 2x1 + 4x3 = 5 4x2 − x4 = 3 Encode into the matrix −−−−−−−−−−−−−−→   −1 3   Next, we get a leading 1 in the ﬁrst column of R1.   −1 3   Replace R1 with − 1 3 R1 −−−−−−−−−−−−−−→   1 3   2Carl also ﬁnds starting with R3 to be more symmetric, in a purely poetic way. 3inﬁnite, in fact 572 Systems of Equations and Matrices Now we eliminate the nonzero entry below our leading 11 3   Replace R2 with −2R1 + R2 −−−−−−−−−−−−−−−−−→    1 − 1 3 2 0 3 4 0 We proceed to get a leading 1 in R2. 0 − 1 2 4 3 0 −1 3 − 2    3 19 1 3 − 2    Replace R2 with 3 2 R2 −−−−−−−−−−−−−→    1 3 19 2 3    3 19 3 3 We now zero out the entry below the leading 1 in R21 3 19 2 3    Replace R3 with −4R2 + R3 −−−−−−−−−
−−−−−−−−→ Next, it’s time for a leading 1 in R3.   1 − 1 3 3 19 1 0 2 0 −24 −5 −35 0 0 − 1 3 − 2 1 6   Replace R3 with − 1 24 R3 −−−−−−−−−−−−−−−→      1 − 1 3 3 19 1 0 2 0 −24 −5 −35 24 3 19 2 35 24    The matrix is now in row echelon form. To get the reduced row echelon form, we start with the last leading 1 we produced and work to get 0’s above it 24 3 19 2 35 24    Replace R2 with −6R3 + R2 −−−−−−−−−−−−−−−−−→ Lastly, we get a 0 above the leading 1 of R2 24 3 − 2 3 3 4 35 24    Replace R1 with 1 3 R2 + R1 −−−−−−−−−−−−−−−−−→    At last, we decode to get 24 12 − 24       3 3 4 35 24 12 3 4 35 24    1 0 0 0 1 0 12 − 24    Decode from the matrix −−−−−−−−−−−−−−→    x1 − 5 x2 − 1 x3 + 5 12 x4 = − 5 4 x4 = 24 x4 = 12 3 4 35 24 12 3 4 35 24 We have that x4 is free and we assign it the parameter t. We obtain x3 = − 5 and x1 = 5 the reader to check. 12. Our solution is 5 4 t + 3 24 t + 35 4, 24, t : −∞ < t < ∞ and leave it to 24, x2 = 1 24 t + 35 12 t − 5 12 12, 1 8.2 Systems of Linear Equ
ations: Augmented Matrices 573 Like all good algorithms, putting a matrix in row echelon or reduced row echelon form can easily be programmed into a calculator, and, doubtless, your graphing calculator has such a feature. We use this in our next example. Example 8.2.3. Find the quadratic function passing through the points (−1, 3), (2, 4), (5, −2). Solution. According to Deﬁnition 2.5, a quadratic function has the form f (x) = ax2 +bx+c where a = 0. Our goal is to ﬁnd a, b and c so that the three given points are on the graph of f. If (−1, 3) is on the graph of f, then f (−1) = 3, or a(−1)2 + b(−1) + c = 3 which reduces to a − b + c = 3, an honest-to-goodness linear equation with the variables a, b and c. Since the point (2, 4) is also on the graph of f, then f (2) = 4 which gives us the equation 4a + 2b + c = 4. Lastly, the point (5, −2) is on the graph of f gives us 25a + 5b + c = −2. Putting these together, we obtain a system of three linear equations. Encoding this into an augmented matrix produces    a − b + c = 3 4 4a + 2b + c = 25a + 5b + c = −2 Encode into the matrix −−−−−−−−−−−−−−→   1 −1 2 4 5 25   1 3 1 4 1 −2 18, b = 13 Using a calculator,4 we ﬁnd a = − 7 9. Hence, the one and only quadratic which ﬁts the bill is f (x) = − 7 18 x + 37 9. To verify this analytically, we see that f (−1) = 3, f (2) = 4, and f (5) = −2. We can use the calculator to check our solution as well by plotting the three data points and the function f. 18 and c = 37 18 x2 + 13 The graph of f
(x) = − 7 18 x + 37 with the points (−1, 3), (2, 4) and (5, −2) 18 x2 + 13 9 4We’ve tortured you enough already with fractions in this exposition! 574 Systems of Equations and Matrices 8.2.1 Exercises In Exercises 1 - 6, state whether the given matrix is in reduced row echelon form, row echelon form only or in neither of those forms. 1 0 3 0 1 3 1.   4. 5. 3 −1 2 −4 1 −1   1 3 3 16. 6   In Exercises 7 - 12, the following matrices are in reduced row echelon form. Determine the solution of the corresponding system of linear equations or state that the system is inconsistent. 1 0 −2 7 0 1 7.   10. 11. 1 0 0 −3 20 0 1 0 0 0 1 19   1 0 0 0 0 −8 1 0 0 1 7 4 −. 124 0 0 9 −3 20 0     In Exercises 13 - 26, solve the following systems of linear equations using the techniques discussed in this section. Compare and contrast these techniques with those you used to solve the systems in the Exercises in Section 8.1. 13. −5x + y = 17 x + y = 5          15. 17. 19. 4x − y + z = 5 2y + 6z = 30 x + z = 5 3x − 2y + z = −5 x + 3y − z = 12 0 x + y + 2z = x − y + z = −4 −3x + 2y + 4z = −5 x − 5y + 2z = −18             14. 16. 18. 20. x + y + z = 3 2x − y + z = 0 −3x + 5y + 7z = 7 x − 2y + 3z = 7 −
3x + y + 2z = −5 3 2x + 2y + z = 2x − y + z = −1 1 4 4x + 3y + 5z = 5y + 3z = 2x − 4y + z = −7 x − 2y + 2z = −2 3 −x + 4y − 2z = 8.2 Systems of Linear Equations: Augmented Matrices 575          21. 23. 25. 2x − y + z = 1 2x + 2y − z = 1 3x + 6y + 4z = 9 x + y + z = 4 2x − 4y − z = −1 x − y = 2 2x − 3y + z = −1 4x − 4y + 4z = −13 6x − 5y + 7z = −25 22. 24. 26.          x − 3y − 4z = 3 3x + 4y − z = 13 2x − 19y − 19z = 2 x − y + z = 8 3x + 3y − 9z = −6 7x − 2y + 5z = 39 x1 − x3 = −2 2x2 − x4 = 0 x1 − 2x2 + x3 = 0 −x3 + x4 = 1 27. It’s time for another meal at our local buﬀet. This time, 22 diners (5 of whom were children) feasted for $162.25, before taxes. If the kids buﬀet is $4.50, the basic buﬀet is $7.50, and the deluxe buﬀet (with crab legs) is $9.25, ﬁnd out how many diners chose the deluxe buﬀet. 28. Carl wants to make a party mix consisting of almonds (which cost $7 per pound), cashews (which cost $5 per pound), and peanuts (which cost $2 per pound.) If he wants to make a 10 pound mix with a budget of $35, what are the possible combinations almonds
, cashews, and peanuts? (You may ﬁnd it helpful to review Example 8.1.3 in Section 8.1.) 29. Find the quadratic function passing through the points (−2, 1), (1, 4), (3, −2) 30. At 9 PM, the temperature was 60◦F; at midnight, the temperature was 50◦F; and at 6 AM, the temperature was 70◦F. Use the technique in Example 8.2.3 to ﬁt a quadratic function to these data with the temperature, T, measured in degrees Fahrenheit, as the dependent variable, and the number of hours after 9 PM, t, measured in hours, as the independent variable. What was the coldest temperature of the night? When did it occur? 31. The price for admission into the Stitz-Zeager Sasquatch Museum and Research Station is $15 for adults and $8 for kids 13 years old and younger. When the Zahlenreich family visits the museum their bill is $38 and when the Nullsatz family visits their bill is $39. One day both families went together and took an adult babysitter along to watch the kids and the total admission charge was $92. Later that summer, the adults from both families went without the kids and the bill was $45. Is that enough information to determine how many adults and children are in each family? If not, state whether the resulting system is inconsistent or consistent dependent. In the latter case, give at least two plausible solutions. 32. Use the technique in Example 8.2.3 to ﬁnd the line between the points (−3, 4) and (6, 1). How does your answer compare to the slope-intercept form of the line in Equation 2.3? 33. With the help of your classmates, ﬁnd at least two diﬀerent row echelon forms for the matrix 1 2 3 4 12 8 576 Systems of Equations and Matrices 8.2.2 Answers 1. Reduced row echelon form 2. Neither 3. Row echelon form only 4. Reduced row echelon form 5. Reduced row echelon form 6. Row echelon form only 7. (−2, 7) 9. (−3t + 4, −6t − 6, 2, t) for all real numbers t 8. (−3, 20, 19) 10
. Inconsistent 11. (8s − t + 7, −4s + 3t + 2, s, t) for all real numbers s and t 12. (−9t − 3, 4t + 20, t) for all real numbers t 13. (−2, 7) 15. (−t + 5, −3t + 15, t) for all real numbers t 17. (1, 3, −2) 19. (1, 3, −2) 14. (1, 2, 0) 16. (2, −1, 1) 18. Inconsistent 20. −3, 1 2, 1 21. 1 3, 2 3, 1 22. 19 13, t 13, − 11 13 t + 51 for all real numbers t 13 t + 4 23. Inconsistent 25. −2t − 35 2, t for all real numbers t 4, −t − 11 24. (4, −3, 1) 26. (1, 2, 3, 4) 27. This time, 7 diners chose the deluxe buﬀet. 28. If t represents the amount (in pounds) of peanuts, then we need 1.5t − 7.5 pounds of almonds and 17.5 − 2.5t pounds of cashews. Since we can’t have a negative amount of nuts, 5 ≤ t ≤ 7. 29. f (x) = − 4 30. T (t) = 20 5 x2 + 1 27 t2 − 50 5 x + 23 9 t + 60. Lowest temperature of the evening 595 5 12 ≈ 49.58◦F at 12:45 AM. 8.2 Systems of Linear Equations: Augmented Matrices 577 31. Let x1 and x2 be the numbers of adults and children, respectively, in the Zahlenreich family and let x3 and x4 be the numbers of adults and children, respectively, in the Nullsatz family. The system of equations determined by the given information is   15x1 + 8x2 = 38 15x3 + 8x4 = 39 15x1 + 8x2 + 15x3 + 8x4 = 77 15x1 + 15x3 = 45  We subtracted the cost of the babysitter in E3 so the constant is 77, not 92
. This system is 5, t. Our variables repreconsistent dependent and its solution is 8 15 t + 2 sent numbers of adults and children so they must be whole numbers. Running through the values t = 0, 1, 2, 3, 4 yields only one solution where all four variables are whole numbers; t = 3 gives us (2, 1, 1, 3). Thus there are 2 adults and 1 child in the Zahlenreichs and 1 adult and 3 kids in the Nullsatzs. 5, −t + 4, − 8 15 t + 13 578 Systems of Equations and Matrices 8.3 Matrix Arithmetic In Section 8.2, we used a special class of matrices, the augmented matrices, to assist us in solving systems of linear equations. In this section, we study matrices as mathematical objects of their own accord, temporarily divorced from systems of linear equations. To do so conveniently requires some more notation. When we write A = [aij]m×n, we mean A is an m by n matrix1 and aij is the entry found in the ith row and jth column. Schematically, we have j counts columns from left to right −−−−−−−−−−−−−−−→ a1n a11 a12 a2n a21 a22......... · · · amn am1 am2 · · · · · ·      counts rows from top to bottom With this new notation we can deﬁne what it means for two matrices to be equal. Deﬁnition 8.6. Matrix Equality: Two matrices are said to be equal if they are the same size and their corresponding entries are equal. More speciﬁcally, if A = [aij]m×n and B = [bij]p×r, we write A = B provided 1. m = p and n = r 2. aij = bij for all 1 ≤ i ≤ m and all 1 ≤ j ≤ n. Essentially, two matrices are equal if they are the same size and they have the same numbers in the same spots.2 For example, the two 2 × 3 matrices below are, despite appearances, equal. 0 −2 9 25 117 −3 = √ 3 ln(1) e2 ln(3) −8 12
