The solution is often represented as a vector: 2xI + 4X2 + 2X3 + 4X4 + 2X5 = 4 2xI + 4X2 + 3X3 + 3X4 + 3X5 = 4. 3xI + 6X2 + 6X3 + 3X4 + 6X5 = 6.


 Gervais Geoffrey Underwood
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1 [~ o o : o o ill] i 1. Mrices, Vecors, nd GussJordn Eliminion 1 x y = =  z= The soluion is ofen represened s vecor: n his exmple, he process of eliminion works very smoohly. We cn elimine ll enries off he digonl nd cn mke ech coefficien on he digonl equl o. The process of eliminion works well unless we encouner zero long he digonl. These zeros represen missing erms in some equions. The following exmple illusres how o solve such sysem: X  X  Xs = x + X + X + X + X5 = x + X + X + X + X5 = x + X + X + X + X5 =  n The ugmened mrix of his sysem is M= l~ As in he previous exmples, we re rying o bring he mrix ino digonl form. To keep rck of our work, we will plce cursor in he mrix, s you migh on compuer screen. niilly, he cursor is plced he op posiion of he firs nonzero column of he mrix: o  i i.  i J Our firs gol is o mke he cursor enry equl o. We cn ccomplish seps s follows: his in wo Sep 1 f he cursor enry is, swp he cursor row wih some row below o mke he cursor enry nonzero.8 8To mke he process unmbiguous, swp he cursor row wih he,firs row below h hs nonzero enry in he cursor column.
2 1 CHAPTER 1 Liner Equions ] XJ  n Sep 1 X.j X5 = x , + we x x x~ re + merely x~ x~ xj + xj xj xj wriing x.j + x.j x.j X5 down + = X5 X5 X5 he = equions in differen order. This will cerinly ir no ffec he XJ soluions  X.j of he X5 sysem: = 1J Now we cn proceed s in he previous exmples. n i] Sep o. Divide he cursor row by he cursor enry o mke he cursor enry equl Sep  1X.j X5  = does X5 no = x x chnge + x~ x~ he + soluions x, xj xj XJ +  x.j of + he X5 X5 sysem, = becuse he equion con'esponding o he cursor row hs he sme soluions before nd fer he operion: [/~ x, XJ + x~ X x.j + x, + x.j + X5 = 1: irx ;] Sep muliples Elimine of he cursor ll oher row enries from heinoher cursor rows.'! column by subrcing suible Jj j, f 1, ]   ()x, + X + xj + xjx.j XJ  + x.j X5 + = X5 =  ()  ~]X X5 X5 X.j  + = XJ X5 X + = + x.j x~ x~ + + XJ XJ  [/~ ldd muliple of row. of course. Think of his s subrcing negive muliple of row. :]  () x + X + xj + x.j + X5 = 1 ()
3 1. Mrices, Vecors, nd GussJordn Eliminion 15 Convince yourself h his operion does no chnge he soluions of he sysem. (See Exercise 8.) Now we hve ken cre of he firs column (he firs vrible), so we cn move he cursor o new posiion. Following he pproch ken in Secion 1.1, we move he cursor down digonlly (i.e., one row down nd one column o he righ): o / o r o Le's summrize. For our mehod o work s before, we need nonzero cursor enry. Since no only he cursor enry, bu lso ll enries below. re zero. we cnno ccomplish his by swpping he cursor row wih some row below, s we did in Sep. would no help us o swp he cursor row wih he row bove; his would ffec he firs column of he mrix. which we hve lredy fixed. Thus, we hve o give up on he second column (he second vrible); we will move he cursor o he nex column:  /1 n r~ /1! Sep! Sep! Sep r~ Sep Move he cursor down digonlly (i.e., one row down nd one column o he righ). f he new cursor enry nd ll enries below re zero. move he cursor o he nex column (remining in he sme row). Repe his sep if necessry. Then reurn o Sep. Here. / () since  i:1  /i ;.  () he cursor enry is. we cn proceed direcly o Sep nd elimine ll oher ~~] enries in he cursor column: ~] :~] r~
4 1 CHAPTER 1 Liner Equions . Sep [~   :1 : i  + () () /1 : / :. i 1 : 1 () When we ry o pply Sep o his mrix, we run ou of columns; he process row reducion comes o n end. We sy h he mrix E is in reduced rowecheloi form, or rref for shor. We wrie E = rref(m), where M is he ugmened mrix of he sysem. Reduced rowechelon form A mrix is in reduced rowechelon form if i sisfies ll of he following condiions:. f row hs nonzero enries, hen he firs nonzero enry is, clled he leding in his row. o b. f column conins leding, hen ll oher enries in h column re zero. c. f row conins leding, hen ech row bove conins leding furher o he lef. E= L A mrix in reduced rowechelon form my conin rows of zeros, s in he pre ceding exmple. However, by condiion c, hese rows mus pper s he ls row, of he mrix. Convince yourself h he procedure jus oulined (repeedly performing Seps hrough ) indeed produces mrix wih hese hree properies. See Exercise. For emphsis, le's circle he leding 's in he mrix E = rref(m) we foun( erlier:  X CD = X =   X+ This mrix represens he following sysem: CD Agin, we sy h his sysem is in reduced rowechelon form. The ledin, vribles correspond o he leding 's in he echelon form of he mrix. We ls 1A leding is lernively referred o s pivo.
5 1. Mrices, Vecors, nd GussJordn Eliminion 17 drw he sircse formed by he leding vribles. Th is where he nme echelon form comes from: According o Webser, n echelon is formion "like series of seps." Now we cn solve ech of he preceding equions for he leding vrible: X =  X  X = + X X5 = . X We cn freely choose he nonleding vribles, X= 5 nd X =, where 5 nd re rbirry rel numbers; he nonleding vribles re lernively referred o s he free vribles. The leding vribles re hen deermined by our choices for 5 nd ; h is, X =  5 , X = +, nd X5 = . This sysem hs infiniely mny soluions, nmely, xl=5, X=5, x=+. X=, x5=, where 5 nd re rbirry rel numbers. We cn represen he soluions s vecors in JR?5: Xl  s (s, rbirry). We will ofen find i helpful o wrie his soluion s x) X = X Xl X51 r1 r For exmple, if we se s = =, we ge he priculr soluion X X X X5  X  Here is summry of he eliminion process jus oulined:
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