Solution Manual for Linear Algebra with Applications, 5th Edition

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SOLUTIONSMANUALLINEARALGEBRAWITHAPPLICATIONSFIFTHEDITIONOtto BretscherColby College

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Section 1.1Chapter 1Section 1.11.1.1[x+ 2y= 12x+ 3y= 1]−2×1st equation→[x+ 2y= 1−y=−1]÷(−1)→[x+ 2y= 1y= 1]−2×2nd equation→[x=−1y= 1], so that (x, y) = (−1,1).1.1.2[4x+ 3y= 27x+ 5y= 3]÷4→[x+34y=127x+ 5y= 3]−7×1st equation→[x+34y=12−14y=−12]×(−4)→[x+34y=12y= 2]−34×2nd equation→[x=−1y=2],so that (x, y) = (−1,2).1.1.3[2x+ 4y= 33x+ 6y= 2]÷2→[x+ 2y=323x+ 6y= 2]−3×1st equation→[x+ 2y=320=−52].So there is no solution.1.1.4[2x+ 4y= 23x+ 6y= 3]÷2→[x+ 2y= 13x+ 6y= 3]−3×1st equation→[x+ 2y= 10= 0]This system has infinitely many solutions:if we choosey=t, an arbitrary real number, then the equationx+ 2y= 1 gives usx= 1−2y= 1−2t.Therefore the general solution is (x, y) = (1−2t, t), wheretis anarbitrary real number.1.1.5[2x+ 3y= 04x+ 5y= 0]÷2→[x+32y= 04x+ 5y= 0]−4×1st equation→[x+32y= 0−y= 0]÷(−1)→[x+32y= 0y= 0]−32×2nd equation→[x= 0y= 0],so that (x, y) = (0,0).1.1.6x+ 2y+ 3z=8x+ 3y+ 3z= 10x+ 2y+ 4z=9−I−I→x+ 2y+ 3z= 8y= 2z= 1−2(II)→x+ 3z= 4y= 2z= 1−3(III)→x= 1y= 2z= 1, so that (x, y, z) = (1,2,1).1.1.7x+ 2y+ 3z= 1x+ 3y+ 4z= 3x+ 4y+ 5z= 4−I−I→x+ 2y+ 3z= 1y+z= 22y+ 2z= 3−2(II)−2(II)→x+z=−3y+z=20=−1This system has no solution.1

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Chapter 11.1.8x+ 2y+ 3z= 04x+ 5y+ 6z= 07x+ 8y+ 10z= 0−4(I)−7(I)→x+ 2y+ 3z= 0−3y−6z= 0−6y−11z= 0÷(−3)→x+ 2y+ 3z= 0y+ 2z= 0−6y−11z= 0−2(II)+6(II)→x−z= 0y+ 2z= 0z= 0+III−2(III)→x= 0y= 0z= 0,so that (x, y, z) = (0,0,0).1.1.9x+ 2y+ 3z= 13x+ 2y+z= 17x+ 2y−3z= 1−3(I)−7(I)→x+ 2y+ 3z= 1−4y−8z=−2−12y−24z=−6÷(−4)→x+ 2y+ 3z= 1y+ 2z=12−12y−24z=−6−2(II)+12(II)→x−z= 0y+ 2z=120= 0This system has infinitely many solutions: if we choosez=t, an arbitrary real number, then we getx=z=tandy=12−2z=12−2t. Therefore, the general solution is (x, y, z) =(t,12−2t, t), wheretis an arbitrary realnumber.1.1.10x+ 2y+ 3z= 12x+ 4y+ 7z= 23x+ 7y+ 11z= 8−2(I)−3(I)→x+ 2y+ 3z= 1z= 0y+ 2z= 5Swap:II↔III→x+ 2y+ 3z= 1y+ 2z= 5z= 0−2(II)→x−z=−9y+ 2z= 5z= 0+III−2(III)→x=−9y= 5z= 0,so that (x, y, z) = (−9,5,0).1.1.11[x−2y= 23x+ 5y= 17]−3(I)→[x−2y= 211y= 11]÷11→[x−2y= 2y= 1]+2(II)→[x= 4y= 1],so that (x, y) = (4,1). See Figure 1.1.Figure 1.1: for Problem 1.1.11.1.1.12[x−2y= 32x−4y= 6]−2(I)→[x−2y= 30= 0]2

