LINEARALGEBRAWWLCHENc WWLChen,1982,2008.ThischapteroriginatesfrommaterialusedbytheauthoratImperialCollege,UniversityofLondon,between1981and1990.Itisavailablefreetoallindividuals,ontheunderstandingthatitisnottobeusedfornancialgain,andmaybedownloadedand/orphotocopied,withorwithoutpermissionfromtheauthor.However,thisdocumentmaynotbekeptonanyinformationstorageandretrievalsystemwithoutpermissionfromtheauthor,unlesssuchsystemisnotaccessibletoanyindividualsotherthanitsowners.Chapter1LINEAREQUATIONS1.1. IntroductionExample1.1.1. Try to draw the two lines3x + 2y = 5,6x + 4y = 5.Itiseasytoseethatthetwolinesareparallelanddonotintersect,sothatthissystemoftwolinearequations has no solution.Example1.1.2. Try to draw the two lines3x + 2y = 5,x + y = 2.It is easy to see that the two lines are not parallel and intersect at the point (1, 1), so that this systemof two linear equations has exactly one solution.Example1.1.3. Try to draw the two lines3x + 2y = 5,6x + 4y = 10.It is easy to see that the two lines overlap completely,so that this system oftwo linearequations hasinnitely many solutions.Chapter1: LinearEquations page1of31LinearAlgebra c WWLChen,1982,2008In these three examples,we have shown that a system of two linear equations on the plane R2mayhave no solution, one solution or innitely many solutions. A natural question to ask is whether therecan be any other conclusion. Well, we can see geometrically that two lines cannot intersect at more thanone point without overlapping completely. Hence there can be no other conclusion.In general, we shall study a system ofm linear equations of the forma11x1 +a12x2 + . . . +a1nxn =b1,a21x1 +a22x2 + . . . +a2nxn =b2,...am1x1 + am2x2 + . . . + amnxn = bm,(1)withn variablesx1, x2, . . . , xn. Here we may not be so lucky as to be able to see geometrically what isgoing on. We therefore need to study the problem from a more algebraic viewpoint. In this chapter, weshall conne ourselves to the simpler aspects of the problem. In Chapter 6, we shall study the problemagain from the viewpoint of vector spaces.If we omit reference to the variables, then system (1) can be represented by the arraya11a12. . . a1na21a22. . . a2n.........am1am2. . . amn

b1b2...bm(2)of all the coecients. This is known as the augmented matrix of the system. Here the rst row of thearray represents the rst linear equation, and so on.We also writeAx = b, whereA =a11a12. . . a1na21a22. . . a2n.........am1am2. . . amnand b =b1b2...bmrepresent the coecients andx =x1x2...xnrepresents the variables.Example1.1.4. The array1 3 1 5 10 1 1 2 12 4 0 7 1

543(3)represents the system of 3 linear equationsx1 + 3x2 + x3 + 5x4 + x5 = 5,x2 + x3 + 2x4 + x5 = 4,2x1 + 4x2+ 7x4 + x5 = 3,(4)Chapter1: LinearEquations page2of31< 02 >LinearAlgebra c WWLChen,1982,2008with 5 variablesx1, x2, x3, x4, x5. We can also write1 3 1 5 10 1 1 2 12 4 0 7 1x1x2x3x4x5=543.1.2. ElementaryRowOperationsLet us continue with Example 1.1.4.Example1.2.1. Consider the array (3). Let us interchange the rst and second rows to obtain0 1 1 2 11 3 1 5 12 4 0 7 1

453.Then this represents the system of equationsx2 + x3 + 2x4 + x5 = 4,x1 + 3x2 + x3 + 5x4 + x5 = 5,2x1 + 4x2+ 7x4 + x5 = 3,(5)essentially the same as the system (4), the only dierence being that the rst and second equations havebeen interchanged. Any solution of the system (4) is a solution of the system (5), and vice versa.Example1.2.2. Consider the array (3). Let us add 2 times the second row to the rst row to obtain1 5 3 9 30 1 1 2 12 4 0 7 1

1343.Then this represents the system of equationsx1 + 5x2 + 3x3 + 9x4 + 3x5 = 13,x2 + x3 + 2x4 + x5 = 4,2x1 + 4x2+ 7x4 + x5 = 3,(6)essentially the same as the system (4), the only dierence being that we have added 2 times the secondequation to the rst equation. Any solution of the system (4) is a solution of the system (6), and viceversa.Example1.2.3. Consider the array (3). Let us multiply the second row by 2 to obtain1 3 1 5 10 2 2 4 22 4 0 7 1

