CHAPTER 10 SYSTEMS, MATRICES, AND DETERMINANTS


 Roderick McDowell
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1 CHAPTER 0 SYSTEMS, MATRICES, AND DETERMINANTS
2 PRECALCULUS: A TEACHING TEXTBOOK Lesson 64 Solving Sstems In this chapter, we re going to focus on sstems of equations. As ou ma remember from algebra, sstems are used to solve problems that have more than one unknown. Let s go through a fairl simple eample. The Retro Theater charges $8 for an adult ticket and $5 for a child s ticket. Last Saturda, Retro sold 405 tickets for a total of $,658. How man adults and how man children bought tickets last Saturda? Notice there are two unknowns that we re tring to find: The number of adults and the number of tickets. This is the kind of problem where a sstem is helpful. The first step is to let A be the number of adults who bought a ticket and C the number of children. That gives us two unknowns (or variables). The net step is to write not one but two equations, each with A and C. Since a total of 405 tickets were sold on Saturda, we know that the number of adults plus the number of children must equal 405. That translates into the equation A + C = 405. To write a second equation, we can use the fact that the total amount of mone collected was $,658. Since each adult ticket sold for $8, the amount of mone from adults must have been 8A. And since each child s ticket sold for $5, the amount of mone from children must have been 5C. That gives us the equation 8A + 5C =, 658. These two equations together make up the sstem below. A + C = 405 8A + 5C =,658 Equation Equation There are lots of pairs of values for A and C that will solve each equation individuall. For instance, A =, B = 404 will solve Equation. But so will A = 0, B = 0 and A = 57, B = 48. And there are man more. The same is true for Equation. There are man pairs that will work. You can find some ourself. But the important point is that onl one pair will solve both equations. And that pair is the solution to the sstem. So how do we find the right pair? There are different methods for solving a sstem. You ma remember them from algebra. To solve with the substitution method, we need to first solve Equation for one of the variables. Let s solve for A. A = 405 C 8A + 5C =,658 Equation Equation Equation sas that A is the same as 405 C, which means that we can substitute that epression for A in Equation. 8(405 C) + 5C =,
3 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS The substitution step creates an equation with just one unknown, C. Now we can solve for C., 40 8C + 5C =, 658 C = 58 C = 94 So C must equal 94. With that information, it s eas to solve for A. All we have to do is go back to either one of the original equations and put 94 in for C. We ll use Equation since it s easier. A + C = 405 A + 94 = 405 substituting A = We end up with A =, C = 94, which means that Retro must have sold tickets to adults and 94 children. You can check these answers ourself b substituting them back into the two original equations. You ll find that A =, C = 94 solves both. Elimination Method Instead of substituting, we could also have solved our sstem b either addition or subtraction. To show ou how it works, we ll solve the sstem again, this time b using addition. The first step is to multipl both sides of Equation b 8. 8( A + C) = ( 8)(405) 8A + 5C =,658 Equation Equation Now we simplif. 8 A + ( 8 C) =,40 8 A + 5 C =,658 Equation Equation Notice that the A terms are opposites. That means if we add the left and right sides of Equation to the same sides of Equation, the A s will cancel out. It s perfectl legal to do this addition step, b the wa. Since the epressions in Equation are equal, we re just adding the same quantit to both sides of Equation. 8 A + ( 8 C) + 8A + 5C =, (, 40) The net step is to simplif b combining like terms and adding numbers. 8C + 5C = 58 Notice that since the A s were opposites, the canceled each other out. We re left with an equation that has just one unknown (C). Finishing the steps, we get the same answer for C as before. C = 58 C = 94 58
4 PRECALCULUS: A TEACHING TEXTBOOK Now all we would have to do is substitute 94 for C in either one of the original equations and solve for A. The steps are eactl the same as before, so we won t bother showing them again. The main point is that instead of substituting to get one of the equations down to a single variable, we can also add the epressions of one equation to both sides of the other equation. But to use the addition method, one pair of terms must be opposites. That s the onl wa the ll cancel each other out, leaving just the other variable. There s also such a thing as a subtraction method. If we had multiplied both sides of Equation b 8 instead of 8, the A terms would have been eactl the same. 8( A + C) = (8)(405) 8A + 5C =,658 8A + 8C =, 40 8A + 5C =,658 Equation Equation Equation Equation We can still get the A terms to cancel out. But we have to do it b subtracting both sides of Equation from both sides of Equation. 8A + 5 C (8A + 8 C) =, 658, 40 8A + 5C 8A 8C =, 658, 40 C = 58 C = 94 See, we end up with the same answer for C as before. The main point is that another wa to eliminate a variable in a sstem is to add or subtract both sides of one equation to (from) the other. These two methods together are sometimes called the elimination method. Multipling Both Equations In some sstems, both equations need to be multiplied b something in order to eliminate a variable. Here s an eample like that. + = 4 5 = 5 Equation Equation There s no wa we can make either the s or the s the same or opposites b multipling just one of the equations. What we can do, though, is multipl both sides of Equation b and both sides of Equation b. This will make the terms opposites. Simplifing gives us this. ( + ) = ( 4) ( 5 ) = ( 5) 6 9 = 6 0 = 50 Equation Equation Equation Equation 58