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Section 1.1This system has infinitely many solutions:If we choosey=t, an arbitrary real number, then the equationx−2y= 3 gives usx= 3 + 2y= 3 + 2t.Therefore the general solution is (x, y) = (3 + 2t, t), wheretis anarbitrary real number. (See Figure 1.2.)Figure 1.2: for Problem 1.1.12.1.1.13[x−2y= 32x−4y= 8]−2(I)→[x−2y= 30= 2], which has no solutions. (See Figure 1.3.)Figure 1.3: for Problem 1.1.13.1.1.14The system reduces tox+ 5z= 0y−z= 00= 1, so that there is no solution; no point in space belongs to all threeplanes.Compare with Figure 2b.1.1.15The system reduces tox= 0y= 0z= 0so the unique solution is (x, y, z) = (0,0,0). The three planes intersect atthe origin.1.1.16The system reduces tox+ 5z= 0y−z= 00= 0, so the solutions are of the form (x, y, z) = (−5t, t, t), wheretis anarbitrary number. The three planes intersect in a line; compare with Figure 2a.1.1.17[x+ 2y=a3x+ 5y=b]−3(I)→[x+ 2y=a−y=−3a+b]÷(−1)→[x+ 2y=ay= 3a−b]−2(II)3

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Chapter 1[x=−5a+ 2by= 3a−b], so that (x, y) = (−5a+ 2b,3a−b).1.1.18x+ 2y+ 3z=ax+ 3y+ 8z=bx+ 2y+ 2z=c−I−I→x+ 2y+ 3z=ay+ 5z=−a+b−z=−a+c−2(II)→x−7z= 3a−2by+ 5z=−a+b−z=−a+c÷(−1)→x−7z= 3a−2by+ 5z=−a+bz=a−c+7(III)−5(III)→x= 10a−2b−7cy=−6a+b+ 5cz=a−c,so that (x, y, z) = (10a−2b−7c,−6a+b+ 5c, a−c).1.1.19The system reduces tox+z= 1y−2z=−30=k−7.a. The system has solutions ifk−7 = 0, ork= 7.b. Ifk= 7 then the system has infinitely many solutions.c. Ifk= 7 then we can choosez=tfreely and obtain the solutions(x, y, z) = (1−t,−3 + 2t, t).1.1.20The system reduces tox−3z=1y+ 2z=1(k2−4)z=k−2This system has a unique solution ifk2−46= 0, that is, ifk6=±2.Ifk= 2, then the last equation is 0 = 0, and there will be infinitely many solutions.Ifk=−2, then the last equation is 0 =−4, and there will be no solutions.1.1.21Letx,y, andzrepresent the three numbers we seek. We set up a system of equations and solve systematically(although there are short cuts):∣∣∣∣∣∣x+y= 24x+z= 28y+z= 30∣∣∣∣∣∣−(I)→∣∣∣∣∣∣x+y= 24−y+z= 4y+z= 30∣∣∣∣∣∣÷(−1)→∣∣∣∣∣∣x+y= 24y−z=−4y+z= 30∣∣∣∣∣∣−(II)−(II)→∣∣∣∣∣∣x+z= 28y−z=−42z= 34∣∣∣∣∣∣÷2→∣∣∣∣∣∣x+z= 28y−z=−4z= 17∣∣∣∣∣∣−(III)+(III)→∣∣∣∣∣∣x= 11y= 13z= 17∣∣∣∣∣∣We see thatx= 11, y= 13,andz= 17.4

