Solution Manual for Elementary Linear Algebra, 7th Edition

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CONTENTSChapter 1Systems of Linear Equations..................................................................1Chapter 2Matrices ................................................................................................27Chapter 3Determinants.........................................................................................64Chapter 4Vector Spaces .......................................................................................88Chapter 5Inner Product Spaces..........................................................................133Chapter 6Linear Transformations......................................................................180Chapter 7Eigenvalues and Eigenvectors ...........................................................217

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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..........................................................................................................21Project Solutions...........................................................................................................26

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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+=itislinear 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, let3xt=and2xs=and obtain113263913.xxt−+=Dividing this equation by 13 you obtain11231123 .xstxst−+==+−So, you can describe the solution set as1123 ,xst=+−2,xs=and3,xt=wheretandsare any real numbers.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 setas232xt=+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 conclude that2.z=Substituting this value into Equation 2 produces 226y+=or2.y=Finally, substituting2y=and2z=into Equation 1, you obtain24x−=or6.x=So, the system has exactly onesolution:6,2,xy==and2.z=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)xyx−2y= 5x−12y+ 23+= 4−361236912(7, 1)xy0.2x−0.5y=−27.80.3x+ 0.4y= 68.750150−50100150250(101, 96)xy−123456−2123454x+y= 4232316x+y=

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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, youobtain 93x=or13.x=The solution is:13x=and12.y= −(e) The solutions in (c) and (d) are the same.36.(a)(b) Because each equation has the same line as a graph,there are infinitely many solutions.(c) All solutions of this system lie on the line53252142.yx=+So let,xt=then the solution set is,xt=53252142,yt=+wheretis any real number.(d) Adding 3 times the first equation to the secondequation you obtain5.32.11.2500.xy−+==Choosingxt=as the free variable, you obtain2.15.31.25yt=+or 215312.5yt=+or53252142.yt=+So, you can describe the solution setas,xt=53252142,yt=+wheretis any realnumber.(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= 0−5.3x+ 2.1y= 1.2515.9x−6.3y=−3.753−32−2

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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.Adding2−times the first equation to the fourthequation, yields14234242343420321463.xxxxxxxxxx+=−−=−=−+−= −Multiplying the fourth equation by1,−and interchangingit with the second equation, yields142342423434463321.20xxxxxxxxxx+=−+=−=−−=Adding3−times the second equation to the third, and2−times the second equation to the fourth, produces1423434343446312208.7136xxxxxxxxx+=−+=−= −−= −Dividing the third equation by 12 yields142343434523334463.7136xxxxxxxxx+=−+=−= −−= −Adding7−times the third equation to the fourth yields142343445233443334463.xxxxxxxx+=−+=−= −=Using back-substitution, the original system has exactlyone solution:11,x=21,x=31,x=and41.x=Answers may vary slightly for Exercises 58–60.58.Using a computer software program or graphing utility,you obtain0.8,x=1.2,y=2.4.z= −60.Using a computer software program or graphing utility,you obtain6.8813,163.3111,xy== −210.2915,z= −59.2913.w= −62.0xyz===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===64.0xyz===is clearly a solution.Dividing the first equation by 12 yields5112120.1240xyzxyz++=+−=Adding12−times the first equation to the second yields511212020.xyzyz++=−−=Lettingzt=be the free variable, you can describe thesolution as34,xt=2 ,yt=−and,zt=wheretis anyreal number.66.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 Equations768.(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 of the text, and will be proved later inTheorem 2.5.(b) True. See definition on page 6 of the text.(c) False. Consider the following system of three linearequations with two variables.236391.xyxyx+= −−−==The solution to this system is:1,5.xy== −70.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.72.Substituting1Ax=and1By=into the original systemyields230.25346ABAB+=−= −Reduce the system to row-echelon form.8120259122812025172ABABABA+=−= −+== −So,2534A= −and25.51B=Because1Ax=and1 ,By=the solution of the original system of equationsis:3425x=−and51.25y=74.Substituting1,Ax=1 ,By=and1Cz=into the original system yields225341230.ABCABABC+−=−= −++=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= −76.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 Equations78.Interchange the two equations and row reduce.()323232646146xykxyxykyk−= −+=−= −+=+So, if23,k= −there will be an infinite number ofsolutions.80.Reduce the system.()22142xkykyk+=−=−If1,k= ±there will be no solution.82.Interchange the first two equations and row reduce.02431xyzkykzkyz++=+=−−=If0,k=then there is an infinite number of solutions.Otherwise,024513.xyzyzz++=+==Because this system has exactly one solution, the answeris all0.k≠84.Reducing the system to row-echelon form produces()()5020102xyzyzaybzc++=−=−+−=()5020222.xyzyzabzc++=−=+−=So, you see that(a) if 2220,ab+−≠then there is exactly onesolution.(b) if 2220ab+−=and0,c=then there is aninfinite number of solutions.(c) if 2220ab+−=and0,c≠there is no solution.86.If1230,ccc===then the system is consistentbecause0xy==is a solution.88.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−≠90.The two lines coincide.23700xy−==Letting,yt=73 .2tx+=The graph does not change.92.212001312120xyxy−=−=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.x21−1−2−4−5145−2−33y