52/3 32 · 13 log(0.001) Now that we have an agreed upon understanding of what it means for two matrices to equal each other, we may begin deﬁning arithmetic operations on matrices. Our ﬁrst operation is addition. Deﬁnition 8.7. Matrix Addition: Given two matrices of the same size, the matrix obtained by adding the corresponding entries of the two matrices is called the sum of the two matrices. More speciﬁcally, if A = [aij]m×n and B = [bij]m×n, we deﬁne A + B = [aij]m×n + [bij]m×n = [aij + bij]m×n As an example, consider the sum below. 1Recall that means A has m rows and n columns. 2Critics may well ask: Why not leave it at that? Why the need for all the notation in Deﬁnition 8.6? It is the authors’ attempt to expose you to the wonderful world of mathematical precision. 8.3 Matrix Arithmetic 579   3 2 4 −1 0 −7    +  −1 4 −5 −3 1 8    =  2 + (−1) 4 + (−5) 0 + 8 3 + 4 (−1) + (−3) (−71 −4 8 −6 It is worth the reader’s time to think what would have happened had we reversed the order of the summands above. As we would expect, we arrive at the same answer. In general, A + B = B + A for matrices A and B, provided they are the same size so that the sum is deﬁned in the ﬁrst place. This is the commutative property of matrix addition. To see why this is true in general, we appeal to the deﬁnition of matrix addition. Given A = [aij]m×n and B = [bij]m×n, A + B = [aij]m×n + [bij]m×n = [aij + bij]m×n = [bij + aij]m×n = [bij]m×n
+ [aij]m×n = B + A where the second equality is the deﬁnition of A + B, the third equality holds by the commutative law of real number addition, and the fourth equality is the deﬁnition of B + A. In other words, matrix addition is commutative because real number addition is. A similar argument shows the associative property of matrix addition also holds, inherited in turn from the associative law of real number addition. Speciﬁcally, for matrices A, B, and C of the same size, (A + B) + C = A + (B + C). In other words, when adding more than two matrices, it doesn’t matter how they are grouped. This means that we can write A + B + C without parentheses and there is no ambiguity as to what this means.3 These properties and more are summarized in the following theorem. Theorem 8.3. Properties of Matrix Addition Commutative Property: For all m × n matrices, A + B = B + A Associative Property: For all m × n matrices, (A + B) + C = A + (B + C) Identity Property: If 0m×n is the m × n matrix whose entries are all 0, then 0m×n is called the m × n additive identity and for all m × n matrices A A + 0m×n = 0m×n + A = A Inverse Property: For every given m × n matrix A, there is a unique matrix denoted −A called the additive inverse of A such that A + (−A) = (−A) + A = 0m×n The identity property is easily veriﬁed by resorting to the deﬁnition of matrix addition; just as the number 0 is the additive identity for real numbers, the matrix comprised of all 0’s does the same job for matrices. To establish the inverse property, given a matrix A = [aij]m×n, we are looking for a matrix B = [bij]m×n so that A + B = 0m×n. By the deﬁnition of matrix addition, we must have that aij + bij = 0 for all i and j. Solving, we get bij = −aij. Hence, given a matrix A
, its additive inverse, which we call −A, does exist and is unique and, moreover, is given by the formula: −A = [−aij]m×n. The long and short of this is: to get the additive inverse of a matrix, 3A technical detail which is sadly lost on most readers. 580 Systems of Equations and Matrices take additive inverses of each of its entries. With the concept of additive inverse well in hand, we may now discuss what is meant by subtracting matrices. You may remember from arithmetic that a − b = a + (−b); that is, subtraction is deﬁned as ‘adding the opposite (inverse).’ We extend this concept to matrices. For two matrices A and B of the same size, we deﬁne A − B = A + (−B). At the level of entries, this amounts to A − B = A + (−B) = [aij]m×n + [−bij]m×n = [aij + (−bij)]m×n = [aij − bij]m×n Thus to subtract two matrices of equal size, we subtract their corresponding entries. Surprised? Our next task is to deﬁne what it means to multiply a matrix by a real number. Thinking back to arithmetic, you may recall that multiplication, at least by a natural number, can be thought of as ‘rapid addition.’ For example. We know from algebra4 that 3x = x + x + x, so it seems natural that given a matrix A, we deﬁne 3A = A + A + A. If A = [aij]m×n, we have 3A = A + A + A = [aij]m×n + [aij]m×n + [aij]m×n = [aij + aij + aij]m×n = [3aij]m×n In other words, multiplying the matrix in this fashion by 3 is the same as multiplying each entry by 3. This leads us to the following deﬁnition. Deﬁnition 8.8. Scalara Multiplication: We deﬁne the product of a real number and a matrix to be the matrix obtained by multiplying each of its entries by said real number. More speci�
�cally, if k is a real number and A = [aij]m×n, we deﬁne kA = k [aij]m×n = [kaij]m×n aThe word ‘scalar’ here refers to real numbers. ‘Scalar multiplication’ in this context means we are multiplying a matrix by a real number (a scalar). One may well wonder why the word ‘scalar’ is used for ‘real number.’ It has everything to do with ‘scaling’ factors.5 A point P (x, y) in the plane can be represented by its position matrix, P : (x, y) ↔ P = x y Suppose we take the point (−2, 1) and multiply its position matrix by 3. We have 3P = 3 −2 1 = 3(−2) 3(1) = −6 3 which corresponds to the point (−6, 3). We can imagine taking (−2, 1) to (−6, 3) in this fashion as a dilation by a factor of 3 in both the horizontal and vertical directions. Doing this to all points (x, y) in the plane, therefore, has the eﬀect of magnifying (scaling) the plane by a factor of 3. 4The Distributive Property, in particular. 5See Section 1.7. 8.3 Matrix Arithmetic 581 As did matrix addition, scalar multiplication inherits many properties from real number arithmetic. Below we summarize these properties. Theorem 8.4. Properties of Scalar Multiplication Associative Property: For every m × n matrix A and scalars k and r, (kr)A = k(rA). Identity Property: For all m × n matrices A, 1A = A. Additive Inverse Property: For all m × n matrices A, −A = (−1)A. Distributive Property of Scalar Multiplication over Scalar Addition: For every m × n matrix A and scalars k and r, (k + r)A = kA + rA Distributive Property of Scalar Multiplication over Matrix Addition: For all m × n matrices A and B scalars k, k(A + B) = kA + kB Zero Product Property: If A is an m × n matrix and k is a scalar,
then kA = 0m×n if and only if k = 0 or A = 0m×n As with the other results in this section, Theorem 8.4 can be proved using the deﬁnitions of scalar multiplication and matrix addition. For example, to prove that k(A + B) = kA + kB for a scalar k and m × n matrices A and B, we start by adding A and B, then multiplying by k and seeing how that compares with the sum of kA and kB. k(A + B) = k [aij]m×n + [bij]m×n = k [aij + bij]m×n = [k (aij + bij)]m×n = [kaij + kbij]m×n As for kA + kB, we have kA + kB = k [aij]m×n + k [bij]m×n = [kaij]m×n + [kbij]m×n = [kaij + kbij]m×n which establishes the property. The remaining properties are left to the reader. The properties in Theorems 8.3 and 8.4 establish an algebraic system that lets us treat matrices and scalars more or less as we would real numbers and variables, as the next example illustrates. Example 8.3.1. Solve for the matrix A: 3A − 2 −1 5 3 using the deﬁnitions and properties of matrix arithmetic. + 5A = −4 2 6 −2 + 1 3 9 12 −3 39 582 Solution. Systems of Equations and Matrices 3A − 3A + − 3A + (−1) 2 −1 3 5 2 −1 5 3 2 −1 3 5 + 5A = + 5A + 5A = = = 2 −1 3 5 2 −1 5 3 (−1)(−1) (−1)(5) −2 3A + 1 −3 −5 + (−1)(5A) + (−1)(5A) = + ((−1)(5))A = + (−5)A = 3A + (−1) 3A + (−1) 3A + (−1)(2) (−1)(3) 3A + (−5)A + −2 1 −3 −5 (3 + (−5))A + −2 1 −
3 −5 + − −2 1 −3 −5 = = (−2)A + 02×2 = (−2)A = (−2)A = −4 2 6 −2 −4 2 6 −2 −4 2 6 −2 1 3 + + + 9 12 −3 39 1 (9) 1 3 3 (−3) 1 1 3 3 4 3 −1 13 (12) (39) −1 6 5 11 −1 6 5 11 −1 6 5 11 −1 6 5 11 −1 6 5 11 −1 6 5 11 + − −2 1 −3 −5 −1 6 5 11 −1 − (−2) − −2 1 −3 −5 6 − 1 5 − (−3) 11 − (−5) − 1 2 ((−2)A) = − 1 2 − 1 2 (−2) A = 1 5 8 16 1 5 8 16 (1) − 1 (5) 2 (16) (8 1A = A = − 1 2 − 5 2 −4 − 16 2 − 1 2 − 5 2 −4 −8 The reader is encouraged to check our answer in the original equation. 8.3 Matrix Arithmetic 583 While the solution to the previous example is written in excruciating detail, in practice many of the steps above are omitted. We have spelled out each step in this example to encourage the reader to justify each step using the deﬁnitions and properties we have established thus far for matrix arithmetic. The reader is encouraged to solve the equation in Example 8.3.1 as they would any other linear equation, for example: 3a − (2 + 5a) = −4 + 1 We now turn our attention to matrix multiplication - that is, multiplying a matrix by another matrix. Based on the ‘no surprises’ trend so far in the section, you may expect that in order to multiply two matrices, they must be of the same size and you ﬁnd the product by multiplying the corresponding entries. While this kind of product is used in other areas of mathematics,6 we deﬁne matrix multiplication to serve us in solving systems of linear equations. To that end, we begin by deﬁning the product of a row and a column. We motivate the general deﬁnition with an example. Consider the two matrices A and B below. 3 (9). A = 2 −10 0 −8 1 8 −5 9 0 −2
