Elementary Linear Algebra, 8th Edition Solution Manual

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Elementary Linear Algebra, 8th Edition Solution Manual - Page 1 preview image

Complete Solutions Manualto AccompanyElementary Linear AlgebraEIGHTH EDITIONRon LarsonThe Pennsylvania State University,State College, PA

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CONTENTSChapter 1Introduction to Systems of Linear Equations........................................1Chapter 2Matrices ................................................................................................29Chapter 3Determinants.........................................................................................76Chapter 4Vector Spaces .....................................................................................104Chapter 5Inner Product Spaces..........................................................................154Chapter 6Linear Transformations......................................................................206Chapter 7Eigenvalues and Eigenvectors ...........................................................244

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C H A P T E R1Systems of Linear EquationsSection 1.1Introduction to Systems of Linear Equations........................................2Section 1.2Gaussian Elimination and Gauss-Jordan Elimination ..........................8Section 1.3Applications of Systems of Linear Equations.....................................14Review Exercises..........................................................................................................22Project Solutions..........................................................................................................27

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2C H A P T E R1Systems of Linear EquationsSection 1.1Introduction to Systems of Linear Equations2.Because the termxycannot be rewritten asaxby+forany real numbersaandb, the equation cannot be writtenin the form12.a xa yb+=So, this equation isnotlinear in the variablesxandy.4.Because the terms2xand2ycannot be rewritten asaxby+for any real numbersaandb, the equationcannot be written in the form12.a xa yb+=So, thisequation isnotlinear in the variablesxandy.6.Because the equation is in the form12,a xa yb+=it islinear in the variablesxandy.8.Choosingyas the free variable, letyt=and obtain12121639393.xtxtxt−==+=+So, you can describe the solution set as163xt=+and,yt=wheretis any real number.10.Choosing2xand3xas free variables, let2xs=and3xt=and obtain112243612.xst+−=11231123 .xstxst+−==−+So, you can describe the solution set as1123 ,xst=−+32, and,xtxs==wheresandtare any real number.12.3223xyxy+=−+=Adding the first equation to the second equationproduces a new second equation, 55y=or1.y=So,( )2323 1 ,xy=−=−and the solution is:1,x= −1.y=This is the point where the two lines intersect.14.The two lines coincide.Multiplying the first equation by 2 produces a new firstequation.2343224xyxy−=−+= −Adding 2 times the first equation to the second equationproduces a new second equation.23200xy−==Choosingyt=as the free variable, you obtain232.xt=+So, you can describe the solution set as232xt=+and,yt=wheretis any real number.16.317437xyxy−+=+=Subtracting the first equation from the second equationproduces a new second equation,510x= −or2.x= −So,()4237y−+=or5,y=and the solution is:2,x= −5.y=This is the point where the two linesintersect.x4−2−2−3−4−3(−1, 1)y−x+ 2y= 3x+ 3y= 2−2x+y=−4x213−34−2−2−3−44y43x−y= 11213xy−2−4−6−8−2268−x+ 3y= 174x+ 3y= 7(−2, 5)

