U.G.Mathematics (Short Questions and Answers) Dr.Madhumanial Pal Department of Applied Mathematics with Oceanology andComputer Programming Vidyasagar University Midnapore - 721102 West Bengal ~ ,Asian"B""ks7JJ'itJatt.t . . i H l i t t . ~ 7/28,Mahavir Lane, VardanHouse, Ansari Road, Darya Ganj,New Delhi-11 0002 Corporate andEditorial office 7/28, MahavirLane, Vardan House, Ansari Road, Darya Ganj, New Delhi-IIO 002. 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PublishedbyKamalJagasiaforAsianBooksPvt.Ltd.,7/28,MahavirLane,VardanHouse, AnsariRoad, Darya Ganj, New Delhi-I I 0 002 Printed atYash Prinlographs Dedicatedtomybelovedson Aniket "This page is Intentionally Left Blank" Preface ThisbooktitledU.C.Mathematics(ShortQuestionsandAnswers)has beenwrittenforthestudentsofB.A.,B.Sc.generalcourseforallIndian Universities.Alltheeffortshavebeenmadetomakethisbookusefulfor othercompetitiveexaminations.Thoughthebookisofgeneralnaturebut anefforthasbeenmadetocoverupmostof thetopicsprescribedforB.A., B.Sc.generalcourses.Itisashortanswertypebookwhichincludesthe topicsclassicalalgebra,linearalgebra,abstractalgebra,differentialcalculus, integralcalculus,differentialequation,LPP,dynamics,probability,statistics and numerical methods.Necessary theoremsand formulaehavebeen outlined inthebeginningofeachchapterwhichmaybealmostessentialinspecific problems. While writing thisbook,the worksof severalauthors havebeen frequently consulted.SoIam really indebted to those authors and express my gratitude to all suchpersons. This book could not havebeen completed without the extreme support and encouragementof mywifeAnita and indirectsupport of my sonAniket. Iwouldliketothankmyteachersandcolleaguesfortheirinvaluablesug-gestionsand support. Iwould alsoliketothank the publisher of Asian Books Private Limited for bringing outthe bookinsuch agoodshape in suchashorttime. It isearnestly hoped that the book will be liked by the students and teachers alike.Comments and suggestions forthe further improvement of the book shall be gratefully accepted. January 2004 MadhumangalPal Contents Preface (v) Classical Algebra I.l 1.1Complex Number 1.1 1.1.1Geometrical representation (,f a complex number:Argand diagram 1.1 1.1.2Representation of complex numbers 1.5 1.1.3De Moiver's theorem and its use1.10 1.1.4Euler's definition of exponential and trigonometrical functions ....1.12 1.2Theory of Equation.1.13 1.2.1Remainder theorem and synthetic division1.13 1.2.2Descartes'rule of signs1.17 1.2.3Relation between roots and coefficients1.20 1.2.4Transformationof equation1.22 2Linear Algebra2.1 2.1Determinant2.1 2.1.1Some properties of determinants2.1 2.1.2Minor, cofactor and adjoint of determinant2.4 2.1.3Cramer's rule2.8 2.3Matrices2.9 2.2.1Symmetric and skew-symmetric matrices2.15 2.2.2Adjoint and inverse of a matrix2.17 2.2.3Orthogonal matrix2.19 2.2.4Rank of a matrix2.20 2.2.5Solution of equations by matrixinverse method2.22 2.3Vector Space2.23 2.3.1Linearly dependence andindependence of vectors2.24 2.3.2Basis of a vector space2.26 2.3.3Eigenvalues2.27 3Abstract Algebra 3.1 3.1Set Theory3.1 3.1.1Basic set operations 3.2 3.1.2Laws of algebra of sets 3.5 3.1.3Cartesian product of sets3.6 3.2Relation 3.9 3.2.1Types of relations 3.9 3.3Mapping 3.11 3.4Permutation 3.15 3.5Group 3.18 3.5.1Cyclic group, sub-group 3.22 3.6Ring,Integral Domain and Field 3.24 4Geometry 4.1 4.1Transformation of Axes4.1 4.2Pair of Straight Lines4.7 4.3Polar Equation4.16 4.4General Equation of SecondDegree4.24 4.5Direction Cosines and Ratios4.27 4.6Plane4.31--4.7Straight Lines4.39 4.8Sphere4.46 4.9Cylinder4.49 4.10Cone4.50 5Vector Algebra5.1 5.1Elementary Operations5.1 5.2Scalar Product or Dot Product5.5 5.3Vector Product or Cross Product5.8 5.4MechanicalProblems5.16 6Differential Calculus6.1 6.1Number System6.1 6.1.1Absolute value of a real number6.3 6.2Sequence6.5 6.3Infinite Series6.13 6.3.1Series of positive terms6.15 6.3.2Alternating Series6.20 6.4Function6.22 6.5Limit and Continuity6.26 6.5.1Continuity of a function6.32 6.6Uniform continuity6.36 6.7Differentiation6.37 6.8Successive Differentiation6.48 6.9PartialDifferentiation6.53 6.10Mean Value Theorem6.61 6.11Maximum and Minimum6.68 6.12Tangent and Normal6.70 6.13Asymptotes6.75 6.14Envelope6.77 6.15Radius of Curvature6.79 7Integral Calculus7.1 7.1Integration7.1 7.1.1Indefinite integration7.1 7.1.2Definite integration7.8 7.2Beta and Gamma Functions7.17 7.3Multiple Integral 7.21 7.4Area,Surface andVolume7.23 8Differential Equation 8.1 8. IIntroduction 8.1 8.2Homogeneous Equation8.10 8.3Linear Equation8.12 8.4First Order Higher Degree.....8.17 8.5Orthogonal