−12   Let R1 denote the ﬁrst row of A and C1 denote the ﬁrst column of B. To ﬁnd the ‘product’ of R1 with C1, denoted R1 · C1, we ﬁrst ﬁnd the product of the ﬁrst entry in R1 and the ﬁrst entry in C1. Next, we add to that the product of the second entry in R1 and the second entry in C1. Finally, we take that sum and we add to that the product of the last entry in R1 and the last entry in C1. Using entry notation, R1·C1 = a11b11 +a12b21 +a13b31 = (2)(3)+(0)(4)+(−1)(5) = 6+0+(−5) = 1. We can visualize this schematically as follows 2 −10 0 −8 1 8 −5 9 0 −2 −12   −−−−−−−−−→ 0 −1 2 a11b11 (2)(3) 3 4 5      −−−−−−−−−→ 0 −1 2 + + a12b21 (0)(4) 3 4 5      −−−−−−−−−→ 0 −1 2 + + a13b31 (−1)(5) 3 4 5      To ﬁnd R2 · C3 where R2 denotes the second row of A and C3 denotes the third column of B, we proceed similarly. We start with ﬁnding the product of the ﬁrst entry of R2 with the ﬁrst entry in C3 then add to it the product of the second entry in R2 with the second entry in C3, and so forth. Using entry notation, we have R2·C3 = a21b13+a22b23+a23b33 = (−10)(2)+(3)(−5)+(5)(−2) = −45. Schematically, 2 −10 0 −8 1 8 −5 9 0 −2
−12   6See this article on the Hadamard Product. 584 Systems of Equations and Matrices −−−−−−−−−→ −10 3 5 2 −5 −2      a21b13 = (−10)(2) = −20 + −−−−−−−−−→ −10 5 3 2 −5 −2      −−−−−−−−−→ −10 3 5 2 −5 −2      a22b23 = (3)(−5) = −15 a23b33 = (5)(−2) = −10 + Generalizing this process, we have the following deﬁnition. Deﬁnition 8.9. Product of a Row and a Column: Suppose A = [aij]m×n and B = [bij]n×r. Let Ri denote the ith row of A and let Cj denote the jth column of B. The product of Ri and Cj, denoted Ri · Cj is the real number deﬁned by Ri · Cj = ai1b1j + ai2b2j +... ainbnj Note that in order to multiply a row by a column, the number of entries in the row must match the number of entries in the column. We are now in the position to deﬁne matrix multiplication. Deﬁnition 8.10. Matrix Multiplication: Suppose A = [aij]m×n and B = [bij]n×r. Let Ri denote the ith row of A and let Cj denote the jth column of B. The product of A and B, denoted AB, is the matrix deﬁned by that is AB = [Ri · Cj]m×r      AB = R1 · C1 R1 · C2 R2 · C1 R2 · C2... R1 · Cr... R2 · Cr......... Rm · C1 Rm · C2... Rm · Cr  
   There are a number of subtleties in Deﬁnition 8.10 which warrant closer inspection. First and foremost, Deﬁnition 8.10 tells us that the ij-entry of a matrix product AB is the ith row of A times the jth column of B. In order for this to be deﬁned, the number of entries in the rows of A must match the number of entries in the columns of B. This means that the number of columns of A must match7 the number of rows of B. In other words, to multiply A times B, the second dimension of A must match the ﬁrst dimension of B, which is why in Deﬁnition 8.10, Am×n is being multiplied by a matrix Bn×r. Furthermore, the product matrix AB has as many rows as A and as many columns of B. As a result, when multiplying a matrix Am×n by a matrix Bn×r, the result is the matrix ABm×r. Returning to our example matrices below, we see that A is a 2 × 3 matrix and B is a 3 × 4 matrix. This means that the product matrix AB is deﬁned and will be a 2 × 4 matrix. A = 2 −10 0 −8 1 8 −5 9 0 −2 −12   7The reader is encouraged to think this through carefully. 8.3 Matrix Arithmetic 585 Using Ri to denote the ith row of A and Cj to denote the jth column of B, we form AB according to Deﬁnition 8.10. AB = R1 · C1 R1 · C2 R1 · C3 R1 · C4 R2 · C1 R2 · C2 R2 · C3 R2 · C4 = 1 7 2 14 −45 6 −4 47 Note that the product BA is not deﬁned, since B is a 3 × 4 matrix while A is a 2 × 3 matrix; B has more columns than A has rows, and so it is not possible to multiply a row of B by a column of A. Even when the dimensions of A and B are compatible such that AB and BA are both deﬁned, the product AB and BA aren’t necessarily equal.8 In other words, AB may not equal
BA. Although there is no commutative property of matrix multiplication in general, several other real number properties are inherited by matrix multiplication, as illustrated in our next theorem. Theorem 8.5. Properties of Matrix Multiplication Let A, B and C be matrices such that all of the matrix products below are deﬁned and let k be a real number. Associative Property of Matrix Multiplication: (AB)C = A(BC) Associative Property with Scalar Multiplication: k(AB) = (kA)B = A(kB) Identity Property: For a natural number k, the k × k identity matrix, denoted Ik, is deﬁned by Ik = [dij]k×k where dij = 1, if i = j 0, otherwise For all m × n matrices, ImA = AIn = A. Distributive Property of Matrix Multiplication over Matrix Addition: A(B ± C) = AB ± AC and (A ± B)C = AC ± BC The one property in Theorem 8.5 which begs further investigation is, without doubt, the multiplicative identity. The entries in a matrix where i = j comprise what is called the main diagonal of the matrix. The identity matrix has 1’s along its main diagonal and 0’s everywhere else. A few examples of the matrix Ik mentioned in Theorem 8.5 are given below. The reader is encouraged to see how they match the deﬁnition of the identity matrix presented there   I2 I3 [1] I1         I4 8And may not even have the same dimensions. For example, if A is a 2 × 3 matrix and B is a 3 × 2 matrix, then AB is deﬁned and is a 2 × 2 matrix while BA is also deﬁned... but is a 3 × 3 matrix! 586 Systems of Equations and Matrices The identity matrix is an example of what is called a square matrix as it has the same number of rows as columns. Note that to in order to verify that the identity matrix acts as a multiplicative identity, some care must be taken depending on the order of the multiplication. For example, take the matrix 2 × 3 matrix A from earlier A = 2 −10 0 −1
5 3 In order for the product IkA to be deﬁned, k = 2; similarly, for AIk to be deﬁned, k = 3. We leave it to the reader to show I2A = A and AI3 = A. In other words, and 1 0 0 1 2 −10 0 −1 5 3 = 2 −10 0 −1 5 3 2 −10 0 −10 0 −1 5 3 While the proofs of the properties in Theorem 8.5 are computational in nature, the notation becomes quite involved very quickly, so they are left to a course in Linear Algebra. The following example provides some practice with matrix multiplication and its properties. As usual, some valuable lessons are to be learned. Example 8.3.2. 1. Find AB for A = −23 −1 46 17 2 −34 and B =   −3 2 1 5 −4 3   2. Find C2 − 5C + 10I2 for C = 1 −2 4 3 3. Suppose M is a 4 × 4 matrix. Use Theorem 8.5 to expand (M − 2I4) (M + 3I4). Solution. 1. We have AB = −23 −1 46 17 2 −34   −3 2 1 5 −. Just as x2 means x times itself, C2 denotes the matrix C times itself. We get 8.3 Matrix Arithmetic 587 C2 − 5C + 10I2 = = = = 1 −2 4 3 1 −2 4 3 2 − 5 1 −2 4 3 1 −2 4 3 + + 10 1 0 0 1 −5 10 −15 −20 10 0 0 10 + + 5 10 −15 −10 −5 −10 10 15 0 0 0 0 3. We expand (M − 2I4) (M + 3I4) with the same pedantic zeal we showed in Example 8.3.1. The reader is encouraged to determine which property of matrix arithmetic is used as we proceed from one step to the next. (M − 2I4) (M + 3I4) = (M − 2I4) M + (M − 2I4) (3I4) = M M − (2I4) M + M (3I4) − (2I4) (3I4) = M 2 − 2 (
I4M ) + 3 (M I4) − 2 (I4 (3I4)) = M 2 − 2M + 3M − 2 (3 (I4I4)) = M 2 + M − 6I4 Example 8.3.2 illustrates some interesting features of matrix multiplication. First note that in part 1, neither A nor B is the zero matrix, yet the product AB is the zero matrix. Hence, the the zero product property enjoyed by real numbers and scalar multiplication does not hold for matrix multiplication. Parts 2 and 3 introduce us to polynomials involving matrices. The reader is encouraged to step back and compare our expansion of the matrix product (M − 2I4) (M + 3I4) in part 3 with the product (x − 2)(x + 3) from real number algebra. The exercises explore this kind of parallel further. As we mentioned earlier, a point P (x, y) in the xy-plane can be represented as a 2 × 1 position matrix. We now show that matrix multiplication can be used to rotate these points, and hence graphs of equations. Example 8.3.3. Let. Plot P (2, −2), Q(4, 0), S(0, 3), and T (−3, −3) in the plane as well as the points RP, RQ, RS, and RT. Plot the lines y = x and y = −x as guides. What does R appear to be doing to these points? 2. If a point P is on the hyperbola x2 − y2 = 4, show that the point RP is on the curve y = 2 x. 588 Systems of Equations and Matrices Solution. For P (2, −2), the position matrix is P = 2 −2, and RP = = √ 2 √ √ √ We have that R takes (2, −2) to (2 2), (0, 3) √ 2 2, 3 is moved to 2). Plotting these in the coordinate plane along with the lines y = x and y = −x, we see that the matrix R is rotating these points counterclockwise by 45◦. 2, 0). Similarly, we ﬁnd (4, 0) is moved to (2 √, and (−3, −3) is moved to (0, −3 − 3 2 RQ RP RS −4 −3 −2
