and then substitute this into the second equation to get 5(11 4 y) 3y

Transcription

1 Math E-b Lecture # Notes The priary focus of this week s lecture is a systeatic way of solvig ad uderstadig systes of liear equatios algebraically, geoetrically, ad logically. Eaple #: Solve the syste y 3y 9. This is easy to solve. We would like to eplore the ay ways of castig this proble geoetrically, algebraically, ad otherwise. Substitutio: We ca rewrite the first equatio as y ad the substitute this ito the secod equatio to get ( y) 3y 9 or 3y 9, so 3y 6 ad therefore y. We the substitute this ito y to get ( ) or 3. This approach should be failiar to all. Geoetry: Each of these liear equatios represets a lie, ad the siultaeous solutio represets the itersectio of these two lies. The geoetric perspective ay ot produce a eplicit solutio, but it s helpful i uderstadig the ature of the solutio. I this case, we see that the lies itersect i a poit, aely (3, ). It could have bee the case that the lies were parallel (with o solutio) or coicidet (with a etire lie of solutios. It should be clear that these are the oly three possibilities i this case. Gauss-Jorda eliiatio: It is possible to work uch ore systeatically to arrive at a algebraic solutio by usig just three basic rules: ) It s OK to scale ay equatio by a ozero costat (this will ot chage the solutios). ) It s OK to iterchage ay pair of equatios (this clearly will ot affect the itersectio). 3) It s OK to add ay scalar ultiple of oe equatio to ay other equatio. [This will ot affect the solutios, though this observatio eeds soe justificatio (ad you are ecouraged to do so).] I our eaple, we ca the produce the followig sequece of equivalet liear systes: y copy st eq. y copy st eq. y st eq. + (d eq.) 3 d eq. (st eq.) (d eq.) copy d eq. 3y 9 3y 6 3 y y This ay ot at first see sipler tha our earlier ethod, but it is systeatic ad geeralizable. I fact, if we agree to aitai the order of the variables ad keep the left ad right sides of the equatios distict, all of the essetial iforatio ca be captured i the followig augeted atri: 3 9. We see the atri of coefficiets to the left of the divider bar ad the costats of the right-had sides to the right of the divider bar. Oce we kow how to iterpret these ubers, we ca easily reproduce the equatios. The algebraic rules that we used above ca ow be traslated ito valid rules for aipulatig the rows of this array. Eleetary Row Operatios: ) It s OK to scale ay row by a ozero costat. ) It s OK to iterchage ay pair of rows. 3) It s OK to add ay scalar ultiple of oe row to ay other row. The previous aipulatios of equatios the becoe aipulatio of rows: R R R R 0 3 RR 3 9 R R 0 0

2 The solutio of this syste thus becoes a strategy for gettig fro the startig augeted atri to the fial, siplified augeted atri that allows us to directly read off the solutio. This fial atri (which is actually uiquely deteried) is called the reduced row-echelo for (RREF) of the give atri. We ll see as we look at other eaples that there s a siple strategy for gettig to the RREF for ay give syste. Basically, it s this: Strategy for gettig to the reduced row-echelo for (RREF): A) Scale or iterchage to produce a Leadig i the upper left positio (if possible). A Leadig refers to where the first ozero etry o a row (as read fro left to right) is equal to. B) Use the first row as a pivot row, leavig it uchaged but addig (or subtractig) appropriate ultiples of it to the other rows to produce 0 s elsewhere i the colu cotai the Leadig (the first colu i this case). We call this cleaig the colu. C) Scale or iterchage to produce a Leadig i the secod row shifted oe colu to the right (possibly ore). D) Usig the secod row as the pivot row to clea this et colu leavig oly the Leadig i the pivot row ad 0 s elsewhere i that colu. E) Cotiue this process to get Leadig s i each of the rows, shiftig to the right as you desced through the rows. F) Ay all-zero rows should appear at the botto of the array. It s possible to use differet steps to get fro the iitial atri to its RREF equivalet. It ca be proved (try it!) that o atter what steps are take, the RREF will be uiquely deteried. Properties of a atri i reduced row-echelo for (RREF): ) Every ozero row has a Leadig. ) The Leadig s shift oe or ore colus to the right as you desced the rows. 3) Ay colu cotaiig a Leadig has 0 s elsewhere i that colu (clea). ) Ay all-zero rows are located at the botto of the atri. Defiitio: The rak of a atri is the uber of Leadig s i the RREF of the atri. As we ll see i the subsequet eaples, the rak for a cosistet syste is the uber of variables that we ca solve for either for a specific value or i ters of oe or ore paraeters. I our iitial eaple with two equatios i two variables, the rak was ad we were able to solve for both variables to get specific values. 3yz Eaple #: Solve the syste yz. Geoetrically, each equatio represets a plae i R 3, ad the siultaeous solutio will correspod to the itersectio of these two plaes. Three possible thigs could happe: (a) the plaes itersect i a lie, (b) the plaes are parallel ad do t itersect at all, or (c) the plaes are coicidet ad itersect everywhere to give a etire plae of solutios. It s ot possible to get a uique solutio, ad it s really ot sufficiet to say that there are ifiitely ay solutios (though there ay be) without akig clear whether these are o a lie or a plae ad eplicitly represetig these solutios (i ters of oe or ore paraeters). Algebraically, we ca proceed usig augeted atrices ad row reductio: 3 RR 6 6 R 6 6 R 6 6 R R R R 0 3 R 6R 0 z z R 0 3 y z 3 y 3 z This syste has rak ad we were able to solve for ad y i ters of the other variable z. I order to epress the solutios i a ore evehaded way, we itroduce a paraeter t ad epress all of the variables i ters of