583.Then this represents the system of equationsx1 + 3x2 + x3 + 5x4 + x5 = 5,2x2 + 2x3 + 4x4 + 2x5 = 8,2x1 + 4x2+ 7x4 + x5 = 3,(7)Chapter1: LinearEquations page3of31< 03 >LinearAlgebra c WWLChen,1982,2008essentiallythesameasthesystem(4), theonlydierencebeingthatthesecondequationhasbeenmultiplied through by 2. Any solution of the system (4) is a solution of the system (7), and vice versa.In the general situation, it is not dicult to see the following.PROPOSITION1A. (ELEMENTARY ROW OPERATIONS) Considerthearray (2)correspondingto the system (1).(a) Interchanging thei-th andj-th rows of (2) corresponds to interchanging thei-th andj-th equationsin (1).(b) Adding a multiple of thei-th row of (2) to thej-th row corresponds to adding the same multiple ofthei-th equation in (1) to thej-th equation.(c) Multiplyingthei-throwof(2)byanon-zeroconstantcorrespondstomultiplyingthei-thequationin (1) by the same non-zero constant.In all three cases, the collection of solutions to the system (1) remains unchanged.Letusinvestigatehowwemayuseelementaryrowoperationstohelpussolveasystemof linearequations. As a rst step, let us continue again with Example 1.1.4.Example1.2.4. Consider again the system of linear equationsx1 + 3x2 + x3 + 5x4 + x5 = 5,x2 + x3 + 2x4 + x5 = 4,2x1 + 4x2+ 7x4 + x5 = 3,(8)represented by the array1 3 1 5 10 1 1 2 12 4 0 7 1

543. (9)Let us now perform elementary row operations on the array (9). At this point, do not worry if you donot understand why we are taking the following steps. Adding 2 times the rst row of (9) to the thirdrow, we obtain1 3 1 5 10 1 1 2 10 2 2 3 1

547.From here, we add 2 times the second row to the third row to obtain1 3 1 5 10 1 1 2 10 0 0 1 1

541. (10)Next, we add 3 times the second row to the rst row to obtain1 0 2 1 20 1 1 2 10 0 0 1 1

741.Next, we add the third row to the rst row to obtain1 0 2 0 10 1 1 2 10 0 0 1 1

641.Chapter1: LinearEquations page4of31< 04 >LinearAlgebra c WWLChen,1982,2008Finally, we add 2 times the third to row to the second row to obtain1 0 2 0 10 1 1 0 10 0 0 1 1