−1 −1 1 2 3 Q x T −2 −3 −4 P RT For a generic point P (x, y) on the hyperbola x2 − y2 = 4, we have RP = = √ which means R takes (x, y) to y = 2 x, we replace x with √, √ 2 2 y and y with √ √ +. To show that this point is on the curve √ 2 2 y and simplify. 8.3 Matrix Arithmetic 589 √ √ = 2 2 x − y2 x2 x2 − y2 Since (x, y) is on the hyperbola x2 − y2 = 4, we know that this last equation is true. Since all of our steps are reversible, this last equation is equivalent to our original equation, which establishes the point is, indeed, on the graph of y = 2 x is a hyperbola, and it is none other than the hyperbola x2 − y2 = 4 rotated counterclockwise by 45◦.9 Below we have the graph of x2 − y2 = 4 (solid line) and y = 2 x. This means the graph of y = 2 x (dashed line) for comparison3 −1 −1 −2 −3 When we started this section, we mentioned that we would temporarily consider matrices as their own entities, but that the algebra developed here would ultimately allow us to solve systems of linear equations. To that end, consider the system    3x − y + z = 8 x + 2y − z = 4 2x + 3y − 4z = 10 In Section 8.2, we encoded this system into the augmented matrix   3 −1 1 2 8 1 2 −1 4 3 −4 10   9See Section 7.5 for more details. 590 Systems of Equations and Matrices Recall that the entries to the left of the vertical line come from the coeﬃcients of the variables in the system, while those on the right comprise the associated constants. For that reason, we may form the coeﬃcient matrix A, the unknowns matrix X and the constant matrix B as below   A = 3 −1 1 2 1 2 −1 3 − 10 x y z We now consider the matrix equation AX = B. AX =
B   3 −1 3 −4 x y z 3x − y + z x + 2y − z 2x + 3y − 4z   =   =         8 4 10 8 4 10   We see that ﬁnding a solution (x, y, z) to the original system corresponds to ﬁnding a solution X for the matrix equation AX = B. If we think about solving the real number equation ax = b, we would simply ‘divide’ both sides by a. Is it possible to ‘divide’ both sides of the matrix equation AX = B by the matrix A? This is the central topic of Section 8.4. 8.3 Matrix Arithmetic 8.3.1 Exercises For each pair of matrices A and B in Exercises 1 - 7, ﬁnd the following, if deﬁned 3A A − 2B −B AB 1. A = 3. A = 2 −3 1 4 −1 3 5 2, B = 5 −3 1 4 A2 BA 2. A = −1 5 −3 6, B = 2 10 1 −7 41 3 −5 11 7 −9 591     5. A = 7. A =  , B = 1 2 3 7 8 9 6. A =   −3 1 −2 4 5 −6  , B = −5 1 8 2 −3 3 −7 5 1 −2 1 −1   , B =  1 2 1 17 33 19 10 19 11   In Exercises 8 - 21, use the matrices 3 2 −5 C = 10 − 11 13 9 0 −5 to compute the following or state that the indicated operation is undeﬁned. 8. 7B − 4A 11. E + D 14. A − 4I2 9. AB 12. ED 15. A2 − B2 10. BA 13. CD + 2I2A 16. (A + B)(A − B) 17. A2 − 5A
− 2I2 18. E2 + 5E − 36I3 19. EDC 20. CDE 22. Let A = a b c d e f E1 = 21. ABCEDI2 0 1 1 0 E2 = 5 0 0 1 E3 = 1 −2 1 0 Compute E1A, E2A and E3A. What eﬀect did each of the Ei matrices have on the rows of A? Create E4 so that its eﬀect on A is to multiply the bottom row by −6. How would you extend this idea to matrices with more than two rows? 592 Systems of Equations and Matrices In Exercises 23 - 29, consider the following scenario. In the small village of Pedimaxus in the country of Sasquatchia, all 150 residents get one of the two local newspapers. Market research has shown that in any given week, 90% of those who subscribe to the Pedimaxus Tribune want to keep getting it, but 10% want to switch to the Sasquatchia Picayune. Of those who receive the Picayune, 80% want to continue with it and 20% want switch to the Tribune. We can express this situation using matrices. Speciﬁcally, let X be the ‘state matrix’ given by X = T P where T is the number of people who get the Tribune and P is the number of people who get the Picayune in a given week. Let Q be the ‘transition matrix’ given by Q = 0.90 0.20 0.10 0.80 such that QX will be the state matrix for the next week. 23. Let’s assume that when Pedimaxus was founded, all 150 residents got the Tribune. (Let’s call this Week 0.) This would mean X = 150 0 Since 10% of that 150 want to switch to the Picayune, we should have that for Week 1, 135 people get the Tribune and 15 people get the Picayune. Show that QX in this situation is indeed QX = 135 15 24. Assuming that the percentages stay the same, we can get to the subscription numbers for Week 2 by computing Q2X. How many people get each paper in Week 2? 25. Explain why the transition matrix does what we want it to do. 26. If the conditions do not change from
week to week, then Q remains the same and we have what’s known as a Stochastic Process10 because Week n’s numbers are found by computing QnX. Choose a few values of n and, with the help of your classmates and calculator, ﬁnd out how many people get each paper for that week. You should start to see a pattern as n → ∞. 27. If you didn’t see the pattern, we’ll help you out. Let Xs = 100 50. Show that QXs = Xs This is called the steady state because the number of people who get each paper didn’t change for the next week. Show that QnX → Xs as n → ∞. 10More speciﬁcally, we have a Markov Chain, which is a special type of stochastic process. 8.3 Matrix Arithmetic 593 28. Now let Show that Qn → S as n → ∞. 29. Show that SY = Xs for any matrix Y of the form Y = y 150 − y This means that no matter how the distribution starts in Pedimaxus, if Q is applied often enough, we always end up with 100 people getting the Tribune and 50 people getting the Picayune. 30. Let z = a + bi and w = c + di be arbitrary complex numbers. Associate z and w with the matrices Z = a b −b a and W = c d −d c Show that complex number addition, subtraction and multiplication are mirrored by the associated matrix arithmetic. That is, show that Z + W, Z − W and ZW produce matrices which can be associated with the complex numbers z + w, z − w and zw, respectively. 31. Let A = 1 2 3 4 and B = 0 −3 2 −5 Compare (A + B)2 to A2 + 2AB + B2. Discuss with your classmates what constraints must be placed on two arbitrary matrices A and B so that both (A + B)2 and A2 + 2AB + B2 exist. When will (A + B)2 = A2 + 2AB + B2? In general, what is the correct formula for (A + B)2? In Exercises 32 - 36, consider the following deﬁnitions. A square matrix is said to be an upper triangular matrix if all of its entries
below the main diagonal are zero and it is said to be a lower triangular matrix if all of its entries above the main diagonal are zero. For example9 0 −5 from Exercises 8 - 21 above is an upper triangular matrix whereas F = 1 0 3 0 is a lower triangular matrix. questions with your classmates. (Zeros are allowed on the main diagonal.) Discuss the following 594 Systems of Equations and Matrices 32. Give an example of a matrix which is neither upper triangular nor lower triangular. 33. Is the product of two n × n upper triangular matrices always upper triangular? 34. Is the product of two n × n lower triangular matrices always lower triangular? 35. Given the matrix A = 1 2 3 4 write A as LU where L is a lower triangular matrix and U is an upper triangular matrix? 36. Are there any matrices which are simultaneously upper and lower triangular? 8.3 Matrix Arithmetic 8.3.2 Answers 1. For A = 2 −3 4 1 and B = 5 −2 8 4 595 3A = 6 −9 12 3 −B = −5 2 −4 −8 A2 = 1 −18 13 6 A − 2B = −8 1 −7 −12 AB = −2 −28 30 21 BA = 8 −23 20 16 2. For A = −1 5 −3 6 and B = 2 10 1 −7 3A = −3 15 −9 18 −B = −2 −10 7 −1 A2 = −14 25 −15 21 A − 2B = −5 −15 4 11 AB = −37 −5 −48 −24 BA = −32 70 4 −29 3. For A = −1 3 5 2 and B = 7 0 8 −3 1 4 3A = −3 9 15 6 −B = −7 0 −8 3 −1 −4 A2 = 16 3 5 19 A − 2B is not deﬁned AB = −16 3 4 29 2 48 BA is not deﬁned 4. For A = 2 4 6 8 and B = −1 3 −5 11 7 −9 3A = 6 12 18 24 −B = 1 −3 5 9 −11 −7 A2 = 28 40 60 88 A − 2B is not deﬁned AB = 26 −30 34 50 −54 58 BA is not deﬁned 596 5. For A =  
3A = 7 8 9     and B = 1 2 3   21 24 27 A2 is not deﬁned AB =   7 14 21 8 16 24 9 18 27   6. For A =   −3 1 −2 4 5 −6   and B = −5 1 8 3A =   3 −6 −9 12 15 −18   A2 is not deﬁned AB is not deﬁned Systems of Equations and Matrices −B = −1 −2 −3 A − 2B is not deﬁned BA = [50] −B = 5 −1 −8 A − 2B is not deﬁned BA = 32 −34 7. For A =   2 −3 3 −7 5 1 −2 1 −1    and B =  1 1 2 17 33 19 10 19 11   3A = A2 =     6 −9 9 −21 15 3 −6 3 −3   −40 −4 23 −10 −4 11 15 21 −36   AB = 8. 7B − 4A =     4 −29 −47 −2 −B =   −1 −2 −1 −17 −33 −19 −10 −19 −11   A − 2B =   0 −7 3 −31 −65 −40 −27 −37 −23   BA =    . AB = −10 1 −20 −1 8.3 Matrix Arithmetic 597 10. BA = 12. ED = −9 −12 1 −2    67 11 3 − 178 3 −72 −30 −40    14. A − 4I2 = −3 2 3 0 16. (A + B)(A − B) = −7 3 46 2 11. E
+ D is undeﬁned 13. CD + 2I2A = 238 3 −126 361 863 5 15 15. A2 − B2 = 17. A2 − 5A − 2I2 = −8 16 3 25 0 0 0 0 18. E2 + 5E − 36I3 =   −30 20 −15 0 −36 0 −36 0 0   19. EDC =    3449 15 − 407 15 − 101 99 6 − 9548 3 −648 −324 −35 −360    20. CDE is undeﬁned 21. ABCEDI2 = − 90749 15 − 156601 15 − 28867 5 − 47033 5 d e f c a b E1 interchanged R1 and R2 of A. 22. E1A = E2A = E3A = d 5a 5b 5c f a − 2d b − 2e c − 2f f e e d E4 = 1 0 0 −6 E2 multiplied R1 of A by 5. E3 replaced R1 in A with R1 − 2R2. 598 Systems of Equations and Matrices 8.4 Systems of Linear Equations: Matrix Inverses We concluded Section 8.3 by showing how we can rewrite a system of linear equations as the matrix equation AX = B where A and B are known matrices and the solution matrix X of the equation corresponds to the solution of the system. In this section, we develop the method for solving such an equation. To that end, consider the system 2x − 3y = 16 3x + 4y = 7 To write this as a matrix equation, we follow the procedure outlined on page 590. We ﬁnd the coeﬃcient matrix A, the unknowns matrix X and constant matrix B to be A = 2 −3 4 3 X = x y B = 16 7 In order to motivate how we solve a matrix equation like AX = B, we revisit solving a similar equation involving real numbers. Consider the equation 3x = 5. To solve, we simply divide both sides by 3 and obtain x = 5 3. How can we go about deﬁning an analogous process for matrices? To answer this question, we solve 3x = 5 again, but this time, we pay
