Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.


 Arline O’Brien’
 4 years ago
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1 This documet was writte ad copyrighted by Paul Dawkis. Use of this documet ad its olie versio is govered by the Terms ad Coditios of Use located at The olie versio of this documet is available at At the above web site you will fid ot oly the olie versio of this documet but also pdf versios of each sectio, chapter ad complete set of otes. Preface Here are my olie otes for my Liear Algebra course that I teach here at Lamar Uiversity. Despite the fact that these are my class otes they should be accessible to ayoe watig to lear Liear Algebra or eedig a refresher. These otes do assume that the reader has a good workig kowledge of basic Algebra. This set of otes is fairly self cotaied but there is eough Algebra type problems (arithmetic ad occasioally solvig equatios) that ca show up that ot havig a good backgroud i Algebra ca cause the occasioal problem. Here are a couple of warigs to my studets who may be here to get a copy of what happeed o a day that you missed.. Because I wated to make this a fairly complete set of otes for ayoe watig to lear Liear Algebra I have icluded some material that I do ot usually have time to cover i class ad because this chages from semester to semester it is ot oted here. You will eed to fid oe of your fellow class mates to see if there is somethig i these otes that was t covered i class.. I geeral I try to work problems i class that are differet from my otes. However, with a Liear Algebra course while I ca make up the problems off the top of my head there is o guaratee that they will work out icely or the way I wat them to. So, because of that my class work will ted to follow these otes fairly close as far as worked problems go. With that beig said I will, o occasio, work problems off the top of my head. Also, I ofte do t have time i class to work all of the problems i the otes ad so you will fid that some sectios cotai problems that were t worked i class due to time restrictios.. Sometimes questios i class will lead dow paths that are ot covered here. I try to aticipate as may of the questios as possible i writig these otes up, but the reality is that I ca t aticipate all the questios. Sometimes a very good questio gets asked i class that leads to isights that I ve ot icluded here. You should always talk to someoe who was i class o the day you missed ad compare these otes to their otes ad see what the differeces are. 4. This is somewhat related to the previous three items, but is importat eough to merit its ow item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!! Usig these otes as a substitute for class is liable to get you i trouble. As already oted ot everythig i these otes is covered i class ad ofte material or isights ot i these otes is covered i class. 005 Paul Dawkis
2 Systems of Equatios ad Matrices Itroductio We will start this chapter off by lookig at the applicatio of matrices that almost every book o Liear Algebra starts off with, solvig systems of liear equatios. Lookig at systems of equatios will allow us to start gettig used to the otatio ad some of the basic maipulatios of matrices that we ll be usig ofte throughout these otes. Oce we ve looked at solvig systems of liear equatios we ll move ito the basic arithmetic of matrices ad basic matrix properties. We ll also take a look at a couple of other ideas about matrices that have some ice applicatios to the solutio to systems of equatios. Oe word of warig about this chapter, ad i fact about this complete set of otes for that matter, we ll start out i the first sectio or to doig a lot of the details i the problems, but towards the ed of this chapter ad ito the remaiig chapters we will leave may of the details to you to check. We start off by doig lots of details to make sure you are comfortable workig with matrices ad the various operatios ivolvig them. However, we will evetually assume that you ve become comfortable with the details ad ca check them o your ow. At that poit we will quit showig may of the details. Here is a listig of the topics i this chapter. Systems of Equatios I this sectio we ll itroduce most of the basic topics that we ll eed i order to solve systems of equatios icludig augmeted matrices ad row operatios. Solvig Systems of Equatios Here we will look at the Gaussia Elimiatio ad GaussJorda Method of solvig systems of equatios. Matrices We will itroduce may of the basic ideas ad properties ivolved i the study of matrices. Matrix Arithmetic & Operatios I this sectio we ll take a look at matrix additio, subtractio ad multiplicatio. We ll also take a quick look at the traspose ad trace of a matrix. Properties of Matrix Arithmetic We will take a more i depth look at may of the properties of matrix arithmetic ad the traspose. 005 Paul Dawkis
3 Iverse Matrices ad Elemetary Matrices Here we ll defie the iverse ad take a look at some of its properties. We ll also itroduce the idea of Elemetary Matrices. Fidig Iverse Matrices I this sectio we ll develop a method for fidig iverse matrices. Special Matrices We will itroduce Diagoal, Triagular ad Symmetric matrices i this sectio. LUDecompositios I this sectio we ll itroduce the LUDecompositio a way of factorig certai kids of matrices. Systems Revisited Here we will revisit solvig systems of equatios. We will take a look at how iverse matrices ad LUDecompositios ca help with the solutio process. We ll also take a look at a couple of other ideas i the solutio of systems of equatios. Systems of Equatios Let s start off this sectio with the defiitio of a liear equatio. Here are a couple of examples of liear equatios. 5 6x 8y+ 0z 7x x 9 I the secod equatio ote the use of the subscripts o the variables. This is a commo otatioal device that will be used fairly extesively here. It is especially useful whe we get ito the geeral case(s) ad we wo t kow how may variables (ofte called ukows) there are i the equatio. So, just what makes these two equatios liear? There are several mai poits to otice. First, the ukows oly appear to the first power ad there are t ay ukows i the deomiator of a fractio. Also otice that there are o products ad/or quotiets of ukows. All of these ideas are required i order for a equatio to be a liear equatio. Ukows oly occur i umerators, they are oly to the first power ad there are o products or quotiets of ukows. The most geeral liear equatio is, ax + ax + ax b () where there are ukows, x, x,, x, ad a, a,, a, b are all kow umbers. Next we eed to take a look at the solutio set of a sigle liear equatio. A solutio set (or ofte just solutio) for () is a set of umbers t, t,, t so that if we set x t, x t,, x t the () will be satisfied. By satisfied we mea that if we plug these umbers ito the left side of () ad do the arithmetic we will get b as a aswer. 005 Paul Dawkis