5 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS See, the s are opposites, which means we can eliminate those terms b adding both sides of Equation to both sides of Equation = = 8 Now we have just one variable in the equation, so we can solve for. 9 = 8 = To find, we just need to substitute for in either one of the original equations. Let s use Equation. 6 9 = 6 9() = 6 8 = 6 = 0 = 5 The solutions to the sstem are = 5, =. And ou can check that for ourself b substituting this pair in for both or the original equations. Solving Sstems b Graphing Another wa to solve a sstem is b graphing both equations on the same coordinate plane. When both of the equations are linear, as with all of our eamples, the graph will be two straight lines. Here s the graph of a linear sstem. += +=5 Figure 64. +=5 (,) += Notice that the two lines intersect at just one point: =, =. That s the solution pair that will solve both equations. Remember, all the points on each line will solve the equation for that line, but onl the point that s on both lines will solve both equations. To solve a sstem b graphing, then, ou have to graph each equation on the same coordinate plane and look for the point(s) where the graphs intersect. Adding both sides of Equation to Equation would have worked just as well. 58
6 PRECALCULUS: A TEACHING TEXTBOOK If the coordinates of the intersection point are fractions or decimals, then it can be hard to see the eact answers to the sstem. One wa to get around this problem is to graph the equations on our calculator. To graph the sstem + = 5 and + = this wa, the first step is to solve both equations for. That gives us this. 5 = + = + Net, we enter both of these into the calculator in the usual wa. The standard window ( Xmin = 0, Xma = 0, Ymin = 0, Yma = 0 ) will work fine for this sstem. So to see the graph of both functions on the same plane, we can just press GRAPH. Now to find the coordinates of the intersection point, we use the calculate function. We just press nd, then TRACE to get to that menu. We need intersect, which is choice 5. 5 Now we select the equation = + as the first curve, and = + as the second curve. The last step is to guess, b moving the cursor onto the intersection point. The answer comes out to =, =, which we know is right. The main point, though, is that graphing a sstem on a calculator can be helpful when the coordinates are fractions or decimals. 584
7 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS Some Special Cases Some sstems are a special case because the don t have an solutions. The easiest wa to see wh is to graph the two equations. Look at the graph of this sstem. Figure = 4= 4= 4= The lines are parallel, so no solutions. Since the lines are parallel, the never intersect. That means the sstem has no solutions. If we were to tr to solve the sstem using algebra, with the substitution or elimination method, we would end up with a false equation. Watch this. 4 = Equation 4 = Equation Subtracting Equation from Equation gives us an obviousl false equation. 4 ( 4 ) = ( ) 0 = 4 That s alwas what happens when a linear sstem has no solutions. Both variables will cancel out, and ou ll be left with a false equation. Sstems with no solutions are said to be inconsistent. There s one other special case we should review. That s when the two equations are actuall the same. Here s a case like that. = 5 Equation = Equation It s kind of hard to tell that these equations are the same, because the re in a different form. But watch what happens when we divide both sides of Equation b and move the term to the left side = = ( + 5) = + 5 = 5 585
8 PRECALCULUS: A TEACHING TEXTBOOK Equation is eactl the same as Equation. A sstem like this actuall has an infinite number of solutions, because an linear equation has an infinite number of solution pairs. Since both equations are the same, all those pairs will solve the sstem. If we had tried to solve this algebraicall, both the  and terms would have canceled out and we would have been left with a true equation like 0 = 0. That alwas happens when both equations of a sstem are actuall the same. Sstems of this kind are called dependent sstems. Practice 64 a. Select the choice that is identical to each trig epression: + csc cot + cos A. csc B. cos C. sec D. E. sin b. Use our knowledge of trig identities (not a calculator) to find the trig value below. If cos θ =, find 8 cos θ sin θ c. Solve each sstem of equations: = + 4 = 6 d. Solve the sstem of equation + = 0 = 7 using a graphing calculator. e. The graph of = can be epressed as a set of parametric equations. If = t, and = f ( t), then what does f ( t ) equal? A. D. 5 t B. t + C. 8t 7 t 4 4 E. 8t + f. Solve the word problem below. A local convenience store recentl ran a sale on their fountain drinks. During the sale, a large fountain drink cost $.50 and a small cost $.00. If 50 drinks were sold altogether and the store made $689 during the sale, how man of each size were sold? Problem Set 64 Tell whether each sentence below is True or False.. Sstems are used to solve problems that have more than one unknown.. Sstems can be solved b the substitution, elimination, or graphing methods. 586
9 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS For each pair of polar coordinates below, select the correct conversion into rectangular coordinates. Don t use the polar conversion function on our calculator.. P (70,5 ) A. (85,85 ) B. ( 85,85 ) C. ( 85, 85) D. (85, 85 ) E. (85, 85 ) 4. Q(50, 60 ) A. ( 5, 5) B. ( 5, 5 ) C. ( 5, 5 ) D. (5, 5 ) E. (5, 5 ) For each pair of rectangular coordinates below, select the correct conversion into polar coordinates. Don t use the polar conversion function on our calculator. 5. M (8,6) A. ( 0, 6.9 ) B. (, 6.9 ) C. (0, 6.9 ) D. ( 0, 6.9 ) E. (0, 6.9 ) 6. N( 4,9) A. ( 5.6, 59.4 ) B. ( 5, 0.6 ) C. (5.6,59.4 ) D. ( 5.6,59.4 ) E. (5.6, 59.4 ) Select the choice that is identical to each trig epression given. 7. cos θ sinθ A. sinθ B. cosθ C. tanθ D. secθ E. cscθ (a) 8. + sec tan + sin A. B. sec C. sin D. csc E. cos Use our knowledge of trig identities (not a calculator) to find each trig value below. 9. If tan( θ ) = 4, find cotθ (b) 0. If 6 cos α =, find 7 cos α sin α 587
