Unit-VII: Linear Algebra-I. To show what are the matrices, why they are useful, how they are classified as various types and how they are solved.

Transcription

1 Unit-VII: Liner lger-i Purpose of lession : To show wht re the mtries, wh the re useful, how the re lssified s vrious tpes nd how the re solved. Introdution: Mtries is powerful tool of modern Mthemtis nd its stud is eoming importnt d d due to its wide pplitions in lmost ever rnh of siene nd espeill in Phsis nd Engineering. Mtries re used Soiologists in the stud of dominne within group, Demogrphers in the stud of irths nd deths, et, Eonomist in the stud of inter-industr eonomists, input-output tles nd for vrious prtil usiness purposes, sttistiins in the stud of design of eperiments nd for vrious prtil usiness purposes, sttistiins in the stud of design of eperiments nd multivrite nlsis Engineers in the stud of network nlsis whih is used in eletril nd ommunition engineering. pplition: The following tle shows the numer of trnsistors nd resistors purhsed mnufturer from suppliers nd B for the first week of Jnur B Trnsistor esistors ) Write the dt in the tle s mtri S ) )Use slr multiplition to find mtri S whose entries re ll % lrge thn the orresponding entries of S ) Suppose tht S is the suppl mtri for the seond week of Jnur. Find S S nd eplin wht its entries represent. Solution: ) The suppl tle n e written s the following mtri S

2 ) If the entries of S re % lrge thn the entries of S, then S S. S. S S. ) The entries of S S give the totl numer of trnsistors nd resistors purhsed from suppliers nd B for the first two weeks of Jnur. s s Emple : The following tle Shows the numer of eonom, mid-size nd lrge rs rented individuls nd orportions t rentl gen in single d Tle Eonom mid-size lrge Individuls Corportions Tle Bonus points Free miles Eonom Mid-size Lrge Tle shows the numer of onus points nd free miles given in promotionl progrm for eh of the three r tpes. Find ) Totl onus points for individuls ) Totl free miles for individuls ) Totl onus points for orportions d) Totl free miles for orportions [ ]

3 The row mtri from the Tle epresents the numer of eonom, medium size nd lrge rented Individuls The olumn mtri From tle represents the onus points given for eh eonom, mid-size nd lrge r tht is rented. The produt of these two mtries is mtri whose entr is the sum of the produts of the orresponding entries [ ] [ ] The produt of ove two mtries, gives the totl numer of onus points given to rentl of the rs. individuls on the Eo Mid Lr Bonus Pts X Free Miles Eo Mid Lr Trffi ontrol in the future: The ove piture shows the intersetions of four one w streets. rs per hour wnt to enter intersetion P from the north on First venue while rs per hour wnt to hed est from intersetion Q on Elm Street. The letters w,,, nd z represent the numer of rs per hour pssing the four points etween these four intersetions, s shown in piture. ) Find vlues for w,, nd z tht would relize this desired trffi flow.

4 B) If onstrution on Ok Street limits z to rs per hour, then how mn rs per hour would hve to pss w,, nd. Solution: The solution to the prolem is sed on the ft tht the numer of rs entering n intersetion per hour must equl the numer leving tht intersetion per hour, if the trffi is to keep flowing Sine rs ( ) enter intersetion P, must leve, w. Writing similr eqution for eh intersetion ields the sstem of equtions. W W X z z w z z w z - z z The sstem is dependent sstem does not hve unique solution.(z is non negtive integer) )If z is limited to euse of onstrution, then the solution is (,,,) To keep trffi flowing when z, the sstem must route rs pst w, pst nd pst.