4 The first thig to otice about the solutio set to a sigle liear equatio that cotais at least two variables with ozero coefficets is that we will have a ifiite umber of solutios. We will also see that while there are ifiitely may possible solutios they are all related to each other i some way. Note that if there is oe or less variables with ozero coefficiets the there will be a sigle solutio or o solutios depedig upo the value of b. Let s fid the solutio set s for the two liear equatios give at the start of this sectio. Example Fid the solutio set for each of the followig liear equatios. 5 (a) 7x x 9 (b) 6x 8y+ 0z Solutio (b) The first thig that we ll do here is solve the equatio for oe of the two ukows. It does t matter which oe we solve for, but we ll usually try to pick the oe that will mea the least amout (or at least simpler) work. I this case it will probably be slightly easier to solve for x so let s do that. 5 7x x 9 5 7x x 9 5 x x 6 7 Now, what this tells us is that if we have a value for x the we ca determie a correspodig value for x. Sice we have a sigle liear equatio there is othig to restrict our choice of x ad so we we ll let x be ay umber. We will usually write this as x t, where t is ay umber. Note that there is othig special about the t, this is just the letter that I usually use i these cases. Others ofte use s for this letter ad, of course, you could choose it to be just about aythig as log as it s ot a letter represetig oe of the ukows i the equatio (x i this case). Oce we ve chose x we ll write the geeral solutio set as follows, 5 x t x t 6 7 So, just what does this tell us as far as actual umber solutios go? We ll choose ay value of t ad plug i to get a pair of umbers x ad x that will satisfy the equatio. For istace pickig a couple of values of t completely at radom gives, 005 Paul Dawkis 4
5 t 0: x, x 0 7 t 7 : 5 x ( 7), x We ca easily check that these are i fact solutios to the equatio by pluggig them back ito the equatio. 5 t 0: 7 ( 0) t 7 : 7( ) ( 7) 9 So, for each case whe we plugged i the values we got for x ad x we got  out of the equatio as we were supposed to. Note that sice there a ifiite umber of choices for t there are i fact a ifiite umber of possible solutios to this liear equatio. (b) We ll do this oe with a little less detail sice it works i essetially the same maer. The fact that we ow have three ukows will chage thigs slightly but ot overly much. We will first solve the equatio for oe of the variables ad agai it wo t matter which oe we chose to solve for. 0z 6x+ 8y 4 z x+ y I this case we will eed to kow values for both x ad y i order to get a value for z. As with the first case, there is othig i this problem to restrict out choices of x ad y. We ca therefore let them be ay umber(s). I this case we ll choose x t ad y s. Note that we chose differet letters here sice there is o reaso to thik that both x ad y will have exactly the same value (although it is possible for them to have the same value). The solutio set to this liear equatio is the, 4 x t y s z t+ s So, if we choose ay values for t ad s we ca get a set of umber solutios as follows. 4 x 0 y z ( 0) + ( ) x y 5 z + ( 5) As with the first part if we take either set of three umbers we ca plug them ito the 005 Paul Dawkis 5
6 equatio to verify that the equatio will be satisfied. We ll do oe of them ad leave the other to you to check ( 5) The variables that we got to choose for values for ( x i the first example ad x ad y i the secod) are sometimes called free variables. We ow eed to start talkig about the actual topic of this sectio, systems of liear equatios. A system of liear equatios is othig more tha a collectio of two or more liear equatios. Here are some examples of systems of liear equatios. x+ y 9 x y 4x 5x + x 9 6x + x 9 x + 0x 5x x 7 7x x 4x 5 x 0x 4 x x + x x + x 4 5 x + x x + 9x 0 4 7x + 0x + x + 6x 9x As we ca see from these examples systems of equatio ca have ay umber of equatios ad/or ukows. The system may have the same umber of equatios as ukows, more equatios tha ukows, or fewer equatios tha ukows. A solutio set to a system with ukows, x, x,, x, is a set of umbers, t, t,, t, so that if we set x t, x t,, x t the all of the equatios i the system will be satisfied. Or, i other words, the set of umbers t, t,, t is a solutio to each of the idividual equatios i the system. For example, x, y 5 is a solutio to the first system listed above, x+ y 9 x y because, ( ) + ( 5) 9 & ( ) ( 5) () However, x 5, y is ot a solutio to the system because, ( 5) + ( ) 9 & ( 5) ( ) We ca see from these calculatios that x 5, y is NOT a solutio to the first equatio, but it IS a solutio to the secod equatio. Sice this pair of umbers is ot a solutio to both of the equatios i () it is ot a solutio to the system. The fact that it s 005 Paul Dawkis 6