621. (11)We remark here that the array (10) is said to be in row echelon form, while the array (11) is said to bein reduced row echelon form precise denitions will follow in Sections 1.51.6. Let us see how we maysolve the system (8) by using the arrays (10) or (11). First consider (10). Note that this represents thesystemx1 + 3x2 + x3 + 5x4 + x5 = 5,x2 + x3 + 2x4 + x5 = 4,x4 + x5 = 1.(12)First of all, take the third equationx4 + x5 = 1.If we letx5 = t, thenx4 = 1 t. Substituting these into the second equation, we obtain (you must dothe calculation here)x2 + x3 = 2 + t.If we letx3 = s, thenx2 = 2 +t s. Substituting all these into the rst equation, we obtain (you mustdo the calculation here)x1 = 6 + t + 2s.Hencex = (x1, x2, x3, x4, x5) = (6 + t + 2s, 2 + t s, s, 1 t, t)is a solution of the system (12) for every s, t R. In view of Proposition 1A, these are also precisely thesolutions of the system (8). Alternatively, consider (11) instead. Note that this represents the systemx12x3x5 = 6,x2 + x3x5 = 2,x4 + x5 = 1.(13)First of all, take the third equationx4 + x5 = 1.If we letx5 = t, thenx4 = 1 t. Substituting these into the second equation, we obtain (you must dothe calculation here)x2 + x3 = 2 + t.If we letx3 = s, thenx2 = 2 +t s. Substituting all these into the rst equation, we obtain (you mustdo the calculation here)x1 = 6 + t + 2s.Hencex = (x1, x2, x3, x4, x5) = (6 + t + 2s, 2 + t s, s, 1 t, t)Chapter1: LinearEquations page5of31< 05 >LinearAlgebra c WWLChen,1982,2008isasolutionofthesystem(13)foreverys, t R. InviewofProposition1A,thesearealsopreciselythe solutions of the system (8). However, if you have done the calculations as suggested, you will noticethat the calculation is easier for the system (13) than for the system (12). This is clearly a case of thearray (11) in reduced row echelon form having more 0s than the array (10) in row echelon form, so thatthe system (13) has fewer non-zero coecients than the system (12).1.3. RowEchelonFormDefinition. A rectangular array of numbers is said to be in row echelon form if the following conditionsare satised:(1) The left-most non-zero entry of any non-zero row has value 1. These are called the pivot entries.(2) All zero rows are grouped together at the bottom of the array.(3) The pivot entry of a non-zero row occurring lower in the array is to the right of the pivot entry ofa non-zero row occurring higher in the array.Next,weinvestigatehowwemayreduceagivenarraytorowechelonform. Weshallillustratetheideas by working on an example.Example1.3.1. Consider the array0 0 5 0 15 50 2 4 7 1 30 1 2 3 0 10 1 2 4 1 2.Step 1: Locate the left-most non-zero column and cover all columns to the left of this column (in ourillustration here, denotes an entry that has been covered). We now have 0 5 0 15 5 2 4 7 1 3 1 2 3 0 1 1 2 4 1 2.Step2: Considerthepartof thearraythatremainsuncovered. Byinterchangingrowsif necessary,ensure that the top-left entry is non-zero. So let us interchange rows 1 and 4 to obtain 1 2 4 1 2 2 4 7 1 3 1 2 3 0 1 0 5 0 15 5.Step 3: If the top entry on the left-most uncovered column is a, then we multiply the top uncovered rowby 1/a to ensure that this entry becomes 1. So let us divide row 1 by 1 to obtain 1 2 4 1 2 2 4 7 1 3 1 2 3 0 1 0 5 0 15 5!Step4: Wenowtrytomakeallentriesbelowthetopentryontheleft-mostuncoveredcolumnzero.This can be achieved by adding suitable multiples of row 1 to the other rows. So let us add 2 timesrow 1 to row 2 to obtain 1 2 4 1 2 0 0 1 1 1 1 2 3 0 1 0 5 0 15 5.Chapter1: LinearEquations page6of31< 06 >LinearAlgebra c WWLChen,1982,2008Then let us add 1 times row 1 to row 3 to obtain 1 2 4 1 2 0 0 1 1 1 0 0 1 1 1 0 5 0 15 5.Step 5: Now cover the top row. We then obtain 0 0 1 1 1 0 0 1 1 1 0 5 0 15 5.Step 6: Repeat Steps 15 on the uncovered array, and as many times as necessary so that eventually thewhole array gets covered. So let us continue. Following Step 1, we locate the left-most non-zero columnand cover all columns to the left of this column. We now have 0 1 1 1 0 1 1 1 5 0 15 5.Following Step 2, we interchanging rows if necessary to ensure that the top-left entry is non-zero. So letus interchange rows 1 and 3 (here we do not count any covered rows) to obtain 5 0 15 5 0 1 1 1 0 1 1 1.FollowingStep3, wemultiplythetoprowbyasuitablenumbertoensurethatthetopentryontheleft-most uncovered column becomes 1. So let us multiply row 1 by 1/5 to obtain 1 0 3 1 0 1 1 1 0 1 1 1.Following Step 4, we do nothing! Following Step 5, we cover the top row. We then obtain 0 1 1 1 0 1 1 1.FollowingStep1, welocatetheleft-mostnon-zerocolumnandcoverall columnstotheleftof thiscolumn. We now have 1 1 1 1 1 1.FollowingStep2,wedonothing! FollowingStep3,wemultiplythetoprowbyasuitablenumbertoensure that the top entry on the left-most uncovered column becomes 1. So let us multiply row 1 by 1to obtain 1 1 1 1 1 1.Chapter1: LinearEquations page7of31< 07 >LinearAlgebra c WWLChen,1982,2008Following Step 4, we now try to make all entries below the top entry on the left-most uncovered columnzero. So let us add row 1 to row 2 to obtain 1 1 1 0 0 0.Following Step 5, we cover the top row. We then obtain 0 0 0.FollowingStep1, welocatetheleft-mostnon-zerocolumnandcoverall columnstotheleftof thiscolumn. We now have .Step . Uncover everything! We then have0 1 2 4 1 20 0 1 0 3 10 0 0 1 1 10 0 0 0 0 0,in row echelon form.In practice, we do not actually cover any entries of the array, so let us repeat here the same argumentwithout covering anything the reader is advised to compare this with the earlier discussion. We startwith the array0 0 5 0 15 50 2 4 7 1 30 1 2 3 0 10 1 2 4 1 2.Interchanging rows 1 and 4, we obtain0 1 2 4 1 20 2 4 7 1 30 1 2 3 0 10 0 5 0 15 5.Adding 2 times row 1 to row 2, and adding 1 times row 1 to row 3, we obtain0 1 2 4 1 20 0 0 1 1 10 0 0 1 1 10 0 5 0 15 5.Interchanging rows 2 and 4, we obtain0 1 2 4 1 20 0 5 0 15 50 0 0 1 1 10 0 0 1 1 1.Chapter1: LinearEquations page8of31< 08 >LinearAlgebra c WWLChen,1982,2008Multiplying row 1 by 1/5, we obtain0 1 2 4 1 20 0 1 0 3 10 0 0 1 1 10 0 0 1 1 1.Multiplying row 3 by 1, we obtain0 1 2 4 1 20 0 1 0 3 10 0 0 1 1 10 0 0 1 1 1.Adding row 3 to row 4, we obtain0 1 2 4 1 20 0 1 0 3 10 0 0 1 1 10 0 0 0 0 0,in row echelon form.Remarks. (1) As already observed earlier, we do not actually physically cover rows or columns. In anypractical situation, we simply copy these entries without changes.(2)Thestepsindicatedthetherstpartofthelastexampleareforguidanceonly. Inpractice,wedo not have to follow the steps above religiously, and what we do is to a great extent dictated by goodcommon sense. For instance, suppose that we are faced with the array