5 Definition Mtries nd Tpes of Mtries: Twent si femle students nd twent-four mle students responded to surve on inome in ollege lger lss. mong the femle students, lssified themselves s low-inome, s middle-inome, nd s high-inome. mong the mle students, were low-inome, were middle-inome, nd were high-inome. Eh student is lssified in two ws, ording to gender nd inome. This informtion n e written in mtri: Femle Mle L M H Squre mtri: In this mtri we n see the lss of mkeup ording to gender nd inome. mtri provides onvenient w to orgnize two w lssifition of dt. mtri is retngulr rr of rel numers. The rows of mtri run horizontll, nd the olumns run vertill. mtri with m rows nd n olumns hs size m n (red m n ). The numer of rows is lws given first. For emple, the mtri used to lssif the students is mtri When m n i. e the numer of rows is the sme s the numer of olumns the rr is lled squre mtri. The others re lled retngulr mtri. ow-mtri: In mtri if there is onl one row it is lled row mtri Column mtri: In mtri if there is onl one olumn it is lled olumn mtri Null mtri : In mtri, if ll the elements of mtri re zero, it is lled null mtri nd is denoted o Equl Mtri: Two mtries re sid to e equl if i) The re of the sme tpe i.e the hve sme numer of rows nd olumns ii) The elements in the orresponding positions of the two mtries re equl. Lotion of n element: To lote n prtiulr element of mtri the elements re usull denoted letter followed two suffies whih respetivel speif the row nd the olumn in whih it ppers. Thus the element ourring in p th row nd q th olumn will e written s pq. Note: The mtries re generll denoted pitl letters. Digonl Elements of Mtri: n element ij of squre mtri is sid to e digonl element if i j.

6 Slr Multiple of mtri: If k is numer nd is mtri, then k is defined s the mtri eh element of whih is k times the orresponding element of. Tringulr Mtri: if ever element ove (or elow ) the digonl is zero, the mtri is lled tringulr mtri Emple: upper tri ngulr lower tri ngulr Trnsposed Mtri: The mtri of order n m otined interhnging the rows nd olumns of mtri of order m n is lled the trnspose mtri of or trnspose of the mtri nd is denoted or t. epresenttion of Points: point is represented in two dimensions its oordintes. These two vlues re speified s the elements of mtri i.e. [, ] or In three-dimensions mtri [ z ] is used or z ow mtri or olumn mtri like [ ] or re frequentl lled position vetors. Elementr Trnsformtions: n elementr trnsformtion, whih is lso known s E-trnsformtion, is n opertion of n one of the following tpes. i) Interhnge of two rows (olumns) ii) iii) The multiplition of the elements of rows (olumns) non-zero numer. The ddition to the elements of row (olumn), the orresponding elements of row ( olumn) multiplied n numer n elementr trnsformtion is sid to e row-trnsformtion or olumn trnsformtion ording s it pplies to rows or olumns.

7 Smols to e used for elementr trnsformtion: i) ij for the interhnge of i th nd j th rows ii) i (k) for the multiplition of i th row k iii) ij (k) for the ddition to the i th row, the produts of the j th row k. Similrl we use the smols C ij, C i (k), C ij (k) for the orresponding olumn opertions. Equivlent Mtries: Two mtries nd B re sid to e equivlent if one n e otined from the other sequene of elementr trnsformtions. Two equivlent mtries hve the sme order. The smol is used for equivlene. Minor : Let e mtri squre or retngulr, from it delete ll rows leving ertin t-rows nd ll olumns leving ertin t-olumn. Now if t > then the elements tht re left, onstitute squre mtri of order t nd the determinnt of this mtri is lled minor of of order t nk of Mtri: Definition: mtri is sid to e rnk r when I ) It hs t lest one non-zero minor of order r nd ii) Ever minor of order higher thn r vnishes. Briefl, the rnk of mtri is the lrgest order of n non-vnishing minor of the mtri For emple:), ρ() ), ρ() Note : i) If mtri hs non-zero minor or order r, its rnk is r ii) If ll minors of mtri of order of r re zero, its rnk is r Emple: () ρ ) ( ρ

8 Emple:. Find the rnk of the mtri Emple:. edue the mtri to unit mtri onl elementr row trnsformtions Solution: Emple:. Find the rnk of the mtri ) ( ; - operte ; ρ () : ρ ; - I

9 Norml form of Mtri: B mens of sequene of elementr trnsformtions ever mtri of order m n n e redued to one of the following form Emple: edue the mtri to its norml form Solution: Consisten of sstem of liner equtions: Consider the following sstem of m liner equtions in n-unknowns,,, n. n n. n n m m m. mn n n The sstem of equtions n e written s r I ) ( ; ; ρ

