Matrices. Introduction


 Bertram Lucas
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1 Mtrices Introduction
2 Mtrices  Introduction Mtrix lgebr hs t lest two dvntges: Reduces complicted systems of equtions to simple expressions Adptble to systemtic method of mthemticl tretment nd well suited to computers Definition: A mtrix is set or group of numbers rrnged in squre or rectngulr rry enclosed by two brckets 4 c b d
3 Properties: Mtrices  Introduction A specified number of rows nd specified number of columns Two numbers (rows x columns) describe the dimensions or size of the mtrix. Exmples: x mtrix x4 mtrix x mtrix 4 4 5
4 Mtrices  Introduction A mtrix is denoted by bold cpitl letter nd the elements within the mtrix re denoted by lower cse letters e.g. mtrix [A] with elements ij A mxn = ma n m m ij ij ij in n mn i goes from to m j goes from to n
5 Mtrices  Introduction TYPES OF MATRICES. Column mtrix or vector: The number of rows my be ny integer but the number of columns is lwys 4 m
6 Mtrices  Introduction TYPES OF MATRICES. Row mtrix or vector Any number of columns but only one row 6 5 n
7 Mtrices  Introduction TYPES OF MATRICES. Rectngulr mtrix Contins more thn one element nd number of rows is not equl to the number of columns m n
8 Mtrices  Introduction TYPES OF MATRICES 4. Squre mtrix The number of rows is equl to the number of columns ( squre mtrix A hs n order of m) m x m The principl or min digonl of squre mtrix is composed of ll elements ij for which i=j
9 Mtrices  Introduction TYPES OF MATRICES 5. Digonl mtrix A squre mtrix where ll the elements re zero except those on the min digonl 9 5 i.e. ij = for ll i = j ij = for some or ll i = j
10 Mtrices  Introduction TYPES OF MATRICES 6. Unit or Identity mtrix  I A digonl mtrix with ones on the min digonl i.e. ij = for ll i = j ij = for some or ll i = j ij ij
11 Mtrices  Introduction TYPES OF MATRICES 7. Null (zero) mtrix  All elements in the mtrix re zero ij For ll i,j
12 Mtrices  Introduction TYPES OF MATRICES 8. Tringulr mtrix A squre mtrix whose elements bove or below the min digonl re ll zero
13 Mtrices  Introduction TYPES OF MATRICES 8. Upper tringulr mtrix A squre mtrix whose elements below the min digonl re ll zero i.e. ij = for ll i > j ij ij ij ij ij ij
14 Mtrices  Introduction TYPES OF MATRICES 8b. Lower tringulr mtrix A squre mtrix whose elements bove the min digonl re ll zero ij ij ij ij ij i.e. ij = for ll i < j ij 5
15 Mtrices Introduction TYPES OF MATRICES 9. Sclr mtrix A digonl mtrix whose min digonl elements re equl to the sme sclr A sclr is defined s single number or constnt i.e. ij = for ll i = j ij = for ll i = j ij ij ij
16 Mtrices Mtrix Opertions
17 Mtrices  Opertions EQUALITY OF MATRICES Two mtrices re sid to be equl only when ll corresponding elements re equl Therefore their size or dimensions re equl s well 5 5 A = B = A = B
18 Mtrices  Opertions Some properties of equlity: IIf A = B, then B = A for ll A nd B IIf A = B, nd B = C, then A = C for ll A, B nd C 5 A = B = b b b b b b b b b If A = B then b ij ij
19 Mtrices  Opertions ADDITION AND SUBTRACTION OF MATRICES The sum or difference of two mtrices, A nd B of the sme size yields mtrix C of the sme size c ij ij b ij Mtrices of different sizes cnnot be dded or subtrcted
20 Mtrices  Opertions Commuttive Lw: A + B = B + A Associtive Lw: A + (B + C) = (A + B) + C = A + B + C A x B x C x
21 Mtrices  Opertions A + = + A = A A + (A) = (where A is the mtrix composed of ij s elements)