attention to the properties of real numbers being used at each step. Recall that dividing by 3 is the same as multiplying by 1 3 = 3−1, the so-called multiplicative inverse 1 of 3. 3x = 5 3−1(3x) = 3−1(5) Multiply by the (multiplicative) inverse of 3 Associative property of multiplication Inverse property Multiplicative Identity 3−1 · 3 x = 3−1(5) 1 · x = 3−1(5) x = 3−1(5) If we wish to check our answer, we substitute x = 3−1(5) into the original equation 3x 3 3−1(5) 3 · 3−1 (5 Associative property of multiplication? = 5 = 5 Multiplicative Identity Inverse property Thinking back to Theorem 8.5, we know that matrix multiplication enjoys both an associative property and a multiplicative identity. What’s missing from the mix is a multiplicative inverse for the coeﬃcient matrix A. Assuming we can ﬁnd such a beast, we can mimic our solution (and check) to 3x = 5 as follows 1Every nonzero real number a has a multiplicative inverse, denoted a−1, such that a−1 · a = a · a−1 = 1. 8.4 Systems of Linear Equations: Matrix Inverses 599 Solving AX = B Checking our answer AX = B A−1(AX) = A−1B A−1A X = A−1B I2X = A−1B X = A−1B AX A A−1B AA−1 B I2B The matrix A−1 is read ‘A-inverse’ and we will deﬁne it formally later in the section. At this stage, we have no idea if such a matrix A−1 exists, but that won’t deter us from trying to ﬁnd it.2 We want A−1 to satisfy two equations, A−1A = I2 and AA−1 = I2, making A−1 necessarily a 2 × 2 matrix.3 Hence, we assume A−1 has the form A−1 = x1 x2 x3 x4 for real numbers x1, x2, x3 and x4. For reasons which will become clear later, we focus our attention on the equation AA
−1 = I2. We have AA−1 = I2 x1 x2 2 −3 x3 x4 4 3 2x1 − 3x3 2x2 − 3x4 3x1 + 4x3 3x2 + 4x4 = = 1 0 0 1 1 0 0 1 This gives rise to two more systems of equations 2x1 − 3x3 = 1 3x1 + 4x3 = 0 2x2 − 3x4 = 0 3x2 + 4x4 = 1 At this point, it may seem absurd to continue with this venture. After all, the intent was to solve one system of equations, and in doing so, we have produced two more to solve. Remember, the objective of this discussion is to develop a general method which, when used in the correct scenarios, allows us to do far more than just solve a system of equations. If we set about to solve these systems using augmented matrices using the techniques in Section 8.2, we see that not only do both systems have the same coeﬃcient matrix, this coeﬃcient matrix is none other than the matrix A itself. (We will come back to this observation in a moment.) 2Much like Carl’s quest to ﬁnd Sasquatch. 3Since matrix multiplication isn’t necessarily commutative, at this stage, these are two diﬀerent equations. 600 Systems of Equations and Matrices 2x1 − 3x3 = 1 3x1 + 4x3 = 0 2x2 − 3x4 = 0 3x2 + 4x4 = 1 Encode into a matrix −−−−−−−−−−−−−→ Encode into a matrix −−−−−−−−−−−−−→ 3 2 −3 1 4 0 2 −3 0 4 1 3 To solve these two systems, we use Gauss-Jordan Elimination to put the augmented matrices into reduced row echelon form. (We leave the details to the reader.) For the ﬁrst system, we get 2 −3 1 4 0 3 Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ 4 1 0 17 0 1 − 3 17 which gives x1 = 4 17. To solve the second system, we use the exact same row operations, in the same order, to put
its augmented matrix into reduced row echelon form (Think about why that works.) and we obtain 17 and x3 = − 3 2 −3 0 4 1 3 Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ 1 0 0 1 3 17 2 17 which means x2 = 3 17 and x4 = 2 17. Hence, A−1 = x1 x2 x3 x4 = 4 17 − 3 17 3 17 2 17 We can check to see that A−1 behaves as it should by computing AA−1 As an added bonus, AA−1 = 2 −3 4 3 4 17 − 3 17 3 17 2 17 1 0 0 1 = = I2 A−1A = 4 17 − 3 17 3 17 2 17 2 −3 4 3 = 1 0 0 1 = I2 We can now return to the problem at hand. From our discussion at the beginning of the section on page 599, we know X = A−1B = 4 17 − 3 17 3 17 2 17 16 7 = 5 −2 so that our ﬁnal solution to the system is (x, y) = (5, −2). As we mentioned, the point of this exercise was not just to solve the system of linear equations, but to develop a general method for ﬁnding A−1. We now take a step back and analyze the foregoing discussion in a more general context. In solving for A−1, we used two augmented matrices, both of which contained the same entries as A 8.4 Systems of Linear Equations: Matrix Inverses 601 3 2 −3 1 4 0 2 − We also note that the reduced row echelon forms of these augmented matrices can be written as 4 1 0 17 0 1 − 3 17 3 17 2 17 1 0 0 1 I2 I2 = = x1 x3 x2 x4 where we have identiﬁed the entries to the left of the vertical bar as the identity I2 and the entries to the right of the vertical bar as the solutions to our systems. The long and short of the solution process can be summarized as A 1 0 A 0 1 Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ I2 I2 x1 x3
x2 x4 Since the row operations for both processes are the same, all of the arithmetic on the left hand side of the vertical bar is identical in both problems. The only diﬀerence between the two processes is what happens to the constants to the right of the vertical bar. As long as we keep these separated into columns, we can combine our eﬀorts into one ‘super-sized’ augmented matrix and describe the above process as A 1 0 0 1 Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ I2 x1 x2 x3 x4 We have the identity matrix I2 appearing as the right hand side of the ﬁrst super-sized augmented matrix and the left hand side of the second super-sized augmented matrix. To our surprise and delight, the elements on the right hand side of the second super-sized augmented matrix are none other than those which comprise A−1. Hence, we have A I2 Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ I2 A−1 In other words, the process of ﬁnding A−1 for a matrix A can be viewed as performing a series of row operations which transform A into the identity matrix of the same dimension. We can view this process as follows. In trying to ﬁnd A−1, we are trying to ‘undo’ multiplication by the matrix A. The identity matrix in the super-sized augmented matrix [A\|I] keeps a running memory of all of the moves required to ‘undo’ A. This results in exactly what we want, A−1. We are now ready 602 Systems of Equations and Matrices to formalize and generalize the foregoing discussion. We begin with the formal deﬁnition of an invertible matrix. Deﬁnition 8.11. An n × n matrix A is said to be invertible if there exists a matrix A−1, read ‘A inverse’, such that A−1A = AA−1 = In. Note that, as a consequence of our deﬁnition, invertible matrices are square, and as such, the conditions in Deﬁnition 8.11 force the matrix A−1 to be same dimensions as A, that is, n × n. Since
not all matrices are square, not all matrices are invertible. However, just because a matrix is square doesn’t guarantee it is invertible. (See the exercises.) Our ﬁrst result summarizes some of the important characteristics of invertible matrices and their inverses. Theorem 8.6. Suppose A is an n × n matrix. 1. If A is invertible then A−1 is unique. 2. A is invertible if and only if AX = B has a unique solution for every n × r matrix B. The proofs of the properties in Theorem 8.6 rely on a healthy mix of deﬁnition and matrix arithmetic. To establish the ﬁrst property, we assume that A is invertible and suppose the matrices B and C act as inverses for A. That is, BA = AB = In and CA = AC = In. We need to show that B and C are, in fact, the same matrix. To see this, we note that B = InB = (CA)B = C(AB) = CIn = C. Hence, any two matrices that act like A−1 are, in fact, the same matrix.4 To prove the second property of Theorem 8.6, we note that if A is invertible then the discussion on page 599 shows the solution to AX = B to be X = A−1B, and since A−1 is unique, so is A−1B. Conversely, if AX = B has a unique solution for every n × r matrix B, then, in particular, there is a unique solution X0 to the equation AX = In. The solution matrix X0 is our candidate for A−1. We have AX0 = In by deﬁnition, but we need to also show X0A = In. To that end, we note that A (X0A) = (AX0) A = InA = A. In other words, the matrix X0A is a solution to the equation AX = A. Clearly, X = In is also a solution to the equation AX = A, and since we are assuming every such equation as a unique solution, we must have X0A = In. Hence, we have X0A = AX0 = In, so that X0 = A−1 and A is invertible. The foregoing discussion
justiﬁes our quest to ﬁnd A−1 using our super-sized augmented matrix approach A In Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ In A−1 We are, in essence, trying to ﬁnd the unique solution to the equation AX = In using row operations. What does all of this mean for a system of linear equations? Theorem 8.6 tells us that if we write the system in the form AX = B, then if the coeﬃcient matrix A is invertible, there is only one solution to the system − that is, if A is invertible, the system is consistent and independent.5 We also know that the process by which we ﬁnd A−1 is determined completely by A, and not by the 4If this proof sounds familiar, it should. See the discussion following Theorem 5.2 on page 380. 5It can be shown that a matrix is invertible if and only if when it serves as a coeﬃcient matrix for a system of equations, the system is always consistent independent. It amounts to the second property in Theorem 8.6 where the matrices B are restricted to being n × 1 matrices. We note that, owing to how matrix multiplication is deﬁned, being able to ﬁnd unique solutions to AX = B for n × 1 matrices B gives you the same statement about solving such equations for n × r matrices − since we can ﬁnd a unique solution to them one column at a time. 8.4 Systems of Linear Equations: Matrix Inverses 603 constants in B. This answers the question as to why we would bother doing row operations on a super-sized augmented matrix to ﬁnd A−1 instead of an ordinary augmented matrix to solve a system; by ﬁnding A−1 we have done all of the row operations we ever need to do, once and for all, since we can quickly solve any equation AX = B using one multiplication, A−1B. Example 8.4.1. Let A =   1 3 0 −1 1 2   2 5 4 1. Use row operations to ﬁnd A−1. Check your answer by ﬁnding A−1A and AA−1. 2.