10 Conditions for Consisten: Here the sstem of equtions X B is lled the non homogeneous set of equtions nd the sstem of equtions X is lled the homogeneous set of equtions. sstem of liner equtions is sid to e onsistent if it possesses solution nd inonsistent ( not onsistent) if it does not possess solution homogeneous sstem lws possesses t lest one solution (nmel the trivil solution ), homogeneous sstem is lws onsistent. But non homogeneous sstem m or m not e onsistent. The eqution X B is onsistent i.e. hs solution (unique or infinite ) iff the mtries nd D re of the sme rnk i.e. ρ() ρ(d). i) If ρ() ρ(d) n, the numer of unknowns, then onl the given sstem of equtions will hve unique solution ii) If ρ() ρ(d) < n, then the given sstem will hve infinite solutions. iii) If ρ() ρ(d) then the given sstem of equtions will e inonsistent or will not hve n solution. Emple: Test the onsisten of the following equtions z z - z Solution: B n X,,,, n mn m m n n n D n mn m m n n n, : - B, z X, D D

11 If ρ() ρ(d) < numer of unknowns, the sstem is onsistent nd hs n infinite numer of solutions Emple: Show tht the following equtions re inonsistent z z -z Soln.: B i.e. D[/B] ; z ; ;

12 ρ(), ρ(d) ρ() ρ(d) Hene the given equtions re inonsistent Emple: Show tht the sstem of equtions z z z - -z is onsistent. Solution: ρ(), ρ(d) ρ() ρ(d) The sstem is inonsistent Emple : Investigte the vlues of nd µ so tht the equtions z z z µ hve i) Unique solution ii) no solution iii) n infinite numer of solutions : B, X B D z ) (

13 Solution: () i) If eqution () hs unique solution then the determinnt of the oeffiient mtri i.e the given sstem of equtions will hve unique solution if whtever m the vlue of µ. If nd µ, then ρ(), ρ(d) ρ() ρ(d) Hene the given sstem of equtions hs no solution ) If in the ugmented D, nd µ then It is ler tht ρ() ρ(d) (< the numer of unknowns ) Therefore their eist infinite numer of solutions to the given set of equtions µ z / µ µ B D D

14 Bsill, there re TWO tpes of methods ville, viz, Diret methods nd Indiret methods or Itertive methods. Diret Method: i) Crmer s rule ii) Gussin Elimintion iii) Guss-Jordn method iv) Choleske Method Indiret Method: i) Joi Method ii) Guss seidl method iii) Suessive Over-eltion Tehnique Guss Elimintion method: Guss Elimintion method pplied to three liner equtions. First we eplin this method pplied to prtiulr sstems of order three given () The ugmented mtri of the ove sstem To eliminte from the seond eqution multipling the first eqution the seond eqution. Similrl to eliminte from the third eqution multipl the first eqution third eqution. Thus we get the mtri. nd dd it to nd dd it to the

15 Where, re lled the multipliers for the first stge of elimintion. The first eqution is lled the pivotl eqution nd is lled the first pivot is the new pivot nd the multiplier is i.e. multipl the seond row nd dd it to the third row. The vlues of, nd n e otined the sustitution emrk: This method fils if one of the pivots s, or Vnishes. We n modif rerrnging the rows so tht the pivot is non-zero. Emple: Solve Guss elimintion Method Fromthelst eqution z z z

16 Solution: The ugmented mtri is Eliminting from seond nd third eqution The ugmented mtri is Eliminting from the third eqution - z, -, ; z z z

17 Guss-Jordn Method: Insted of eliminting onl in the third eqution, we ould hve eliminted it from the first eqution lso, so tht t the end of the seond stge, the ugmented mtri eomes. Guss-Jordn Method: onsider () The ugmented mtri of the ove sstem () This modifition of Gussin elimintion is lled the Guss-Jordn method Emple: Solve Guss-Jordn method Solution: The ugmented mtri is z z z ;