22 Mtrices  Opertions SCALAR MULTIPLICATION OF MATRICES Mtrices cn be multiplied by sclr (constnt or single element) Let k be sclr quntity; then ka = Ak Ex. If k=4 nd A 4
23 Mtrices  Opertions Properties: k (A + B) = ka + kb (k + g)a = ka + ga k(ab) = (ka)b = A(k)B k(ga) = (kg)a
24 Mtrices  Opertions MULTIPLICATION OF MATRICES The product of two mtrices is nother mtrix Two mtrices A nd B must be conformble for multipliction to be possible i.e. the number of columns of A must equl the number of rows of B Exmple. A x B = C (x) (x) (x)
25 Mtrices  Opertions B x A = Not possible! (x) (4x) A x B = Not possible! (6x) (6x) Exmple A x B = C (x) (x) (x)
26 Mtrices  Opertions b b b b b b c c c c ( ( ( ( b b b b ) ) ) ) ( ( ( ( b b b b ) ) ) ) ( ( ( ( b b b b Successive multipliction of row i of A with column j of B row by column multipliction ) ) ) ) c c c c
27 Mtrices  Opertions ) (7 ) ( 8) (4 5) (7 6) ( 4) (4 ) ( ) ( 8) ( 5) ( 6) ( 4) ( Remember lso: IA = A
28 Mtrices  Opertions Assuming tht mtrices A, B nd C re conformble for the opertions indicted, the following re true:. AI = IA = A. A(BC) = (AB)C = ABC  (ssocitive lw). A(B+C) = AB + AC  (first distributive lw) 4. (A+B)C = AC + BC  (second distributive lw) Cution!. AB not generlly equl to BA, BA my not be conformble. If AB =, neither A nor B necessrily =. If AB = AC, B not necessrily = C
29 Mtrices  Opertions AB not generlly equl to BA, BA my not be conformble ST TS S T
30 Mtrices  Opertions If AB =, neither A nor B necessrily =
31 Mtrices  Opertions TRANSPOSE OF A MATRIX If : A A x T T A A Then trnspose of A, denoted A T is: T ij ji For ll i nd j
32 Mtrices  Opertions To trnspose: Interchnge rows nd columns The dimensions of A T re the reverse of the dimensions of A A A T T A A x x
33 Mtrices  Opertions Properties of trnsposed mtrices:. (A+B) T = A T + B T. (AB) T = B T A T. (ka) T = ka T 4. (A T ) T = A
34 Mtrices  Opertions. (A+B) T = A T + B T
35 Mtrices  Opertions (AB) T = B T A T 8 8 8
36 Mtrices  Opertions SYMMETRIC MATRICES A Squre mtrix is symmetric if it is equl to its trnspose: A = A T A b A T b b d b d
37 Mtrices  Opertions When the originl mtrix is squre, trnsposition does not ffect the elements of the min digonl d b c A d c b A T The identity mtrix, I, digonl mtrix D, nd sclr mtrix, K, re equl to their trnspose since the digonl is unffected.
38 Mtrices  Opertions INVERSE OF A MATRIX Consider sclr k. The inverse is the reciprocl or division of by the sclr. Exmple: k=7 the inverse of k or k  = /k = /7 Division of mtrices is not defined since there my be AB = AC while B = C Insted mtrix inversion is used. The inverse of squre mtrix, A, if it exists, is the unique mtrix A  where: AA  = A  A = I
39 Mtrices  Opertions Exmple: A A A Becuse:
40 Mtrices  Opertions Properties of the inverse: ) ( ) ( ) ( ) ( ) ( A k ka A A A A A B AB T T A squre mtrix tht hs n inverse is clled nonsingulr mtrix A mtrix tht does not hve n inverse is clled singulr mtrix Squre mtrices hve inverses except when the determinnt is zero When the determinnt of mtrix is zero the mtrix is singulr
41 Mtrices  Opertions DETERMINANT OF A MATRIX To compute the inverse of mtrix, the determinnt is required Ech squre mtrix A hs unit sclr vlue clled the determinnt of A, denoted by det A or A If then A 6 A 6 5 5
42 Mtrices  Opertions If A = [A] is single element (x), then the determinnt is defined s the vlue of the element Then A =det A = If A is (n x n), its determinnt my be defined in terms of order (n) or less.