3 t this paraeter. To iiize fractios, let s take z t. The y 3t. If you are failiar with the z t t R paraeterizatio of lies, you ight rewrite this i vector for as y 3t. This represets a lie z 0 passig through the poit (, 3,0) i the directio of the vector v,,. You do t eed to kow this to be able to produce solutios, but it does ake eplicit the earlier observatio that we epected the two plaes to itersect i a lie. 3y Eaple #3: Solve the syste y 7. y Geoetrically, we are seekig the itersectio of three lies i R. The likelihood is that there is o coo itersectio of all three lies. For a overdeteried syste such as this (ore coditios tha there are variables) we would typically epect to get o solutios. However, it is possible that we could have a uique solutio or eve that the three lies ight coicide to yield a etire lie of solutios. Algebraically, we proceed as before usig augeted atrices ad a row reductio strategy: , where soe steps have bee erged This fial array cotais a clear cotradictio. The botto row says that 0 0y. The oly logical coclusio is that the preise that a coo solutio eisted ust have bee false. There are o solutios. It s iportat to ote that the fial array above is actually ot i reduced row-echelo for. Techically we should still clea the last colu: This is worth otig because if a calculator is used to fid the RREF, it will carry out this process to copletio. Note: If a Leadig appears to the right of the divider bar i the RREF for a give augeted atri, this idicates that the syste ust be icosistet, i.e. there are o solutios to the liear syste. y3z Eaple #: Solve the syste 3 y z 7. 3yz Here we have a syste of 3 equatios i 3 ukows. Each equatio represets a plae i R 3. The itersectio of these plaes could give either (a) a sigle poit (uique solutio), (b) o solutios (either due to parallel plaes or the itersectio of two plaes givig a lie parallel to the third plae), (c) a lie of solutios if the three plaes itersect i that way, or (d) a etire plae of solutios (if all three plaes coicide). Algebraically, proceed with augeted atrices ad row reductio: y. (, y, z) (.,.,3.) z 3. 3

4 Note that the rak i this case is 3 ad we were able to solve for specific values for all 3 variables. There was o eed for paraeters ad the solutio is uique Eaple #: Solve the syste 733. Geoetrically, we are a little beyod the cofort zoe. There are ow variables, so we are lookig at the itersectio of two objects i R, but what are these objects? We ca proceed by aalogy. If we were i R with coordiates (, ) ad had the relatio 0, we would recogize this as a lie (oe degree of freedo). If we were i R 3 with coordiates (,, 3) ad had the relatio 33 0, we would recogize this as a plae (two degrees of freedo). So, by aalogy, if we are i R with coordiates (,, 3, ) ad had the relatio 33 0, we ight ituitively thik of this as a hyperplae (three degrees of freedo) i this four-diesioal space. This visualizatio ay ot get us to a eplicit solutio, but it s helpful for uderstadig the possibilities. The itersectio of two hyperplaes could yield o solutios (if they are parallel), but they will ost likely yield a set of solutios with two degrees of freedo (thus requirig two idepedet paraeters). Typically, the ipositio of each additio costrait (equatio) reduces by oe the uber of degrees of freedo of the itersectio, though this is defiitely ot always the case. It should be ephasized that we ca forally solve this syste ad paraetrically represet all solutios without ay geoetric visualizatio. Algebraically, we proceed with augeted atrices ad row reductio: The correspodig equatios are If we ow itroduce two idepedet paraeters s ad t ad, i order to iiize fractios, let 3 s ad 6 st 8 7st t, we ca epress all solutios paraetrically as. I ters of vectors i R, this 3 s t st, R ca be epressed i the vector for s t. This agrees with the geoetric discussio above i the sese that we have a iitial vector to get to a poit i the solutio space, ad we ca the ove i two idepedet directios fro there to spa the epected two degrees of freedo. Rak vs. paraeters I each of the previous eaples where solutios eisted (cosistet systes), there was a siple relatioship betwee the uber of variables, the rak of the syste, ad the uber of paraeters ecessary to epress all solutios. The rak represets the uber of variables we ca solve for i ters of the other variables, ad we ll eed to itroduce a idepedet paraeter for each of the variables that we caot otherwise solve for. Therefore, [Rak] + [Nuber of paraeters] = [Nuber of Variables]. That is, if the syste is cosistet with rak k ad if there are variables, we will require k paraeters to epress all solutios. The greater the rak, the fewer the uber of paraeters. If the rak is (full rak), the o paraeters are eeded ad the syste will have a uique solutio. Keep i id that a give syste ay also be icosistet i which case there are o solutios.