10 PRECALCULUS: A TEACHING TEXTBOOK Solve each sstem of equations below.. + = 8 + = 7. + = = 0 (c). + 0 = = 4 Solve each sstem of equations below using a graphing calculator = 6 + = (d) 5. = 8 = 7 Select the answer for each question below. 6. The graph of 4 = + 9. (Hint: Use our graphing calculator.) A. intersects the ais at eactl one point. B. intersects the ais at eactl two points. C. intersects the ais at eactl three points. D. intersects the ais at eactl four points. E. does not intersect the ais. 7. Which of the following is equivalent to the equation = 0 in polar form? A. D. r = 7r cosθ B. r = 7r cosθ C. r + 7r sinθ = 0 E. r + 7r cosθ = 0 r = 7r sinθ (e) 8. The graph of = + 4 can be epressed as a set of parametric equations. If = t, and = f ( t), then what does f ( t ) equal? A. 4t 5 B. t + C. t + D. 4 t E. t Answer each question below. Estimate an irrational answers to two decimal places. 9. If = 4cosθ and = 4sinθ, then + =? 0. What is the magnitude of vector v with initial point (, 7) and terminal point ( 4, 5)?. If sin α 4sinα = 0 over the interval 0 α 80, then α =? (All angles are in degrees.) Solve the word problem below. (f). Cit of Fun, an amusement park, charges $5 per child and $45 per adult. On an average summer da, the sell a total of,0 tickets and make $75,450. Based on this data, how man of each ticket tpe do the sell on an average summer da? 588
11 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS Lesson 65 More Complicated Sstems We started learning about sstems in the last lesson. All of our eamples involved two linear equations. But, as ou probabl remember from algebra, some sstems have higherdegree equations. Here s an eample. = = Equation Equation Equation is a parabola, and Equation is a line. A graph of this sstem shows that the graphs intersect at not one but two points. Figure 65. = = = +8+9 (6,) (,) =  Higherdegree sstems can have more then one pair of solutions. This is one of the big differences between linear sstems and higherdegree or nonlinear sstems. When nonlinear equations are involved, it s possible for the graphs to intersect more than once. That makes the sstem have more than one pair of solutions. We can find both solution pairs without ever looking at the graph. We can solve the sstem algebraicall b the same basic method we use for linear sstems. Since Equation is alread solved for, the substitution method is probabl the easiest. So we substitute for in Equation and then simplif. ( ) = 8 9 = = = We end up with a quadratic equation with just one variable. The equation can be solved b factoring. 0 = ( + 6)( + ) = 6, 589
12 PRECALCULUS: A TEACHING TEXTBOOK As with most quadratic equations, we end up with two solutions. We re not finished et, though, because we still have to find matching values for. What we do is substitute each of our answers for into either of the original equations. Equation is a lot simpler, so we ll use it. for =6 for = 6 = = = = We end up with the solution pairs ( 6, ) and (, ). And those were the intersection points on the sstem s graph. That s how a nonlinear sstem can be solved algebraicall. Here s another eample of a nonlinear sstem. This one has a thirddegree (cubic) function. = = Equation Equation See, Equation is third degree and Equation is second degree. To solve the sstem algebraicall, we should use the substitution method, since both equations are solved for. Let s substitute for in Equation. That will eliminate. Now we move everthing to the left and simplif. = = 0 This equation can be solved b factoring. ( ) = 0 = 0 = 0 = 0 = We end up with two answers for. To find the matching values for, we substitute these into either one of the original equations. Equation is simpler. for =0 for = = (0) = () = 0 = 8 590
13 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS So this sstem also has two solution pairs: (0,0) and (,8). The graph of the sstem confirms these answers. Figure 65. = = =  =  (,8) 55 (0,0) Obviousl, there are lots of other kinds of nonlinear sstems besides our two eamples. Some have even more than two pairs of solutions. Here are a couple of other eamples. +5 =98  = (,4) (,4) (,4)   = +5 =98 (,4) Figure =5 + =5  =5 (,4) (,4) + =5 (0,5) Some nonlinear sstems are so hard that the can t be solved with algebra. That s when the graphing calculator becomes essential. To find the solutions, ou can use the intersect function, as we discussed in the last lesson. Three Variables, Three Equations Linear sstems can also get more complicated. Instead of having just two equations with two variables, it s possible for a sstem to have three equations with three variables. As ou ma remember, this kind of sstem would be useful in solving a problem with three unknowns. Here s an eample of a threevariable linear sstem z = z = z = 6 Equation Equation Equation We solve threevariable, threeequation sstems using basicall the same method as we do sstems having onl two variables. The first step is to eliminate a variable. We can make the terms in Equations and opposites b multipling both sides of Equation b. ( z) = () multipling b 59
14 PRECALCULUS: A TEACHING TEXTBOOK That gives us this z = z = z = 6 Equation Equation Equation Now we can eliminate the s in those two equations b adding them. We ll add the sides of Equation to the sides of Equation. Added from Equation z + (5 4 + z) = z = 4 Instead of getting an answer for or z, we end up with a twovariable equation. There s no wa to pin down specific answers for either of those variables right now. So the net step is to go back to the original sstem and tr to eliminate the terms in two other equations. We used Equations and before, so this time let s eliminate the s in Equations and. To make those terms opposites, we can multipl both sides of Equation b. That gives us this. ( z) = () multipling b z = z = z = 6 Equation Equation Equation Now we can add Equation to Equation and simplif. Added from Equation z + ( z) = z = We end up with another twovariable equation. We can t get a specific value for or with this equation either. But what we can do is put together the twovariable equations that we ve created in order to create another sstem. We ll call the first twovariable equation, 7 + z = 4, Equation 4, and the second twovariable equation, z =, Equation z = 4 z = Equation 4 Equation 5 59