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Section 1.1Introduction to Systems of Linear Equations318.5216521xyxy−=+=Adding the first equation to the second equationproduces a new second equation,742x=or6.x=So, 6521y−=or3,y= −and the solution is:6,x=3.y= −This is the point where the two lines intersect.20.1242325xyxy−++=−=Multiplying the first equation by 6 produces a new firstequation.322325xyxy+=−=Adding the first equation to the second equationproduces a new second equation,428x=or7.x=So,725y−=or1,y=and the solution is:7,x=1.y=This is the point where the two lines intersect.22.0.20.527.80.30.468.7xyxy−= −+=Multiplying the first equation by 40 and the secondequation by 50 produces new equations.820111215203435xyxy−= −+=Adding the first equation to the second equationproduces a new second equation,232323x=or101.x=So,()8 101201112y−= −or96,y=and the solutionis:101,x=96.y=This is the point where the twolines intersect.24.21236344xyxy+=+=Adding 6 times the first equation to the second equationproduces a new second equation,00.=Choosingxt=as the free variable, you obtain44 .yt=−So,you can describe the solution asxt=and44 ,yt=−wheretis any real number.26.From Equation 2 you have23.x=Substituting this value into Equation 1 produces12126x−=or19.x=So, the system has exactly one solution:19x=and23.x=28.From Equation 3 you have2.z=Substituting this value into Equation 2 produces3211y+=or3.y=Finally, substituting3y=into Equation 1, you obtain35x−=or8.x=So, the system has exactlyone solution:8,3, and2.xyz===30.From the second equation you have20.x=Substituting this value into Equation 1 produces130.xx+=Choosing3xas the free variable, you have3xt=and obtain10xt+=or1.xt= −So, you can describe thesolution set as1,xt= −20,x=and3.xt=xy−3369−3−6−93x−5y= 216x+ 5y= 21(6,−3)xy0.2x−0.5y=−27.80.3x+ 0.4y= 68.750150−50100150250(101, 96)xy−123456−2123454x+y= 4232316x+y=xyx−2y= 5x−12y+ 23+= 4−361236912(7, 1)

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4Chapter 1Systems of Linear Equations32.(a)(b) This system is inconsistent, because you see twoparallel lines on the graph of the system.34.(a)(b) Two lines corresponding to two equations intersectat a point, so this system is consistent.(c) The solution is approximately13x=and12.y= −(d) Adding18−times the second equation to the firstequation, you obtain105y−=or12.y= −Substituting12y= −into the first equation, youobtain93x=or13.x=The solution is:13x=and12.y= −(e) The solutions in (c) and (d) are the same.36.(a)(b) Because the lines coincide, the system is consistent.(c) All solutions of this system lie on the line127.yx=+So, let,xt=then the solution set is12,7,xtyt==+wheretis any real number.(d) Adding 3 times the first equation to the secondequation you obtain44.16.33.1500.xy−+==Choosingxt=as a free variable, you obtain14.72.11.05ty−+=or14721105ty−+=or127.yt=+So, you can describe the solution set as12,7,xtyt==+wheretis any real number.(e) The solutions in (c) and (d) are the same.38.Adding2−times the first equation to the secondequation produces a new second equation.322010xy+==Because the second equation is a false statement, theoriginal system of equations has no solution.40.Adding6−times the first equation to the secondequation produces a new second equation.12220140xxx−==Now, using back-substitution, the system has exactly onesolution:10x=and20.x=42.Multiplying the first equation by32produces a new firstequation.121214040xxxx+=+=Adding4−times the first equation to the secondequation produces a new second equation.1214000xx+==Choosing2xt=as the free variable, you obtain114.xt= −So you can describe the solution set as114xt= −and2,xt=wheretis any real number.44.To begin, change the form of the first equation.1212532632xxxx+= −−= −Multiplying the first equation by 3 yields a new firstequation.1212352232xxxx+= −−= −Adding –3 times the first equation to the second equationproduces a new second equation.1223522111122xxx+= −−=Multiplying the second equation by211−yields a newsecond equation.12235221xxx+= −= −Now, using back-substitution, the system has exactly onesolution:121 and1.xx= −= −4−4−664x−5y= 3−8x+ 10y= 142−2−339x−4y= 51213x+y= 0−3−23244.1x−6.3y=−3.15−14.7x+ 2.1y= 1.05