18 Eliminting from seond nd third eqution -z (½) (/) z -z - z, -, Emple :Solve the following sstem of liner equtions z z z Using Guss-Jordn method. Solution: Let the given sstem of equtions e written in the mtri form X B where The ugmented mtri is ; i.e ; ndb z,x : B : B -

19 We get the following sstem of equtions X, nd z Then required solution is,, z - - / /

20 Unit-VIII: Liner lger-ii Liner Trnsformtions: De trnsformtion ( or funtion or m trnsformtion T from n to m is The set n is lled the domin of T tht the domin of T in n nd the For in n, the vetor T() in m is The set of ll imges T() is lled th For emple: X nd u o S tht multiplition trnsfo (zero) vetor. Mtri Trnsformtion : T: n mtri, For simpliit, we sometime Emple: Let nd define trnsformtion T: Find T(u), the imge of u under th,,u efinition: mpping): rule tht ssigns to eh vetor in n vetor T( T, nd m is lled the o-domin of T. The nottion T o-domin is m. lled the imge of (under the tion of T). he rnge of T. nd orms X into nd Similrl multiplition trns m for eh in n, T() is omputed s (), where es denote suh mtri trnsformtion T() so tht T() () he trnsformtion T. -, () in m. T: n m indites sforms u into the o is n m n.

21 Definition: trnsformtion ( or mpping ) T is liner if i) T(u v) T(u) T(v) for ll u, in the domin of T ii) T(u) T(u) for ll u nd ll slrs. Chrteristi Eqution of Mtri : If ( ij ) is n squre mtri of order n nd e n vrile, then the mtri [ - I] is lled the hrteristi mtri of. The determinnt of this mtri i.e. Is lled the hrteristi funtion. The eqution is the hrteristi eqution of nd of degree n. The roots of the eqution re lled the hrteristis or Eigen vlues of. Emple: Determine the hrteristi roots of Solution: The hrteristi eqution is ie B sntheti division method ) (-)(-), re the require roots or Eigen vlues Emple: Find the eigen vlues nd the eigenvetor of the mtri Solution: The hrteristi eqution of is i.e epnding we get The roots of this eqution re, whih re the eigen vlues of. To find the eigen vetors, onsider the mtri eqution is ) ( u u T I I

22 This gives homogeneous liner equtions (-) (-) i) For ove equtions eomes - - Both represent the sme eqution - giving the eigenvetor ii) gin for, Both represent the sme giving the eigenvetor Emple: Find the eigen vlue nd the eigen vetors of the mtri Solution: The hrteristi eqution of the mtri is. X where e i X I X X Let I

23 i.e Epnding, we get - : - ( -) i.e or ( -) i.e or ( -) ( -) i.e,, These re the eigen vlues of To find the eigen vetor onsider the mtri eqution is i.e This gives the three homogeneous equtions (-) - - (-) - () - (-) i) eqution () eome () - Solving n two non-identil equtions of () for,, the rule of ross multiplition. We get. wherex X I

24 Therefore the eigenvetor orresponding to is ii) eqution () eome () - Solving the rule of ross multiplition We get Therefore the eigenvetor orresponding to is iii) eqution () eome () - - Solving the rule of ross multiplition We get Therefore the eigenvetor orresponding to is Orthogonl Mtri: squre mtri is sid to e orthogonl if I Singulr mtri nd Nonsingulr mtri: squre mtri is sid to e singulr if its determinnt is zero nd is sid to e nonsingulr if its determinnt is not equl to zero. Inverse of squre mtri: If nd B re two squre mtries of the sme order suh tht B-B I then B is lled the inverse of denoted -. Similrit of Mtries: Two squre mtries nd B of the sme order re sid to e similr if there eists non singulr mtri P suh tht B P - P Here B is sid to e similr to.

25 Emple: edue the mtri to the digonl form. Solution: The hrteristi eqution of is re the eigen vlues of. Consider Csei) let, we get - or or / / X (,) is the eigen vetor orresponding to Cseii) let, we get - or or / / X (,) is the eigen vetor orresponding to. X I ) (- -.. ie ie I. nd e i djp p nd we X X P p - P hve mtri digonl the is p- p- D P Thus P P

26 edue the following mtries into digonl mtries : ns. : ns.