15 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS We alread know how to solve twovariable, twoequation sstems. So now we can find solutions for and. We can make the s opposites b multipling both sides of Equation 4 b and both sides of Equation 5 b 7. (7 + z) = (4) 7( z) = 7( ) 87 + z = z = 56 Equation 4 Equation 5 Equation 4 Equation 5 Now we can eliminate the s b adding and simplifing. Added from Equation z + (87 + z) = z = 0 z = We have a value for z. To find a matching value for, we just substitute equations. Using Equation 4, we get this. 7 + = 4 z = into either one of the twovariable Now we solve for = 4 7 = 4 = With values for both z and, we can find the matching value for b substituting into an of the original threevariable equations. Using Equation, we get this. The solutions to the sstem are =, =, and 6() = + + = = = z =. 59
16 PRECALCULUS: A TEACHING TEXTBOOK Graphing in D What does the graph of a threevariable, threeequation sstem look like? Well, ou ma remember from algebra that graphs of threevariable equations don t fit on a flat coordinate plane. The re dimensional. The coordinate sstem has an ais, a ais and a zais, with the zais etending above and below the other two aes. Here s what the point =, =, z = 4 looks like in a D coordinate sstem. Figure 65.4 z 4 (,,4) origin In a D graph, ever point in space is represented b not two but three numbers: (,, z ). Graphs of threevariable equations are also dimensional. The have all sorts of interesting shapes. Here are a few eamples. z Figure 65.5 z A linear threevariable equation has the simplest of all D graphs. It s just a plane (like a flat piece of paper) floating in space. The graph of our sstem, then, is three planes in space. Our three answers ( =, =, z = ) represent the point (in space) where all three of those intersect. 594
17 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS Practice 65 a. For the pair of rectangular coordinates K( 5, 8), select the correct conversion into polar coordinates. Don t use the polar conversion function on our calculator. A. (9.4, 07. ) B. ( 9.4, 07. ) C. (9.4, 07. ) D. ( 9.4, 07. ) E. ( 5, 7. ) Solve each sstem of equations below. b. = + 4 = c. + 7z = 0 + z = z = 7 + = 6 = d. Solve the sstem of equations irrational answers to two decimal places. using a graphing calculator. Estimate an π π e. f ( ) = sin(arctan ) and g( ) = tan(arcsin ). If 0, then g f =? f. Solve the word problem below. Big Time Cinema offers three different ticket prices. Children tickets cost $4.50, adult tickets cost $9.00, and senior tickets cost $5.50. On a certain Frida night, the theater sold twice as man adult tickets as it did senior and child tickets combined. If the theater sold a total of, tickets and made $6,58.50, how man of each ticket tpe did the theater sell? Problem Set 65 Tell whether each sentence below is True or False.. Higherdegree sstems can have more than one pair of solutions.. Graphs of threevariable, threeequation sstems can be graphed on a flat coordinate plane. Answer each question below. Estimate an irrational numbers to one decimal place.. Find the components of a vector h with magnitude 5 and direction angle θ = Find the magnitude and direction angle of k = 4i 7j. 595
18 PRECALCULUS: A TEACHING TEXTBOOK Answer each question below. Estimate an irrational answers to one decimal place. 5. An airplane is fling 50 kilometers per hour and heading 4 south of east. If the wind is blowing at a rate of 75 kilometers per hour 70 south of west, what is the groundspeed and direction of the airplane? 70 4 v =75 km/h v =50 km/h 6. Cars that slide down ic hills often have to be pulled up b tow trucks. If a small car weighing 8,900 Newtons is held still b a cable from a tow truck on an ice slide inclined at an angle of 8, what is the pull on the cable? 8 F =8,900 N For each pair of parametric equations below, eliminate the parameter and select the direct relationship between and. 7. = 4 t, = t A. = 8 B. = 4 C. = 4 6 D. = 6 4 E. = 8 8. = t, = 8t + 6t A. D. = B. = E. = C. = 54 8 = + For each pair of rectangular coordinates below, select the correct conversion into polar coordinates. Don t use the polar conversion function on our calculator. 9. E( 4,5) (a) 0. G( 0, ) A. (4., 7. ) B. (4.,7. ) C. ( 4.,7. ) D. ( 4., 7. ) E. ( 7, 6.8 ) A. (., 0.8 ) B. ( 4,.8 ) C. (., 0.8 ) D. (., 0.8 ) E. (., 0.8 ) 596
19 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS Solve each sstem of equations below.. 4 = 5 =. + 4 = 6 = (b). = + = (c) z = z = 6 6 z = 8 Solve each sstem of equations below using a graphing calculator. Estimate an irrational answers to two decimal places. 5. = = 5 (d) 6. + = 9 4 = + + Select the answer for each question below. 7. The graph of which of the following functions is smmetric with respect to the origin? A. D. g( ) = ( 5) B. f ( ) = e C. F( ) = 4sin h( ) = ( + ) E. G( ) = 8. What is the domain of f ( ) =? 9 A. > B. < or > C. ± D. { } < E. { < < } Answer each question below. Estimate an irrational answers to two decimal places. 9. The sides of a triangle are 7, 8, and 9 centimeters. What is the measure in degrees of the angle opposite the 7 centimeters side? 0. If + =, then =? 7 8 π π (e). f ( ) = sin(arctan ) and g( ) = tan(arcsin ). If 0, then g f =? 9 Solve the word problem below. (f). For a fundraiser to go on a church mission trip, Penelope sold three different tpes of cand bars. Choconut Bars cost $, AlmondBliss Bars cost $.50, and FudgeBrownie Bars cost $. She ended up selling three times as man FudgeBrownie bars than the other two combined. If Penelope sold 500 cand bars and made $587.50, how man of each tpe of bar did she sell? 597