Use A−1 to solve the following systems of equations    (a) 3x + y + 2z = 26 −y + 5z = 39 2x + y + 4z = 117    (b) 3x + y + 2z = 4 −y + 5z = 2 2x + y + 4z = 5    (c) 3x + y + 2z = 1 −y + 5z = 0 2x + y + 4z = 0 Solution. 1. We begin with a super-sized augmented matrix and proceed with Gauss-Jordan elimination.   Replace R1 −−−−−−−→ with 1 3 R1 Replace R3 with −−−−−−−−−−→ −2R1 + R3  1 1 3 0 −1 1 2  1 1 3 0 −1 1 0 3  1 0 0  Replace R2 −−−−−−−−→ with (−1)R2  1 0 0 Replace R3 with −−−−−−−−−−→ 3 R2 + R3 − 1  Replace R3 −−−−−−−→ with 3 13 R3   1 1 2 1 1 3 0 −5 13 1 0 3 1 0 3 0 −5 0 1 0 3 0 −1 1 13 1 − 2 13       0 0 1 0 0 1 0 0 3 13 5 1 8 3 1 0 3 0 −5 13 5 0 1 − 2 13 1 0 3 0 −1 1 3 3 1 0 3 0 −1 1 13   13 Replace R1 with − 2 3 R3 + R1 −−−−−−−−−−−−→ Replace R2 with 5R3 + R2    1 0 0 39 − 2 39 − 2 13 − 8 17 1 0 3 1 0 − 10 0 1 − 2 13 13 1 13    13 15 13 3 13 604 Systems of Equations and Matrices    1 0 0 39 − 2 39 − 2 13 − 8 17 1
0 3 1 0 − 10 0 1 − 2 13 13 1 13    Replace R1 with −−−−−−−−−−→ 3 R2 + R1 − 1    2 13 − 7 9 1 0 0 13 0 1 0 − 10 0 0 1 − 2 13 13 − 8 13 1 13    13 15 13 3 13 13 15 13 3 13 We ﬁnd A−1 =    2 13 − 7 13 15 13 3 13 9 13 − 10 − 2 13  13 − 8 13 1 13 A−1A =   and   . To check our answer, we compute 2 13 − 7 9 13 − 10 − 2 13 13 − 8 13 1 13 13 15 13 3 13       1 3 0 −   = I3  AA−1 =    1 3 0 − 13 − 10 − 2 13 13 − 8 13 1 13 2 13 −   = I3  13 15 13 3 13 2. Each of the systems in this part has A as its coeﬃcient matrix. The only diﬀerence between the systems is the constants which is the matrix B in the associated matrix equation AX = B. We solve each of them using the formula X = A−1B. (a) X = A−1B = (b) X = A−1B = (c) X = A−1B =          2 13 − 7 13 − 8 13 − 7 13 − 8 13 − 7 13 − 8 9 13 − 10 − 2 13 9 13 − 10 − 2 13 9 13 − 10 − 2 13 13 1 13 2 13 1 13 2 13 1 13                   26 39 117  13 15 13
3 13 13 15 13 3 13 13 15 13 3 13     =   −39 91   . Our solution is (−39, 91, 26). 26  . We get 5  13, 19 13, 9 13. 5 13 19 13 9 13  . We ﬁnd 9  13, − 10 13, − 2 13.6 9 13 − 10 13 − 2 13 In Example 8.4.1, we see that ﬁnding one inverse matrix can enable us to solve an entire family of systems of linear equations. There are many examples of where this comes in handy ‘in the wild’, and we chose our example for this section from the ﬁeld of electronics. We also take this opportunity to introduce the student to how we can compute inverse matrices using the calculator. 6Note that the solution is the ﬁrst column of the A−1. The reader is encouraged to meditate on this ‘coincidence’. 8.4 Systems of Linear Equations: Matrix Inverses 605 Example 8.4.2. Consider the circuit diagram below.7 We have two batteries with source voltages VB1 and VB2, measured in volts V, along with six resistors with resistances R1 through R6, measured in kiloohms, kΩ. Using Ohm’s Law and Kirchhoﬀ’s Voltage Law, we can relate the voltage supplied to the circuit by the two batteries to the voltage drops across the six resistors in order to ﬁnd the four ‘mesh’ currents: i1, i2, i3 and i4, measured in milliamps, mA. If we think of electrons ﬂowing through the circuit, we can think of the voltage sources as providing the ‘push’ which makes the electrons move, the resistors as obstacles for the electrons to overcome, and the mesh current as a net rate of ﬂow of electrons around the indicated loops. The system of linear equations associated with this circuit is    (R1 + R3) i1 − R3i2 −
R1i4 = VB1 −R3i1 + (R2 + R3 + R4) i2 − R4i3 − R2i4 = 0 −R4i2 + (R4 + R6) i3 − R6i4 = −VB2 −R1i1 − R2i2 − R6i3 + (R1 + R2 + R5 + R6) i4 = 0 1. Assuming the resistances are all 1kΩ, ﬁnd the mesh currents if the battery voltages are (a) VB1 = 10V and VB2 = 5V (b) VB1 = 10V and VB2 = 0V (c) VB1 = 0V and VB2 = 10V (d) VB1 = 10V and VB2 = 10V 2. Assuming VB1 = 10V and VB2 = 5V, ﬁnd the possible combinations of resistances which would yield the mesh currents you found in 1(a). 7The authors wish to thank Don Anthan of Lakeland Community College for the design of this example. VB1R5R1R2R6VB2R3R4i1i2i3i41 606 Solution. Systems of Equations and Matrices 1. Substituting the resistance values into our system of equations, we get    2i1 − i2 − i4 = VB1 −i1 + 3i2 − i3 − i4 = 0 −i2 + 2i3 − i4 = −VB2 −i1 − i2 − i3 + 4i4 = 0 This corresponds to the matrix equation AX = B where     A = 2 −1 −1 0 −1 0 −1 3 −1 −1 2 −1 4 −1 −1 −1     X =         i1 i2 i3 i4 B =     VB1 0 −VB2 0     When we input the
matrix A into the calculator, we ﬁnd from which we have A−1 =     1.625 1.25 1.125 1 1.25 1.125 1.5 1.25 1.25 1.625 To solve the four systems given to us, we ﬁnd X = A−1B where the value of B is determined by the given values of VB1 and VB2 1 (a) B =         10 0 −5 0, 1 (b) B =         10 0 0 0, 1 (c10 0, 1 (d) B =         10 0 10 0 (a) For VB1 = 10V and VB2 = 5V, the calculator gives i1 = 10.625 mA, i2 = 6.25 mA, i3 = 3.125 mA, and i4 = 5 mA. We include a calculator screenshot below for this part (and this part only!) for reference. 8.4 Systems of Linear Equations: Matrix Inverses 607 (b) By keeping VB1 = 10V and setting VB2 = 0V, we are removing the eﬀect of the second battery. We get i1 = 16.25 mA, i2 = 12.5 mA, i3 = 11.25 mA, and i4 = 10 mA. (c) Part (c) is a symmetric situation to part (b) in so much as we are zeroing out VB1 and making VB2 = 10. We ﬁnd i1 = −11.25 mA, i2 = −12.5 mA, i3 = −16.25 mA, and i4 = −10 mA, where the negatives indicate that the current is ﬂowing in the opposite direction as is indicated on the diagram. The reader is encouraged to study the symmetry here, and if need be, hold up a mirror to the diagram to literally ‘see’ what is happening. (d) For V
B1 = 10V and VB2 = 10V, we get i1 = 5 mA, i2 = 0 mA, i3 = −5 mA, and i4 = 0 mA. The mesh currents i2 and i4 being zero is a consequence of both batteries ‘pushing’ in equal but opposite directions, causing the net ﬂow of electrons in these two regions to cancel out. 2. We now turn the tables and are given VB1 = 10V, VB2 = 5V, i1 = 10.625 mA, i2 = 6.25 mA, i3 = 3.125 mA and i4 = 5 mA and our unknowns are the resistance values. Rewriting our system of equations, we get    1.25R2 − 4.375R3 + 3.125R4 = 5.625R1 + 4.375R3 = 10 0 −3.125R4 − 1.875R6 = −5 0 −5.625R1 − 1.25R2 + 5R5 + 1.875R6 = The coeﬃcient matrix for this system is 4 × 6 (4 equations with 6 unknowns) and is therefore not invertible. We do know, however, this system is consistent, since setting all the resistance values equal to 1 corresponds to our situation in problem 1a. This means we have an underdetermined consistent system which is necessarily dependent. To solve this system, we encode it into an augmented matrix 0 4.375 1.25 −4.375     5.25 0 0 0 −5.625 −1.25 0 3.125 0 −3.125 0 0 0 0 0 10 0 0 0 −1.875 −5 0 1.875 5     and use the calculator to write in reduced row echelon form 608 Systems of Equations and Matrices     1 0 0 0 0.7 0 1 −3..7 0 0 −1.5 −4 0.6 1.6 0 1 1 0     Decoding this system from the matrix, we get �
�   R1 + 0.7R3 = 1.7 R2 − 3.5R3 − 1.5R6 = −4 R4 + 0.6R6 = 1.6 1 R5 = We can solve for R1, R2, R4 and R5 leaving R3 and R6 as free variables. Labeling R3 = s and R6 = t, we have R1 = −0.7s + 1.7, R2 = 3.5s + 1.5t − 4, R4 = −0.6t + 1.6 and R5 = 1. Since resistance values are always positive, we need to restrict our values of s and t. We know R3 = s > 0 and when we combine that with R1 = −0.7s + 1.7 > 0, we get 0 < s < 16 7. Similarly, R6 = t > 0 and with R4 = −0.6t + 1.6 > 0, we ﬁnd 0 < t < 8 In order visualize the inequality R2 = 3.5s + 1.5t − 4 > 0, we graph the 3. line 3.5s + 1.5t − 4 = 0 on the st-plane and shade accordingly.8 Imposing the additional conditions 0 < s < 16 3, we ﬁnd our values of s and t restricted to the region depicted on the right. Using the roster method, the values of s and t are pulled from the region (s, t) : 0 < s < 16 3, 3.5s + 1.5t − 4 > 0. The reader is encouraged to check that the solution presented in 1(a), namely all resistance values equal to 1, corresponds to a pair (s, t) in the region. 7 and = 16 7 −2 −1 1 2 4 t −2 −1 1 2 4 t −1 −1 The region where 3.5s + 1.5t − 4 > 0 The region for our parameters s and t. t = 8 3 8See Section 2.4 for a review of this procedure. 8.4 Systems of Linear Equations: Matrix Inverses 609 8.4.1 Exercises In Exercises 1 - 8, ﬁnd the inverse of the matrix or state
that the matrix is not invertible. 1. A = 3. C = 1 2 3 4 6 15 14 35 2. B = 4. D = 12 −7 −5 3 2 −1 16 −9     5. E = 7. G = 0 3 2 −1 4 3 2 −5   3 2 3 11 4 19 −3 1 2 3   6. F = 83 4 −3 6 2 1 2 −2 0 0 0 −3 8 16 4 − In Exercises 9 - 11, use one matrix inverse to solve the following systems of linear equations. 3x + 7y = 26 5x + 12y = 39 9. 10. 3x + 7y = 0 5x + 12y = −1 11. 3x + 7y = −7 5 5x + 12y = In Exercises 12 - 14, use the inverse of E from Exercise 5 above to solve the following systems of linear equations.    12. 3x + 4z = 1 2x − y + 3z = 0 −3x + 2y − 5z = 0    13. 3x + 4z = 0 2x − y + 3z = 1 −3x + 2y − 5z = 0    14. 3x + 4z = 0 2x − y + 3z = 0 −3x + 2y − 5z = 1 15. This exercise is a continuation of Example 8.3.3 in Section 8.3 and gives another application of matrix inverses. Recall that given the position matrix P for a point in the plane, the matrix RP corresponds to a point rotated 45◦ counterclockwise from P where a) Find R−1. (b) If RP rotates a point counterclockwise 45◦, what should R−1P do? Check your answer by ﬁnding R−1P for various points on the coordinate axes and the lines y = ±x. (c) Find R−1P where P corresponds to a generic point P (x, y). Verify that this takes points on the curve y = 2 x to points on the curve x2 − y2 = 4. 610 Systems of Equations and Matrices 16. A Sasquatch