2 3 2 13 2 0 2

.If wefollowthestepsreligiously, thenweshall multiplyrow1by1/2. However, notethatthiswillintroducefractionstosomeentriesof thearray, andanysubsequentcalculationwill becomerathermessy. Instead, let us multiply row 1 by 3 to obtain

6 9 6 33 2 0 2

.Then let us multiply row 2 by 2 to obtain

6 9 6 36 4 0 4

.Adding 1 times row 1 to row 2, we obtain

6 9 6 30 5 6 1

.In this way, we have avoided the introduction of fractions until later in the process. In general, if we startwith an array with integer entries, then it is possible to delay the introduction of fractions by omittingStep 3 until the very end.Chapter1: LinearEquations page9of31< 09 >LinearAlgebra c WWLChen,1982,2008Example1.3.2. Consider the array2 1 3 2 51 3 2 4 13 2 0 0 2.Try following the steps indicated in the rst part of the previous example religiously and try to see howcomplicated the calculations get. On the other hand, we can modify the steps with some common sense.First of all, we interchange rows 1 and 2 to obtain1 3 2 4 12 1 3 2 53 2 0 0 2.The reason for taking this step is to put an entry 1 at the top left without introducing fractions anywhere.When we next add multiples of row 1 to the other rows to make 0s below this 1, we do not introducefractions either. Now adding 2 times row 1 to row 2, we obtain1 3 2 4 10 5 1 6 33 2 0 0 2.Adding 3 times row 1 to row 3, we obtain1 3 2 4 10 5 1 6 30 7 6 12 1.Next, multiplying row 2 by 7, we obtain1 3 2 4 10 35 7 42 210 7 6 12 1.Multiplying row 3 by 5, we obtain1 3 2 4 10 35 7 42 210 35 30 60 5.Note that here we are essentially covering up row 1. Also, we have multiplied rows 2 and 3 by suitablemultiplestothattheirleadingnon-zeroentriesarethesame, inpreparationfortakingthenextstepwithout introducing fractions. Now adding 1 times row 2 to row 3, we obtain1 3 2 4 10 35 7 42 210 0 23 18 26.Here, the array is almost in row echelon form, except that the leading non-zero entries in rows 2 and 3arenotequalto1. However,wecanalwaysmultiplyrow2by1/35androw3by1/23ifwewanttoobtain the row echelon form1 3 2 4 10 1 1/5 6/5 3/50 0 1 18/23 26/23.Ifthisdiersfromtheansweryougotwhenyoufollowedthestepsindicatedinthepreviousexamplereligiously, do not worry. row echelon forms are not unique!Chapter1: LinearEquations page10of31< 10 >LinearAlgebra c WWLChen,1982,20081.4. ReducedRowEchelonFormDefinition. A rectangular array of numbers is said to be in reduced row echelon form if the followingconditions are satised:(1) The left-most non-zero entry of any non-zero row has value 1. These are called the pivot entries.(2) All zero rows are grouped together at the bottom of the array.(3) The pivot entry of a non-zero row occurring lower in the array is to the right of the pivot entry ofa non-zero row occurring higher in the array.(4) Each column containing a pivot entry has 0s everywhere else in the column.We now investigate how we may reduce a given array to reduced row echelon form. Here, we basicallytakeanextrasteptoconvertanarrayfromrowechelonformtoreducedrowechelonform. Weshallillustrate the ideas by continuing on an earlier example.Example1.4.1. Consider again the array0 0 5 0 15 50 2 4 7 1 30 1 2 3 0 10 1 2 4 1 2.We have already shown in Example 1.3.1 that this array can be reduced to row echelon form0 1 2 4 1 20 0 1 0 3 10 0 0 1 1 10 0 0 0 0 0.Step 1: Cover all zero rows at the bottom of the array. We now have0 1 2 4 1 20 0 1 0 3 10 0 0 1 1 1 .Step 2: We now try to make all the entries above the pivot entry on the bottom row zero (here again wedo not count any covered rows). This can be achieved by adding suitable multiples of the bottom rowto the other rows. So let us add 4 times row 3 to row 1 to obtain0 1 2 0 3 20 0 1 0 3 10 0 0 1 1 1 .Step 3: Now cover the bottom row. We then obtain0 1 2 0 3 20 0 1 0 3 1 .Step4: RepeatSteps23ontheuncoveredarray, andasmanytimesasnecessarysothateventuallythe whole array gets covered. So let us continue. Following Step 2, we add 2 times row 2 to row 1 toobtain0 1 0 0 9 40 0 1 0 3 1 .Chapter1: LinearEquations page11of31< 11 >LinearAlgebra c WWLChen,1982,2008Following Step 3, we cover row 2 to obtain0 1 0 0 9 4 .Following Step 2, we do nothing! Following Step 3, we cover row 1 to obtain .Step . Uncover everything! We then have0 1 0 0 9 40 0 1 0 3 10 0 0 1 1 10 0 0 0 0 0,in reduced row echelon form.Again, inpractice, wedonotactuallycoveranyentriesof thearray, soletusrepeatherethesameargument without covering anything the reader is advised to compare this with the earlier discussion.We start with the row echelon form0 1 2 4 1 20 0 1 0 3 10 0 0 1 1 10 0 0 0 0 0.Adding 4 times row 3 to row 1, we obtain0 1 2 0 3 20 0 1 0 3 10 0 0 1 1 10 0 0 0 0 0.Adding 2 times row 2 to row 1, we obtain0 1 0 0 9 40 0 1 0 3 10 0 0 1 1 10 0 0 0 0 0,in reduced row echelon form.1.5. SolvingaSystemofLinearEquationsLetusrstsummarizewhatwehavedonesofar. Westudyasystem(1)of mlinearequationsinnvariablesx1, . . . , xn. Ifweomitreferencetothevariables, thenthesystem(1)canberepresentedbythearray(2),withmrowsandn + 1columns. Wenextreducethearray(2)torowechelonformorreduced row echelon form by elementary row operations.ByProposition1A,thesystemoflinearequationsrepresentedbythearrayinrowechelonformorreduced row echelon form has the same solution set as the system (1). It follows that to solve the systemChapter1: LinearEquations page12of31< 12 >LinearAlgebra c WWLChen,1982,2008(1), it remains to solve the system represented by the array in row echelon form or reduced row echelonform. We now describe a simple way to obtain all solutions of this system.Definition. Any column of an array (2) in row echelon form or reduced row echelon form containing apivot entry is called a pivot column.First of all, let us eliminate the situation when the system has no solutions. Suppose that the array(2) has been reduced to row echelon form, and that this contains a row of the form0 . . . 0