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Section 1.1Introduction to Systems of Linear Equations546.Multiplying the first equation by 20 and the secondequation by 100 produces a new system.12120.64.27217xxxx−=+=Adding7−times the first equation to the secondequation produces a new second equation.1220.64.26.212.4xxx−== −Now, using back-substitution, the system has exactly onesolution:13x=and22.x= −48.Adding the first equation to the second equation yields anew second equation.2431044xyzyzxy++=+=+=Adding4−times the first equation to the third equationyields a new third equation.24310344xyzyzyz++=+=−−= −Dividing the second equation by 4 yields a new secondequation.35422344xyzyzyz++=+=−−= −Adding 3 times the second equation to the third equationyields a new third equation.354277422xyzyzz++=+=−=Multiplying the third equation by47−yields a new thirdequation.354222xyzyzz++=+== −Now, using back-substitution the system has exactly onesolution:0,4, and2.xyz=== −50.Interchanging the first and third equations yields a newsystem.12312312311432475323xxxxxxxxx−+=+−=−+=Adding2−times the first equation to the secondequation yields a new second equation.12323123114326915323xxxxxxxx−+=−=−+=Adding5−times the first equation to the third equationyields a new third equation.123232311432691521812xxxxxxx−+=−=−= −At this point, you realize that Equations 2 and 3 cannotboth be satisfied. So, the original system of equations hasno solution.52.Adding4−times the first equation to the secondequation and adding2−times the first equation to thethird equation produces new second and third equations.1323234132154521545xxxxxx+=−−= −−−= −The third equation can be disregarded because it is thesame as the second one. Choosing3xas a free variableand letting3,xt=you can describe the solution as12451522134xtxt=−=−3,xt=wheretis any real number.54.Adding3−times the first equation to the secondequation produces a new second equation.123232528168xxxxx−+=−= −Dividing the second equation by 8 yields a new secondequation.1232325221xxxxx−+=−= −Adding 2 times the second equation to the first equationyields a new first equation.1323021xxxx+=−= −Letting3xt=be the free variable, you can describe thesolution as1,xt= −221,xt=−and3,xt=wheretisany real number.

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6Chapter 1Systems of Linear Equations56.Adding 3 times the first equation to the fourth equationyields1423424234214202367.xxxxxxxxxx−+=−−=−=−++=Interchanging the second equation with the thirdequation yields1424234234210422367.xxxxxxxxxx−+=−=−−=−++=Adding4−times the second equation to the thirdequation, and adding2−times the second equation tothe fourth equation yields1424343421032347.xxxxxxxx−+=−=−+=+=Adding 3 times the second equation to the third equationyields1424344210321313.xxxxxxx−+=−=−+==Using back-substitution, the original system has exactlyone solution:12341,1,1, and1.xxxx====Answers may vary slightly for Exercises 58–62.58.Using a software program or graphing utility, you obtain0.8,x=1.2,y=2.4.z= −60.Using a software program or graphing utility, you obtain10,20,40,12.xyzw== −== −62.Using a software program or graphing utility, you obtain6.8813,163.3111,xy== −210.2915,z= −59.2913.w= −64.0xyz===is clearly a solution.Dividing the first equation by 2 produces320430.8330xyxyzxyz+=+−=++=Adding4−times the first equation to the second equation,and8−times the first equation to the third, yields32030.930xyyzyz+=−−=−+=Adding3−times the second equation to the thirdequation yields3203060.xyyzz+=−−==Using back-substitution, you conclude there is exactlyone solution:0.xyz===66.0xyz===is clearly a solution.Dividing the second equation by 2 yields a new secondequation.12163080xyzxyz++=+−=Adding3−times the second equation to the firstequation produces a new first equation.52128080xzxyz−+=+−=Lettingzt=be the free variable, you can describe thesolution as516,2 ,xtyt== −and,zt=wheretis anyreal number.68.Letx=the speed of the plane that leaves first andy=the speed of the plane that leaves second.3280Equation 123200Equation 2yxxy−=+=327222160232003360960xyxyyy−+=+===96080880xx−==Solution: First plane: 880 kilometers per hour; secondplane: 960 kilometers per hour