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Section 1.2Gaussian Elimination and Gauss-Jordan Elimination910.1232123225170971154760684−−−−−−⎡⎤⎡⎤⎢⎥⎢⎥−−⇒−−⎢⎥⎢⎥⎢⎥⎢⎥−−−⎣⎦⎣⎦Add 2 times Row 1 to Row 2.Add 5 times Row 1 to Row 3.12.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.211010101212121210102110101010100202010121102110101010100101010101100011⎡⎤⎡⎤⎢⎥⎢⎥−−⇒−−⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎡⎤⎡⎤⎢⎥⎢⎥⇒−−⇒⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎡⎤⎡⎤⎢⎥⎢⎥⇒⇒⎢⎥⎢⎥⎢⎥⎢⎥−⎣⎦⎣⎦Because the matrix is in row-echelon form, convert backto a system of linear equations.1323011xxxx+===By back-substitution,131.xx= −= −So, the solutionis:121,1,xx= −=and31.x=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 matrixisin 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 conclude that the system has no solution.

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10Chapter 1Systems of Linear Equations32.The augmented matrix for this system is2132402114 .7506−⎡⎤⎢⎥−⎢⎥⎢⎥−⎣⎦Gaussian elimination produces the following.33311122222212345135212242213241121121120211402114021140177506750607800⎡⎤⎡⎤⎡⎤−−−−⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥−⇒−⇒−⇒−⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥−−−−−−−⎣⎦⎣⎦⎣⎦⎣⎦Back-substitution now yields( )( )()32313211223311222267761012126108.xxxxxx==+=+==−+=−+=So, the solution is:128,10,xx==and36.x=34.The augmented matrix for this system is11531021 .2110−⎡⎤⎢⎥−⎢⎥⎢⎥−−⎣⎦Subtracting the first row from the second row yields a new second row.115301322110−⎡⎤⎢⎥−−⎢⎥⎢⎥−−⎣⎦Adding2−times the first row to the third row yields a new third row.115301320396−⎡⎤⎢⎥−−⎢⎥⎢⎥−−⎣⎦Multiplying the second row by1−yields a new second row.115301320396−⎡⎤⎢⎥−⎢⎥⎢⎥−−⎣⎦Adding 3 times the second row to the third row yields a new third row.115301320000−⎡⎤⎢⎥−⎢⎥⎢⎥⎣⎦Adding1−times the second row to the first row yields a new first row.102101320000−⎡⎤⎢⎥−⎢⎥⎢⎥⎣⎦Converting back to a system of linear equations produces13232132.xxxx−=−=Finally, choosing3xt=as the free variable, you can describe the solution as112 ,xt=+223 ,xt=+and3,xt=wheretis any real number.

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Section 1.2Gaussian Elimination and Gauss-Jordan Elimination1136The augmented matrix for this system is1218 .36321⎡⎤⎢⎥−−−−⎣⎦Gaussian elimination produces the following matrix.12180003⎡⎤⎢⎥⎣⎦Because the second row corresponds to the equation 03,=there is no solution to the original system.38.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 computer software program or graphing utility,you obtain123456112021.xxxxxx== −=== −=42.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.44.The corresponding equations are all 00.=So, there arethree free variables. So,1,xt=2,xs=and3,xr=where,, andtsrare any real numbers.

