MEI Mthemtics i Eductio d Idustry MEI Structured Mthemtics Module Summry Sheets FP, Further Cocepts for Advced Mthemtics Topic : Mtrices Topic : Comple Numbers Topic : Curve Sketchig Topic 4: Algebr Purchsers hve the licece to mke multiple copies for use withi sigle estblishmet November, 4 MEI, Ok House, 9 Epsom Cetre, White Horse Busiess Prk, Trowbridge, Wiltshire. BA4 XG. Compy No. 549 Egld d Wles Registered with the Chrity Commissio, umber 589 Tel: 5 7777. F: 5 775755.
Summry FP Topic : Mtrices Chpter Pges - Eercise A Q., 5 Mtrices re rectgulr rrys of umbers which c be used to covey iformtio. A mtri with rows d m colums is kow s m mtri. A mtri with equl umbers of rows d colums is squre. The etries i mtri re clled elemets. If ll elemets re the mtri is kow s the zero mtri. The Idetity Mtri for y sized squre mtri is oe where ll elemets i the ledig digol re d ll other elemets re. i.e., for mtri I = If mtri is multiplied by umber the every elemet is multiplied by tht umber. Mtrices re coformble if they re the sme size. Coformble mtrices my be dded or subtrcted. Mtri dditio is commuttive i.e. A + B = B + A Mtri dditio is ssocitive. i.e. (A + B) + C = A + (B + C) 5 E.g. A=, B=, C = 4 7 8 4 5 (i) Fid, if possible, A+ B, A+ C (ii) Show tht A+ B= B+ A. + 5 + 8 (i) A+ B= = + 7 4+ 8 A+ Cis ot possible s Ad Cre ot coformb le. 5 8 (ii) B + A= + + = = + 7+ 8+ 4 A B 4 E.g. A =, 4 A+A = A = = 4 8 E.g. I = E.g. If A d B re s give bove, d D= the show tht A + B + D = A + ( B + D ) ( + ) 8 8 A B + D = + = 9 7 4 8 = A + ( B + D ) = + = 4 8 9 Chpter Pges - Eercise B Q. (i), 4(i) Trsformtios my be represeted by mtrices. If X, represets the positio vector of the poit ( y, ) y ' d X', represets the positio vector of the poit ', ' y' c d if M= the X is trsformed to X' b d by M if X' = MX. c is trsformed to, is trsformed to. b d is the mtri for reflectio i the lie y =. cosθ -siθ is the mtri for rottio of θ siθ cosθ ticlockwise bout the origi. is the mtri for elrgemet, cetre the origi d scle fctor. Nottio A trsformtio is described by bold itlic letter; e.g. T. The mtri tht represets this trsformtio is deoted by bold upright cpitl letter; e.g.t. ( y ) E.g. Describe the trsformtio give by M =., M represets stretch prllel to the is. E.g. Describe the trsformtio give by.8. M =..8.8.,..8 i.e. rottio of 5. ticlockwise bout the origi. FP; Further Cocepts for Advced Mthemtics pge Competece sttemets FP, m, m, m4
Summry FP Topic : Mtrices Chpter Pges 4- Eercise C Q. (i), (i), 5 Chpter Pges -5 Eercise D Q., 7 Multiplictio of mtrices mtrices, A d B my be multiplied provided the umber of colums i A is equl to the umber of rows i B i.e. If A is m d B is p the AB is mtri of size m p i.e. Size(A) =, Size(B) = 5 The size(ab) = 5. Note tht BA cot be clculted s the umber of colums of B does ot equl the umber of rows i A. If A d B re squre the AB d BA c be clculted but they re ot, i geerl, equl. So mtri multiplictio is ot commuttive. i.e. AB BA Mtri multiplictio is ssocitive. i.e. A(BC) = (AB)C Nottio The poit P hs positio vector p. The imge of P uder the trsformtio T c be deoted T(P) or P'. The imge, T(P), hs positio vector p' = T(p). T(p) is foud by evlutig the mtri product Tp. Compositio of trsformtios The trsformtio MN represets the trsformtio N followed by M. E.g. If N is reflectio i the y is, the N =. If M is reflectio i the is, the M =. MN = =. This represets rottio through 8 bout the origi. 