n1corresponding to the last column of the array being a pivot column. This row represents the equation0x1 + . . . + 0xn = 1;clearly the system cannot have any solution.Definition. Suppose that the array (2) in row echelon form or reduced row echelon form satises theconditionthatitslastcolumnisnotapivotcolumn. Thenanyvariablexicorrespondingtoapivotcolumn is called a pivot variable. All other variables are called free variables.Example1.5.1. Consider the array0 1 0 0 90 0 1 0 30 0 0 1 10 0 0 0 0

4110,representing the systemx29x5 = 4,x3+ 3x5 = 1,x4 + x5 = 1.Note that the zero row in the array represents an equation which is trivial! Here the last column of thearray is not a pivot column. Now columns 2, 3, 4 are the pivot columns, so thatx2, x3, x4are the pivotvariables andx1, x5are the free variables.To solve the system, we allow the free variables to take any values we choose, and then solve for thepivot variables in terms of the values of these free variables.Example1.5.2. Consider the system of 4 linear equations5x3+ 15x5 = 5,2x2 + 4x3 + 7x4 + x5 = 3,x2 + 2x3 + 3x4= 1,x2 + 2x3 + 4x4 + x5 = 2,(14)in the 5 variablesx1, x2, x3, x4, x5. If we omit reference to the variables, then the system can be repre-sented by the array0 0 5 0 150 2 4 7 10 1 2 3 00 1 2 4 1

5312. (15)Chapter1: LinearEquations page13of31< 13 >LinearAlgebra c WWLChen,1982,2008As in Example 1.3.1, we can reduce the array (15) to row echelon form0 1 2 4 10 0 1 0 30 0 0 1 10 0 0 0 0