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12Chapter 1Systems of Linear Equations46.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.48.(a) IfAis theaugmentedmatrix of a system of linear equations, then the number of equations in this system is three(because it is equal to the number of rows of the augmented matrix). The number of variables is two because it isequal to the number of columns of the augmented matrix minus one.(b) Using Gaussian elimination on the augmented matrix of 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 linear equations, then the number of equations is three, because it is equalto the number of rows of the coefficient matrix. The number of variables is also three, because it is equal to the numberof columns of the coefficient matrix.Using Gaussian elimination onAyou obtain the following coefficient matrix of an equivalent system.31221006000k⎡⎤−⎢⎥+⎢⎥⎢⎥⎣⎦Because the homogeneous system is always consistent, the homogeneous system with the coefficient matrixAisconsistent for any value ofk.50.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.52.Because the system composed of Equations 1 and 2 is consistent, but has a free variable, this system must have an infinitenumber of solutions.54.Use Gauss-Jordan elimination as follows.1231231231014560360120127890612000000−⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⇒−−⇒⇒⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥−−⎣⎦⎣⎦⎣⎦⎣⎦

Solution Manual for Elementary Linear Algebra, 7th Edition - Page 15 preview image

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Section 1.2Gaussian Elimination and Gauss-Jordan Elimination1356.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⎡⎤⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦⎣⎦⎣⎦58.(a) False. A 47×matrix has 4 rows and 7 columns.(b) True. Reduced row-echelon form of a given matrix is unique while row-echelon form is not. (See also exercise 64of this section.)(c) True. See Theorem 1.1 on page 21.(d) False. Multiplying a row by anonzeroconstant is one of the elementary row operations. However, multiplying a rowof a matrix by a constant0c=isnotan elementary row operation. (This would change the system by eliminating theequation corresponding to this row.)60.No, the row-echelon form is not unique. For instance,1201⎡⎤⎢⎥⎣⎦and10 .01⎡⎤⎢⎥⎣⎦The reduced row-echelon form is unique.62.First, you need0a≠or0.c≠If0,a≠then you have.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≠64.Row reduce the augmented matrix for this system.()2101201002010120λλλλλλλ⎡⎤−⎡⎤⎡⎤⇒⇒ ⎢⎥⎢⎥⎢⎥−++−⎢⎥⎣⎦⎣⎦⎣⎦To have a nontrivial solution you must have()()220210.λλλλ−−=−+=So, if1λ= −or2,λ=the system will have nontrivial solutions.66.Answers will vary.Sample answer:Because the thirdrow consists of all zeros, choose a third equation that is amultiple of one of the other two equations.3241282xzyzyz+= −+=+=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.

Solution Manual for Elementary Linear Algebra, 7th Edition - Page 16 preview image

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14Chapter 1Systems of Linear EquationsSection 1.3Applications of Systems of Linear Equations2.(a) Because there are three points, choose a second-degree polynomial,()2012.p xaa xa x=++Then substitute0, 2,x=and 4 into( )p xand equate the results to0,2,y=−and 0, respectively.( )( )( )( )( )( )201202012012201201200022242444160aaaaaaaaaaaaaaaa++==++=++= −++=++=Use Gauss-Jordan elimination on the augmented matrix for this system.12100010001242010214160001⎡⎤⎡⎤⎢⎥⎢⎥−⇒−⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦So,( )2122.p xxx= −+(b)4.(a) Because there are three points, choose a second-degree polynomial,()2012.pxaa xa x=++Then substitute2, 3,x=and 4 into( )p xand equate the results to4, 4,y=and 4, respectively.( )( )( )( )( )( )2012012201201220120122224433394444164aaaaaaaaaaaaaaaaaa++=++=++=++=++=++=Use Gauss-Jordan elimination on the augmented matrix for this system.1244100413940100141640010⎡⎤⎡⎤⎢⎥⎢⎥⇒⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦So,( )4.p x=(b)xy−2246−2−44(0, 0)(2,−2)(4, 0)x(2, 4)(4, 4)(3, 4)112345235y

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