’s diet consists of three primary foods: Ippizuti Fish, Misty Mushrooms, and Sun Berries. Each serving of Ippizuti Fish is 500 calories, contains 40 grams of protein, and has no Vitamin X. Each serving of Misty Mushrooms is 50 calories, contains 1 gram of protein, and 5 milligrams of Vitamin X. Finally, each serving of Sun Berries is 80 calories, contains no protein, but has 15 milligrams of Vitamin X.9 (a) If an adult male Sasquatch requires 3200 calories, 130 grams of protein, and 275 milligrams of Vitamin X daily, use a matrix inverse to ﬁnd how many servings each of Ippizuti Fish, Misty Mushrooms, and Sun Berries he needs to eat each day. (b) An adult female Sasquatch requires 3100 calories, 120 grams of protein, and 300 milligrams of Vitamin X daily. Use the matrix inverse you found in part (a) to ﬁnd how many servings each of Ippizuti Fish, Misty Mushrooms, and Sun Berries she needs to eat each day. (c) An adolescent Sasquatch requires 5000 calories, 400 grams of protein daily, but no Vitamin X daily.10 Use the matrix inverse you found in part (a) to ﬁnd how many servings each of Ippizuti Fish, Misty Mushrooms, and Sun Berries she needs to eat each day. 17. Matrices can be used in cryptography. Suppose we wish to encode the message ‘BIGFOOT LIVES’. We start by assigning a number to each letter of the alphabet, say A = 1, B = 2 and so on. We reserve 0 to act as a space. Hence, our message ‘BIGFOOT LIVES’ corresponds to the string of numbers ‘2, 9, 7, 6, 15, 15, 20, 0, 12, 9, 22, 5, 19.’ To encode this message, we use an invertible matrix. Any invertible matrix will do, but for this exercise, we choose   A = 2 −3 3 −7 5 1 −2 1 −1   Since A is 3 × 3 matrix, we encode our message string into a matrix M with 3 rows. To do this, we take the ﬁrst three numbers, 2 9 7,
and make them our ﬁrst column, the next three numbers, 6 15 15, and make them our second column, and so on. We put 0’s to round out the matrix.   M = 6 20 2 9 15 7 15 12 9 19 0 0 0 22 5   To encode the message, we ﬁnd the product AM AM =   2 −3 3 −7 5 1 −2 1 −1     6 20 2 9 15 7 15 12 9 19 0 0 0 22 5    =  12 1 42 3 100 −23 38 57 39 36 −12 −42 −152 −46 −133   9Misty Mushrooms and Sun Berries are the only known ﬁctional sources of Vitamin X. 10Vitamin X is needed to sustain Sasquatch longevity only. 8.4 Systems of Linear Equations: Matrix Inverses 611 So our coded message is ‘12, 1, −12, 42, 3, −42, 100, 36, −152, −23, 39, −46, 38, 57, −133.’ To decode this message, we start with this string of numbers, construct a message matrix as we did earlier (we should get the matrix AM again) and then multiply by A−1. (a) Find A−1. (b) Use A−1 to decode the message and check this method actually works. (c) Decode the message ‘14, 37, −76, 128, 21, −151, 31, 65, −140’ (d) Choose another invertible matrix and encode and decode your own messages. 18. Using the matrices A from Exercise 1, B from Exercise 2 and D from Exercise 4, show AB = D and D−1 = B−1A−1. That is, show that (AB)−1 = B−1A−1. 19. Let M and N be invertible n × n matrices. Show that (M N )−1 = N −1M −1 and compare your work to Exercise 31 in Section 5.2. 612 Systems of Equations and Matrices 8.4.2 Answers 1. A−1 = −2 3 1 2 − 1 2
3. C is not invertible 5. E−1 =   −1 8 4 1 −3 −1 1 −6 −3   2. B−1 = 3 7 5 12 4. D−1 = 9 2 − 1 2 8 −1 6. F −    16 3 0 2 − 35 −90 − 1 2 0 1 5 0 −7 −36      0 7 2 0 1 7. G is not invertible 8. H −1 = The coeﬃcient matrix is B−1 from Exercise 2 above so the inverse we need is (B−1)−1 = B. 12 −7 −5 3 26 39 9. = 39 −13 10. 11. 12 −7 −5 3 0 −1 12 −7 −5 3 −7 5 = = 7 −3 So x = 39 and y = −13. So x = 7 and y = −3. −119 50 So x = −119 and y = 50. The coeﬃcient matrix is E =   3 0 2 −1 4 3 2 −5   from Exercise 5, so E−1 =   −1 8 4 1 −3 −1 1 −6 −3   12. 13. 14.       −1 8 4 1 −3 −1 1 −6 −3 −1 8 4 1 −3 −1 1 −6 −3 −1 8 4 1 −3 −1 1 −6 −1 1 1 8 −3 −6 4 −1 −3   So x = −1, y = 1 and z = 1.   So x = 8, y = −3 and z = −6.   So x = 4, y = −1 and z = −3. 8.4 Systems of Linear Equations: Matrix Inverses 613 16. (a) The adult male Sasquatch needs: 3 servings of Ippizuti Fish, 10 servings of Misty Mush- rooms, and 15 servings of Sun Berries daily. (b) The adult female Sasquatch needs: 3 servings
of Ippizuti Fish and 20 servings of Sun Berries daily. (No Misty Mushrooms are needed!) (c) The adolescent Sasquatch requires 10 servings of Ippizuti Fish daily. (No Misty Mush- rooms or Sun Berries are needed!) 17. (a) A−1 =     1 2 1 17 33 19 10 19 11     (b) 1 1 2 17 33 19 10 19 11   42 3 12 1 100 −23 38 57 39 36 −12 −42 −152 −46 −133    =  6 20 2 9 15 7 15 12 0 22 5 9 19 0 0   (c) ‘LOGS RULE’ 614 Systems of Equations and Matrices 8.5 Determinants and Cramer’s Rule 8.5.1 Definition and Properties of the Determinant In this section we assign to each square matrix A a real number, called the determinant of A, which will eventually lead us to yet another technique for solving consistent independent systems of linear equations. The determinant is deﬁned recursively, that is, we deﬁne it for 1 × 1 matrices and give a rule by which we can reduce determinants of n × n matrices to a sum of determinants of (n − 1) × (n − 1) matrices.1 This means we will be able to evaluate the determinant of a 2 × 2 matrix as a sum of the determinants of 1 × 1 matrices; the determinant of a 3 × 3 matrix as a sum of the determinants of 2 × 2 matrices, and so forth. To explain how we will take an n × n matrix and distill from it an (n − 1) × (n − 1), we use the following notation. Deﬁnition 8.12. Given an n × n matrix A where n > 1, the matrix Aij is the (n − 1) × (n − 1) matrix formed by deleting the ith row of A and the jth column of A. For example, using the matrix A below, we ﬁnd the matrix A23 by deleting the second row and third column of A1 5 1 4 2  
Delete R2 and C3 −−−−−−−−−−−→ A23 = 3 1 2 1 We are now in the position to deﬁne the determinant of a matrix. Deﬁnition 8.13. Given an n × n matrix A the determinant of A, denoted det(A), is deﬁned as follows If n = 1, then A = [a11] and det(A) = det ([a11]) = a11. If n > 1, then A = [aij]n×n and det(A) = det [aij]n×n = a11 det (A11) − a12 det (A12) + −... + (−1)1+na1n det (A1n) ‘det(A)’ and ‘\|A\|’ There are two commonly used notations for the determinant of a matrix A: We have chosen to use the notation det(A) as opposed to \|A\| because we ﬁnd that the latter is often confused with absolute value, especially in the context of a 1 × 1 matrix. In the expansion a11 det (A11)−a12 det (A12)+−...+(−1)1+na1n det (A1n), the notation ‘+−...+(−1)1+na1n’ means that the signs alternate and the ﬁnal sign is dictated by the sign of the quantity (−1)1+n. Since the entries a11, a12 and so forth up through a1n comprise the ﬁrst row of A, we say we are ﬁnding the determinant of A by ‘expanding along the ﬁrst row’. Later in the section, we will develop a formula for det(A) which allows us to ﬁnd it by expanding along any row. Applying Deﬁnition 8.13 to the matrix A = 4 −3 1 2 we get 1We will talk more about the term ‘recursively’ in Section 9.1. 8.5 Determinants and Cramer’s Rule 615 det(A) = det 4 −3 1 2 = 4 det (A11) − (−3) det (A12) = 4 det([
1]) + 3 det([2]) = 4(1) + 3(2) = 10 For a generic 2 × 2 matrix A = a b c d we get det(A) = det a b c d = a det (A11) − b det (A12) = a det ([d]) − b det ([c]) = ad − bc This formula is worth remembering Equation 8.1. For a 2 × 2 matrix, det a b c d = ad − bc Applying Deﬁnition 8.13 to the 3 × 3 matrix A =   3 1 0 −1 1 2 2 5 4   we obtain det(A) = det     3 1 0 − det (A11) − 1 det (A12) + 2 det (A13) = 3 det −1 5 1 4 − det 0 5 2 4 + 2 det 0 −1 1 2 = 3((−1)(4) − (5)(1)) − ((0)(4) − (5)(2)) + 2((0)(1) − (−1)(2)) = 3(−9) − (−10) + 2(2) = −13 To evaluate the determinant of a 4 × 4 matrix, we would have to evaluate the determinants of four 3 × 3 matrices, each of which involves the ﬁnding the determinants of three 2 × 2 matrices. As you can see, our method of evaluating determinants quickly gets out of hand and many of you may be reaching for the calculator. There is some mathematical machinery which can assist us in calculating determinants and we present that here. Before we state the theorem, we need some more terminology. 616 Systems of Equations and Matrices Deﬁnition 8.14. Let A be an n × n matrix and Aij be deﬁned as in Deﬁnition 8.12. The ij minor of A, denoted Mij is deﬁned by Mij = det (Aij). The ij cofactor of A, denoted Cij is deﬁned by Cij = (−1)i+jMij = (−1)i+j det (Aij). We note that in Deﬁnition 8.13, the sum a11 det (A11)