5 The situatio of a icosistet syste ca also be characterized by the fact that the rak of its augeted atri ( rak A b ) is greater that the rak of its coefficiet atri ( raka ) because of the Leadig appearig to the right of the divider bar. [See Eaple #3.] Observatios about rak For ay atri A ( rows, colus): ) rak( A ) because each colu ca have o ore tha oe Leadig i the RREF of the atri. ) rak( A ) because each row ca have o ore tha oe Leadig i the RREF of the atri. 3) rak( A ) i(, ) follows fro the observatios above. ) If A b is a augeted atri for a liear syste ad if RREF A b has a Leadig i its last colu (to the right of the divider bar), the the syste is icosistet. [This is equivalet to havig RREF A b RREF A.] ) If a syste with augeted atri A b is cosistet ad if rak( A ), the the syste will have a uique solutio. A b is cosistet ad if rak( A ) k, the the syste will require k paraeters to epress all solutios. 6) If a syste with augeted atri Vector for of a liear syste a a b Give a syste of liear equatios i variables, we ca use vector additio ad a a b a a b scalar ultiplicatio to epress this i the vector for:. If we let a a b a a b v,, v, ad let b, this ca be ore succictly epressed as v v b. a a b I this perspective, the eistece of solutios ca be uderstood i ters of whether the give vector b o the right-had-side ca be built out of the collectio of vectors v,, v. We will soo re-epress this by sayig that b v v spa,,. If this is ot the case, the syste will be icosistet. If it is the case, there ay be oly oe way to do this (uique solutio) or ay ways to do this (paraeterized faily of solutios). y Lookig back at Eaple # with 3y 9, we could write this i vector for as y 3 9. If you draw the vectors v ad v 3 as well as the vector b 9, you ll see that you ll eed to take 3 ad y to get ad this is uique i this case. Matri for of a liear syste If we take the vector for above ad asseble the vectors v,, v side-by-side to for a atri a a Av v, ad if we write, we ca defie the product of this atri ad the a a

6 vector as A v v v v. Usig this defiitio, we ca epress the liear syste a a b siply as A b. This is called the atri for of the liear syste. a a b This ca also be uderstood i ters of (liear) fuctios. Note that if we write L( ) A, we have the iput vector R ad the output vector L( ) AbR. We ca therefore uderstad such a syste of liear equatios i ters of the (liear) fuctio L : R L R R or R. We also soeties represet this by writig either: R A R or We will eplore this perspective i detail i the et lecture. A R AbR Hoogeeous systes a a 0 A syste of liear equatios of the for is called a hoogeeous syste (the righthad-sides are all equal to 0). It ca easily be see that every hoogeeous syste is cosistet because a a a 0 solutio is (,, ) (0,,0). If you look back at all of the cosistet eaples we previously cosidered ad replace all the right-had-sides by 0, you ll also fid that the solutios will be quite siilar ad require the sae uber of paraeters, but the costat ters will all be 0. The geoetric iterpretatio of this is that the solutios for a give o-hoogeeous syste ad the correspodig hoogeeous syste will siply be parallel traslates with the hoogeeous solutios passig through the origi (we ll soo call this a subspace) ad the o-hoogeeous solutios parallel to the hoogeeous solutios (kow as a affie space). Notes by Robert Witers 6