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Section 1.1Introduction to Systems of Linear Equations770.(a) False. Any system of linear equations is eitherconsistent, which means it has a unique solution,or infinitely many solutions; or inconsistent, whichmeans it has no solution. This result is stated onpage 5, and will be proved later in Theorem 2.5.(b) True. See definition on page 6.(c) False. Consider the following system of three linearequations with two variables.236391.xyxyx+= −−−==The solution to this system is:1,5.xy== −72.Because1xt=and2,xs=you can write32133.xstxx=+−=+−One system could be12312333xxxxxx−+=−+−= −Letting3xt=and2xs=be the free variables, you candescribe the solution as13,xst=+−2,xs=and3,xt=wheretandsare any real numbers.74.Substituting1Ax=and1By=into the original systemyields3211723.6ABAB+=−−= −Reduce the system to row-echelon form.27189121817ABAB+=−−= −271893926ABA+=−= −Using back substitution,23A= −and1.2B=Because1Ax=and1 ,By=the solution of the original systemof equations is:32x= −and2.y=76.Substituting1,Ax=1 ,By=and1Cz=into the original system yields225.341230ABCABABC+−=−= −++=Reduce the system to row-echelon form.22534155ABCABC+−=−= −= −3411161755ABBCC−=−−+= −=−So,1.C= −Using back-substitution,()116117,B−+−= −or1B=and( )34 11,A−= −or1.A=Because1,Ax=1,By=and1,Cz=the solution of the original system of equations is:1,x=1,y=and1.z= −78.Multiplying the first equation bysinθand the second bycosθproduces()()()()22sincossinsinsincoscoscos.xyxyθθθθθθθθ+=−+=Adding these two equations yields()22sincossincossincos.yyθθθθθθ+=+=+So,()()()()cossincossinsincos1xyxθθθθθθ+=++=and()()221sinsincoscossincoscossin.coscosxθθθθθθθθθθ−−−===−Finally, the solution iscossinxθθ=−andcossin.yθθ=+

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8Chapter 1Systems of Linear Equations80.Reduce the system to row-echelon form.()2010xkyky+=−=200, 10xkyyk+==−≠200, 10xyk==−≠If210,k−≠that is if1,k≠ ±the system will haveexactly one solution.82.Reduce the system to row-echelon form.()268314xykzk z++=−= −This system will have no solution if 830,k−=that is,83.k=84.Reduce the system to row-echelon form.()16430kxykx+=+=The system will have an infinite number of solutionswhen34430.kk+== −86.Reducing the system to row-echelon form produces()()5020102xyzyzaybzc++=−=−+−=()5020222.xyzyzabzc++=−=+−=So, you see that(a) if2220,ab+−≠then there is exactly onesolution.(b) if2220ab+−=and0,c=then there is aninfinite number of solutions.(c) if2220ab+−=and0,c≠there is no solution.88.If1230,ccc===then the system is consistentbecause0xy==is a solution.90.Multiplying the first equation byc, and the second bya,produces.acxbcyecacxdayaf+=+=Subtracting the second equation from the first yields().acxbcyecadbc yafec+=−=−So, there is a unique solution if0.adbc−≠92.The two lines coincide.23700xy−==Letting,yt=73 .2tx+=The graph does not change.94.212001312120xyxy−=−=Subtracting 5 times the second equation from 3 times thefirst equation produces a new first equation,2600,x−= −or300.x=So,()21 300200y−=or315,y=and the solution is:300,x=315.y=Thegraphs are misleading because they appear to be parallel,but they actually intersect at()300, 315 .Section 1.2Gaussian Elimination and Gauss-Jordan Elimination2.Because the matrix has 4 rows and 1 column, it has size41.×4.Because the matrix has 1 row and 1 column, it has size11.×6.Because the matrix has 1 row and 5 columns, it has size15.×8.314314437505−−−−−−Add 3 times Row 1 to Row 2.10.1232123225170971154760684−−−−−−−−−−−−−Add 2 times Row 1 to Row 2. Then add 5 times Row 1to Row 3.x21−1−2−4−5145−2−33y