7 8 9 A= 4, B= 4 5 4 AB = 4 7+ 8+ 9+ + 4 AB = 7+ 4 8+ 4 9+ 4 + 4 4 5 7 5 8 5 9 5 4 + + + + 9 5 8 AB = 5 7 79 8 4 C=, D=, 4 5 CD=, DC = 9 5 E.g. If A d B re s give bove, d D= the show tht AB D = A( BD ) 9 5 8 AB D = 5 7 79 8 = 4 ( = A BD ) 4 E.g. M =, N = (i) Fid MN. (ii) Fid the imge of the poit (,) uder the trsformtio N followed by M. (i) MN = =. 4 (ii) p =, MNp = = 4 Chpter Pges 8- Determits b If A =, the umber d bc is c d clled the Determit of A. The vlue is writte A or Det( A). FP; Further Cocepts for Advced Mthemtics pge Competece sttemets FP, m, m, m5 E.g. A = A = ( ) 4 = 4 4 B = B = = Note: 4 8 AB = D et( AB) = = Det( A)Det( B)
Summry FP Topic : Mtrices Chpter Pges 8- Iverse Mtrices The iverse of squre mtri A is writte A - d is such tht AA - = A - A = I b - d b If A=,the A = c d A c Note tht the iverse of the mtri A c be writte s follows: Iterchge the two elemets of the ledig digol Chge the sig of ech elemet i the trilig digol Divide ech term by the determit of A. E.g. Give tht A = d B = 4 fid A d B - - A = A = 4 4 - A = = 4 4 4 4 = = - B B B does ot eist. Eercise: Tke y A d B d show tht (AB) - = B - A -. Eercise E Q. (v),(vii) Chpter Pges -4 Eercise F Q., 7 Chpter Pges -9 Eercise G Q. (i), 5 Chpter Pges 4-4 Eercise H Q. (i), (iii), 4(i) Oly squre mtrices hve iverses. A iverse of mtri oly eists if A. If A = the the mtri is sid to be sigulr. Use of determits The determit of the mtri A is the (siged) re scle fctor of the trsformtio represeted by A. b i.e. the mtri M = trsforms the uit squre c d ito prllelogrm with re d bc. A sigulr mtri trforms ll poits ito lie through the origi. Simulteous Equtios The simulteous equtios + by = c d + ey = f b c c be writte = d e y f As mtri equtio this c be writte AX = B - - Multiply both sides by the iverse of A A AX=A B - i.e. X = A B. A solutio c be foud if A Ivrit poits d lies If mtri represets reflectio i lie, the y poit o the lie remis o the lie. If ll poits o lie AB re mpped oto poits i AB (ot ecessrily the sme poit) the the lie AB is kow s ivrit lie. If poit is trsformed oto itself the the poit is kow s ivrit poit. FP; Further Cocepts for Advced Mthemtics pge 4 Competece sttemets FP m, m7, m8, m9, m, m, m, m E.g. A trsformtio is sher prllel to the y is such tht ech poit is moved by times its distce from the is. Give the mtri tht represets this trsformtio d show tht re is preserved. (i.e. the re of the prllelogrm tht results from trsformig squre hs the sme re. M = M = E.g. Solve + 5y = 7, 4 y = 5 AX = B where A =, X =, 4 y 7 B = - 5 A = = A = - 4 5 7 - AX = B X = A B= - 4 = = - i.e. =, y= 4 M = (i) Show tht (, ) is ivrit poit. (ii) Fid the equtio of the ivrit lie. 4 4 (i) = = 4 (ii) For y poit ( y, ), = y y 4+ y = y = Note tht (, ) lies o this lie. Altertively, we hve lredy foud tht (, ) is ivrit poit; so lso is (,) d so the ivrit lie is the lie joiig these two poits.