Trajectories.....8.19 8.6Higher Order First Degree8.20 9Linear Programming Problem9.1 9.1GraphicalMethod9.1 9.2MathematicalPreliminaries9.5 9.3Simplex Method9.20 9.4Duality9.26 9.5Transportation Problem9.31 9.6Assignment Problem9.37 toDynamics of Particles10.1 10.1Principle of Dynamics10.1 10.2Work,Power andEnergy10.5 10.3Impulse and Impulsive Forces....10.10 10.4Collision of Elastic Bodies....10.12 10.5Motionina Straight Line....10.15 10.6Simple Harmonic Motion....10.17 10.7Motion inaPlane....10.19 10.8Central Orbit....10.24 10.9Planetary Motion ....10.28 10.10Artificial Satellites ....10.29 11Probability and Statistics 11.1 11.1Measure of Central Tendency 11.1 11.2Measure of Dispersion 11.5 11.3Theory of Probabil ity 1.15 I 1.4Distribution Function 1.24 11.5Mathematical Expectation 1.34 , 1.6 Correlation and Regression 1.42 11.7Sampling Distribution 1.47 11.8Testing of Hypothesis 1.48 12Numerical Methods 12.1 12.1Errors 12.1 12.1.1Sources of error 12.1 12.1.2Absolute, relative and percentage error 12.2 12.1.3Generation of round-off errors .12.3 12.2Interpolation 12.2.1Lagrange Interpolation 12.2.2 Finite differences 12.2.3 Properties of Lland V 12.2.4 Shift operator E 12.2.5 Newtoninterpolation formulae 12.3Numerical Differentiation 12.3.1Differentiation based on Newtonforwardinterpolation 12.3.2 Differentiation based on Newtonbackwardinterpolation 12.4NumericalIntegration 12.4.1Euler-Maclaurin sum formula 12.5Solution of Algebraic andTranscedental Equations 12.5.1Bisection method 12.5.2 Regula-Falsi method or Method offalse position 12.5.3Iteration method 12.5.4 Newton-Raphson method 12.6Solution of System of Linear Equations 12.6.1Gauss-Elimination method 12.6.2 Gauss-Seidaliteration method 12.7Solution of Differential Equation 12.7.1Taylor's series method 12.7.2Euler method 12.7.3Modified Euler method 12.7.4 Runge-Kutter methods 12.5 12.5 12.7 12.9 12.10 12.12 12.15 12.15 12.15 12.18 12.21 12.22 12.22 12.23 12.24 12.25 12.28 12.28 12.29 12.31 12.31 12.33 12.33 12.34 Ch.1IIClassicalAlgebra 1.1ComplexNumber 1.1.1Geometricalrepresentationof acomplexnumber:Ar-ganddiagram Letz=x+ iybe acomplexnumber. Let (r, B)be the polarof Pwhose cartesian coordinates be (x, y). imaginary axis P y ox real axis Figure:1.1.1 Then x= rcosB,y= rsinO.Hencer= vx2 +y2 and B =. . '.z= x + iy = r( cos B + i sin B)iscalled the polar form ornormal form or modulusamplitude form of thecomplexnumber. r=lengthof OP,calledthemodulusormagnitudeofthecomplex number z and denotedbyjzjorbymod z,i.e.,jzl=vx2 + y2. The angleB isthe anglebetweenthelinesegment0 Pand the positivedi-rection of x-axis,called the amplitude or argument of the complex number zandisdenotedbyampz= B = tan-1

There are infinite number of valuesof B forwhichcos B =sinO=But thereisoneand onlyonevalueaof 0suchthat-7r < a7r.The valueof 0 whichlieswithin(-7r, 7rJisknownastheprincipalvalueof theamplitude and it isunique. Note1Theprincipalamplitudeofthecomplexnumbersa + ib,-a + ib, -a - ib,a- ib,wherea, b ;:::0maybe calculatedusingthe formulaetan-1

7r- tan-1 ,-7r + tan-1 aIlcl- tan-1 respectively. Ex.1(i)Represent'-i' in the Argand plane and finditsamplitude. (ii)Find theprincipal amplitude of thecomplexnumber-1 + iV3. 1.2U.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) (iii)Find the principleamplitude of the complex numberl. (iv)Findthemodulusandtheprincipal amplitudeof thecomplexnumber 1/(1 + i). (v)Find the modulusand the principal amplitude of the followingcomplex numbers (a)sin/3 +i cos /3, (b)1+itan!1I". (vi)Showthat the points3 - 4i, -3 +4iand 6 - 8iin the Argand plane are collinear. (vii)Find the principal amplitude of -1- i . SOLUTION: (i)Representation of -i in Argandplane isshownbelow: Imaginary axis + I IIIRealaxis Origin t-i Figure:1.1. 2 The principal amplitude of -i is- tan-1 ~=- ~ . (ii)The principal amplitude of the complex number-1 +iV3 is 11"- tan-1 ( ~ )=11"- i =2;. (iii)The principal amplitude of 1 istan-l (i) =o. (iv) 11-i1-i11 Wehave1 +i= (1- i)(1 +i)= -2- = i - ii. The modulusisJ ~+~=~ . The principal amplitude is-tan-1 (*) =-tan-l 1 =- ~ . (v)(a)Modulusof sin/3 +i cos /3isJSin2/3+cos2 /3=1 and amplitude is 1(cos/3)1111"11" tan- sin /3= tan-cot /3= tan- {tan( 2"- /3)}= 2"- /3. (b)ModulusisJ1 +tan2 i1l"=sec 3;andprincipalamplitudeof 1 + itan 3;=1 - itan 2;is - tan 211'211"211" tan-1 (15) = - tan-l tan 5= -5' CH.l:CLASSICALALGEBRA1.3 (vi)The adjacentfigureshowsthat the givenpointsare collinear. Imaginaryaxis -3+4i4 Real axis -4 -6 -S6-Si Figure1.1. 