7 a solutio to oe of them is t material. I order to be a solutio to the system the set of umbers must be a solutio to each ad every equatio i the system. It is completely possible as well that a system will ot have a solutio at all. Cosider the followig system. x 4y 0 () x 4y It is clear (hopefully) that this system of equatios ca t possibly have a solutio. A solutio to this system would have to be a pair of umbers x ad y so that if we plugged them ito each equatio it will be a solutio to each equatio. However, sice the left side is idetical this would mea that we d eed a x ad a y so that x 4y is both 0 ad  for the exact same pair of umbers. This clearly ca t happe ad so () does ot have a solutio. Likewise, it is possible for a system to have more tha oe solutio, although we do eed to be careful here as we ll see. Let s take a look at the followig system. x+ y 8 (4) 8x 4y We ll leave it to you to verify that all of the followig are four of the ifiitely may solutios to the first equatio i this system. x 0, y 8 x, y, x 4, y 0 x 5, y 8 Recall from our work above that there will be ifiitely may solutios to a sigle liear equatio. We ll also leave it to you to verify that these four solutios are also four of the ifiitely may solutios to the secod equatio i (4). Let s ivestigate this a little more. Let s just fid the solutio to the first equatio (we ll worry about the secod equatio i a secod). Followig the work we did i Example we ca see that the ifiitely may solutios to the first equatio i (4) are x t y t+ 8, t is ay umber Now, if we also fid just the solutios to the secod equatio i (4) we get x t y t+ 8, t is ay umber These are exactly the same! So, this meas that if we have a actual umeric solutio (foud by choosig t above ) to the first equatio it will be guarateed to also be a solutio to the secod equatio ad so will be a solutio to the system (4). This meas that we i fact have ifiitely may solutios to (4). Let s take a look at the three systems we ve bee workig with above i a little more detail. This will allow us to see a couple of ice facts about systems. 005 Paul Dawkis 7
8 Sice each of the equatios i (),(), ad (4) are liear i two ukows (x ad y) the graph of each of these equatios is that of a lie. Let s graph the pair of equatios from each system o the same graph ad see what we get. 005 Paul Dawkis 8
9 From the graph of the equatios for system () we ca see that the two lies itersect at the poit (,5) ad otice that, as a poit, this is the solutio to the system as well. I other words, i this case the solutio to the system of two liear equatios ad two ukows is simply the itersectio poit of the two lies. Note that this idea is validated i the solutio to systems () ad (4). System () has o solutio ad we ca see from the graph of these equatios that the two lies are parallel ad hece will ever itersect. I system (4) we had ifiitely may solutios ad the graph of these equatios shows us that they are i fact the same lie, or i some ways the itersect at a ifiite umber of poits. Now, to this poit we ve bee lookig at systems of two equatios with two ukows but some of the ideas we saw above ca be exteded to geeral systems of equatios with m ukows. First, there is a ice geometric iterpretatio to the solutio of systems with equatios i two or three ukows. Note that the umber of equatios that we ve got wo t matter the iterpretatio will be the same. If we ve got a system of liear equatios i two ukows the the solutio to the system represets the poit(s) where all (ot some but ALL) the lies will itersect. If there is o solutio the the lies give by the equatios i the system will ot itersect at a sigle poit. Note i the o solutio case if there are more tha two equatios it may be that ay two of the equatios will itersect, but there wo t be a sigle poit were all of the lies will itersect. If we ve got a system of liear equatios i three ukows the the graphs of the equatios will be plaes i Dspace ad the solutio to the system will represet the 005 Paul Dawkis 9
10 poit(s) where all the plaes will itersect. If there is o solutio the there are o poit(s) where all the plaes give by the equatios of the system will itersect. As with lies, it may be i this case that ay two of the plaes will itersect, but there wo t be ay poit where all of the plaes itersect at that poit. O a side ote we should poit out that lies ca itersect at a sigle poit or if the equatios give the same lie we ca thik of them as itersectig at ifiitely may poits. Plaes ca itersect at a poit or o a lie (ad so will have ifiitely may itersectio poits) ad if the equatios give the same plae we ca thik of the plaes as itersectig at ifiitely may places. We eed to be a little careful about the ifiitely may itersectio poits case. Whe we re dealig with equatios i two ukows ad there are ifiitely may solutios it meas that the equatios i the system all give the same lie. However, whe dealig with equatios i three ukows ad we ve got ifiitely may solutios we ca have oe of two cases. Either we ve got plaes that itersect alog a lie, or the equatios will give the same plae. For systems of equatios i more tha three variables we ca t graph them so we ca t talk about a geometric iterpretatio, but we ca still say that a solutio to such a system will represet the poit(s) where all the equatios will itersect eve if we ca t visualize such a itersectio poit. From the geometric iterpretatio of the solutio to two equatios i two ukows we ow that we have oe of three possible solutios. We will have either o solutio (the lies are parallel), oe solutio (the lies itersect at a sigle poit) or ifiitely may solutios (the equatios are the same lie). There is simply o other possible umber of solutios sice two lies that itersect will either itersect exactly oce or will be the same lie. It turs out that this is i fact the case for a geeral system. Theorem Give a system of equatios ad m ukows there will be oe of three possibilities for solutios to the system.. There will be o solutio.. There will be exactly oe solutio.. There will be ifiitely may solutios. If there is o solutio to the system we call the system icosistet ad if there is at least oe solutio to the system we call it cosistet. Now that we ve got some of the basic ideas about systems take care of we eed to start thikig about how to use liear algebra to solve them. Actually that s ot quite true. We re ot goig to do ay solvig util the ext sectio. I this sectio we just wat to get some of the basic otatio ad ideas ivolved i the solvig process out of the way before we actually start tryig to solve them. 005 Paul Dawkis 0