2110, (16)representing the systemx2 + 2x3 + 4x4 + x5 = 2,x3+ 3x5 = 1,x4 + x5 = 1.(17)Alternatively, as in Example 1.4.1, we can reduce the array (15) to reduced row echelon form0 1 0 0 90 0 1 0 30 0 0 1 10 0 0 0 0

4110, (18)representing the systemx29x5 = 4,x3+ 3x5 = 1,x4 + x5 = 1.(19)By Proposition 1A, the three systems (14), (17) and (19) have exactly the same solution set. Now, weobservefrom(16)or(18)thatcolumns2, 3, 4arethepivotcolumns, sothatx2, x3, x4arethepivotvariables andx1, x5are the free variables. Ifwe assignvaluesx1=s andx5=t,thenwe have,from(17) (harder) or (19) (easier), that(x1, x2, x3, x4, x5) = (s, 9t 4, 3t + 1, t + 1, t). (20)It follows that (20) is a solution of the system (14) for everys, t R.Example 1.5.3. Let us return to Example 1.2.4, and consider again the system (8) of 3 linear equations inthe 5 variables x1, x2, x3, x4, x5. If we omit reference to the variables, then the system can be representedby the array (9). We can reduce the array (9) to row echelon form (10), representing the system (12).Alternatively, wecanreducethearray(9)toreducedrowechelonform(11), representingthesystem(13). By Proposition 1A, the three systems (8), (12) and (13) have exactly the same solution set. Now,we observe from (10) or (11) that columns 1, 2, 4 are the pivot columns, so thatx1, x2, x4are the pivotvariables andx3, x5are the free variables. Ifwe assignvaluesx3=s andx5=t,thenwe have,from(12) (harder) or (13) (easier), that(x1, x2, x3, x4, x5) = (6 + t + 2s, 2 + t s, s, 1 t, t). (21)It follows that (21) is a solution of the system (8) for everys, t R.Example1.5.4.Inthisexample, wedonotbothereventoreducethematrixtorowechelonform.Consider the system of 3 linear equations2x1 + x2 + 3x3 + 2x4 = 5,x1 + 3x2 + 2x3 + 4x4 = 1,3x1 + 2x2= 2,(22)Chapter1: LinearEquations page14of31< 14 >LinearAlgebra c WWLChen,1982,2008in the 4 variables x1, x2, x3, x4. If we omit reference to the variables, then the system can be representedby the array2 1 3 21 3 2 43 2 0 0

512. (23)As in Example 1.3.2, we can reduce the array (23) to the form1 3 2 40 35 7 420 0 23 18

12126, (24)representing the systemx1 + 3x2 + 2x3 + 4x4 = 1,35x2 + 7x3 + 42x4 = 21,23x3 + 18x4 = 26.(25)Notethatthearray(24)isalmostinrowechelonform, exceptthatthepivotentriesarenot1. ByProposition1A, thetwosystems(22)and(25)haveexactlythesamesolutionset. Now, weobservefrom (24) that columns 1, 2, 3 are the pivot columns, so thatx1, x2, x3are the pivot variables andx4isthe free variable. If we assign valuesx4 = s, then we have, from (25), that(x1, x2, x3, x4) =

1623s + 2823, 2423s 1923, 1823s + 2623, s

. (26)It follows that (26) is a solution of the system (22) for everys R.1.6. HomogeneousSystemsConsider a homogeneous system ofm linear equations of the forma11x1 +a12x2 + . . . +a1nxn = 0,a21x1 +a22x2 + . . . +a2nxn = 0,...am1x1 + am2x2 + . . . + amnxn = 0,(27)with n variables x1, x2, . . . , xn. If we omit reference to the variables, then system (27) can be representedby the arraya11a12. . . a1na21a22. . . a2n.........am1am2. . . amn