3 (vii)Wehave tan-l =tan-II = but the point isin the third quad-h..Ird'71"371" rant,soteprmClpaamp Itue IS-71"+4'=-4"' Exercise1Findthemodulusandtheprincipalamplitudeof thefollowing complexnumbers(i)3 + 4i,(ii)4 - 8i,(iii)1 - i, (iv)3 +iv's,(v)-2 +i,(vi)i,(vii)-1. [Ans.(i)5, tan-l t, (ii)4V5, - tan-l 2,(iii)V2, (iv)2v's,(v)V5,11'-tan-l !, (vi)1, (vii)1,1I'.J Ex.2(i)Showthat the points-2 +i, 1 +2i, 4 +5i, 1 +4iarethe vertices of aparallelogram. (ii)Showthatthepoints2 - 2i, 5 +i, 2 +4i, -1 +iaretheverticesofa square. (iii)Namethenatureof thefigureyouget,if thepointsrepresentingthe numbers2 3iinArgandplaneare joined. (iv)Showthatthe points-1, i, -i, !( V2 +iV2) are concyclic . SOLUTION: (i)LetA( -2+i), B(1 + 2i), C(4+5i), D(l +4i) be the vertices of a parallel-ogram.The middlepoints of the diagonalAC and BD arerespectively 1 +3i,both areequal.AlsoIABI= 1(-2 +i)- (1+2i)1=I - 3 - il= J9TI = v'iO and ICDI=:;1{4+5i) - (1+4i)1=13+il= J9TI = v'iO, i.e.,twoopposite sidesare equal. Hencethe givenpointsrepresentthe verticesof aparallelogram. (ii)LetA(2 - 2i), B(5 + i), C(2 +4i), D( -1 +i)be the verticesof asquare. The middle points of the diagonal AC and BD are respectively 2+i, both are equal.AlsoIABI=1(2- 2i)- (5+i)1=1- 3 - 3il=J9+9 =3V2 1.4U.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) andIBGI=1(5+i)- (2+4i)1=13- 3il=V9 +9=3V2,i.e.,two adjacentsidesare equal. IAGI=1(2- 2i)- (2+4i)1= 1 - 6il= J36 = 6. .AB2 + BC2 =18 + 18=36=AG2. Thus one angleisrightangle. Hencethe givenpoints representthe verticesof asquare. (iii)Representation of the complex numbers2 3ior,-2 - 3i, -2 +3i, 2-3i,2 +3iinArgandplane isshownintheadjacentfigure.From figure itisobviousthatthegivenpoints representarectanglewhosesidesare 4and 6units. Imaginary axis -2+3i2+3i Real axis -2-3i2-3i Figure:1.1.4 (iv)Letz =0 be the centre of the circle. Now,Iz - 11=1, Iz- il =1, Iz + il =1, Iz- !(V2+iV2)1 = 1- ~- i ~ 1= J!+! = 1. 2v'2V222 That is,the distancesbetweenz =0afidthe givenpointsare1. Hencethegivenpointsare concyclic. Ex.3(i)Find the area of the rectanglewhose fourverticesarea bi. (ii)Find the area of the rectanglewhose fourverticesare3 2i . SOLUTION: (i)Theverticesof therectangleare-a - ib, -a + ib, a- ib, a+ ib.The lengthalongtherealaxisis2aandthat of alongtheimaginaryaxisis 2b.Hencethe area of the rectangleis2ax2b= 4absquare units. (ii)Same asabove.Put a=3, b =2. CH.1:CLASSICALALGEBRA 1.1.2Representation of complexnumbers Ex.4Expressthe followingcomplexnumbersintopolar form (i)-1 + iv'3,(ii)1 - iv'3,(iii)1 + iv'3,(iv)-1 - i,(v)1 - i . SOLUTION: 1.5 (i)Let-1 = r cos 8, v'3 = r sin8.Thenr2= (_1)2+ (v'3)2= 4 or,r= 2 and tan 8 = -V3 = - tan i= tan(1I"- i) = tan(31f) or,8 =2;. Hence-1 + iV3 =2(cos2;+ i sin 2;). (ii)Let1 = rcos8, -V3 = r sin8.Thenr2= (1)2+ (-v'3)2= 4or,r= 2 and tan 8 =- v'3,butthepointisin the fourthquadrant, 8 =-i (astan i=v'3) Hence1- iV3 =2{cos(-i) + iSin(-i)} =2(cos i- isin i) (iii)Let1 =rcos8, V3= rsin8.Then r2= (1)2+ (y'3)2=4 or,r= 2 and tan8 = v'3 = tan ior,8 =i Hence1 + iV3 =2(cos i+ i sin i). (iv)Let-1 = r cos 8, -1 =r sin8.Then r2= (_1)2 + (_1)2= 2 or,r= y'2 and tan 8 == ~=1,butthe pointisin the third quadrant, 8 = -11"+ tan-II = -11"+ 1 = _3;. Hence-1- i=y'2{cos( - 3;) + i sine _3;)} =y'2(cos 3;- isin ~ ) . (v)Let1 = r cos 9, -1 =r sin9.Then r2=(1)2 + (_1)2=2 or,r=y'2 and tan 9 =11=. -1.The pointisinthe fourthquadrant,so9 =- 1. Hence1 - i=y'2{ cos( -i) + i sine -1)}ii\ y'2(cos 1 - i sin i) Ex.5Expressthe followingexpressions in the formA + iB 1(1 ')211'5'9'14 (.)(.. )+ ,(... ) z +(.)( )z+ z+ z 11 + .,113., 111--1'IV(1')3'v'6+ '11+ '13 z-zz- -zzzz SOLUTION: 1l-il-i1-i11 (i)1 + i= (1- i)(l + i)= 1 - i 2=-2- ='2- '2i. (ii)(1+ i)21 + 2i + i2 1 + 2i - 1=~ 3-i3-i == 3-i3-i 6i + 2i26i - 213 . 9 - i 2= 10 = -5 + 5z. 2i(3 + i) = = (3- i)(3 + i) 1.6U.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) (iii)Letz=x+ iy. Thenz + 1= (x + iy) + 1= (x+ 1)+ iy z - 1(x + iy) - 1(x- 1)+ iy _{(x + 1) + iy}{(x - 1)- iy} - {(x -1) + iy}{(x - 1)- iy} (x- 1)(x + 1)+ y2+ iy{(x - 1)- (x + 1)} = (x_1)2+y2 _(x2 +y2 -1) - 2ixy - .(x- 1)2+ y2 x2 + y2- 1- 2xy. =(x- 1)2+ y2+ (x_1)2+ y2 z. (iv)Weknow-i3 = -i2.i= -(-1).i = i. 111 -:----:--::- = ----::----(1- i)31 - 3i + 3i2 - i3 1 - 3i - 3 + i 1-2 + 2i-2 + 2i == -:-------:--:------,- = -2 - 2i(-2 - 2i)( -2 + 2i)(_2)2 + (2)2 -2 + 2i11. =8=-4 + 4z. (v)i5 = (i2)2.i=(_1)2.i = i,i9 = (i2)4.i= (_1)4.i= i, i14= (i2)7= (_1)7=-1, i6 = (i2)3=(-1)3 = -1, ill=(i2)5.i=(-1)5.i =-i, i13 = (i2)6.i= (-1)6.i =i. i5 + i9 + i14i + i- 12i - 1 Hence'6 '11'13=.. =--=-2i + 1. z+z+z-1-z+z-1 Exercise2Express in the formA + iB (.)1( .. )1( ... )2 + 3i(.)a + iba - ib 1--'b'11 11l-2-"IV--'b - -+'b' a-zz-z-za-zaz (v)(vi)(a+ib)2+(b+ia)2. [A(.)a.b( .. )11.( ... )18.