Summry FP Topic : Comple Numbers Chpter Pges 4-49 Eercise A Q. (ii), (vi), (ii), (vi) Chpter Pge 49 Eercise A Q. 5 Comple Numbers The comple umber, j, is such tht j =.Thus j=. A comple umber tkes the geerl form + bj. Comple umbers c be dded d subtrcted i the usul lgebric wy: E.g. ( + bj) +(c + dj) = ( + c) + (b + d)j Comple umbers c be multiplied i the usul wy, usig j = -. E.g. ( + bj)(c + dj) = c + dj + bcj + bdj = (c bd) + (d + bc)j. A qudrtic c ow be described s lwys hvig two roots, though they my be comple. (Note tht if oe is comple the so is the other.) 4± 5 4± E.g. z + 4z+ = z= = 4± j = = ± j Note tht two comple umbers c be dded to give rel umber E.g. ( + j) + ( j) = Two comple umbers my lso be multiplied to give rel umber. E.g. ( + j) ( j) = 9 + 4 = Comple cojugtes If z = + yj, the z* = yj is clled the comple cojugte of z. Comple roots of qudrtic equtio occur i cojugte pirs. E.g. If + j is root the the other oe is j ( + ) + ( ) = ( + ) + ( ) E.g. j 4 j 4 j ( + ) ( ) = ( ) + ( + ) ( + j) ( j) E.g. Solve the equtio 4 7 4 4 4 7 4 ± ± z = = = ± j Solutio by the method of completig the squre: z z z z + 4 + 7 = + 4 = 7 z + z+ = (i) Oe root is 5 jso the other is 5+ j ( z ( j) )( z + ( j) ) = ( 5 j) ( 5 j) ( 5 j )( j) z (ii) The equtio is 5 5 z z z + + z + + + = 5 9 z + z+ = 4 = 7 j E.g. j 4 j 4 j E.g. 4 = + 5j = 4+ ( ) j + j 4+ j ( ) j = + 8j 9j j = j+ = 8 j z + z+ = = z+ = 4 4 4 7 z+ = ± z = ± j E.g. A root of qudrtic equtio is 5 j (i) Stte the other root, (ii) Fid the equtio. 5+ = Chpter Pges 49-5 Eercise B Q. (ii), (ii) 4(iii),9 Divisio Remember tht ( + yj )( yj ) = + y y y The = j j == + yj + yj yj + y Therefore frctio with comple umber i the deomitor my be rtiolised by multiplyig top d bottom of the frctio by the comple cojugte of the deomitor. N.B. The process is the sme s for surds: FP; Further Cocepts for Advced Mthemtics pge 5 Competece sttemets FP, j, j, j, j4 4 4 The = == = 4 4+ 4+ 4 + j E.g. R tiolise + 4j ( j)( j) j j 4 j j = = + 4+ = 4 5 ( j) ( + j)( 4j) + j = + 4j + 4j 4j 8 j + 9j + = 9 + = ( 8 + j) 5 E.g. Solve the equtio + j= j + j= j + j = j j ( + j) = ( j) = + j
Summry FP Topic : Comple Numbers Chpter Pges 55-57 Eercise C Q (iii), (ii),(v) The Argd digrm The comple umber + yj c be represeted geometriclly o coordite system by the poit (, y). Rel umbers (i.e. (, )) form the is which is therefore clled the rel is. Imgiry umbers (i.e. (, y)) form the y is which is therefore clled the imgiry is. Such represettio is clled Argd digrm. Comple cojugtes re reflectios i the imgiry (y ) is. + yj c be described by the vector. The sum y of comple umbers c therefore be see s the sum of vectors. The modulus is the legth of the comple umber, + y E.g. Show the umber r = + j o rgd digrm. Im r (, ) z = + j, z = j z z z + z Re Chpter Pges 58-59 Eercise D Q. (v), Chpter Pges -5 Loci The distce betwee z d z is z z. z z = k is stisfied by ll poits, z, which re k uits from z. i.e. the locus of z is circle, cetre z d rdius k Modulus d Argumet form for comple umbers. A comple umber c be described i Crtesi form. i.e. z = + yj describes the positio vector from the origi to the poit (, y). E.g. The locus of ll poits stisfyig z j = 5 is circle, cetre + j rdius 5 E.g. the poit (, )c be writte i π modulus-rgumet form s,. It c lso be described by mes of the legth of the umber together with the gle it mkes with the positive rel is. θ r (, y) π Eercise E Q (ii), (i), (iii) The distce, r, is the modulus of z. The gle is mesured ticlockwise from the rel is d is mesured i rdis. Without restrictig the rge of θ this is