43 Mtrices  Opertions MINORS If A is n n x n mtrix nd one row nd one column re deleted, the resulting mtrix is n (n) x (n) submtrix of A. The determinnt of such submtrix is clled minor of A nd is designted by m ij, where i nd j correspond to the deleted row nd column, respectively. m ij is the minor of the element ij in A.
44 Mtrices  Opertions eg. A Ech element in A hs minor Delete first row nd column from A. The determinnt of the remining x submtrix is the minor of m
45 Mtrices  Opertions Therefore the minor of is: m And the minor for is: m
46 Mtrices  Opertions COFACTORS The cofctor C ij of n element ij is defined s: i j C ij ( ) mij When the sum of row number i nd column j is even, c ij = m ij nd when i+j is odd, c ij =m ij c c c ( i ( i ( i,,, j j j ) ) ) ( ) ( ) ( ) m m m m m m
47 Mtrices  Opertions DETERMINANTS CONTINUED The determinnt of n n x n mtrix A cn now be defined s A det A c c c n n The determinnt of A is therefore the sum of the products of the elements of the first row of A nd their corresponding cofctors. (It is possible to define A in terms of ny other row or column but for simplicity, the first row only is used)
48 Mtrices  Opertions Therefore the x mtrix : A Hs cofctors : c m And: c m And the determinnt of A is: A c c
49 Mtrices  Opertions Exmple : A A ( )() ()() 5
50 Mtrices  Opertions For x mtrix: A The cofctors of the first row re: c c c ( )
51 Mtrices  Opertions The determinnt of mtrix A is: A c c Which by substituting for the cofctors in this cse is: A ( ) ( ) ( )
52 Mtrices  Opertions Exmple : A 4 ) ()( ) ()( ) )( ( A ) ( ) ( ) ( A
53 ADJOINT MATRICES Mtrices  Opertions A cofctor mtrix C of mtrix A is the squre mtrix of the sme order s A in which ech element ij is replced by its cofctor c ij. Exmple: If A 4 The cofctor C of A is C 4
54 Mtrices  Opertions The djoint mtrix of A, denoted by dj A, is the trnspose of its cofctor mtrix T dja C It cn be shown tht: A(dj A) = (dja) A = A I Exmple: A A dja 4 ()(4) ()( ) T C 4
55 Mtrices  Opertions I dja A 4 4 ) ( I A dja 4 4 ) (
56 Mtrices  Opertions USING THE ADJOINT MATRIX IN MATRIX INVERSION Since AA  = A  A = I nd A(dj A) = (dja) A = A I then A dja A
57 Mtrices  Opertions Exmple A 4 A = To check AA  = A  A = I I A A I AA
58 Mtrices  Opertions Exmple A The determinnt of A is A = ()()()()+()(4) =  The elements of the cofctor mtrix re c c c ( ), ( ), ( ), c c c ( ), ( 4), ( ), c c c (), (7), (5),
59 Mtrices  Opertions C The cofctor mtrix is therefore so T C dja nd A dja A
60 Mtrices  Opertions The result cn be checked using AA  = A  A = I The determinnt of mtrix must not be zero for the inverse to exist s there will not be solution Nonsingulr mtrices hve nonzero determinnts Singulr mtrices hve zero determinnts
61 Mtrix Inversion Simple x cse
62 Simple x cse Let d c b A nd z y x w A Since it is known tht A A  = I then z y x w d c b
63 Simple x cse Multiplying gives w by x bz cw dy cx dz It cn simply be shown tht A d bc
64 Simple x cse thus w y b cw y d w cw b d d w d bc d A
65 Simple x cse x z b cx z d x cx b d b x d bc b A
66 Simple x cse by w dy w c by dy c c y d cb c A
67 Simple x cse bz x dz x c bz dz c z d bc A