20 PRECALCULUS: A TEACHING TEXTBOOK Lesson 66 Solving a Sstem with Determinants We ve been reviewing how to solve sstems of equations. And, as the eamples have shown, the process can be prett tedious. It would be nice if we had a shortcut for solving sstems. A Nice Shortcut Well, the mathematicians agreed and came up with a prett neat shortcut for solving certain kinds of sstems. The quadratic formula is a shortcut for solving quadratic equations. Instead of having to factor or complete the square to solve a tough quadratic equation, we can just plug in the numbers on the right side of the formula, which is b ± b 4ac =. The quadratic formula basicall turns an quadratic equation into just an arithmetic problem, a because we just have to do the arithmetic on the right side. Remember, the quadratic formula is derived b solving the general equation a + b + c = 0 for in terms of the coefficients a, b, and c. Since an quadratic equation can fit into the form a + b + c = 0, this is kind of like solving ever possible quadratic equation at once. The shortcut for solving sstems is derived in prett much the same wa. We start with a twovariable, twoequation sstem with letters in place of the coefficients. a + b = c d + e = f Equation Equation It s possible to solve this sstem for and b elimination, even though the coefficients aren t known. We ll just end up with answers in terms of a, b, c, d, e, and f. To eliminate the s in both equations, we first must make their coefficients the same. We can do that b multipling both sides of Equation b e and both sides of Equation b b. Simplifing gives us this. e( a + b) = ce b( d + e) = b( f ) ae + be = ce bd + be = bf Equation Equation Equation Equation Now the s are opposites, so we can add the sides of Equation to the sides of Equation. Added from Equation. bd be + (ae + be) = bf + ce bd + be + ( ae + be) = bf + ce Now we can simplif. Of course, be and be cancel out. bd + ae = bf + ce 598
21 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS This looks complicated, but we can think of it as an equation with one unknown,. The rest of the letters are just the constants. We can combine the two terms on the left side b reversing the distributing process. We just take an outside parentheses. ( bd + ae) = bf + ce This gets the left side down to a single term, with a coefficient of bd + ae. We can even put the coefficient in front of the in the usual wa. ( bd + ae) = bf + ce Now we finish solving b undoing the multiplication b of bd + ae. All we have to do is divide both sides of the equation b bd + ae. bf + ce = or bd + ae ce bf = ae bd Our answer for, then, is ce bf ae bd. To find, we need to substitute ce bf ae bd for in either one of the original equations. We ll use Equation. a + b = c ce bf a + b = c ae bd substituting for We multipl the fraction b a and then move the result to the right side. ace abf ae bd + b = c ace abf b = c ae bd Net, we subtract the fraction from c. The common denominator is ae bd. c( ae bd) ace abf b = ae bd ae bd Now we distribute the c in the top of the first fraction and subtract the tops. ace bcd ace abf b = ae bd ae bd ace bcd ( ace abf ) b = ae bd 599
22 PRECALCULUS: A TEACHING TEXTBOOK Simplifing on top gives us this. ace bcd ace + abf b = ae bd abf bcd b = ae bd To finish solving for, we need to get rid of the b on the left. The easiest wa to do it is to multipl both sides b b. abf bcd b = b ae bd b abf bcd = b( ae bd) There s one last simplification step. We can factor out a b on top and cancel one pair of b s. b( af cd) = b( ae bd) af cd = ae bd So our answers for the sstem are ce bf = ae bd and af cd =. ae bd Make sure ou understand what these answers mean. For a specific twovariable sstem, with actual numbers in for the letters a through f, we could find the sstem s solutions for and just b substituting the value of the letters into the fractions ce bf and af cd. It s just like substituting the values or a, b, and c in the quadratic ae bd ae bd formula, ago. b ± b 4ac. To show ou how it works, let s take one of the sstems we solved a couple of lessons a Now let s look at an actual sstem. This one is from a couple of lessons ago. + = 4 5 = 5 Equation Equation 600
23 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS Instead of having to go through the long process of having to solve this sstem with the elimination method, all we ce bf af cd have to do is put the values for a, b, c, d, e, and f into the formulas = and =. For our sstem, ae bd ae bd a =, b =, c = 4, d =, e = 5, and f = 5. Plugging those numbers in gives us this. ( 4)( 5) ()( 5) = ()( 5) ()() ()( 5) ( 4)() = ()( 5) ()() These are the eact same answers we got before. = 5 = Using Determinants There s a simple and important wa to remember the formulas ce bf af cd = and =. Notice that the ae bd ae bd denominators are the same, and that the contain the letters a, b, d, and e, which represent the coefficients for and in the equations a + b = c and d + e = f. To remember the formula for the denominators, what we do is write these coefficients down in the shape of a rectangle, and list them in the same order that the appear in the equations. a d b e Now to find the epression for the denominators, all we do is cross multipl and subtract the products. a d b e =ae bd This fournumber smbol with vertical bars is called a determinant. The numbers inside are called elements. And the value of the determinant is calculated b cross multipling and subtracting the products. We ll call the a b determinant for the denominator D, so we have D = ae bd d e =. The numerators of our and formulas can also be written as determinants. The numerator is ce bf. This doesn t include the coefficients of the terms in the equations a + b = c and d + e = f. To set up the determinant for the numerator, then, all ou have to do is start with the letters on the right side of both equations. In other words, replace a and d with c and f. a b D = and replace the coefficients with d e a + b = c d + e = f 60
24 PRECALCULUS: A TEACHING TEXTBOOK This gives us the determinant c f b e, which we ll call N, since it s the determinant for the value of the numerator. Using the same rule to calculate the value of this determinant cross multipling and subtracting the products gives us the correct epression for the numerator of the value. c b N = = ce bf f e To find the numerator for the value, we just start with a b D = again. Onl this time, we replace the  d e coefficients with the letters in the right side of the equations a + b = c and d + e = f. The coefficients are b and e. So we take out b and e and put in c and f. a + b = c d + e = f That gives us a new determinant, a d c f, which we ll call N, since it s the determinant for the value of the numerator. Using the same rule to calculate the value of this determinant cross multipling and subtracting the products gives us the correct epression for the numerator of the value. a c N = = af cd d f So the formulas for the solutions to an twovariable, twoequation sstem determinants. c b a c f e d f N = and = or = and = a b a b D D d e d e a + b = c d + e = f N can be written using a + b = c To solve an sstem of the form, all ou have to do is remember that the denominators equal a d + e = f determinant with the coefficients of and (in the order the appear in the equations), and the numerators are determinants that replace the coefficients of the particular variable ( or ) with the numbers on the right side of the equals sign. Then ou just set up those determinants and calculate their value b cross multipling and subtracting. This method of solving a sstem is also called Cramer s Rule, after a famous mathematician. To give ou one last quick eample, let s solve the sstem below with determinants. 6 7 = = 60
