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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. LU-Decompositios I this sectio we ll itroduce the LU-Decompositio 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 LU-Decompositios 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 o-zero 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 o-zero 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 D-space ad the solutio to the system will represet the 005 Paul Dawkis 9

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

14 (b) This is similar to the first oe. We will multiply each elemet of the secod row by oe-half or each coefficiet of the secod equatio by oe-half. 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 oe-half. (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 o-zero etry (i.e. the left most o-zero 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 row-echelo form if it satisfies items of the reduced row-echelo form defiitio. 005 Paul Dawkis 5

16 Notice from these defiitios that a matrix that is i reduced row-echelo form is also i row-echelo form while a matrix i row-echelo form may or may ot be i reduced row-echelo form. Example The followig matrices are all i row-echelo form Noe of the matrices i the previous example are i reduced row-echelo form. The etries that are prevetig these matrices from beig i reduced row-echelo form are highlighted i red ad uderlied (for those without color priters...). I order for these matrices to be i reduced row-echelo 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 o-zero etry) we do t eed the umbers above it to be zero i order for the matrix to be i reduced row-echelo form. Example The followig matrices are all i reduced row-echelo 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 row-echelo form, the fact that it does t have ay o-zero 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

22 R + R R R R R R 8 R Okay, we re ow i row-echelo 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

25 R + 5R R R R R R 8R Okay, we re i row-echelo 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 row-echelo 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 row-echelo form R R Paul Dawkis 5

26 R + R R R R 4R R R + R R We are ow i row-echelo 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 Gauss-Jorda Elimiatio. So with that i mid let s go all the way to reduced row-echelo 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 row-echelo 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 o-zero) 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 o-zero solutios i additio to the trivial solutio. Note that whe we say o-zero 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 o-zero 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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