ot uique, so it is usul to defie θ s the pricipl gle i the rge π < θ π. Chpter Pges -7 Eercise F Q. (ii), Chpter Pges 9-7 Eercise G Q., 9 Loci rgz = cost. is stright lie. Equtios O pge it ws see tht if qudrtic equtio hs o rel roots the the two comple roots occur i comple cojugte pirs. i.e. if oe root is + bj the the other is bj. This pplies more geerlly to polyomil equtio of degree. If polyomil equtio hs rel coefficiets the comple roots occur i comple cojugte pirs. i.e. cubic equtio hs either three rel roots or oe rel root d two comple roots which re cojugte pir. E.g. Solve the equtio. z + z = By ispectio f() = so oe root is z = + + = ( z )( z z ) ± The roots of z + z+ = re z = z = ( ± j) d z = FP; Further Cocepts for Advced Mthemtics pge Competece sttemets FP, j5, j, j7, j8, j9
Summry FP Topic : Curve Sketchig Chpter Pges 74-7 Chpter Pges 77-85 Eercise A Q., 7, 5 Chpter Pges 87-9 Eercise B Q., (iii) Discotiuities d Asymptotes If fuctio cotis frctio with epressio for i the deomitor such tht for give vlue of the deomitor is zero, the there is discotiuity t tht vlue of. E.g. y = + 5 hs discotiuity t =. The curve pproches, but does ot cross, the lie =. This lie is clled symptote. The power of frctio is the differece betwee the powers i the umertor d deomitor. E.g. y = + 5 hs power =. y = + 5 hs power =. y = + 5 hs power =. If the frctio hs egtive power the the curve will pproch y = If the frctio hs power the it will pproch the lie y =. If the power is the the curve will pproch the lie y = m + c. Sketchig Curves See lso the topic i C d C.. Fid where the curve cuts the es.. Fid the verticl symptotes d emie the behviour of the grph either side of them.. Emie the behviour s teds to ifiity. 4. Complete the sketch. Iequlities There re two wys to solve iequlity f() > (i) Drw the grph of y = f(), idetifyig the regio stisfied by the iequlity (ii) Mipulte the iequlity lgebriclly. Iequlities my be mipulted like equtios ecept: Ay umber multiplyig or dividig both sides must be positive Cre is eeded with subtrctio. E.g. Sketch the curve y = + ( )( ) (i) =, y =.; y =, = (ii) Discotiuities t =, = (iii) > y > ; < < y<, < < y > ; < y< (iv) As, y from below As, y from bove E.g. Sketch the grph of y =. + (i) Whe =,y =. (ii) There re o discotiuities. (iii) As,y from bove As,y from below 4 E.g. Solve the iequlity < by lgebric mes d lso by drwig grphs. Providig >, < 4 4 < 4 + < < < 4 i.e. < < 4 If <, > 4 4 > 4 + > > 4 d < i.e. < i.e. < < 4 d < From the two grphs the iequlities derived bove c be deduced. FP; Further Cocepts for Advced Mthemtics pge 7 Competece sttemets FP, c, c, c, 4
Summry FP Topic 4: Algebr Chpter 4 Pges 97-99 Eercise 4A Q. 4, 5 Idetities Two differet wys of writig the sme epressio re ideticlly equl. The epressios will hve the sme vlue for ll vlues of. E.g. ( + ) + 4+ ( )( ) I idetity the coefficiet of equl powers o ech side will be the sme. E.g. If + 5 + b, fid d b. R.H.S + + ( + b) Compre with coefficiets o L.H.S + = =, + b= 5 b= + + i.e. 5 Chpter 4 Pges -4 Eercise 4B Q., 4 Chpter 4 Pges 5-7 Eercise 4C Q, 8, 9 Chpter 4 Pges - Eercise 4D Q., 4 Roots of polyomils. A qudrtic equtio hs two roots. They my be rel d distict, rel d coicidet, or comple (i which cse they re comple cojugte pir). If the two roots of + b + c = re α d b c α + β =, αβ =. A cubic hs roots, t lest oe of which is rel. The other two roots my be rel d coicidet or comple (i which cse they re comple cojugte pir.) If the three roots of re, d d αβγ = = You should be fmilir with the process of fidig equtio whose roots re relted to give equtio. E.g. If the roots of equtio re α, β d γ d the roots of secod equtio re α', β'd γ ' where there is symmetric reltioship (e.g. α'= α+, etc) the you eed to be ble to fid α' + β ' + γ ', α' β ' + β ' γ ' + γ ' α' d αβ ' ' γ' i terms of α, β d γ.. A qurtic equtio hs 4 roots. Comple roots occur i cojugte pirs, so there re either, or 4 rel roots (some of which my be coicidet). 