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Section 1.11.1.22Letx= the number of male children andy= the number of female children.Then the statement “Emile has twice as many sisters as brothers” translates intoy= 2(x−1) and “Gertrude has as many brothers as sisters” translates intox=y−1.Solving the system[−2x+y=−2x−y=−1]givesx= 3 andy= 4.There are seven children in this family.1.1.23a Note that the demandD1for product 1 increases with the increase of priceP2; likewise the demandD2for product 2 increases with the increase of priceP1. This indicates that the two products are competing; somepeople will switch if one of the products gets more expensive.b SettingD1=S1andD2=S2we obtain the system[70−2P1+P2=−14 + 3P1105 +P1−P2=−7 + 2P2], or[−5P1+P2=−84P1−3P2= 112],which yields the unique solutionP1= 26 andP2= 46.1.1.24The total demand for the product of Industry A is 1000 (the consumer demand) plus 0.1b (the demand fromIndustry B). The outputamust meet this demand:a= 1000 + 0.1b.Setting up a similar equation for Industry B we obtain the system[a= 1000 + 0.1bb= 780 + 0.2a]or[a−0.1b= 1000−0.2a+b= 780],which yields the unique solutiona= 1100 andb= 1000.1.1.25The total demand for the products of Industry A is 310 (the consumer demand) plus 0.3b(the demand fromIndustry B). The outputamust meet this demand:a= 310 + 0.3b.Setting up a similar equation for Industry B we obtain the system[a= 310 + 0.3bb= 100 + 0.5a]or[a−0.3b= 310−0.5a+b= 100],which yields the solutiona= 400 andb= 300.1.1.26Sincex(t) =asin(t) +bcos(t) we can computedxdt=acos(t)−bsin(t) andd2xdt2=−asin(t)−bcos(t).Substituting these expressions into the equationd2xdt2−dxdt−x= cos(t) and simplifyinggives (b−2a) sin(t) + (−a−2b) cos(t) = cos(t). Comparing the coefficients of sin(t) and cos(t) on both sides ofthe equation then yields the system[−2a+b= 0−a−2b= 1], so thata=−15andb=−25. See Figure 1.4.Figure 1.4: for Problem 1.1.26.5

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Chapter 11.1.27a Substitutingλ= 5 yields the system[7x−y= 5x−6x+ 8y= 5y]or[2x−y= 0−6x+ 3y= 0]or[2x−y= 00= 0].There are infinitely many solutions, of the form (x, y) =(t2, t), wheretis an arbitrary real number.b Proceeding as in part (a), we find (x, y) =(−13t, t).c Proceedings as in part (a), we find only the solution (0,0).1.1.28Use the distance from Stein to Schaffhausen (and from Stein to Constance) as the unit of length.Let the speed of the boat and the speed of the river flow bevbandvs, respectively.Using the formula speed =distancetimeand measuring time in hours we get:vb−vs= 1vb+vs=32We findvb=54andvs=14. The time it takes to travel from Stein to Constance is time =distancespeed=1vb=45hours,or 48 minutes.1.1.29Letvbe the speed of the boat relative to the water, andsbe the speed of the stream; then the speed of theboat relative to the land isv+sdownstream andv−supstream. Using the fact that (distance) = (speed)(time),we obtain the system[8= (v+s)138= (v−s)23]←downstream←upstreamThe solution isv= 18 ands= 6.1.1.30The thermal equilibrium condition requires thatT1=T2+200+0+04, T2=T1+T3+200+04, andT3=T2+400+0+04.We can rewrite this system as−4T1+T2=−200T1−4T2+T3=−200T2−4T3=−400The solution is (T1, T2, T3) = (75,100,125).1.1.31To assure that the graph goes through the point (1,−1), we substitutet= 1 andf(t) =−1 into the equationf(t) =a+bt+ct2to give−1 =a+b+c.Proceeding likewise for the two other points, we obtain the systema+b+c=−1a+ 2b+ 4c= 3a+ 3b+ 9c= 13.The solution isa= 1, b=−5, andc= 3, and the polynomial isf(t) = 1−5t+ 3t2. (See Figure 1.5.)6