68 Simple x cse So tht for x mtrix the inverse cn be constructed in simple fshion s c b d A A A c A b A d Exchnge elements of min digonl Chnge sign in elements off min digonl Divide resulting mtrix by the determinnt z y x w A
69 Simple x cse Exmple A A Check inverse A  A=I I 4 4
70 Mtrices nd Liner Equtions Liner Equtions
71 Liner Equtions Liner equtions re common nd importnt for survey problems Mtrices cn be used to express these liner equtions nd id in the computtion of unknown vlues Exmple n equtions in n unknowns, the ij re numericl coefficients, the b i re constnts nd the x j re unknowns n n nn n n n n n n b x x x b x x x b x x x
72 Liner Equtions The equtions my be expressed in the form AX = B where,, n nn n n n n x x x X A nd b n b b B n x n n x n x Number of unknowns = number of equtions = n
73 Liner Equtions If the determinnt is nonzero, the eqution cn be solved to produce n numericl vlues for x tht stisfy ll the simultneous equtions To solve, premultiply both sides of the eqution by A  which exists becuse A = Now since We get A  AX = A  B A  A = I X = A  B So if the inverse of the coefficient mtrix is found, the unknowns, X would be determined
74 Liner Equtions Exmple x x x x x x x x The equtions cn be expressed s x x x
75 Liner Equtions When A  is computed the eqution becomes X A B Therefore x x x,, 7
76 Liner Equtions The vlues for the unknowns should be checked by substitution bck into the initil equtions x x x x x x x x 7) ( ) ( () ) ( () 7) ( ) ( () 7,, x x x
77
78 Complex Numbers Lesson 5.
79 The Imginry Number i It's ny number you cn imgine By definition i i Consider i powers if i i i i i 4 i i i 5 4 i i i i i...
80 Using i Now we cn hndle quntities tht occsionlly show up in mthemticl solutions i Wht bout 49 8
81 Complex Numbers Combine rel numbers with imginry numbers + bi Rel prt Exmples 4i Imginry prt 6 i 4.5 i 6
82 Try It Out Write these complex numbers in stndrd form + bi
83 Opertions on Complex Numbers Complex numbers cn be combined with ddition subtrction multipliction division Consider 9 i 7 5i i 8 i 4i 4 i
84 Opertions on Complex Numbers Division technique Multiply numertor nd denomintor by the conjugte of the denomintor i 5 i i 5 i 5 i 5 i 5i 6i 5 4i 6 5i 6 5 i 9 9 9
85 Complex Numbers on the Clcultor Possible result Reset mode Complex formt to Rectngulr Now clcultor does desired result
86 Complex Numbers on the Clcultor Opertions with complex on clcultor Mke sure to use the correct chrcter for i. Use nd i
87 Wrning Consider 6 49 It is tempting to combine them The multiplictive property of rdicls only works for positive vlues under the rdicl sign Insted use imginry numbers i 7i 47i 8
88 Try It Out Use the correct principles to simplify the following:
89 Assignment Lesson 5. Pge 4 Exercises 69 EOO
90 sttistics STATISTICS
91 Defintions: Sttistics ; Mesure of centrl tendency Defintion of sttistics: sttistics my be defined s the science of collection, orgniztion, nlysis nd interprettion of numericl dt. Mesures of Centrl Tendency: An verge is clled mesure of centrl tendency, becuse it tends to lie centrlly with the vlues of the vrible rrnged ccording to mgnitude.