11 We re goig to start off with a simplified way of writig the system of equatios. For this we will eed the followig geeral system of equatios ad m ukows. a x + a x + + a x b m m a x + a x + + a x b m m a x + a x + + a x b m m (5) I this system the ukows are x, x,, xm ad the a ij ad b i are kow umbers. Note as well how we ve subscripted the coefficiets of the ukows (the a ij). The first subscript, i, deotes the equatio that the subscript is i ad the secod subscript, j, deotes the ukow that it multiples. For istace, a 6 would be i the coefficiet of x 6 i the third equatio. Ay system of equatios ca be writte as a augmeted matrix. A matrix is just a rectagular array of umbers ad we ll be lookig at these i great detail i this course so do t worry too much at this poit about what a matrix is. Here is the augmeted matrix for the geeral system i (5). a a a m b a a am b a a am b Each row of the augmeted matrix cosists of the coefficiets ad costat o the right of the equal sig form a give equatio i the system. The first row is for the first equatio, the secod row is for the secod equatio etc. Likewise each of the first colums of the matrix cosists of the coefficiets from the ukows. The first colum cotais the coefficiets of x, the secod colum cotais the coefficiets of x, etc. The fial colum (the + st colum) cotais all the costats o the right of the equal sig. Note that the augmeted part of the ame arises because we tack the b i s oto the matrix. If we do t tack those o ad we just have a a a m a a a m a a am ad we call this the coefficiet matrix for the system. Example Write dow the augmeted matrix for the followig system. 005 Paul Dawkis
12 x 0x + 6x x4 x+ 9x 5x4 4x+ x 9x+ x4 7 Solutio There really is t too much to do here other tha write dow the system Notice that the secod equatio did ot cotai a x ad so we cosider its coefficiet to be zero. Note as well that give a augmeted matrix we ca always go back to a system of equatios. Example For the give augmeted matrix write dow the correspodig system of equatios Solutio So sice we kow each row correspods to a equatio we have three equatios i the system. Also, the first two colums represet coefficiets of ukows ad so we ll have two ukows while the third colum cosists of the costats to the right of the equal sig. Here s the system that correspods to this augmeted matrix. 4x x 5x 8x 4 9x + x There is oe fial topic that we eed to discuss i this sectio before we move oto actually solvig systems of equatio with liear algebra techiques. I the ext sectio where we will actually be solvig systems our mai tools will be the three elemetary row operatios. Each of these operatios will operate o a row (which should t be too surprisig give the ame ) i the augmeted matrix ad sice each row i the augmeted matrix correspods to a equatio these operatios have equivalet operatios o equatios. Here are the three row operatios, their equivalet equatio operatios as well as the otatio that we ll be usig to deote each of them. Row Operatio Equatio Operatio Notatio Multiply row i by the costat c Multiply equatio i by the costat c cr i 005 Paul Dawkis
13 Iterchage rows i ad j Iterchage equatios i ad j Ri R j Add c times row i to row j Add c times equatio i to equatio j R j + cri The first two operatios are fairly self explaatory. The third is also a fairly simple operatio however there are a couple thigs that we eed to make clear about this operatio. First, i this operatio oly row (equatio) j actually chages. Eve though we are multiplyig row (equatio) i by c that is doe i our heads ad the results of this multiplicatio are added to row (equatio) j. Also, whe we say that we add c time a row to aother row we really mea that we add correspodig etries of each row. Let s take a look at some examples of these operatios i actio. Example 4 Perform each of the idicated row operatios o give augmeted matrix (a) R (b) R (c) R R (d) R + 5R (e) R R Solutio I each of these we will actually perform both the row ad equatio operatio to illustrate that they are actually the same operatio ad that the ew augmeted matrix we get is i fact the correct oe. For referece purposes the system correspodig to the augmeted matrix give for this problem is, x+ 4x x 6x x 4x 0 7x + x x 5 Note that at each part we will go back to the origial augmeted matrix ad/or system of equatios to perform the operatio. I other words, we wo t be usig the results of the previous part as a startig poit for the curret operatio. (a) Okay, i this case we re goig to multiply the first row (equatio) by . This meas that we will multiply each elemet of the first row by  or each of the coefficiets of the first equatio by . Here is the result of this operatio x x + x xx 4x x+ x x Paul Dawkis
14 (b) This is similar to the first oe. We will multiply each elemet of the secod row by oehalf or each coefficiet of the secod equatio by oehalf. Here are the results of this operatio. x+ 4x x 4 5 x x x x+ x x 5 Do ot get excited about the fractio showig up. Fractios are goig to be a fact of life with much of the work that we re goig to be doig so get used to seeig them. Note that ofte i cases like this we will say that we divided the secod row by istead of multiplied by oehalf. (c) I this case were just goig to iterchage the first ad third row or equatio x+ x x xx 4x 0 4 x+ 4x x (d) Okay, we ow eed to work a example of the third row operatio. I this case we will add 5 times the third row (equatio) to the secod row (equatio). So, for the row operatio, i our heads we will multiply the third row times 5 ad the add each etry of the results to the correspodig etry i the secod row. Here are the idividual computatios for this operatio. st etry : 6+ ( 5)( 7) 4 d etry : + ( 5)( ) 4 rd etry : 4+ ( 5)( ) 9 th 4 etry : ( )( ) For the correspodig equatio operatio we will multiply the third equatio by 5 to get, 5x+ 5x 5x 5 the add this to the secod equatio to get, 4x 4x 9x 5 Puttig all this together gives ad rememberig that it s the secod row (equatio) that we re actually chagig here gives, 4 x+ 4x x x4 x 9x x+ x x Paul Dawkis 4