00...0(28)of all the coecients.Note that the system (27) always has a solution, namely the trivial solutionx1 = x2 = . . . = xn = 0.Indeed, if wereducethearray(28)torowechelonformorreducedrowechelonform, thenitisnotdicult to see that the last column is a zero column and so cannot be a pivot column.Chapter1: LinearEquations page15of31< 15 >LinearAlgebra c WWLChen,1982,2008On the other hand,if the system (27) has a non-trivial solution,then we can multiply this solutionby any non-zero real number dierent from 1 to obtain another non-trivial solution. We have thereforeproved the following simple result.PROPOSITION 1B. The homogeneous system (27) either has the trivial solution as its only solutionor has innitely many solutions.The purpose of this section is to discuss the following stronger result.PROPOSITION 1C. Suppose that the system (27) has more variables than equations; in other words,suppose thatn > m. Then there are innitely many solutions.To see this, let us consider the array (28) representing the system (27). Note that (28) hasm rows,correspondingtothenumberof equations. Also(28)has n + 1columns, wherenisthenumberofvariables. However, thecolumnof (28)ontheextremerightisazerocolumn, correspondingtothefactthatthesystemishomogeneous. Furthermore,thiscolumnremainsazerocolumnifweperformelementary row operations on the array (28). If we now reduce (28) to row echelon form by elementaryrow operations, then there are at most m pivot columns, since there are only m equations in (27) and mrows in (28). It follows that if we exclude the zero column on the extreme right, then the remainingncolumns cannot all be pivot columns. Hence at least one of the variables is a free variable. By assigningthis free variable arbitrary real values, we end up with innitely many solutions for the system (27).1.7. ApplicationtoNetworkFlowSystemsof linearequationsarisewhenweinvestigatetheowof somequantitythroughanetwork.Such networks arise in science, engineering and economics. Two such examples are the pattern of tracow through a city and distribution of products from manufacturers to consumers through a network ofwholesalers and retailers.A network consists of a set of points, called the nodes, and directed lines connecting some or all of thenodes. The ow is indicated by a number or a variable. We observe the following basic assumptions:The total ow into a node is equal to the total ow out of a node.The total ow into the network is equal to the total ow out of the network.Example1.7.1. The picture below represents a system of one way streets in a particular part of somecity and the trac ow along the streets between the junctions:x1200200 A x2B 300x3x4300 C x5D 500400 300Chapter1: LinearEquations page16of31< 16 >LinearAlgebra c WWLChen,1982,2008We now have a system of three linear equationsI1 + I2 I3 = 0,13I16I2= 20,13I1+ 8I3 = 50.(30)The augmented matrix is given by1 1 113 6 013 0 8

02050, with reduced row echelon form1 0 00 1 00 0 1

213.HenceI1 = 2,I2 = 1 andI3 = 3. Note here that we have not considered the right hand loop. Supposeagainthatwetakethepositivedirectiontobeclockwise. ByOhmslaw, thevoltagedropacrossthe8 resistor is 8I3, while the voltage drop across the 6 resistor is 6I2. On the other hand, the voltagedropacross the30V electrical sourceis negative. TheVoltagelawappliedtothis loopthengives8I3 + 6I230 = 0. But this equation can be obtained by combining the last two equations in (30).1.9. ApplicationtoEconomicsInthissection, wedescribeasimpleexchangemodel duetotheeconomistLeontief. Aneconomyisdividedintosectors. Weknowthetotaloutputforeachsectoraswellashowoutputsareexchangedamong the sectors. The value of the total output of a given sector is known as the price of the output.Leontief has shown that there exist equilibrium prices that can be assigned to the total output of thesectors in such a way that the income for each sector is exactly the same as its expenses.Example 1.9.1. An economy consists of three sectors A, B, C which purchase from each other accordingto the table below:proportion of output from sectorA B Cpurchased by sectorA 0.2 0.6 0.1purchased by sectorB 0.4 0.1 0.5purchased by sectorC 0.4 0.3 0.4LetpA, pB, pCdenote respectively the value of the total output of sectorsA, B, C. For the expense tomatch the value for each sector, we must have0.2pA + 0.6pB + 0.1pC =pA,0.4pA + 0.1pB + 0.5pC = pB,0.4pA + 0.3pB + 0.4pC = pC,leading to the homogeneous linear equations0.8pA0.6pB0.1pC = 0,0.4pA0.9pB + 0.5pC = 0,0.4pA + 0.3pB0.6pC = 0,giving rise to the augmented matrix0.8 0.6 0.10.4 0.9 0.50.4 0.3 0.6

000, or simply8 6 14 9 54 3 6

000.Chapter1: LinearEquations page21of31< 17 >LinearAlgebra c WWLChen,1982,2008This can be reduced by elementary row operations to16 0 130 12 110 0 0

000,leading to the solution (pA, pB, pC) = t(1316,1112, 1) if we assign the free variable pC = t, or to the solution(pA, pB, pC) = t(39, 44, 48) if we assign the free variablepC= 48t, wheret is a real parameter. For thelatter,the choicet = 106gives rise to the prices of 39,44 and 48 million for the three sectorsA, B, Crespectively.1.10. ApplicationtoChemistryChemical equations consist of reactants and products. The problem is to balance such equations so thatthe following two rules apply:Conservation of mass: No atoms are produced or destroyed in a chemical reaction.Conservation of charge: The total charge of reactants is equal to the total charge of the products.Example1.10.1. Considertheoxidationof ammoniatoformnitricoxideandwater, givenbythechemical equation(x1)NH3 + (x2)O2 (x3)NO + (x4)H2O.Here the reactants are ammonia (NH3) and oxygen (O2), while the products are nitric oxide (NO) andwater (H2O). Our problem is to nd the smallest positive integer values ofx1, x2, x3, x4such that theequation balances. To do this, the technique is to equate the total number of each type of atoms on thetwo sides of the chemical equation:atomN: x1 = x3,atomH: 3x1 = 2x4,atomO: 2x2 = x3 + x4.These give rise to a homogeneous system of 3 linear equationsx1x3= 0,3x12x4 = 0,2x2x3 x4 = 0,in the 4 variablesx1, . . . , x4, with augmented matrix1 0 1 03 0 0 20 2 1 1