92 Arithmetic Men(A.M.): The rithmetic men of n individul series is defined s the quotient of the sum of ll the vlues of the vrible by the totl number of items. x n n i x i
93 Exmple Exmple : The blood pressure of seven middle ged men were s follows: 5, 4,, 7, 46, 4 nd. The men is x
94 In cse of discrete frequency distribution A.M. is clculted s: A.M. (X) = (f i x i ) / f i OR (f x) / f Where f i is the frequency of x i ( I n)
95 Exmple Sol. Clculte the A.M. for the following dt: Income (in rs.): No. of emp: s.no. Income in rs. (x i ) No. of employees (f i ) (f i x i ) Men = (f i x i ) / f i = 79/ = f = (f i x i ) = 79
96 For Grouped or continuous frequency distribution, Arithmtic men is clculted s: Arithmtic Men = A + f i u i * h f i This method is clled STEP DEVIATION METHOD
97 EXAMPLE Clculte the Arithmetic men of the mrks scored by the students of clss in clss test from the following dt : Mrks Number of students Mid Point (x i ) ui= (xi A)/h fiui A = Totl By step Devition method, Arithmtic men = A+( fi ui / fi) * h = 5 + (/)* = 8
98 Medin : The medin of sttisticl series is defined s the size of the middle most item (or the A.M. of two middle most items), provided the items re in the order of mgnitude. For n individul series, to find medin we proceed s follow: () Arrnge the observtions in scending or descending order of mgnitude. (b) If n is odd; then medin = (n+/) th observtion. If n is even; them medin = A.M. of (n/) th & (n/) + th observtion.
99 Exmple : Find the medin of the vlues: Sol: We rrnge the vlues in scending order n = 8 (even); Medin = A.M. of (8/) th & (8/) + th observtions = (+5)/ =
100 In cse of discrete frequency distribution Medin is clculted s: Step I: Find the cumultive frequency (C.F.). Step : Find N/ where N= f i Step : See the C.F. just greter thn N/. Step 4: The vlue of x corresponding to C.F. just greter thn N/ is medin.
101 Exmple : Clculte the Medin of the following frequency distribution : x f c.f N = Here N is ; N/ is 6; C.F. just greter thn N/ is 65; So corresponding vlue of x 5 is medin.
102 For Grouped or continuous frequency distribution, Medin is clculted s: Step I: Find the cumultive frequency (C.F.). Step : Find N/ where N= f i Step : The clss corresponding to C.F. just greter thn N/ is medin clss nd the vlue of the medin is clculted by formul: Medin = l + ((N/F)/f) *h Where l= lower limit of medin clss F = C.F. of clss preceding the medin clss f is the frequency of medin clss h is the width of medin clss.
103 Exmple Clculte the medin from the following distribution: Clss Freuency Cumtive frequency (F) 5 (Medin clss) 5 (f) 6 (C.F. just greter thn N/) N= 49 Here N = 49 N/ = 4.5 C.F. just greter thn N/ is 6 Corresponding clss 5 is medin clss Medin = l +((N/F)/f) *h Medin = 5 + (4.5)/5 * 5 = 9.5
104 MODE The mode of distribution of the vrible is tht vlue of the vrible for which the frequency is mximum. In cse of n individul series, mode is clculted s by counting the number of times the vrious vlues repet themselves nd the vlue which occurs mximum no. of times is the modl vlue.
105 Exmple Find the Mode of the following dt: 4 4 Sol: Since the vlue occurs the mximum no. of times. Hence the modl vlue is.
106 In cse of discrete frequency distribution mode is clculted s: For discrete frequency distribution, generlly mode is clculted by finding the vlue for which frequency is mximum.
107 Exmple Find the mode of the following distribution: Size in inches No of shirts sold shirts of size 6 hve the mximum sle. So mode of distribution is 6.