15 It is importat to remember that whe multiplyig the third row (equatio) by 5 we are doig it i our head ad do t actually chage the third row (equatio). (e) I this case we ll ot go ito the detail that we did i the previous part. Most of these types of operatios are doe almost completely i our head ad so we ll do that here as well so we ca start gettig used to it. I this part we are goig to subtract times the secod row (equatio) from the first row (equatio). Here are the results of this operatio x+ 7x + x xx 4x x+ x x 5 It is importat whe doig this work i our heads to be careful of mius sigs. I operatios such as this oe there are ofte a lot of them ad it easy to lose track of oe or more whe you get i a hurry. Okay, we ve ot got most of the basics dow that we ll eed to start solvig systems of liear equatios usig liear algebra techiques so it s time to move oto the ext sectio. Solvig Systems of Equatios I this sectio we are goig to take a look at usig liear algebra techiques to solve a system of liear equatios. Oce we have a couple of defiitios out of the way we ll see that the process is a fairly simple oe. Well, it s fairly simple to write dow the process ayway. Applyig the process is fairly simple as well but for large systems ca take quite a few steps. So, let s get the defiitios out of the way. A matrix (ay matrix, ot just a augmeted matrix) is said to be i reduced rowechelo form if it satisfies all four of the followig coditios.. If there are ay rows of all zeros the they are at the bottom of the matrix.. If a row does ot cosist of all zeros the its first ozero etry (i.e. the left most ozero etry) is a. This is called a leadig.. I ay two successive rows, either of which cosists of all zeroes, the leadig of the lower row is to the right of the leadig of the higher row. 4. If a colum cotais a leadig the all the other etries of that colum are zero. A matrix (agai ay matrix) is said to be i rowechelo form if it satisfies items of the reduced rowechelo form defiitio. 005 Paul Dawkis 5
16 Notice from these defiitios that a matrix that is i reduced rowechelo form is also i rowechelo form while a matrix i rowechelo form may or may ot be i reduced rowechelo form. Example The followig matrices are all i rowechelo form Noe of the matrices i the previous example are i reduced rowechelo form. The etries that are prevetig these matrices from beig i reduced rowechelo form are highlighted i red ad uderlied (for those without color priters...). I order for these matrices to be i reduced rowechelo form all of these highlighted etries would eed to be zeroes. Notice that we did t highlight the etries above the i the fifth colum of the third matrix. Sice this is ot a leadig (i.e. the leftmost ozero etry) we do t eed the umbers above it to be zero i order for the matrix to be i reduced rowechelo form. Example The followig matrices are all i reduced rowechelo form I the secod matrix o the first row we have all zeroes i the etries. This is perfectly acceptable ad so do t worry about it. This matrix is i reduced rowechelo form, the fact that it does t have ay ozero etries does ot chage that fact sice it satisfies the coditios. Also, i the secod matrix of the secod row otice that the last colum does ot have zeroes above the i that colum. That is perfectly acceptable sice the i that colum is ot a leadig for the fourth row. 005 Paul Dawkis 6
17 Notice from Examples ad that the oly real differece betwee rowechelo form ad reduced rowechelo form is that a matrix i rowechelo form is oly required to have zeroes below a leadig while a matrix i reduced rowechelo from must have zeroes both below ad above a leadig. Okay, let s ow start thikig about how to use liear algebra techiques to solve systems of liear equatios. The process is actually quite simple. To solve a system of equatios we will first write dow the augmeted matrix for the system. We will the use elemetary row operatios to reduce the augmeted matrix to either rowechelo form or to reduced rowechelo form. Ay further work that we ll eed to do will deped upo where we stop. If we go all the way to reduced rowechelo form the i may cases we will ot eed to do ay further work to get the solutio ad i those time where we do eed to do more work we will geerally ot eed to do much more work. Reducig the augmeted matrix to reduced rowechelo form is called GaussJorda Elimiatio. If we stop at rowechelo form we will have a little more work to do i order to get the solutio, but it is geerally fairly simple arithmetic. Reducig the augmeted matrix to rowechelo form ad the stoppig is called Gaussia Elimiatio. At this poit we should work a couple of examples. Example Use Gaussia Elimiatio ad GaussJorda Elimiatio to solve the followig system of liear equatios. x+ x x 4 x+ x + x x+ x Solutio Sice we re asked to use both solutio methods o this system ad i order to for a matrix to be i reduced rowechelo form the matrix must also be i rowechelo form. Therefore, we ll start off by puttig the augmeted matrix i rowechelo form, the stop to fid the solutio. This will be Gaussia Elimiatio. After doig that we ll go back ad pick up from rowechelo form ad further reduce the matrix to reduced row echelo form ad at this poit we ll have performed GaussJorda Elimiatio. So, let s start off by gettig the augmeted matrix for this system. 4 0 As we go through the steps i this first example we ll mark the etry(s) that we re goig to be lookig at i each step i red so that we do t lose track of what we re doig. We should also poit out that there are may differet paths that we ca take to get this matrix ito rowechelo form ad each path may well produce a differet rowechelo 005 Paul Dawkis 7