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Section 1.2Gaussian Elimination and Gauss-Jordan Elimination912.Because the matrix is in reduced row-echelon form, youcan convert back to a system of linear equations1223.xx==14.Because the matrix is in row-echelon form, you canconvert back to a system of linear equations123320.1xxxx++== −Using back-substitution, you have31.x= −Letting2xt=be the free variable, you can describe thesolution as112 ,xt=−2,xt=and31,x= −wheretis any real number.16.Gaussian elimination produces the following.31 15101212101210101231 15101210120202020231120121101210120121012102020044101201210011−−−−−−−−−−− Because the matrix is in row-echelon form, convert backto a system of linear equations.132332211xxxxx+=+==By back-substitution,1231,1, and1.xxx== −=18.Because the fourth row of this matrix corresponds to theequation 02,=there is no solution to the linearsystem.20.Because the leading 1 in the first row is not farther to theleft than the leading 1 in the second row, the matrix isnotin row-echelon form.22.The matrix satisfies all three conditions in the definitionof row-echelon form. However, because the third columndoes not have zeros above the leading 1 in the third row,the matrix isnotin reduced row-echelon form.24.The matrix satisfies all three conditions in the definitionof row-echelon form. Moreover, because each columnthat has a leading 1 (columns one and four) has zeroselsewhere, the matrix is in reduced row-echelon form.26.The augmented matrix for this system is2616 .2616−−−Use Gauss-Jordan elimination as follows.261613813826162616000−−−−−−Converting back to a system of linear equations, you have38.xy+=Choosingyt=as the free variable, you can describethe solution as83xt=−and,yt=wheretis anyreal number.28.The augmented matrix for this system is210.1 .321.6−−Gaussian elimination produces the following.1122085112207724111220511221210.132321.6101100101−−−−  −− −−Converting back to a system of equations, the solution is:15x=and12.y=30.The augmented matrix for this system is120116 .328−Gaussian elimination produces the following.1201201160163280881201200160160880040−−−−−−−Because the third row corresponds to the equation040,= −you can conclude that the system hasno solution.

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10Chapter 1Systems of Linear Equations32.The augmented matrix for this system is3232203124 .67022−−−−Gaussian elimination produces the following.2223313222331322233132223313113232203124018670226702211018036661101800742110180016−−−−−−−−−−−−−−−−−−−Back-substitution now yields( )()( )323123113322222233336886101068.xxxxxx==+=+==+−=+−=So, the solution is:1238,10, and6.xxx===34.The augmented matrix for this system is11531021 .2110−−−−Subtracting the first row from the second row yields anew second row.115301322110−−−−−Adding2−times the first row to the third row yields anew third row.115301320396−−−−−Multiplying the second row by1−yields a new secondrow.115301320396−−−−Adding 3 times the second row to the third row yields anew third row.115301320000−−Adding1−times the second row to the first row yields anew first row.102101320000−−Converting back to a system of linear equations produces13232132.xxxx−=−=Finally, choosing3xt=as the free variable, you candescribe the solution as112 ,xt=+223 ,xt=+and3,xt=wheretis any real number.36.The augmented matrix for this system is1218 .36321−−−−Gaussian elimination produces the following matrix.12180003Because the second row corresponds to the equation03,=there is no solution to the original system.

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Section 1.2Gaussian Elimination and Gauss-Jordan Elimination1138.The augmented matrix for this system is2112634011 .1526352113−−−−−Gaussian elimination produces the following.617101111111526315263152633401101161710012112609510009510052113023113118023113118−−−−−−−−−−−−−−−−−−−−−−−6171061710111111111111439011111111750321750321111111111461710617101111111111178115621111152631526301010000143900000152631526301010014390000−−−−−−−−−−−−−−−1001439000012−−Back-substitution now yields()( )()( )()( )( )()106171061711111111111129043904324420.352635 02 4621.wzwyzwxyzw= −=+=+−== −−−= −−−−== −−−−= −−−−−=So, the solution is:1,0,4,xyz===and2.w= −40.Using a software program or graphing utility, theaugmented matrix reduces to100002010001.001003000104000011−So, the solution is:123452,1,3,4, and1.xxxxx== −===42.Using a computer software program or graphing utility,you obtain123456112021.xxxxxx== −=== −=44.The corresponding equations are1230.0xxx=+=Choosing4xt=and3xt=as the free variables, youcan describe the solution as10,x=2,xs= −3,xs=and4,xt=wheresandtare any real numbers.46.The corresponding equations are all00.=So, there arethree free variables. So,1,xt=2,xs=and3,xr=where,, andtsrare any real numbers.