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Section 1.1Figure 1.5: for Problem 1.1.31.1.1.32Proceeding as in the previous exercise, we obtain the systema+b+c=pa+ 2b+ 4c=qa+ 3b+ 9c=r∣∣∣∣∣∣.The unique solution isa= 3p−3q+rb=−2.5p+ 4q−1.5rc= 0.5p−q+ 0.5r.Only one polynomial of degree 2 goes through the three given points, namely,f(t) = 3p−3q+r+ (−2.5p+ 4q−1.5r)t+ (0.5p−q+ 0.5r)t2.1.1.33f(t) is of the format2+bt+c.Sof(1) =a(12) +b(1) +c= 3, andf(2) =a(22) +b(2) +c= 6.Also,f′(t) = 2at+b, meaning thatf′(1) = 2a+b= 1.So we have a system of equations:a+b+c= 34a+ 2b+c= 62a+b= 1which reduces toa= 2b=−3c= 4.Thus,f(t) = 2t2−3t+ 4 is the only solution.1.1.34f(t) is of the format2+bt+c.So,f(1) =a(12) +b(1) +c= 1 andf(2) = 4a+ 2b+c= 0.Also,∫21f(t)dt=∫21(at2+bt+c)dt=a3t3+b2t2+ct|21=83a+ 2b+ 2c−(a3+b2+c)=73a+32b+c=−1.So we have a system of equations:a+b+c= 14a+ 2b+c= 073a+32b+c=−1which reduces toa= 9b=−28c= 20.7

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Chapter 1Thus,f(t) = 9t2−28t+ 20 is the only solution.1.1.35f(t) is of the format2+bt+c.f(1) =a+b+c= 1,f(3) = 9a+ 3b+c= 3, andf′(t) = 2at+b, sof′(2) = 4a+b= 1.Now we set up our system to bea+b+c= 19a+ 3b+c= 34a+b= 1.This reduces toa−c3= 0b+43c= 10 = 0.We write everything in terms ofa, revealingc= 3aandb= 1−4a.So,f(t) =at2+ (1−4a)t+ 3afor an arbitrarya.1.1.36f(t) =at2+bt+c, sof(1) =a+b+c= 1, f(3) = 9a+ 3b+c= 3. Also,f′(2) = 3, so 2(2)a+b= 4a+b= 3.Thus, our system isa+b+c= 19a+ 3b+c= 34a+b= 3.When we reduce this, however, our last equation becomes 0 = 2, meaning that this system is inconsistent.1.1.37f(t) =ae3t+be2t, sof(0) =a+b= 1 andf′(t) = 3ae3t+ 2be2t, sof′(0) = 3a+ 2b= 4.Thus we obtain the system[a+b= 13a+ 2b= 4],which reveals[a= 2b=−1].Sof(t) = 2e3t−e2t.1.1.38f(t) =acos(2t) +bsin(2t) and 3f(t) + 2f′(t) +f′′(t) = 17 cos(2t).f′(t) = 2bcos(2t)−2asin(2t) andf′′(t) =−4bsin(2t)−4acos(2t).So, 17 cos(2t) = 3(acos(2t) +bsin(2t)) + 2(2bcos(2t)−2asin(2t)) + (−4bsin(2t)−4acos(2t)) = (−4a+ 4b+3a) cos(2t) + (−4b−4a+ 3b) sin(2t) = (−a+ 4b) cos(2t) + (−4a−b) sin(2t).So, our system is:[−a+ 4b= 17−4a−b= 0].This reduces to:[a=−1b= 4].So our function isf(t) =−cos(2t) + 4 sin(2t).8