18 form of the matrix. Keep this i mid as you work these problems. The path that you take to get this matrix ito rowechelo form should be the oe that you fid the easiest ad that may ot be the oe that the perso ext to you fids the easiest. Regardless of which path you take you are oly allowed to use the three elemetary row operatios that we looked i the previous sectio. So, with that out of the way we eed to make the leftmost ozero etry i the top row a oe. I this case we could use ay three of the possible row operatios. We could divide the top row by  ad this would certaily chage the red  ito a oe. However, this will also itroduce fractios ito the matrix ad while we ofte ca t avoid them let s ot put them i before we eed to. Next, we could take row three ad add it to row oe, or we could take three times row ad add it to row oe. Either of these would also chage the red  ito a oe. However, this row operatio is the oe that is most proe to arithmetic errors so while it would work let s ot use it uless we eed to. This leaves iterchagig ay two rows. This is a operatio that wo t always work here to get a ito the spot we wat, but whe it does it will usually be the easiest operatio to use. I this case we ve already got a oe i the leftmost etry of the secod row so let s just iterchage the first ad secod rows ad we ll get a oe i the leftmost spot of the first row pretty much for free. Here is this operatio. 4 R R Now, the ext step we ll eed to take is chagig the two umbers i the first colum uder the leadig ito zeroes. Recall that as we move dow the rows the leadig MUST move off to the right. This meas that the two umbers uder the leadig i the first colum will eed to become zeroes. Agai, there are ofte several row operatios that ca be doe to do this. However, i most cases addig multiples of the row cotaiig the leadig (the first row i this case) oto the rows we eed to have zeroes is ofte the easiest. Here are the two row operatios that we ll do i this step. R + R 4 R R Notice that sice each operatio chaged a differet row we wet ahead ad performed both of them at the same time. We will ofte do this whe multiple operatios will all chage differet rows. We ow eed to chage the red 5 ito a oe. I this case we ll go ahead ad divide the secod row by 5 sice this wo t itroduce ay fractios ito the matrix ad it will give us the umber we re lookig for. 005 Paul Dawkis 8
19 R Next, we ll use the third row operatio to chage the red 6 ito a zero so the leadig of the third row will move to the right of the leadig i the secod row. This time we ll be usig a multiple of the secod row to do this. Here is the work i this step R + R Notice that i both steps were we eeded to get zeroes below a leadig we added multiples of the row cotaiig the leadig to the rows i which we wated zeroes. This will always work i this case. It may be possible to use other row operatios, but the third ca always be used i these cases. The fial step we eed to get the matrix ito rowechelo form is to chage the red  ito a oe. To do this we do t really have a choice here. Sice we eed the leadig oe i the third row to be i the third or fourth colum (i.e. to the right of the leadig oe i the secod colum) we MUST retai the zeroes i the first ad secod colum of the third row. Iterchagig the secod ad third row would defiitely put a oe i the third colum of the third row, however, it would also chage the zero i the secod colum which we ca t allow. Likewise we could add the first row to the third row ad agai this would put a oe i the third colum of the third row, but this operatio would also chage both of the zeroes i frot of it which ca t be allowed. Therefore, our oly real choice i this case is to divide the third row by . This will retai the zeroes i the first ad secod colum ad chage the etry i the third colum ito a oe. Note that this step will ofte itroduce fractios ito the matrix, but at this poit that ca t be avoided. Here is the work for this step. 0 6 R At this poit the augmeted matrix is i rowechelo form. So if we re goig to perform Gaussia Elimiatio o this matrix we ll stop ad go back to equatios. Doig this gives, x+ x + x 0 6 x + x x 005 Paul Dawkis 9
20 At this poit solvig is quite simple. I fact we ca see from this that x. Pluggig this ito the secod equatio gives x 4. Fially, pluggig both of these ito the first equatio gives x. Summarizig up the solutio to the system is, x x 4 x This substitutio process is called back substitutio. Now, let s pick back up at the rowechelo form of the matrix ad further reduce the matrix ito reduced rowechelo form. The first step i doig this will be to chage the umbers above the leadig i the third row ito zeroes. Here are the operatios that will do that for us. R R R R The fial step is the to chage the red above the leadig oe i the secod row ito a zero. Here is this operatio R R We are ow i reduced rowechelo form so all we eed to do to perform GaussJorda Elimiatio is to go back to equatios. 0 0 x x x We ca see from this that oe of the ice cosequeces to GaussJorda Elimiatio is that whe there is a sigle solutio to the system there is o work to be doe to fid the solutio. It is geerally give to us for free. Note as well that it is the same solutio as the oe that we got by usig Gaussia Elimiatio as we should expect. Before we proceed with aother example we eed to give a quick fact. As was poited out i this example there are may paths we could take to do this problem. It was also oted that the path we chose would affect the rowechelo form of the matrix. This will ot be true for the reduced rowechelo form however. There is oly oe reduced rowechelo form of a give matrix o matter what path we chose to take to get to that poit. If we kow ahead of time that we are goig to go to reduced rowechelo form for a matrix we will ofte take a differet path tha the oe used i the previous example. I the previous example we first got the matrix i rowechelo form by gettig zeroes uder the leadig s ad the wet back ad put the matrix i reduced rowechelo form by gettig zeroes above the leadig s. If we kow ahead of time that we re goig to wat 005 Paul Dawkis 0