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Elementary Linear Algebra, 8th Edition Solution Manual - Page 15 preview image

12Chapter 1Systems of Linear Equations48.number of $1 billsnumber of $5 billsnumber of $10 billsnumber of $20 billsxyzw====5102095264021xyzwxyzwyzxy+++=+++=−=−= −151020951000151111260100801400001021200100011−−−15821xyzw====The server has 15 $1 bills, 8 $5 bills, 2 $10 bills, and one$20 bill.50.(a) IfAis theaugmentedmatrix of a system of linearequations, then the number of equations in thissystem is three (because it is equal to the number ofrows of the augmented matrix). The number ofvariables is two because it is equal to the number ofcolumns of the augmented matrix minus one.(b) Using Gaussian elimination on the augmented matrixof a system, you have the following.21342426k−−−213006000k−+This system is consistent if and only if60,k+=so6.k= −IfAis thecoefficientmatrix of a system of linearequations, then the number of equations is three,because it is equal to the number of rows of thecoefficient matrix. The number of variables is alsothree, because it is equal to the number of columnsof the coefficient matrix.Using Gaussian elimination onAyou obtain thefollowing coefficient matrix of an equivalent system.31221006000k−+Because the homogeneous system is alwaysconsistent, the homogeneous system with thecoefficient matrixAis consistent for any value ofk.52.Using Gaussian elimination on the augmented matrix, you have the following.()()1100110011001100011001100110011001100020101000100000000000bacabcabc−−−+From this row reduced matrix you see that the original system has a unique solution.54.Because the system composed of Equations 1 and 2 is consistent, but has a free variable, this system must have an infinitenumber of solutions.56.Use Gauss-Jordan elimination as follows.1231231231014560360120127890612000000−−−−−

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Elementary Linear Algebra, 8th Edition Solution Manual - Page 16 preview image

Section 1.2Gaussian Elimination and Gauss-Jordan Elimination1358.Begin by finding all possible first rows[][][][][][][][]000 ,001 ,010 ,01, 100 , 10, 1, 10 ,aaabawhereaandbare nonzero real numbers. For each of these examine the possible remaining rows.00000101001001000 ,000 ,000 ,001 ,000 ,000000000000000a100100100100100000 ,010 ,010 ,001 ,01,000000001000000a101011010000 ,001 ,000 ,000 ,010000000000000000aaabaa60.(a) False. A47×matrix has 4 rows and 7 columns.(b) True. Reduced row-echelon form of a given matrixis unique while row-echelon form is not. (See alsoexercise 64 of this section.)(c) True. See Theorem 1.1 on page 21.(d) False. Multiplying a row by anonzeroconstant isone of the elementary row operations. However,multiplying a row of a matrix by a constant0c=isnotan elementary row operation. (This wouldchange the system by eliminating the equationcorresponding to this row.)62.No, the row-echelon form is not unique. For instance,1201and10 .01The reduced row-echelon form isunique.64.First, you need0a≠or0.c≠If0,a≠then youhave.00abababcbcdadbcba−−+So,0adbc−=and0,b=which implies that0.d=If0,c≠then you interchange rows and proceed.00cdabcdadcdadbcbc−−+Again,0adbc−=and0,d=which implies that0.b=In conclusion,abcdis row-equivalent to1000if and only if0,bd==and0a≠or0.c≠66.Row reduce the augmented matrix for this system.22950101010295002950λλλλλλλ+−−−−+−+−To have a nontrivial solution you must have the following.()()229505210λλλλ+−=+−=So, if5λ= −or12,λ=the system will have nontrivial solutions.68.A matrix is in reduced row-echelon form if every columnthat has a leading 1 has zeros in every position above andbelow its leading 1. A matrix in row-echelon form mayhave any real numbers above the leading 1’s.70.(a) When a system of linear equations is inconsistent,the row-echelon form of the correspondingaugmented matrix will have a row that is all zerosexcept for the last entry.(b) When a system of linear equations has infinitelymany solutions, the row-echelon form of thecorresponding augmented matrix will have a rowthat consists entirely of zeros or more than onecolumn with no leading 1’s. The last column will notcontain a leading 1.

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