(.)4ab. ns.1a2+b2 +za2+b2'11-2"+2"z,III5+5Z'IVa2+b2z, (v)275 +i,(vi)4abi.] Ex.6Find the square rootof the followingcomplexnumbers (i)5 - 12i,(ii)i,(iii)a2 - 1 + 2ai . SOLUTION: (i)5 - 12i =9 - 2.3.2i + 4i2 =(3- 2i)2. Hencey'5 - 12i =(3 - 2i). CH.1:CLASSICALALGEBRA (ii)i=~ ( 2 i )=~ ( 1+ 2i+ i2)=~ ( 1+ i)2. HenceVi =*(1 + i). (iii)a2 - 1 + 2ai= a2 + 2ai + i2 = (a+ i)2. Hencev'a2 - 1 + 2ai=(a + i). Exercise3Find the square rootof the followingcomplex numbers (i)16i,(ii)-4i. [Ans.(i)2J2(1 + i),(ii)J2(1 - i).] Ex.7(i)Factorisea2 + ab + b2, usingw,acube rootof unity. (ii)Factorisex6 - 1 intolinearfactors . SOLUTION: (i)Asw2+w+1=Oor,_1=w+w2. Now,a2 + ab + b2 = a2 - (w+ w2)ab + b2 =a2 - wab - w2ab + b2 =a(a- wb)- w2b(a- wb) =(a- wb)(a- w2b). (ii)x6 - 1 = (x3)2- 12= (x3 -1)(x3 + 1) =(x- 1)(x2 + X+ 1)(x + 1)(x2 - X+ 1). Now,x2 + x+ 1 = x2 - (w+ w2)x + 1[since w + w2 = -1] =x2 - xw - xw2 + w3 =x(x - w)- w2(x- w) =(x- w)(x - w2). Similarly,x2 - x + 1 =(x + w)(x + w2). Hence,x6 - 1 =(x- 1)(x + 1)(x - w)(x - w2)(x + w)(x + w2) . . Exercise4Factoriseinto linearfactors (i)x2 + 1,(ii)x3 - a3,(iii)x4- a4, (iv)x6 _y6. 1.7 [Ans.(i)(x + i)(x - i),(ii)(x - a)(x - wa)(x - w2a),(iii)(x - a)(x + a)(x -ia)(x + ia),(iv)(x - y)(x + y)(x - wy)(x + wy)(x - w2y)(x + w2y).] Ex.8(i)If P(z) is a variable point in the complex plane, such that Iz-il = 2,findthelocusof P. (ii)In argand plane,whatarethe points representingIzl::;1 ? (iii)On the complex plane, let p(z)be a variable point, on the complex plane, suchthatIz- 2il= 3;provethatthepointszlieonthecirclewhose centreis(0,2)and radius3. 1.8V.G.(SHORTQUESTIONSANDANSWERS) SOLUTION: (i)Letz= x+ iy.Then Iz- il = 2 becomesIx + iy - il = 2. or,Jx2 + (y- 1)2= 2 or,x2 + (y- 1)2= 4. Which isthe locusof thepointP. (ii)Letz=x+ iy.ThenIzl::;1 becomesIx + iYI::;1. or,Jx2 + y2::;1or,x2 + y2::;1. Which representsacirculardiscof radiusunity and centreat(0,0). (iii)Letz=x + iy.Putting this value inIz - 2il=3 wegetIx + iy - 2il=3, or,Ix + i(y - 2)1=3 or,Jx2 + (y- 2)2=3 or,x2 + (y- 2)2=9, which representsacirclecentre at(0,2)and radiusis3. Ex.9FortwocomplexnumbersZ1,Z2provethat IZ1+ z212 + IZ1- z212.= 2(lzll2 + IZ212). SOLUTION:L.H.S.=1:::1+ z212 + IZ1- z212 =(Z1+ Z2)(Z1+ Z2)+ (Z1- Z2)(z1- Z2) =(Z1Z1 + Z1Z2 + z2ii+ z2zd + (Z1Z1 - Z1Z2 - z2z1+ Z2Z2) =2(lz112+ IZ212). Exercise5If z=x+ iy,then provethat (i)3(x2 + y2)=4x,if 21z- 11=Iz- 21. (ii)Iz- 11=2fz- il representsacircle. (iii)thepointzliesonthecirclewhosecentreisat(-3,0)andradiusis2, if Iz + 31=2. (iv)Iz- 81+ Iz + 81=representsan ellipse. Ex. (i)If x=2 + 3i,findthe valueof x3 - 4x2 + 13x + 1. (ii)Find thevalueof x2 + xy + y2forx=1 + i, y=1 - i. SOLUTION: (i)Givenx=2 + 3ior,x- 2 =3i. Squaring both sidesweget (x- 2)2=(3i)2 or,x2 - 4x + 4 =-9 or,x2 - 4x + 13= Now,x3 - 4x2 + 13x + 1 =x(x2 - 4x + 13) + 1 =x.O+ 1 =1. (ii)x+ y= 2 andxy =(1+ i)(1 - i) = 1 - i2 = 2. Then,x2 + xy + y2=(x + y)2- 2xy + xy = (x + y)2- xy = 4- 2= 2. CH.l:CLASSICALALGEBRA Ex.11(i)If wi-I be acube rootof unity,showthat (1- w)(I- w2)(1- w4)(1- w8) =9. (ii)If wbe an imaginary cube rootof unity,showthat x+wy +w2z =w y+wz +w2x. 1.9 (iii)If wn + w2n + 1=0wherewisan imaginarycuberootof urtityandn an integer,isndivisibleby3?, SOLUTION: (i)Since wisacube rootof unity,w3 =1 and1 + w + w2 =o. Now,(1- w)(1- w2)(1- w4)(1- w8) =(1- w)(1- w2)(1- w3.w)(I- w3.w3.w2) =(1- w)(1- w2)(1- w)(l - w2) ={(I - w)(1- w2)F = {I - (w+ w2) + w3}= (1+ 1 + 1)2= 9. x+ wy + w2 ZX+ wy + w2 Z (ii)L.H.S.== y+wz+w2xw3y+w4z+w2x [sincew3 = 1,sow3y= y,w4z= wz] X+ wy + w2 Z= ~= ~= w=R.H S w2(x + wy + w2z)w2 w3 (iii)If possiblelet,nisdivisibleby3,i.e.,n=3p,pisanypositiveinteger. . ..wn + w2n + 1 =w3p + w6p + 1 =(w3)P+ (w3)2p + 1 =(I)P+ (1)2p + 1= 1 + 1 + 1 = 3 i- O. Hencenisnotdivisibleby3. Exercise6Provethat(i)(1+ w4)4=w2,(ii)(1- w + w2)(1+ w - w2) =4, (iii)(1+ w)(1+ w2)(1+ w4)(1+ w8)= 1, (iv)(k+ kw- w2)3=(k + kw2 - w)3. Ex.12If x= a + b,y= aw+ bw2 and z=aw2 + bw,showthat (i)xyz =a3 + b3, (ii)x2 + y2+ z2=6ab. SOLUTION:(i)Givenx=a + b, y=aw + bw2 and z=aw2 + bw. xyz=(a+ b)(aw + b w 2 ~ ( a w 2+ bw) =(a+ b)(a2w3 + ab(w + w2) + b2w3) =(a+ b)(a2 - ab + b2)(since w3 =1 and w + w2 =-1) =a3 + b3. 