108 For grouped or continuous frequency distribution mode is clcultes s: To find the mode of continuous frequency distribution, we follow the following steps: STEP : Determine the clss of mximum frequency, this clss is modl clss. STEP : Determine the vlue of mode by pplying the formul: Mode = l + (ff / ff f ) * h Where l is the lower limit of modl clss f is the frequency of modl clss h is the width of modl clss f is the frequency of clss preceding the modl clss f is the frequency of clss following the modl clss
109 Exmple Clculte the mode from the following dt: Rent (in Rs.) No. of houses ( f )  (Modl clss) (f) 4 5 (f ) 46 Highest frequency is Hence Modl clss is  Mode = l + (ff / ff f ) * h Here l= ; h= ; f= ; f = 4; f = 5 Mode = + (4)/(*45) * =.9
110 Mesures of dispersion Men Devition: It is defined s the A.M. of the bsolute devitions of ll the vlues tken from ny centrl vlue. Stndrd Devition: The stndrd devition of sttisticl dt is defined s the positive squre root of the A.M. of the squred devitions of items from the A.M. of the series under considertion.
111 For individul series Men devition is clculted s: Men Devition ( x ) = x i  x n Men Devition ( Medin ) = x i  Medin n Cofficient of Men Devition (x) = M.D.(X) x Cofficient of Men Devition (Medin) = M.D.(Medin)/Medin
112 Exmple Clculte the men devition bout men nd its coefficient for the following dt: S.No. Xi xi  x x i  x Men = 7/ =.7 M.D. = 86.4/ = 8.64 Coeff. = 8.64/.7 = 6 x i  x = 86.4
113 For frequency distribution & Grouped Dt M.D.is clculted s: Men Devition ( x ) = f i x i  x n Men Devition ( Medin )= f i x i  Medin N
114 Exmple Clculte the men devition bout men nd its coefficient for the following frequency distribution: X f fx Xx x  x f x  x N = 6 f x = 4 f x  x =88 Men = 4/6=9 M.D. = 88/6=.8 Coefficient of M.D. =.8/9=.8
115 Note: Sme method will be used for finding the men devition bout medin. Insted of men, we re to find medin first then find men devition by following the sme procedure nd sme formuls.
116 For Individul series & frequency distribution Stndrd Devition is clculted s:
117 Exmple Find the S.D. nd C.V. for the following dt: 4,6,,,8 S.No. x x x (x x) x64 Men = x / n Men = 5/5 = C.V. = (4.899/) * = %
118 Exmple Clculte the S.D. nd C.V. for the following dt: X f fx X  x (x x) f (x x ) f = fx = 8 f (x x ) = 99 Men = 8/ = 8 C.V. = S.D. / Men * = 5.9%
119 For Grouped dt S.D. is clculted s: S.D. = f i ( u i ) /N ( f i u i /N) /N * h Where u i = x i  A nd N is the sum of frequency. h
120 Exmple Find the S.D. & C.V. for the following dt: clss f x U = xa/h fu u fu f= fu = 66 fu = 9 Men = A + fu/ f *h =.85 S.D. = f i ( u i ) /N ( f i u i /N) /N * h =. C.V. = C.V. = S.D. / Men * = 58.%
121 OTHER FORMULAS (S.D.) Cofficient of S.D. = S.D./Men Cofficient of vrition = (S.D./Men) * Vrince = Squre of S.D.
122 Rnk Correltion Coefficient: Rnk Cooreltion Coefficient is given by the formul: r = 6 d n (n ) Where n is no. of items d is the difference of rnks.