25 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS We just put the numbers for a through f into the determinants in their proper places. = and = Now cross multipling and subtracting the products for each determinant gives us the answers. 47(5) ( 7)( ) = 6(5) ( 7)() and 6( ) 47() = 6(5) ( 7)() 88 = = and 44 0 = = 5 44 The solutions to the sstem are = and = 5. Practice 66 Solve each sstem of equations below. a. = 4 + = 9 b z = z = z = 9 c. Solve the sstem using determinants: 5 + = + 8 = d. Select the epression below that is identical to cos 4 4 α sin α. A. D. cos α B. C. cos α 4 cos α E. sin α e. Which of the following quadratic equations has roots 7 + i and 7 i? A. D = 0 B = 0 E = 0 C = = 0 60
26 PRECALCULUS: A TEACHING TEXTBOOK f. Solve the word problem below. An observation tower that is 75 feet tall sits on top of a hill. From a point at the base of the hill, the angles of elevation to the top and bottom of the antenna are and 0, respectivel. To the nearest whole number of feet, how high is the hill? 75 ft. 0 Problem Set 66 Tell whether each sentence below is True or False.. The value of a determinant can be found b cross multipling and adding the products.. Using determinants to solve sstems of equations is called Cramer s Rule. If u = 7i + 5j and v = 4i 9j, calculate the value of each epression below. Estimate an irrational answers to two decimal places.. u v 4. u + v For each pair of coordinates below, use the polar conversion function of our calculator to select the correct conversion. 5. Convert H (,5 ) to rectangular coordinate A. ( 7.4,4.6) B. ( 7.4, 4.6) C. (7.4,4.6) D. (4.6, 7.4) E. (7.4, 4.6) 6. Convert I(6, 9) to polar coordinate A. ( 8.4, 9.4 ) B. (8.4, 9.4 ) C. (8.4, 9.4 ) D. ( 8.4, 9.4 ) E. ( 9.4,8.4 ) Solve each sstem of equations below. 7. = + 7 = (a) 8. = + = 8 604
27 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS (b) z = 9 z = z = 5 Solve each sstem of equations below using a graphing calculator. Estimate an irrational answers to one decimal place. 0. = = 5. + = 5 = Solve the sstems below using determinants.. + = = 0 (c). + 0 = = 4 Answer each question below. 4. Select the epression below that is identical to cscα cot α. A. sinα B. secα C. cscα D. tanα secα E. cotα cscα (d) 5. Select the epression below that is identical to cos α sin α. A. cosα B. C. sin α D. cos α E. cos α 6. If = sec θ and = sin θ, what is in terms of? Solve each trig equation below for 0 π. (All angles are in radians.) 7. cos = 0 8. sin + cos = 0 Select the answer for each question below. 9. If a point has polar coordinates (5, π ), then what are its rectangular coordinates? A. ( 5, 5) B. ( 5,0) C. (0, 5) D. (5,0) E. (0,5) 605
28 PRECALCULUS: A TEACHING TEXTBOOK (e) 0. Which of the following quadratic equations has roots 4 + i and 4 i? A. D = 0 B = 0 E = 0 C = = 0 ( 4). Which of the following lines are asmptotes of the graph of f ( ) =? 9 I. = ± II. = III. = A. I onl B. I and II onl C. II onl D. I and III onl E. I, II, and III Solve the word problem below. (f). A tree sits on top of a hill that is 00 feet tall. From a point at the base of the hill, the angles of elevation to the top and bottom of the tree are 45 and 4, respectivel. To the nearest whole number of feet, how tall is the tree? ft
29 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS Lesson 67 More on Determinants In the last lesson, we learned a shortcut for solving an twovariable, twoequation sstem. The shortcut is kind of like the quadratic formula, and the easiest wa to remember it is to use determinants. For instance, in an a + b = c ce bf af cd sstem of the form, the solutions have to be = and =. But to help remember d + e = f ae bd ae bd these solutions, we write them using the determinants below. c b a c f e d f = = a b a b d e d e To find the numerators and denominators, all we have to do is calculate the value of each determinant b cross ce bf af cd multipling and subtracting the products. That gives = and =. ae bd ae bd Determinants A determinant with four numbers inside the vertical bars (two rows and two columns) is actuall called a twobtwo (written ) determinant determinant 5 This kind of determinant is used to write the formula for solving an linear sstem with two variables and two equations. But what about threevariable, threeequation sstems? As it turns out, we can write a formula for those as well, but it requires that we use determinants, which are determinants that have 9 numbers inside the vertical bars: three rows and three columns. Here s an eample determinant We can t just cross multipl and subtract the products to calculate the value of a determinant; there are too man numbers for that. The method for calculating a determinant is a little more complicated. We start b writing down the number in the upper left, which is 47. Then we cross out all the other numbers in the same row and in the same column What s left is a determinant, 5. So we write 47 in front of this new determinant. Actuall, it s called a 4 607
30 PRECALCULUS: A TEACHING TEXTBOOK minor determinant. And what we re doing is multipling 47 b the minor determinant. We ll show ou how the multiplication works in a second The net step is to write down the number to the immediate right of 47, which is 7. Then we cross out all the other numbers in the same row and in the same column again This leaves another minor determinant, from So we multipl this minor determinant b 7, and subtract it 5 47 ( 7) 4 9 Continuing the process, we move across the top to the net number,. Crossing out the same row and same column 5 leaves the minor determinant. We multipl this determinant b and add the result to the others ( 7) Now we calculate each of the minor determinants, b cross multipling and subtracting the products. Simplifing gives us this. The value of the determinant is, [5( ) (4)] ( 7)[ ( ) (9)] + [ (4) 5(9)] 47( 5 ) + 7( 7) + ( 84 45) 47( 7) + 7( 6) + ( 9) ,099 Actuall, this same method will work when calculating a b determinant. After crossing out the row and column and multipling the chosen number b what s left, ou still end up cross multipling. 608