1.10V,G,MATHEMATICS(SHORTQUESTIONSANDANSWERS) (ii)Wehavex=a + b,y=aw+ bw2 andz=aw2 + bw, x2 + y2+ z2=(a+ b)2+ (aw+ bw2)2+ (aw2 + bw)2 =(a2 + 2ab + b2) +(a2w2 + 2abw3 +b2w4)+ (a2w4 + 2abw3 +b2w2) =a2(1+w +w4)+2ab(1+w3 +w3)+b2(1+ w4 +w2) =a2(1+w +w2) +2ab(1+1 + 1)+b2(1+w +w2) = a2,O+ 6ab -f' b2,O= 6ab, 1.1.3DeMoiver'stheorem and its use Theorem 1De Moiver's theorem:The value of (cos 8+i sin 8)n is (cos n8+ i sin n8)when nisinteger and(cos n8 +i sin n8)isone of the value of (cos 8 + i sin8)n,when nisfraction, Q.1State DeMoiver'stheorem, Ex.13(i)Find the cube rootof (1+i), (ii)Solvex3 +1 =0byDeMoivre'stheorem, (l'I'I')F'dhlIt"'t'f(1+i)n1 10tesmaesposItIve10egern,I--,=, 1- 2 SOLUTION: (i)Let1 =r cos 8, 1 =r sin 8, . '.r2= 1 +1 = 2 or,r= v'2 and tan 8 =1or,8 =l Then 1 + i=rcos 8 +irsin8 =v'2(cos ~+i.sin ~ ) , Henceone cube rootof 1 + iis {v'2(cos ~+i sin ~)}1/3=21/6(cos~+i sin ~ ) [byDeMoivre'stheorem] (ii)x3 =-1 =cos 1l'+i sin 1l'=cos( 1l'+ 2k1l')+i sin( 1l'+ 2k1l') k=0,1,2",. 1l'+ 2k1l',,1l' + 2k1l' or,x=cos3+2sm3,k=0,1,2 1l',,1l',,51l',,51l' or,x=cos '3+ 2sm '3' cos 1l'+ 2sm 1l', cos 3+ 2sm 3' N51l'.,51l'(21l' )"(21l') ow,cos 3+2sm 3=cos1l'- '3+2sm1l'- '3 1l',,1l' = cos '3- 2sm '3' 1600'6001iv'3 x=- ,cos smor,-1, 2'-2-' CH.1:CLASSICALALGEBRA1.11 (l+i)n(1+i)2)n (iii)Wehave1 _i= lor,1 _i2 = 1 ( 1 + i2 + 2i) n=1(1 - 1 + 2i) n=1 or,1 + 1or,2 or,in= 1 Weknowi2 =-1, i3 =-i, i4= 1. Hencethe requiredsmallestpositiveintegernis4. Ex.14(i)Findthemodulusandtheprincipleamplitudeof thecomplex number(1+ i)5. (ii)Find the modulus and principal value of the complex number(cos 50+ i sin 50)6. SOLUTION: (i)Let 1 = r cos f),1 =r sin f).Then r2= 1 + 1 = 2 or,r= y'2 and tan f)=1 or,f)=l' ...(1+ i)5=(rcos f)+ i sinf))5= (y'2)5(cos 1+ i sin 1)5 =(y'2) 5(cos5;+ i sin 5;) =(y'2)5{cos(271'- 3;) + isin(271'_3;)} =(y'2)5{cos( _3;) + i sin( _ 3;)}. Hencethe modulusis(y'2)5=25/2 andprincipal amplitude is_ 3: .. (ii)(cos 50+ i sin 50)6=cos 300+ i sin 300=cos(3600- 600)+ i sin(3600- 60)=cos 60- i sin 60=1{ cos( -60) + i sin( -60)}. Herer=1 and f)=-60.Hencethe modulusis1 and principal ampli-tude is-60. Ex.15Showthat(1+ V3i)30=230. SOLUTION:Let1 =rcosf), V3 =rsinf). '.r2= 1 + 3 =4 or,r=2 and tan f)=V3 = tan jor,f)= j, Hence(1+ iV3)30=(r cos f)+ ir sin f)) 30 =r30 (cos 30f) + i sin 30f)) =230{cos(30.J) + isin(30.j)} =230{cos(107l')+ isin(107I')}=230. [sincecos 1071'=1 and sin 1071'=OJ Ex.16(i)(sin f)+ i cos f))n=f:.(sin nf) + i cos nO)?(n> 1)Explain. (ii)Showthat,(sinf) + icosf))n =cosn(7I'/2 - 0) + i sin n(7I'/2 - 0),wheren isan integer. 1.12D.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) SOLUTION: (i)Whenn=2then (sine +i cos e)2=sin2 e - cos2 e +2isinecose =- cos 2e +i sin 2e=(sin 2e +i cos 2e). Hence(sine +i cos e)=(sinne +i cos n8)forn=2,i.e.,forn> 1. (ii)(sine +i cos e)n={cos(I - e)+isin(I - e)}n = cosn(I - e)+isinn(I - e) [by theorem] Ex.17Find the values11/3. SOLUTION:11/3 =(cosO +isinO)1/3=(cos2k71'+isin2k1r)1/3 k=0,1,2, ... =cos2k7r+i sin 2k7rk=012 33'". the required valuesare cos 0 +i sin 0cos211"+i sin 211"cos411"+i sin 411" '33'33. 1 211"..211" or,,cos 3Z SIll 3' [sincecos 4;+i sin 4311"=cos(271'- 2;) +i sin(271'_2;) =cos 2;- i sin 2;] or,1,cos 1200 i sin 1200 or,1, i{}. 1.1.4Euler's definition of exponential and trigonometrical func-tions z2z3 (i)eZ =1 +z +- +- +...... 2!3! (ii)eix =cosx+isinx eix + e-ix eix _e-ix (iii)cos x=2'sinx =2i Ex.18(i)ByEuler'sdefinitionof sinzandcosz,wherezisacomplex quantity,provethatcos2 z - sin2 z =cos 2z. (ii)ByEuler'sexponentialvaluesofsinzandcosz,wherezisacomplex number,establish thatcos 2z=1 - 2 sin2 z. SOLUTION: (i)cos2 z _sin2 z=(etZ e-iZ) 2_(eiZ 2 e2iz +e-2iz +2e2iz +e-2iz - 2 =+------44 e2iz + e-2iz ----- =cos2z. 2 CR.l:CLASSICALALGEBRA e2iz + e-2iz e2iz + e-2iz - 2 (ii)Cos 2z=2= 1 +2 e2iz + e-2iz - 2 =1- 24i2 (eiZ _e-iZ)2 =1 - 22i= 1 - 2 Sin2 z. 1.2Theoryof Equation Ex.1(i)Form thecubicequation whoserootsare 0,2,3. (ii)Form the polynomial whosezerosare1, -2, 3,-4 . SOLUTION: 1.13 (i)Sincethe rootsof the equation are0,2,and 3thecubic equationis (x- O)(x- 2)(x - 3)=0 or,x(x2 - 5x + 6)=0 or,x3 - 5x2 + 6x=O. (ii)Sincethe zerosof the polynomial are1,-2,3,and -4 the polynomial is (x- 1)(x + 2)(x - 3)(x + 4)=(x2 + X- 2)(x2 + X- 12) =x4+ 2x3 - 13x2 - 14x + 24. Exercise1Form theequation whoserootsare (i)1,-1/2,5;(ii)0,0,3,3;(iii)2 + 3i, 1;(iv)2 J3, 5. [Ans.(i)!(2x3 -llx2 + 4x + 5),(ii)x4 - 6x3 + 9x2,(iii)x2 - 3(1 + i)x + 2 + 3i, (iv)x3 - 9x2 + 21x- 5] 1.2.1Remaindertheoremandsyntheticdivision Ex.2If thepolynomialf (x)bedividedbyx - h,showthattheremainder isf(h) . SOLUTION:LetQ bethequotientandRbe theremainder independentof x,whenf(x)isdividedby(x- h). Thenf(x)=(x- h)Q + R. Putting x= h wehavef(h)= O.Q+ R = R. Thus f (h)isthe remainder. Ex.3(i)Usethemethodof syntheticdivisiontodeterminethequotient and remainder when 6x4 + 5x3 - 13x + 2 isdividedbyx + 2. 1.14V.