123 Exmple Find the coefficient of rnk correltion for the following dt: x y R R d=rr d r = (6* 6)/() =.64 d = 6
124 Exmple Seven Competitors in music competition re rnked by the judges x & y in the following order. Wht is the degree of greement between the judges. Also find the coefficient of correltion. S.No. Competitors R R d = RR d A B  4 C D 45 E F G d = r = (7/7*48) =.7857 (Agreement between the judges is high)
125 Binomil Theorem
126 Session Objective. Binomil theorem for positive integrl index. Binomil coefficients Pscl s tringle. Specil cses (i) Generl term (ii) Middle term (iii) Gretest coefficient (iv) Coefficient of x p (v) Term dependent of x (vi) Gretest term
127 Binomil Theorem for positive integrl index Any expression contining two terms only is clled binomil expression eg. +b, + b etc For positive integer n n n n n n n n n n n n b c b c b c b... c b c b n n n n n r r cr b Binomil theorem r n! n! n c cn r for r n r! n r! n r!r! n where r n re clled binomil coefficients. n n... n r C r,...r!.9.8 7!!.. numertor contins r fctors C7 C 7 C
128 Pscl s Tringle b b b b b b b b b b b 4 b 6 b 4b b b 5 b b b 5b b 5 C 4 C C 5 C C 4 C C C C C C 5 C 4 C C 5 C C 4 C C 5 C 4 4 C 4 5 C 5 n n n cr c r cr
129 Observtions from binomil theorem. (+b) n hs n+ terms s r n. Sum of indeces of nd b of ech term in bove expnsion is n. Coefficients of terms equidistnt from beginning nd end is sme s n c r = n c nr n n n n n n n n n n n b c b c b c b... c b c b n n
130 Specil cses of binomil theorem n n n n n n n n n n n x y c x c x y c x y... c y n r r n n r r r c x y n n n n n n n n r n r r x c c x c x... c x c x in scending powers of x n n n n n n x c x c x... c n n n n r cr x r n x in descending powers of x
131 Illustrtive Exmple Expnd (x + y) 4 +(x  y) 4 nd hence 4 4 find the vlue of Solution : x y C x y C x y C x y C x y C x y x 4x y6x y 4xy y Similrly x y x 4x y 6x y 4xy y 4 4 x y x y x 4 6x y y Hence 6 =4
132 Generl term of ( + b) n n nr r r Tr c b,r,,,...,n n n r, First Term T c b r, Second Term T n n c b n nr r r r T c b,r,,,...,n r 4 n n T T T T4 T5 Tn Tn n+ terms kth term from end is (nk+)th term from beginning
133 Illustrtive Exmple Find the 6th term in the expnsion of 9 4x 5 nd its 4th term from the x 5 end. Solution : 9r r T 9 4x 5 r C r 5 x x 5 9! 4 5 T6 T5 C5 5 x 4!5! x x 54 x
134 Illustrtive Exmple Find the 6th term in the 9 expnsion of 4x 5 nd its 4th term from the x 5 end. Solution : 9r r T 9 4x 5 r C r 5 x 4th term from end = 94+ = 7th term from beginning i.e. T x 5 9! 4 5 T7 T6 C6 5 x!6! 5 6 x 5 x x
135 Middle term CseI: n is even, i.e. number of terms odd only one middle term n th term n n n n n n T T c b CseII: n is odd, i.e. number of terms even, two middle terms n n th th term term n n n n n n T T c b n n n n n n T T c b Middle term n x =? x
136 Gretest Coefficient r CseI: n even n c, r n Coefficient of middle n Cn term T n is mx i.e. for r n CseII: n odd n n Coefficient of middle term T or T is mx i.e. for r or n n n n n n C or C
137 Illustrtive Exmple Find the middle term(s) in the expnsion of nd 7 x x 6 hence find gretest coefficient in the expnsion Solution : Number of terms is 7 + = 8 hence middle terms, (7+)/ = 4th nd (7+)/ = 5th x 7! x T4 T C x 6 4!! x 5 x.. 8
138 Illustrtive Exmple Find the middle term(s) in the expnsion 7 of x nd hence find gretest x 6 coefficient in the expnsion Solution : x 7! x T5 T4 C4 x 6!4! x 5 5 x Hence Gretest coefficient is 7 7 7! C4 or C or 5!4!..