21 reduced rowechelo form we ca just take care of the matrix i a colum by colum basis i the followig maer. We first get a leadig i the correct colum the istead of usig this to covert oly the umbers below it to zero we ca use it to covert the umbers both above ad below to zero. I this way oce we reach the last colum ad take care of it of course we will be i reduced rowechelo form. We should also poit out the differeces betwee GaussJorda Elimiatio ad Gaussia Elimiatio. With GaussJorda Elimiatio there is more matrix work that eeds to be performed i order to get the augmeted matrix ito reduced rowechelo form, but there will be less work required i order to get the solutio. I fact, if there s a sigle solutio the the solutio will be give to us for free. We will see however, that if there are ifiitely may solutios we will still have a little work to do i order to arrive at the solutio. With Gaussia Elimiatio we have less matrix work to do sice we are oly reducig the augmeted matrix to rowechelo form. However, we will always eed to perform back substitutio i order to get the solutio. Which method you use will probably deped o which you fid easier. Okay let s do some more examples. Sice we ve doe oe example i excruciatig detail we wo t be botherig to put as much detail ito the remaiig examples. All operatios will be show, but the explaatios of each operatio will ot be give. Example 4 Solve the followig system of liear equatios. x x + x x+ x x x x + x Solutio First, the istructios to this problem did ot specify which method to use so we ll eed to make a decisio. No matter which method we chose we will eed to get the augmeted matrix dow to rowechelo form so let s get to that poit ad the see what we ve got. If we ve got somethig easy to work with we ll stop ad do Gaussia Elimiatio ad if ot we ll proceed to reduced rowechelo form ad do GaussJorda Elimiatio. So, let s start with the augmeted matrix ad the proceed to put it ito rowechelo form ad agai we re ot goig to put i quite the detail i this example as we did with the first oe. So, here is the augmeted matrix for this system. ad here is the work to put it ito rowechelo form. 005 Paul Dawkis
22 R + R R R R R R 8 R Okay, we re ow i rowechelo form. Let s go back to equatio ad see what we ve got. x x + x x x 0 Hmmmm. That last equatio does t look correct. We ve got a couple of possibilities here. We ve either just maaged to prove that 0 (ad we kow that s ot true), we ve made a mistake (always possible, but we have t i this case) or there s aother possibility we have t thought of yet. Recall from Theorem i the previous sectio that a system has oe of three possibilities for a solutio. Either there is o solutio, oe solutio or ifiitely may solutios. I this case we ve got o solutio. Whe we go back to equatios ad we get a equatio that just clearly ca t be true such as the third equatio above the we kow that we ve got ot solutio. Note as well that we did t really eed to do the last step above. We could have just as easily arrived at this coclusio by lookig at the secod to last matrix sice 08 is just as icorrect as 0. So, to close out this problem, the official aswer that there is o solutio to this system. I order to see how a simple chage i a system ca lead to a totally differet type of solutio let s take a look at the followig example. Example 5 Solve the followig system of liear equatios. x x + x x+ x x x x + x 7 Solutio The oly differece betwee this system ad the previous oe is the 7 i the third equatio. I the previous example this was a. Here is the augmeted matrix for this system. 005 Paul Dawkis
23 7 Now, sice this is essetially the same augmeted matrix as the previous example the first few steps are idetical ad so there is o reaso to show them here. After takig the same steps as above (we wo t eed the last step this time) here is what we arrive at For some good practice you should go through the steps above ad make sure you arrive at this matrix. I this case the last lie coverts to the equatio 0 0 ad this is a perfectly acceptable equatio because after all zero is i fact equal to zero! I other words, we should t get excited about it. At this poit we could stop covert the first two lies of the matrix to equatios ad fid a solutio. However, i this case it will actually be easier to do the oe fial step to go to reduced rowechelo form. Here is that step. 0 4 R R We are ow i reduced rowechelo form so let s covert to equatios ad see what we ve got. x+ x 4 x x Okay, we ve got more ukows tha equatios ad i may cases this will mea that we have ifiitely may solutios. To see if this is the case for this example let s otice that each of the equatios has a x i it ad so we ca solve each equatio for the remaiig variable i terms of x as follows. x 4 x x + x So, we ca choose x to be ay value we wat to, ad hece it is a free variable (recall we saw these i the previous sectio), ad each choice of x will give us a differet solutio to the system. So, just like i the previous sectio whe we ll reame the x ad write the solutio as follows, 005 Paul Dawkis