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) (ii)Find the remainder and quotient by synthetic division wherex4 + x-11 isdivisiblebyx+ 2. (iii)For what values of p,2x3 - px2 + 6x - 3piscompletely divisible by x + 2? (iv)Applythemethodofsyntheticdivisiontofindpinorderthat2x4-7x3 + p2x+ 15maybe exactlydivisiblebyx- 3. (v)Find the relationbetweenaandb if 4x3 - 3x2 + 2ax + b isdivisibleby x+ 2? (vi)Showthat2x4- 7x3 + 8ax2 - 3bx+ 17= 0isdivisiblebyx- 2when 32a- 6b- 7 =O. (vii)Findtheremainder,bysyntheticdivision,whenthepolynomial3x4 -4x3 + 2x2- 9x + 1isdividedby2x + 1. SOLUTION: (i)Herethemultiplieris(-2) and wehave -21650-132 .-1214-2882 "-6-0------::7----:-14.,-----4-:-:1-8-4:--The quotientis6i3 - 7x2 + 14x - 41and the remainder\s84. (ii)Herethemultiplieris(-2)and wehave -211001-11 -24-814 1-24-73 The quotientisx3 - 2x2+ 4x - 7 and the remainder is3. (iii)Herethe multiplieris(-2) andwehave (iv) -212-p6-3p .-48 + 2p- 28- 4p '-2-=-----4c---p--,I:-:4-+-:O-=2p'"----=2-=-8---:7-=-p-Forcompletedivisibilitythe remaindermust be zero. Therefore,-28 - 7p=0or,p =-4. Herethemultiplieris3 312-70 p2 15 6-3-9 3p2- 27 2-1 -3 p2_9 3p2-12 CH.1:CLASSICALAI fiEBRA1.15 In order to have exact divisibility the remainder must be zero,3p2 -12 = o or,p2= 4 or,p= 2. (v)If the givenexpression be exactlydivisibleby(x + 2),thenf( -2) must be zerowhere f(x)=4x3 - 3x2 + 2ax + b. This gives-32 - 12- 4a + b =0or,-44 - 4a + b =0 or,-4a + b =44whichistherequiredrelation. (vi)Heref(x)==2x4- 7x3 + 8ax2 - 3bx + 17. (vii) f(2)mustbe zeroforexactdivisibility,this gives 32- 56+ 32a- 6b+ 17=0 or,-7 + 32a- 6b=0 or,32a - 6b- 7 =O. Herewefirstdivideby(x + 1/2)asshownbelow 1 3 -42-9 -2 31119 - --248 1 91 -16 111991107 3 -- --24816 Th ..1 (31121991)dh.d.107 e quobent IS-"23x- 2x + 4x - 8ante remamer IS16' Exercise2(i)Find theremainderwhenx5 - 4x4+ 8x2 - 1 isdividedby x- 3. (ii)Findtherelationbetweenaandbif x3 + 2x2+ 2ax + bisdivisibleby x+2? (iii)Find the quotientandremainderwhenx3 + 3x2 + 3x + 2isdividedby x-I, bysyntheticdivision. (iv)Find the remainderwhenx3 + 5x2 + 1 isdivisiblebyx+ 3. (v)Find the remainder and quotient when 3x2 +4x -11 isdivided byx-I. [Ans.(i)-4,(ii)b - 4a =0,(iii)Quotientisx2 + 4x + 7,remainder is9, (iv)19,(v)-4.] Ex.4Showthat if nisodd,then x+ 1 isafactorof xn + 1. SOLUTION:If nisodd,then(_I)n + 1 =-1 + 1 =O. ...(x + 1)isafactorof xn + 1 whennisodd. 1 ~ 1 6D.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) Ex.5(i)Expressx4+ 5x2 - 3x + 2asapolynomial inx+ 2. (ii)If f(x)=2x4- x3 - 2x2+ 5x- 1,thenshowthatf(x + 3)=2x4+ 23x3 + 97x2 + 182x + 131. SOLUTION: (i)Let us divide x4+5x2 -3x+2 by x+2 by synthetic method in succession. -2105-32 -24-1842 1-2944 -2 1-4 1 1 Hencef(x)=x4+ 5x2 - 3x + 2 =(x + 2)4- 8(x + 2)3+ 29(x + 2)2- 55(x + 2)+ 44. (ii)Letusdivide2x4- x3 - 2x2 + 5x - 1 byx- 3bysyntheticmethod in succession. 32-1-2 6 2 5-1 39132 131 Hencef(x + 3)=2x4 + 23x3 + 97x2 + 182x + 131. Exercise3(i)Expressthepolynomialx4+ 4x3 - 2x2+ X+ 2=0asa polynomial inx+ 2. (ii)Express3x3 - 4x2 + 5x + 6 asapolynomial in x- 1. (iii)Express3x4 + 5x3 - 2x2+ 4x + 6 asapolynomial in x- 1. CH.l:CLASSICA ....ALGEBRA (iv)Expressx4- x3 + 2x2- 3x + 1 asapolynomial in x- 3. [Ans.(i)(x + 2)4- 4(x + 2)3- 2(x + 2)2+ 25(x + 2)- 24, (ii)3(x - 113 + 5(x - 1)2+ 6(x - 1)+ 10,(iii)3(x - 1)4 +17(x - 1)3+ 31(x - 1)2+ 27(x - 1)+ 16,(iv)(x - 3)4 +l1(x - 3)3+ 47(x - 3)2+ 90(x - 3)+ 64] 1.17 Theorem1Fundamental theorem of classical algebra:Every algebraic equation hasatleastonerootreal orimaginary. Q.1State the fundamentaltheorem of classical algebra. 1.2.2Descartes' ruleof signs Theorem2Descartes'ruleof signs:Anequationf (x)=0withreal coefficientscannothavemorepositiverealrootsthantherearechangesof sign in f (x)and cannothavemorenegativereal rootsthan therearechanges ofsigninf (- x ).If thenumberofrealrootsbelessthanthenumberof changesof sign,then itwillbe by an evennumber. Q.2State Descartes'ruleof signs. Ex.6(i)UseDescartes'ruleof signstodeterminethenature of the roots of the equationx6 - 3x2 - X+ 1 =O. (ii)UsingDescartes'rulesignsfindthe leastnumberof imaginaryrootsof the equationx5 + x3 - 2x2+ X- 2 = O. (iii)Find the least number of imaginary roots of the equation 3x5 - 4x2 + 8 = o. (iv)UseDescartes'ruleof signstodeterminethenature of therootsof the equationx8 - 1 =O. (v)UseDescartes'ruleof signtoshowthattheequationx8 + x4+ 1=0 hasno real roots. (vi)Howmanypositiverootsanalgebraicequationcanhaveif thesignof thetermsof the equation be allpositive? SOLUTION: (i)Letf(x)=x6 - 3x2 - X+ 1. There are twochanges of sign in f (x).Hence there cannot be more than twopositiverealroots. Againf( -x) =x6 - 3x2 + X+ 1 Therearetwochangesof signinf( -x).Hencetherecannotbemore than twonegativerealroots. 1.18U.