139 Coefficient of x p in the expnsion of (f(x) + g(x)) n Algorithm Step: Write generl term T r+ Step: Simplify i.e. seprte powers of x from coefficient nd constnts nd equte finl power of x to p Step: Find the vlue of r
140 Term independent of x in (f(x) + g(x)) n Algorithm Step: Write generl term T r+ Step: Simplify i.e. seprte powers of x from coefficient nd constnts nd equte finl power of x to Step: Find the vlue of r
141 Illustrtive Exmple Find the coefficient of x 5 in the expnsion of nd term independent of x x Solution : x r Tr Cr x x r r r C r r r x For coefficient of x 5,  5r = 5 r = T 5 C x Coefficient of x 5 = 85
142 Solution Cont. r Tr Cr x x r r r C r r r x For term independent of x i.e. coefficient of x,  5r = r = 4 T 4 4 C4 4 Term independent of x
143
144 Introduction In this chpter you will lern to dd frctions with different denomintors ( recp) You will lern to work bckwrds nd split n lgebric frction into components clled Prtil Frctions
145
146 A You cn dd nd subtrct severl frctions s long s they shre common denomintor Prtil Frctions Clculte: You will hve seen this plenty of times lredy! If you wnt to combine frctions you must mke the denomintors equivlent Multiply brckets Group terms
147
148 B Prtil Frctions You cn split frction with two liner fctors into Prtil Frctions For exmple: when split up into Prtil Frctions when split up into Prtil Frctions You need to be ble to clculte the vlues of A nd B
149 B Prtil Frctions You cn split frction with two liner fctors into Prtil Frctions Split Split the Frction into its liner prts, with numertors A nd B Crossmultiply to mke the denomintors the sme into Prtil Frctions Group together s one frction This hs the sme denomintor s the initil frction, so the numertors must be the sme If x = : If x = : You now hve the vlues of A nd B nd cn write the nswer s Prtil Frctions
150
151 C Prtil Frctions You cn lso split frctions with more thn liner fctors in the denomintor For exmple: when split up into Prtil Frctions
152 C Prtil Frctions You cn lso split frctions with more thn liner fctors in the denomintor Split Split the Frction into its liner prts Cross Multiply to mke the denomintors equl into Prtil frctions Put the frctions together The numertors must be equl If x = If x = If x = .5 You cn now fill in the numertors
153 C Prtil Frctions You cn lso split frctions with more thn liner fctors in the denomintor Split into Prtil frctions You will need to fctorise the denomintor first Therefore (x + ) is fctor Try substituting fctors to mke the expression Divide the expression by (x + ) You cn now fctorise the qudrtic prt
154 C Prtil Frctions You cn lso split frctions with more thn liner fctors in the denomintor Split the frction into its liner prts Split Cross multiply into Prtil frctions Group the frctions The numertors must be equl If x = If x = If x =  Replce A, B nd C
155
156 D Prtil Frctions You need to be ble to split frction tht hs repeted liner roots into Prtil Frction For exmple: when split up into Prtil Frctions The repeted root is included once fully nd once broken down
157 Prtil Frctions You need to be ble to split frction tht hs repeted liner roots into Prtil Frction Split Split the frction into its prts Mke the denomintors equivlent into Prtil frctions Group up The numertors will be the sme If x =  At this point there is no wy to cncel B nd C to leve A by substituting vlue in Choose ny vlue for x (tht hsn t been used yet), nd use the vlues you know for B nd C to leve A If x = .5 If x = Sub in the vlues of A, B nd C D
158
159 E Prtil Frctions You cn split n improper frction into Prtil Frctions. You will need to divide the numertor by the denomintor first to find the whole prt A regulr frction being split into components A top hevy (improper) frction will hve whole number prt before the frctions
160 E Prtil Frctions You cn split n improper frction into Prtil Frctions. You will need to divide the numertor by the denomintor first to find the whole prt Divide the numertor by the denomintor to find the whole prt Split Now rewrite the originl frction with the whole prt tken out into Prtil frctions Remember, Algebriclly n improper frction is one where the degree (power) of the numertor is equl to or exceeds tht of the denomintor Split the frction into prts (ignore the whole prt for now) Mke denomintors equivlent nd group up The numertors will be the sme If x = If x =
161 Summry We hve lernt how to split Algebric Frctions into Prtil frctions We hve lso seen how to do this when there re more thn components, when one is repeted nd when the frction is improper
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