24 x 4 t x + t x t t is ay umber We therefore get ifiitely may solutios, oe for each possible value of t ad sice t ca be ay real umber there are ifiitely may choices for t. Before movig o let s first address the issue of why we used GaussJorda Elimiatio i the previous example. If we d used Gaussia Elimiatio (which we defiitely could have used) the system of equatios would have bee. x x + x 4 x x To arrive at the solutio we d have to solve the secod equatio for x first ad the substitute this ito the first equatio before solvig for x. I my mid this is more work ad work that I m more likely to make a arithmetic mistake tha if we d just goe to reduced rowechelo form i the first place as we did i the solutio. There is othig wrog with usig Gaussia Elimiatio o a problem like this, but the back substitutio is defiitely more work whe we ve got ifiitely may solutios tha whe we ve got a sigle solutio. Okay, to this poit we ve worked othig but systems with the same umber of equatios ad ukows. We eed to work a couple of other examples where this is t the case so we do t get too locked ito this kid of system. Example 6 Solve the followig system of liear equatios. x 4x 0 5x+ 8x 7 x+ x Solutio So, let s start with the augmeted matrix ad reduce it to rowechelo form ad see if what we ve got is ice eough to work with or if we should go the extra step(s) to get to reduced rowechelo form. Let s start with the augmeted matrix Notice that this time i order to get the leadig i the upper left corer we re probably goig to just have to divide the row by ad deal with the fractios that will arise. Do ot go to great legths to avoid fractios, they are a fact of life with these problems ad so while it s okay to try to avoid them, sometimes it s just goig to be easier to deal with it ad work with them. So, here s the work for reducig the matrix to rowechelo form. 005 Paul Dawkis 4
25 R + 5R R R R R R 8R Okay, we re i rowechelo form ad it looks like if we go back to equatios at this poit we ll eed to do oe quick back substitutio ivolvig umbers ad so we ll go ahead ad stop here at this poit ad do Gaussia Elimiatio. Here are the equatios we get from the rowechelo form of the matrix ad the back substitutio x x x + 4 x 4 So, the solutio to this system is, x x 4 Example 7 Solve the followig system of liear equatios. 7x+ x x 4x4 + x5 8 x x + x4 + x5 4x x 8x+ 0x5 Solutio First, let s otice that we are guarateed to have ifiitely may solutios by the fact above sice we ve got more equatios tha ukows. Here s the augmeted matrix for this system I this example we ca avoid fractios i the first row simply by addig twice the secod row to the first to get our leadig i that row. So, with that as our iitial step here s the work that will put this matrix ito rowechelo form R R Paul Dawkis 5
26 R + R R R R 4R R R + R R We are ow i rowechelo form. Notice as well that i several of the steps above we took advatage of the form of several of the rows to simplify the work somewhat ad i doig this we did several of the steps i a differet order tha we ve doe to this poit. Remember that there are o set paths to take through these problems! Because of the fractios that we ve got here we re goig to have some work to do regardless of whether we stop here ad do Gaussia Elimiatio or go the couple of extra steps i order to do GaussJorda Elimiatio. So with that i mid let s go all the way to reduced rowechelo form so we ca say that we ve got aother example of that i the otes. Here s the remaiig work R R R+ 4R We re ow i reduced rowechelo form ad so let s go back to equatios ad see what we ve got. 8 8 x x4 x5 x + x4 + x5 5 5 x x x4 x5 x + x4 + x5 So, we ve got two free variables this time, 4 x ad 5 x, ad otice as well that ulike ay of the other ifiite solutio cases we actually have a value for oe of the variables here. That will happe o occasio so do t worry about it whe it does. Here is the solutio for this system. 005 Paul Dawkis 6
27 8 8 x + t+ s x x + t+ s 5 5 x t x s s ad t are ay umbers 4 5 Now, with all the examples that we ve worked to this poit hopefully you ve gotte the idea that there really is t ay oe set path that you always take through these types of problems. Each system of equatios is differet ad so may eed a differet solutio path. Do t get too locked ito ay oe solutio path as that ca ofte lead to problems. Homogeeous Systems of Liear Equatios We ve got oe more topic that we eed to discuss briefly i this sectio. A system of liear equatios i m ukows i the form ax + ax + + a mxm 0 ax + ax + + amxm 0 a x+ ax+ + amxm 0 is called a homogeeous system. The oe characteristic that defies a homogeeous system is the fact that all the equatios are set equal to zero ulike a geeral system i which each equatio ca be equal to a differet (probably ozero) umber. Hopefully, it is clear that if we take x 0 x 0 x 0 x m 0 we will have a solutio to the homogeeous system of equatios. I other words, with a homogeeous system we are guarateed to have at least oe solutio. This meas that Theorem from the previous sectio ca the be reduced to the followig for homogeeous systems. Theorem Give a homogeeous system of equatios ad m ukows there will be oe of two possibilities for solutios to the system. 4. There will be exactly oe solutio, x 0, x 0, x 0,, x m 0. This solutio is called the trivial solutio. 5. There will be ifiitely may ozero solutios i additio to the trivial solutio. Note that whe we say ozero solutio i the above fact we mea that at least oe of the x i s i the solutio will ot be zero. It is completely possible that some of them will still be zero, but at least oe will ot be zero i a ozero solutio. We ca make a further reductio to Theorem from the previous sectio if we assume that there are more ukows tha equatios i a homogeeous system as the followig theorem shows. Theorem Give a homogeeous system of liear equatios i m ukows if m> (i.e. there are more ukows tha equatios) there will be ifiitely may solutios to 005 Paul Dawkis 7
28 the system. Matrices I the previous sectio we used augmeted matrices to deote a system of liear equatios. I this sectio we re goig to start lookig at matrices i more geerality. A matrix is othig more tha a rectagular array of umbers ad each of the umbers i the matrix is called a etry. Here are some examples of matrices [ 0 9] [ ] The size of a matrix with rows ad m colums is deoted by m. I deotig the size of a matrix we always list the umber of rows first ad the umber of colums secod. Example Give the size of each of the matrices above. Solutio size : size : 9 0 I this matrix the umber of rows is equal to the umber of colums. Matrices that have the same umber of rows as colums are called square matrices. 4 size : 4 7 This matrix has a sigle colum ad is ofte called a colum matrix. [ 0 9] size : Paul Dawkis 8
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