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Section 1.11.1.39Plugging the three points (x, y) into the equationa+bx+cy+x2+y2= 0, leads to a system of linearequations for the three unknowns (a, b, c).a+ 5b+ 5c+ 25 + 25=0a+ 4b+ 6c+ 16 + 36=0a+ 6b+ 2c+ 36 + 4=0.The solution isa=−20, b=−2, c=−4.−20−2x−4y+x2+y2= 0 is a circle of radius 5 centered at (1,2).1.1.40Plug the three points into the equationax2+bxy+cy2= 1. We obtain a system of linear equationsa+ 2b+ 4c=14a+ 4b+ 4c=19a+ 3b+c=1.The solution isa= 3/20, b=−9/40, c= 13/40. This is the ellipse (3/20)x2−(9/40)xy+ (13/40)y2= 1.1.1.41The given system reduces tox−z=−5a+2b3y+ 2z=4a−b30=a−2b+c.This system has solutions (in fact infinitely many) ifa−2b+c= 0.The points (a, b, c) with this property form a plane through the origin.1.1.42ax1=−3x2= 14 + 3x1= 14 + 3(−3) = 5x3= 9−x1−2x2= 9 + 3−10 = 2x4= 33 +x1−8x2+ 5x3−x4= 33−3−40 + 10 = 0,so that (x1, x2, x3, x4) = (−3,5,2,0).bx4= 0x3= 2−2x4= 2x2= 5−3x3−7x4= 5−6 =−1x1=−3−2x2+x3−4x4=−3 + 2 + 2 = 1,so that (x1, x2, x3, x4) = (1,−1,2,0).1.1.43a The two lines intersect unlesst= 2 (in which case both lines have slope−1).To draw a rough sketch ofx(t), note thatlimt→∞x(t) = limt→−∞x(t) =−1(the linex+t2y=tbecomes almost horizontal)andlimt→2−x(t) =∞, limt→2+x(t) =−∞.9

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Chapter 1Figure 1.6: for Problem 1.1.43a.Figure 1.7: for Problem 1.1.43a.Also note thatx(t) is positive iftis between 0 and 2, and negative otherwise.Apply similar reasoning toy(t). (See Figures 1.6 and 1.7.)bx(t) =−tt−2, andy(t) =2t−2t−2.1.1.44We can think of the line through the points (1,1,1) and (3,5,0) as the intersection of any two planes throughthese two points; each of these planes will be defined by an equation of the formax+by+cz=d.It is requiredthat 1a+ 1b+ 1c=dand 3a+ 5b+ 0c=d.Now the system[a+b+c−d=03a+5b−d=0]reduces to[a+52c−2d=0b−32c+d=0].We can choose arbitrary real numbers forcandd; thena=−52c+ 2dandb=32c−d.For example, if we choosec= 2 andd= 0,thena=−5 andb= 3, leading to the equation−5x+ 3y+ 2z= 0.If we choosec= 0 andd= 1,thena= 2 andb=−1,giving the equation 2x−y= 1.10

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Section 1.1We have found one possible answer:[−5x+3y+2z=02x−y=1].1.1.45To eliminate the arbitrary constantt, we can solve the last equation fortto givet=z−2, and substitutez−2 fortin the first two equations, obtaining[x= 6 + 5(z−2)y= 4 + 3(z−2)]or[x−5z=−4y−3z=−2].This system does the job.1.1.46Letb= Boris’ money,m= Marina’s money, andc= cost of a chocolate bar.We are told that[b+12m=c12b+m= 2c], with solution (b, m) = (0,2c).Boris has no money.1.1.47Let us start by reducing the system:x+ 2y+ 3z= 39x+ 3y+ 2z= 343x+ 2y+z= 26−I−3(I)→x+ 2y+ 3z= 39y−z=−5−4y−8z=−91Note that the last two equations are exactly those we get when we substitutex= 39−2y−3z: either way, we end up with the system[y−z=−5−4y−8z=−91].1.1.48a We set up two equations here, with our variables:x1= servings of rice,x2= servings of yogurt.So our system is:[3x1+12x2= 6030x1+20x2= 300].Solving this system reveals thatx1= 8, x2= 3.b Again, we set up our equations:[3x1+12x2=P30x1+20x2=C],and reduce them to find thatx1=−P15+C25, whilex2=P10−C100.1.1.49Letx1= number of one-dollar bills,x2= the number of five-dollar bills, andx3= the number of ten-dollarbills. Then our system looks like:[x1+x2+x3= 32x1+ 5x2+ 10x3= 100],which reduces to give us solutions that fit:x1= 15 +54x3,x2= 17−94x3, wherex3can be chosen freely. Nowlet’s keep in mind thatx1,x2, andx3must be positive integers and see what conditions this imposes on thevariablex3. We see that sincex1andx2must be integers,x3must be a multiple of 4. Furthermore,x3must bepositive, andx2= 17−94x3must be positive as well, meaning thatx3<689. These constraints leave us with onlyone possibility,x3= 4, and we can compute the corresponding valuesx1= 15 +54x3= 20 andx2= 17−94x3= 8.Thus, we have 20 one-dollar bills, 8 five-dollar bills, and 4 ten-dollar bills.1.1.50Letx1, x2, x3be the number of 20 cent, 50 cent, and 2 Euro coins, respectively. Then we need solutions tothe system:[x1+x2+x3= 1000.2x1+.5x2+2x3= 1000]11