000,which can be simplied by elementary row operations to1 0 1 00 2 1 10 0 3 2

000,leadingtothegeneral solution(x1, . . . , x4)=t(23,56,23, 1)if weassignthefreevariablex4=t. Thechoicet = 6 gives rise to the smallest positive integer solution (x1, . . . , x4) = (4, 5, 4, 6), leading to thebalanced chemical equation4NH3 + 5O2 4NO + 6H2O.Chapter1: LinearEquations page22of31LinearAlgebra c WWLChen,1982,2008ProblemsforChapter11. Consider the system of linear equations2x1 + 5x2 + 8x3 = 2,x1 + 2x2 + 3x3 = 4,3x1 + 4x2 + 4x3 = 1.a) Write down the augmented matrix for this system.b) Reduce the augmented matrix by elementary row operations to row echelon form.c) Use your answer in part (b) to solve the system of linear equations.2. Consider the system of linear equations4x1 + 5x2 + 8x3 = 0,x1+ 3x3 = 6,3x1 + 4x2 + 6x3 = 9.a) Write down the augmented matrix for this system.b) Reduce the augmented matrix by elementary row operations to row echelon form.c) Use your answer in part (b) to solve the system of linear equations.3. Consider the system of linear equationsx1 x2 7x3 + 7x4 = 5,x1 + x2 + 8x3 5x4 = 7,3x12x217x3 + 13x4 =14,2x1 x211x3 + 8x4 = 7.a) Write down the augmented matrix for this system.b) Reduce the augmented matrix by elementary row operations to row echelon form.c) Use your answer in part (b) to solve the system of linear equations.4. Solve the system of linear equationsx+3y2z = 4,2x+7y +2z =10.5. For each of the augmented matrices below, reduce the matrix to row echelon or reduced row echelonform, and solve the system of linear equations represented by the matrix:a)1 1 2 13 2 1 34 3 1 42 1 3 2

56111 b)1 2 3 32 5 3 127 1 8 5

1276. Reduce each of the following arrays by elementary row operations to reduced row echelon form:a)1 2 3 4 50 2 3 4 50 0 3 4 50 0 0 4 5 b)1 1 0 0 00 1 1 0 00 0 1 1 00 0 0 1 1c)1 11 21 31 41 512 12 22 32 42 523 13 23 33 43 53Chapter1: LinearEquations page26of31< 19 >LinearAlgebra c WWLChen,1982,20087. Consider a system of linear equations in ve variables x = (x1, x2, x3, x4, x5) and expressed in matrixformAx = b, where x is written as a column matrix. Suppose that the augmented matrix (A|b)can be reduced by elementary row operations to the row echelon form1 3 2 0 60 0 1 1 20 0 0 1 10 0 0 0 0

4170.a) Which are the pivot variables and which are the free variables?b) Determine all the solutions of the system of linear equations.8. Consider a system of linear equations in ve variables x = (x1, x2, x3, x4, x5) and expressed in matrixformAx = b, where x is written as a column matrix. Suppose that the augmented matrix (A|b)can be reduced by elementary row operations to the row echelon form1 2 0 3 10 1 3 1 20 0 0 1 10 0 0 0 0

5340.a) Which are the pivot variables and which are the free variables?b) Determine all the solutions of the system of linear equations.9. Consider the system of linear equationsx1 + x2 x3 = 1,2x1 + x2 + 2x3 = 5 + 1,x1 x2 + 3x3 = 4 + 2,x12x2 + 7x3 = 10 1.a) Reduce its associated augmented matrix to row echelon form.[Hint: Afteroneortwosteps, wewillndthecalculationsextremelyunpleasant, particularlysincewedonotknowwhetheriszeroornon-zero. Tryrewritingthesystemofequationsasasysteminthevariablesx1, x3, x2,sothatcolumns2and3oftheaugmentedmatrixarenowswapped.]b) Find a value of for which the system is soluble.c) Solve the system.10. Find the minimum value forx4in the following system of one way streets:120 15050 A x1B 80x2x3100 C x4D 100Chapter1: LinearEquations page27of31< 20 >