4 If the four roots of + b + c + d + e = re α, β, γ d δ b α+ β + γ + δ = α =, c αβ + βγ + γδ + αγ + αδ + βδ = αβ =, d αβγ + αβδ + αγδ + βγδ = αβγ =, e αβγδ = = αβγδ FP; Further Cocepts for Advced Mthemtics pge 8 Competece sttemets FP, p, p, p, 5, α + β + γ = α = + b + c + d = b αβ + βγ + γα = αβ = αβγ, c, β α β γ E.g. The roots of 5 re αd β. Fid the equtio with roots d. α+ β =, αβ = 5, α β α β αβ + = + = 4 = αβ = 5 i.e. 5 + + = + + = E.g. If the roots of the equtio + + 4= re αβ, d γ the fid the equtio whose roots re α+, β + d γ+. α+ β + γ =, αβ + βγ + γα =, αβγ = 4 α+ + β+ + γ + = α ( α+ )( β + ) + ( β + )( γ + ) + ( γ + )( α+ ) ( αβ βγ γα ) ( α β γ ) = + + + + + + = + = 5 ( α+ ) ( β + )( γ + ) = αβγ + ( αβ + βγ + γα ) + ( α + β + γ ) + = 4+ + = + 5+ = Notice tht this my hve lso bee chieved by the trsformtio w= +. β + + 4= + + 4= w w + + + 4= w w w w + 5 + = w w w E.g. If the roots of the equtio + + 8 = re α d the fid the equtio whose roots re d. α β Substitute =. The equtio i z will hve roots z d. α β z 8z + + =
Summry FP Topic 4: Algebr Chpter 5 Pges 5-8 Eercise 5A Q., 8, 9 Iductio Give cojecture, it is proved by iductio s follows:. Show tht it is true for prticulr vlue of, typiclly, = or.. Assume it to be true for = k.. Bsed o this ssumptio prove tht it is lso true for = k +. A cojecture c be disproved by sigle couter-emple. The cojecture might be formul for the sum of fiite series. E.g. Prove tht + 4 +... + + = + + 7 It is true for =, for =...9 Assume true for = k. i.e. Sk = k( k + )( k + 7) The Sk + = k( k + )( k + 7) + ( k + )( k + ) = ( k + )( k(k + 7) + ( k + ) ) = ( k + )( k + k + 8) = ( k + )( k + )( k + 9) which is of the sme form s the formul give with k replced by k +. So if true for = k the lso true for = k + But sice it is true for =, it is true for ll positive itegers,. Chpter 5 Pge Chpter 5 Pge Eercise 5B Q., 5 Chpter 5 Pges -4 Eercise 5C Q., 5 Sequeces If sequece is defied iductively i.e. if ech term is defied by the term before it the formul my be derived, or if give my be proved by iductio. Summtio of Series to ifiity. A series is sid to coverge if, s pproches, the sum pproches fiite umber. You met the coditio for geometric series to coverge i C. Other series lso coverge. The method of differeces If the terms of oe series c be epressed s the differece of cosecutive terms of other series the the series c be summed by the ccelltio of middle terms. E.g. If t = s s r r r the t = ( s s ) + ( s s ) +... + ( s s ) r = s s FP; Further Cocepts for Advced Mthemtics pge 9 Competece sttemets FP, p, p4,,, S = + r + r +... + r = S S = providig r <. r ( r ) r E.g. A sequece is defied by u + = u + for positive itegers, with u =. Prove tht u = It is true for =, sice u =. Assume true for = k. i.e. u k = k. The u k+ = u k + = ( k ) + = k + + = k + + So if true for = k the it is lso true for = k + But sice it is true for =, the formul is true for ll positive itegers,. E.g. Show tht =. r r r+ Hece fid S = d deduce the sum to ifiity. r r+ ( r ) = = r r+ r+ r r S = = r r r+ = + + +..... + + 4 5 + + = + = + ( + ) S = 4 Stdrd results my sometimes be used. ( + ) E.g. Fid r. r= ( r+ ) = r+ Sice r = ( + ) r= r= r= ( r+ ) = ( + ) + = ( + ) r=