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) Asthe equation isof degree6,hence the othertworootsmustbe imag-inary. (ii)Letf(x)=x5+x3-2x2+x-2. Asnumber of changesof sign inf(x)is3,there cannotbe morethan 3 positiverealroots. Againfe-x)=_x5 - x3 - 2x2- X- 2 The equation hasnonegativereal rootsastherearenochangesof sign inf( -x). Astheequationisof degree5,theothertworootsmustbe imaginary. Therefore the leastnumber of imaginary rootsistwo. (iii)Letf(x)=3x5 - 4x2 + 8. Since number of changes of sign in f (x) is 2,hence there can be maximum twopositiverealroots. Againfe-x)=-3x5 - 4x2 + 8 Herenumberof changesof signisone.The equationhasonenegative teal root. Asthe degreeof theequationis5,the leastnumber of imaginary roots is2. (iv)Letf(x)=x8 - 1. Aschangeof signinf (x)isone,hencetherecannotbemorethan one positivereal root. Againf( -x) =x8 - 1. Here also number of changes of sign is one.Thus the given equation has onenegativerealroot. As the degree of the equation is 8,hence the other six roots are imaginary. (v)Letf(x)=x8 + x4+ 1. Asthere isnochange of sign,sothe givenequation hasnopositive real roots. Alsof( -x)=x8 + x4+ 1hasnochangesofsign.Hencethegiven equation hasnonegativerealroots . . '.f(x)=0hasnoreal roots. (vi)If thesignof thetermsof theequationbeallpositivei.e.,thereisno changeof signbetweentheterms,.thentheequationcannothaveany positiverealroot. Exercise4(i)Using Descartes' rule signs find the least number of complex roots of the equation x6 - 3x2 - x+ 1 =O. CR.l:CLASSICALALGEBRA1.19 (ii)UsingDescartes'rulesignsfindtheleastnumberof imaginary rootsof theequationx6 - 5x2 - 2x + 1 =O. (iii)ApplyDescartes'ruleof signsto findthenumber of imaginary rootsof 3x4 + 4x2 - 3x - 12=O. (iv)Isittruethatalltherootsof theequation2x3 - llx2 + 28x- 24=0 arenonreal. (v)Find the least number of imaginary roots of the equation 3x5 - 4x2 + 8 = O. (vi)Show that the equation 2x 7 - x5 + 4x3 - 5 =0 has at least two imaginary roots. (vii)Is it possible that all the roots of the equation 2x3 -llx2 + 28x - 24=0 be complex?Justify youranswer. (viii)Provethatx4+ x2 + x-I =0hasonepositive,onenegativeandtwo imaginary roots. [Ans.(i)2,(ii)2,(iii)2,(iv)No,(v)2,(vii)No.] Theorem 3If tworealnumbersaand b be substituted forxin anypolyno-mialf(x)and if theyhaveoppositesigns,thentheequationf(x)=0must haveatleastonerealrootliesbetweenaandb. Ex.7Showthatarootof x3 - 4x + 2 =0liesbetween1 and2 . SOLUTION:Letf(x)=x3 - 4x + 2. f(l)= 1- 4 + 2 = -1 O. Sincef{l)andf(2)areof oppositesigns,theremusthaveatleastonereal rootlyingbetween1 and2. Exercise5Showthatthe equation (i)lOx3 -17x2 + X+ 6 =0hasarootbetween0 and-1. (ii)x4 -12x2 + 12x - 3 =0 has a root between -3 and -4 and another between 2and3. Ex.8If 1,0:, (3, 1', ... be the roots of the equation xn -1 =0,then showthat (1- 0:)(1- (3)(1- 1') ... =n. SOLUTION:Since1, a, (3, 1', ...aretherootsof theequationxn- 1=0, hence xn- 1 =(x- l)(x - o:)(x- (3)(x- 1') ... Dividing both sidesbyx-I wehave xn-1 + xn-2 + ... + x+ 1 =(x - o:)(x- (3)(x- 1') ... Putting x=1 tothisrelation.Then n=(1- 0:)(1- (3)(1- 1') .... 1.20D.G.MATHEMATICS(SHORT(.lUESTIONSANDANSWERS) 1.2.3Relation between rootsandcoefficients Note1If a, 13"betherootsof theequationax3 + bx2 + cx + d=0 then a+ 13+, = -bfa, af3+ 13, +,a = cia, af3, = -d/a. Note2If a, 13", Jbe the roots of the equation ax4 + bx3 + cx2 + dx + e= 0 then a+ 13+, + J=-bfa, af3+ a, + aJ + 13,+ 136+,6 =cia or,(a + f3)(T+ J)+ af3+,6 =cia af3, + af3J+ 13,6+ a,6 =-d/a or,af3(T+ J)+ ,J(a + 13)=-d/a, af3,6 = e/a. Ex.9(i)Solvethe equation x3 - 3x2 + 4 =0,twoof its roots being equal. (ii)Solvethe equation2x3 - 21x2+ 42x - 16 =0the rootsbeingin G.P. (iii)Solve the equation x3 - 3x2 - 6x + 8 =0,given that the roots are in A.P. SOLUTION: (i)Letthe rootsof the equation be a, a, 13. Then 2a + 13= 3and a2 + 2af3= O. These givea=2,13=-1. .'.the roots are2,2, -1. (ii)Lettherootsof the cubicbe a/f3, a, af3. (1)21 Then a-g+ 1 + 13=2and a3 =8. Solvingthesewegeta=2 and 13=4,~ . 11 the rootsare8,2, 2or2,2,8. (iii)Lettherootsof the equation be (a - 13), a, (a + 13). Then wehave3a = 3 or,a= 1. Also3a2 - 132 = -6 or,3 - 132 = -6 or,132 = 9 or,13= 3. the rootsare-2,1,4 or4,1, -2. Ex.10(i)If the roots of the equation x3 - 3x2 + ax + b =0 be in arithmetic progression,provethat a+ b =2. (ii)The sum of tworootsof theequation x3 + alx2 + a2x + a3=0iszero, showthat al a2- a3=O. (iii)Find the condition that the rootsof the equation x3 + px2 + qx + r= 0 arein A.P. CH.1:CLASSICALALGEBRA1.21 (iv)If the roots of the equation x3 +ax2 +bx +c =0 are in G.P.,prove that a3c = b3 . SOLUTION: (i)Letthe rootsbe(a - 13), a, (a +13). Then 3a = 3 ora= 1. Also3a2 - 132 = aor,3 - 132 = aor,132 = 3 - a and a(a2 - 132) =-b or,(1- 3 +a)=-b or,a- 2= -b or,a+b =2whichisthe requiredrelation. (ii)Letthe rootsbe a, -a, 13. Then 13= -aI,_a2 = a2and_a2j3=-a3' Substituting the valuesof 13and a2 weget-a2al=-a3 or,ala2- a3=0,whichisthe requiredrelation. (iii)Letthe rootsbe(a - 13), a, (a +13). p Then 3a =-p or,a=-3"' p2 Again3a2 - 132 =qor,3- 132 = q p2 or,132 =3- qanda(a2 - 132)= -r or,_ P:+ q)=-r 2p23r or,-g+q= p or,-2p3+9pq=27r or,9pq - 2p3=27r. (iv)Letthe rootsbe aj3. =-a+ =b a3 =-c Substituting(1)in(2),weget b -a.a =b or,a=--. a (1) (2) (3) 1.22U.