31 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS Determinants and Solving a Sstem Since determinants have enough room for 9 numbers, the can be used to solve threevariable, three equation sstems. That s enough room for a coefficient for,, and z for each of the three equations. Let s solve the threeequation sstem from a couple of lessons ago using determinants z = z = z = 6 Equation Equation Equation The process is basicall the same as the one we use for two equation sstems. The first step is to create a determinant containing the coefficients for,, and z in each equation in the same order the appear in the a + b + cz = p equations. Our sstem is in the form d + e + fz = q, so the first determinant should have the coefficients in the g + h kz = s following order. a b c D = d e f g h k This is the determinant for the denominator of all three solutions of the sstem. Now we put the numbers in. 5 4 D = Net, we calculate the value of the determinant, following the rules eplained above Now we multipl each number b its minor determinant. The second term is subtracted and the third term is added D = Net, we cross multipl and subtract the products in each minor determinant and simplif the entire epression. D = 5[( 6) 9(6)] 4[6( 6) 9(7)] + [6(6) (7)] D = 5( 66) + 4( 99) + () D = The denominator for,, and z equal 660. D = 660 What about the numerators? The process is basicall the same one we use for twovariable, twoequation sstems. Starting with the determinant for the denominator, to find the numerator for, we replace the coefficients 609
32 PRECALCULUS: A TEACHING TEXTBOOK of with the numbers on the right side of the equals sign in the three equations of the sstem. Here are the equations for the sstem and the determinant for the denominator again 5 4 D = z = z = z = 6 The coefficients are in the first column (5, 6, and 7). We need to replace those with 5,, and 6. N 5 4 = Now we calculate this determinant b the usual method. We multipl 5, 4, and b three minor determinants N = Net, we cross multipl and subtract the products for each minor determinant. N = 5[( 6) 9(6)] 4[( 6) 9(6)] + [(6) (6)] N = 5( 66) + 4( ) + (66) N =,0 The last step for finding is to put together the fraction with N =,0 on top and D = 660 on bottom. N,0 = = = D 660 So the solution for of our sstem is. This is the same answer we got a couple of lessons ago. We need to go through the same steps to find the answers for and z. Onl since the denominator for these will be the same as, we can skip that step. We alread know that D = 660. The find the determinant for the numerator of, we start with D and then replace the coefficients with the numbers on the right side of our three equations z = 5 D = z = z = N =
33 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS Now we calculate this determinant according to our method N = N N = 5[( 6) 9(6)] 5[6( 6) 9(7)] + [6(6) (7)] N = 5( ) 5( 99) + ( 55) N = So = = =. Finall, we go through the same process for the numerator of z. We start with D and then D 660 replace the zcoefficients with the numbers on the right side of our three equations. 5 4 D = Net, we calculate this determinant N z = z = z = z = N z = N = 5[(6) (6)] 4[6(6) (7)] + 5[6(6) (7)] z N = 5( 66) + 4( 55) + 5() z N = 0 Putting this numerator on top of the denominator, D = 660 the sstem, then, are =, =, and z N z 0, gives us = = =. The final answers to D 660 z =, which are the same answers we got before. 6
34 PRECALCULUS: A TEACHING TEXTBOOK Practice 67 a. Solve the sstem of equations: + = 5 = 9 b. Find the value of the determinant: c. Solve the sstem of equations + 5 4z = z = z = 7 using determinants. d. Use our knowledge of trig identities (not a calculator) to find each trig value below. 7 If secθ =, find cosθ + cosθ tan θ 4 e. The amount of potential energ stored in a compressed string varies jointl (directl) with onehalf of the square of the compression of the spring (). If the 80 joules of energ are stored when the spring is compressed b 0.5 centimeters, how man joules of energ would be stored b the same spring compressed b centimeters? Problem Set 67 Tell whether each sentence below is True or False.. Determinants can be used to solve threevariable, threeequation sstems.. The method for evaluating a determinant is the same as evaluating a determinant. Answer each question below. Estimate an irrational answers to one decimal place.. Find the magnitude and direction angle of m = 57i + 4j. 4. Find the unit vector that s in the same direction as u = 4i 7j. Solve each sstem of equations below = 6 + = 9 6. = = 6 (a) 7. + = 4 = 5 6
35 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS Find the value of each determinant below (b) Solve each sstem of equations below using determinants = 6 5 = 4. + = = (c) z = 6 + 4z = z = 6 Use our knowledge of trig identities (not a calculator) to find each trig value below.. If π 5 cos( θ ) =, find sin( θ ) (d) 4. If cscα =, find sinα + sinα cot α Select the correct graph of the pair of parametric equations and polar equation below. 5. = t, = t 6 A. B. C D. E
36 PRECALCULUS: A TEACHING TEXTBOOK (e) 6. r = cos 4θ (The angle is in radians) A. B. C D. E Select the answer for each question below. 7. The figure below shows one period of the graph of = 4sin + for 0 < π. What are the coordinates of the point where the minimum value of the function 6 occurs on this interval? 4 π A.,5 B. ( π, ) C. E. π, π, 4 D. 5 π, In ABC, cot A cos B =? csc B A. a bc B. C. D. a b E. c 9. What is the range of f ( ) = 8sin? 5 A. All real numbers greater than or equal to 0 and less than or equal to 5 B. All real numbers greater than or equal to 8 and less than or equal to 8. C. All real numbers greater than or equal to and less than or equal to 5 5. D. All real numbers greater than or equal to 0 and less than or equal to 5. E. All real numbers greater than or equal to 8 and less than or equal to 0. a c b c b A C c a B 64