− a12 det (A12) + −... + (−1)1+na1n det (A1n) can be rewritten as a11(−1)1+1 det (A11) + a12(−1)1+2 det (A12) +... + a1n(−1)1+n det (A1n) which, in the language of cofactors is a11C11 + a12C12 +... + a1nC1n We are now ready to state our main theorem concerning determinants. Theorem 8.7. Properties of the Determinant: Let A = [aij]n×n. We may ﬁnd the determinant by expanding along any row. That is, for any 1 ≤ k ≤ n, det(A) = ak1Ck1 + ak2Ck2 +... + aknCkn If A is the matrix obtained from A by: – interchanging any two rows, then det(A) = − det(A). – replacing a row with a nonzero multiple (say c) of itself, then det(A) = c det(A) – replacing a row with itself plus a multiple of another row, then det(A) = det(A) If A has two identical rows, or a row consisting of all 0’s, then det(A) = 0. If A is upper or lower triangular,a then det(A) is the product of the entries on the main diagonal.b If B is an n × n matrix, then det(AB) = det(A) det(B). det (An) = det(A)n for all natural numbers n. A is invertible if and only if det(A) = 0. In this case, det A−1 = 1 det(A). aSee Exercise 8.3.1 in 8.3. bSee page 585 in Section 8.3. Unfortunately, while we can easily demonstrate the results in Theorem 8.7, the proofs of most of these properties are beyond the scope of this text. We could prove these properties for generic 2 × 2 8.5 Determinants and Cramer’s Rule 617 or even 3 × 3 matrices by brute force computation, but this manner of proof belies the elegance and symmetry of the determinant. We will prove
what few properties we can after we have developed some more tools such as the Principle of Mathematical Induction in Section 9.3.2 For the moment, let us demonstrate some of the properties listed in Theorem 8.7 on the matrix A below. (Others will be discussed in the Exercises.) A =   3 1 0 −1 1 2   2 5 4 We found det(A) = −13 by expanding along the ﬁrst row. To take advantage of the 0 in the second row, we use Theorem 8.7 to ﬁnd det(A) = −13 by expanding along that row.   det   1 3 0 −1 1 2 2 5 4     = 0C21 + (−1)C22 + 5C23 = (−1)(−1)2+2 det (A22) + 5(−1)2+3 det (A23) = − det 3 2 2 4 − 5 det 3 1 2 1 = −((3)(4) − (2)(2)) − 5((3)(1) − (2)(1)) = −8 − 5 = −13 In general, the sign of (−1)i+j in front of the minor in the expansion of the determinant follows an alternating pattern. Below is the pattern for 2 × 2, 3 × 3 and 4 × 4 matrices, and it extends naturally to higher dimensions. + − − +   + − + − + − + − +       + − + − − + − + + − + − − + − +     The reader is cautioned, however, against reading too much into these sign patterns. In the example above, we expanded the 3 × 3 matrix A by its second row and the term which corresponds to the second entry ended up being negative even though the sign attached to the minor is (+). These signs represent only the signs of the (−1)i+j in the formula; the sign of the corresponding entry as well as the minor itself determine the ultimate sign of the term in the expansion of the determinant. To illustrate some of the other properties in Theorem 8.7, we use row operations to
transform our 3 × 3 matrix A into an upper triangular matrix, keeping track of the row operations, and labeling 2For a very elegant treatment, take a course in Linear Algebra. There, you will most likely see the treatment of determinants logically reversed than what is presented here. Speciﬁcally, the determinant is deﬁned as a function which takes a square matrix to a real number and satisﬁes some of the properties in Theorem 8.7. From that function, a formula for the determinant is developed. 618 Systems of Equations and Matrices each successive matrix.3   3 1 0 −1 1 2 A   2 5 4 Replace R3 −−−−−−−−−−→ with − 2 3 R1 + R3   3 1 0 − Replace R3 with −−−−−−−−−−→ 1 3 R2 + R3   3 1 0 −1 0 0 C   2 5 13 3 Theorem 8.7 guarantees us that det(A) = det(B) = det(C) since we are replacing a row with itself plus a multiple of another row moving from one matrix to the next. Furthermore, since C is upper triangular, det(C) is the product of the entries on the main diagonal, in this case det(C) = (3)(−1) 13 = −13. This demonstrates the utility of using row operations to assist in 3 calculating determinants. This also sheds some light on the connection between a determinant and invertibility. Recall from Section 8.4 that in order to ﬁnd A−1, we attempt to transform A to In using row operations A In Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ In A−1 As we apply our allowable row operations on A to put it into reduced row echelon form, the determinant of the intermediate matrices can vary from the determinant of A by at most a nonzero multiple. This means that if det(A) = 0, then the determinant of A’s reduced row echelon form must also be nonzero, which, according to Deﬁnition 8.4 means that all the main diagonal entries on A’s
reduced row echelon form must be 1. That is, A’s reduced row echelon form is In, and A is invertible. Conversely, if A is invertible, then A can be transformed into In using row operations. Since det (In) = 1 = 0, our same logic implies det(A) = 0. Basically, we have established that the determinant determines whether or not the matrix A is invertible.4 It is worth noting that when we ﬁrst introduced the notion of a matrix inverse, it was in the context of solving a linear matrix equation. In eﬀect, we were trying to ‘divide’ both sides of the matrix equation AX = B by the matrix A. Just like we cannot divide a real number by 0, Theorem 8.7 tells us we cannot ‘divide’ by a matrix whose determinant is 0. We also know that if the coeﬃcient matrix of a system of linear equations is invertible, then system is consistent and independent. It follows, then, that if the determinant of said coeﬃcient is not zero, the system is consistent and independent. 8.5.2 Cramer’s Rule and Matrix Adjoints In this section, we introduce a theorem which enables us to solve a system of linear equations by means of determinants only. As usual, the theorem is stated in full generality, using numbered unknowns x1, x2, etc., instead of the more familiar letters x, y, z, etc. The proof of the general case is best left to a course in Linear Algebra. 3Essentially, we follow the Gauss Jordan algorithm but we don’t care about getting leading 1’s. 4In Section 8.5.2, we learn determinants (speciﬁcally cofactors) are deeply connected with the inverse of a matrix. 8.5 Determinants and Cramer’s Rule 619 Theorem 8.8. Cramer’s Rule: Suppose AX = B is the matrix form of a system of n linear equations in n unknowns where A is the coeﬃcient matrix, X is the unknowns matrix, and B is the constant matrix. If det(A) = 0, then the corresponding system is consistent and independent and the solution for unknowns x1, x
2,... xn is given by: xj = det (Aj) det(A), where Aj is the matrix A whose jth column has been replaced by the constants in B. In words, Cramer’s Rule tells us we can solve for each unknown, one at a time, by ﬁnding the ratio of the determinant of Aj to that of the determinant of the coeﬃcient matrix. The matrix Aj is found by replacing the column in the coeﬃcient matrix which holds the coeﬃcients of xj with the constants of the system. The following example ﬂeshes out this method. Example 8.5.1. Use Cramer’s Rule to solve for the indicated unknowns. 1. Solve 2x1 − 3x2 = 4 5x1 + x2 = −2 for x1 and x2 2. Solve    2x − 3y + z = −1 1 0 x − y + z = 3x − 4z = for z. Solution. 1. Writing this system in matrix form, we ﬁnd A = 2 −3 1 5 X = x1 x2 B = 4 −2 To ﬁnd the matrix A1, we remove the column of the coeﬃcient matrix A which holds the coeﬃcients of x1 and replace it with the corresponding entries in B. Likewise, we replace the column of A which corresponds to the coeﬃcients of x2 with the constants to form the matrix A2. This yields A1 = 4 −3 1 −2 A2 = 2 4 5 −2 Computing determinants, we get det(A) = 17, det (A1) = −2 and det (A2) = −24, so that x1 = det (A1) det(A) = − 2 17 x2 = det (A2) det(A) = − 24 17 The reader can check that the solution to the system is − 2 17, − 24 17. 620 Systems of Equations and Matrices 2. To use Cramer’s Rule to ﬁnd z, we identify x3 as z. We have   A = 2 −3 1 −1 3 1 1 0 −1 1 0  
 A3 = Az =  2 −3 −1 1 −1 1 0 0 3   Expanding both det(A) and det (Az) along the third rows (to take advantage of the 0’s) gives z = det (Az) det(A) = −12 −10 = 6 5 The reader is encouraged to solve this system for x and y similarly and check the answer. Our last application of determinants is to develop an alternative method for ﬁnding the inverse of a matrix.5 Let us consider the 3 × 3 matrix A which we so extensively studied in Section 8.5.1 A =   3 1 0 −1 1 2   2 5 4 We found through a variety of methods that det(A) = −13. To our surprise and delight, its inverse below has a remarkable number of 13’s in the denominators of its entries. This is no coincidence.   2   A−1 = 13 − 7 9 13 − 10 − 2 13 Recall that to ﬁnd A−1, we are essentially solving the matrix equation AX = I3, where X = [xij]3×3 is a 3 × 3 matrix. Because of how matrix multiplication is deﬁned, the ﬁrst column of I3 is the product of A with the ﬁrst column of X, the second column of I3 is the product of A with the second column of X and the third column of I3 is the product of A with the third column of X. In other words, we are solving three equations6 13 − 8 13 15 13 3 13 13 1 13    A  x11 x21 x31    =    1 0 0  A  x12 x22 x32    =    0 1 0  A  x13 x23 x33    =    0 0 1 We can solve each of these systems using Cramer’s Rule. Focusing on the ﬁrst system, we have

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