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Chapter 1this system reduces to:[x1−5x3=−50003x2+6x3=80003].Our solutions are then of the formx1x2x3=5x3−50003−6x3+80003x3. Unfortunately for the meter maids, there are nointeger solutions to this problem. Ifx3is an integer, then neitherx1norx2will be an integer, and no one willever claim the Ferrari.Section 1.21.2.111−2...5234...2−2(I)→11−2...5018...−8−II→10−10...13018...−8[x−10z= 13y+ 8z=−8]−→[x= 13 + 10zy=−8−8z]xyz=13 + 10t−8−8tt, wheretis an arbitrary real number.1.2.234−1...868−2...3÷3→143−13...8368−2...3−6(I)→143−13...83000...−13This system has no solutions, since the last row represents the equation 0 =−13.1.2.3x= 4−2y−3zyandzare free variables; lety=sandz=t.xyz=4−2s−3tst, wheresandtare arbitrary real numbers.1.2.411...12−1...534...2−2(I)−3(I)→11...10−3...301...−1÷(−3)→11...101...−101...−1−II−II→10...201...−100...0, so thatx=2y=−1 .12

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Section 1.21.2.50011...00110...01100...01001...0swap:I↔III→1100...00110...00011...01001...0−I→1100...00110...00011...00−101...0−II+II→10−10...00110...00011...00011...0+III−III−III→1001...0010−1...00011...00000...0x1+x4=0x2−x4=0x3+x4=0−→x1=−x4x2=x4x3=−x4x1x2x3x4=−tt−tt, wheretis an arbitrary real number.1.2.6The system is in rref already.x1= 3 + 7x2−x5x3= 2 + 2x5x4= 1−x5Letx2=tandx5=r.x1x2x3x4x5=3 + 7t−rt2 + 2r1−rr1.2.712023...000132...00014−1...000001...0−II→12023...000132...00001−3...000001...0−2(III)−3(III)→12009...0001011...00001−3...000001...0−9(IV)−11(IV)+3(IV)→12000...000100...000010...000001...0x1+ 2x2= 0x3= 0x4= 0x5= 0−→x1=−2x2x3= 0x4= 0x5= 013

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Solution Manual for Linear Algebra with Applications, 5th Edition - Page 16 preview image

Chapter 1Letx2=t.x1x2x3x4x5=−2tt000, wheretis an arbitrary real number.1.2.8[01023...000048...0]÷4→[01023...000012...0]−2(II)0100−1...000012...0[x2−x5= 0x4+ 2x5= 0]−→[x2=x5x4=−2x5]Letx1=r, x3=s, x5=t.x1x2x3x4x5=rts−2tt, wherer, tandsare arbitrary real numbers.1.2.900012−1...212001−1...01220−11...2swap:I↔II→12001−1...000012−1...21220−21...2−I→12001−1...000012−1...20020−22...2swap:II↔III→12001−1...00020−22...200012−1...2÷2→12001−1...00010−11...100012−1...2x1+ 2x2+x5−x6= 0x3−x5+x6= 1x4+ 2x5−x6= 2−→x1=−2x2−x5+x6x3= 1 +x5−x6x4= 2−2x5+x6Letx2=r, x5=s, andx6=t.x1x2x3x4x5x6=−2r−s+tr1 +s−t2−2s+tst, wherer, sandtare arbitrary real numbers.1.2.10The system reduces tox1+x4=1x2−3x4=2x3+2x4=−3−→x1= 1−x4x2= 2 + 3x4x3=−3−2x414

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