37 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS Answer each question below. Estimate an irrational answers to two decimal places. 0. If f ( ) = e and g( ) 4 =, then f ( g ()) =?. If f ( ) = + 5 and g( ) = 7. What is the maimum value of g( f ( ))? Solve the word problem below. (f). According to Bole s Law, the pressure that the gas at a constant temperature eerts on a container is indirectl proportional (varies inversel) with the volume of the container. If a gas eerts. atmospheres of pressure on a container with a volume of liters, how much pressure would the gas eert on a container with a volume of 4 liters? 65
38 PRECALCULUS: A TEACHING TEXTBOOK Lesson 68 Solving Sstems with Matrices We ve been learning how to solve sstems using determinants. The nice thing about determinants is that the allow us to solve an linear sstem with just arithmetic. No algebra is needed. The drawback of determinants is that as the number of variables and equations in a sstem goes up (beond three), the amount of arithmetic required increases hugel. Since complicated real world problems can sometimes involve sstems with thousands of variables, we need a better shortcut method than determinants. Making a Triangle But before we can eplain a better shortcut, we need to talk a little about sstems with lots of variables. A threevariable, threeequation sstem is eas to solve once it s been worked down into a form like this. + + z = + z = 8 4z = 6 Since the bottom equation just has one unknown, all we have to do is solve it for z and then substitute the result in for z in the middle equation to get an answer for. 4z = 6 z = 4 + z = 8 + ( 4) = 8 = 0 = 0 And once we know that z = 4 and = 0, we can substitute those numbers into the top equation to solve for. + + = ( 4) = So the solutions are =, = 0, and z = 4. = = An eas sstem like this is said to be in triangular form, because the equations are shaped a little like a triangle. + + z = + z = 8 4z = 6 See, since the equations get shorter as ou go from top to bottom, the sstem is a little like an upside down triangle. The second shortcut for solving sstems that we re about to learn involves turning a sstem into triangular form. 66
39 CHAPTER 0: SYSTEMS, MATRICES, AND DETERMINANTS From Sstem to Matri In order to make the steps as fast and simple as possible, this shortcut uses something called a matri. A matri is basicall just a group of numbers laid out in rows and columns in the shape of a rectangle. More technicall, we could sa that a matri is a rectangular arra of numbers. Here s an eample of a matri The numbers are in rows and columns. It s similar to a determinant written with vertical bars. Onl with a matri, instead of plain vertical bars, there are brackets on either side. And notice that the matri has more columns than rows. A matri can have a different number of rows than columns; all sorts of combinations are possible. That s different from determinants, which are square. Here s how we represent a threevariable, threeequation sstem with a matri. We ll set up a matri for the sstem below. + z = z = 0 6z = 5 Inside the matri, we put the coefficients of,, and z, and the numbers on the right side of the equals sign for each equation. And these numbers should be in the same order as the appear in the sstem We also draw a vertical line inside the matri that separates the coefficients from the numbers on the right. The vertical line serves as a simplified equals signs for the three equations. This matri is a simplified wa to write the sstem. It includes onl the essential information: the coefficients and the numbers on the right. The s, s, z s, and equals signs are left out. We can do the same kinds of things to the matri that we can to a sstem. Since it s legal to multipl all the terms of an equation in a sstem b the same number (that s the same as multipling both sides b the same number), we can multipl all of the numbers in an row of our matri b the same number. And since it s legal to add or subtract the like terms of one equation in a sstem to the like terms of one of the other equations, we re allowed to add or subtract all of the numbers in one row of the matri to the corresponding numbers of an other row. When we multipl all of the terms of a row b some number, we just replace the old numbers with the new ones. For instance, if we multiplied the top row b, the matri would change from the original form to the form on the right below What we reall did with this step was to multipl both sides of the first equation b to create a new equation. ( + z) = ( ) z = 0 6z = z = z = 0 6z = 5 67
40 PRECALCULUS: A TEACHING TEXTBOOK The new sstem is equivalent to the original. It has the eact same solutions as the original sstem. But notice that the new matri above has the same coefficients and numbers as the new sstem. So when we multipl all the terms of a row in a matri b the same number, the result is an equivalent matri. When adding (or subtracting) the terms of one row of a matri to those of another, we also just replace the old row with the new. For eample, we add row to row b just replacing row with the sum of the two rows This step is the same as adding the sides of Equation to the sides of Equation and creating a new equation. Replacing Equation with that new equation creates a sstem that s equivalent to the original. + z = z = 0 6z = 5 + z = z = z = 5 Here are the rules for what can be done to a matri. Table 68. Arithmetic that can be done to matrices.... Multipl each number in a row b the same number and replace the old row with the new one. Add (or subtract) the numbers of one row to (from) the corresponding numbers of another row and replace the old row with the new one. Interchange an two rows. An Actual Eample Now let s go through an actual eample. We ll solve the sstem below using matrices. The matri form is on the right. + z = z = z = The basic strateg is to change the rows of the matri so that the sstem ends up in triangular form. Remember, in that form the sstem is reall eas to solve. Here s what a matri looks like when it s in triangular form
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