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) Exercise6(i)Find the condition that the equation x3 + px2 + qx + r=0 mayhavetworootsequalbutof oppositesigns. (ii)Solvethe equation x3 - 6x2 + 3x + 10=0,the rootsbeing in A.P. (iii)Solvetheequationx3 - 7x2 + 36=0 oneof therootbeingdoublethe other. (iv)Determine rsothat one rootof the equation x3 - rx2 + rx - 4= 0shall be reciprocal of another. [Ans.(i)pq - r=0,(ii)5,2,-1 or,-1,2,5,(iii)3,6,-2,(iv)r=5.] Exercise7(i)Solve the equation x3 - 3x2 + 4=0,tworootsbeing equal. (ii)Solvetheequation4x3 - 24x2+ 23x+ 18=0,havinggiventhatthe rootsarein A.P. (iii)Solvethe equation27x3 + 42x2- 28x - 8=0whoserootsare inG.P. [Ans.(i)2,2,-1,(ii)-1/2,2,9/2,(iii)-2/9,2/3,-2] 1.2.4Transformationof equation Ex.11(i)Transformtheequationx3 - 6x2 + 5x + 8 =0intoanotherin which the secondtermismissing. (ii)Find theequationeachof whoserootsisgreaterby2 thantherootsof the equation x4+ 8x3 + X- 5 =O. (iii)Find the equation whose roots are squares of the roots of x3 - 2x cos 0 + 1 =O. (iv)Obtainanequationwhoserootsaretwicetherootsoftheequation x3 + 3x2 + 4x + 5 =O. (v)Form the equation whose roots are reciprocal to those of x4 - 4x3 + 5x2-8x + 7 =O. (vi)Findtheequationwhoserootsareequalinmagnitudebutoppositein signto the rootsof the equation x4+ 3x3 - 7x2 + 2x + 1 =O. SOLUTION: (i)Letustransform the equation by putting x=y + h . ...(y + h)3- 6(y + h)2+ 5(y + h) + 8 =0 or,y3+ (3h- 6)y2+ (3h2 - 12h + 5)y + (h3 - 6h2 + 5h + 8)=0 Toremovethesecondtermweput3h- 6=0or,h=2,thusthe transformed equation isy3+ (12- 24+ 5)y + (8- 24 + 10 + 8)= 0or, y3-7y + 2 =O. CR.1:CLASSICALALGEBRA1.23 (ii)Letabe arootof the given equation. .a4 + 8a3 + a- 5 =0 Lety=a+ 2 or,0:=Y - 2. Putting this valuein(1)weget (y- 2)4+ 8(y - 2)3+ (y- 2)- 5=0 or,y4- 24y2+ 65y - 55=O. (1) (iii)Leta, b, cbetherootsof thegivenequation.Sinceaisarootof the equation . '.a3 - 2a cos 0 + 1 =0or,a (a 2 - 2 cos 0)=-1 Squaring both sidesweget a2(a2 - 2cosO)2=1. Lety= a2..'.y(y - 2 cos 0)2= 1 or,y3- 4y2 cos 0 + 4y cos2 0 =1. (iv)Leta, (3,'Y~ etherootsof the givenequation .0:3 + 3a2 + 4a + 5= 0 Lety =2a or,a=y/2Substituting in(1)weget y3+ 3y2+ 4!+ 5 =0 842 or,y3+ 6y2+ 16y + 40=O. (v)The givenequationisx4- 4x3 + 5x2 - 8x + 7=0 Sb1 ustltutmg x=-, weget y 1458 ---+---+7=0 y4y3y2Y or,7y4- 8y3 + 5y2- 4y + 1 =0whichisthe required equation. (vi)Putting x=-y in thegivenequation.Thetransformed equation is (_y)4 +3( _y)3- 7( _y)2+2( -y) +1 =0 or,y4- 3y3 - 7y2- 2y + 1 =O. (1) Ex.12(i)If a, (3,'Yaretherootsof the equationx3 - 5x2 - 4x + 20=0, determine thevalueof "L..a2(3. (ii)If a, (3,'Ybetherootsof the equation x3 - px2 + qx - r=0 findthevalueof "L..0:2 (3. (iii)If a, (3, 'Ybetherootsof theequationx3 + px + q =0,findthevalueof 1 L(3+'Y' 1.24V.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) (iv)If a, /3"betherootsof x3 - px2 + qx - r=0,findthevalueof(/3+ ,)(, + a)(a + (3). (v)If a, /3"betherootsof the equationx3 + x+ 1=0findthevalueof a2 + /32 + ,2. (vi)If a, /3"betherootsof theequationax3 + bx2 + ex + d =0,findthe valueof E a2 SOLUTION: (i)The givenequation isx3 - 5x2 - 4x + 20=O. Wehavea+ /3+, = 5,a/3 + /3, + a, = -4, a/3, = -20. WeknowE a2/3= (a + /3+ ,)(a/3 + /3, + a,) - 3a/3, = 5x (-4) - 3 x (-20)=40. (ii)Herea+ /3+, =p,a/3+ /3, + a, =q,a/3, =r. WeknowE a2/3=(a + /3+ ,)(a/3 + /3, + a,) - 3a/3, =pq - 3r. (iii)Sincea, (3, ,are theroots of x3 + px + q =0,wehave a+ (3+, = 0,a(3+ /3, + a, = p,a(3, = -q. 1111 Now,L /3+ ,=a+ /3+ (3+ ,+ a+, = ~+ _1_+ ~= _ (.!. + .!.+ .!.) -,-a-/3a(3, (3, + a, + a(3pp === a/3,-qq (iv)Sincea, (3"arethe rootsof x3 - px2 + qx - r=0,wehave a+ /3+ ,=p,a(3+ (3, + a, =q,a(3, =r. Now,(a + (3)(/3 + ,)(, + a)=(a + (3 + ,)(a/3 + /3, -t a,) - ap, =pq - r. (v)Sincea, (3"arethe roots of x3 + x+ 1 =0, a+ /3+ ,= 0,a/3+ (3, + a, = 1,a/3, = -1. Wehavea2 + (32+ ,2 =(a + (3+ ,)2 - 2(a/3 + /3, + a,) = 0 - 2.1= -2. (vi)Here a+ /3+, =- ~ ,a(3+ /3, + a, = !!..,a/3, = - ~ . aaa NowE a2 = a2 + /32 + ,2 = (a + /3+ ,)2 - 2(a/3 + /3, + a,) b2 eb2 - 2ae =- - 2 - =----::---a2 aa2 Ex.13(i)Findtheequationwhoserootsaretherootsoftheequation x4+ 5x3 - 6x2 + 8x - 9 =0with their sign changed. CH.1:CLASSICALALGEBRA1.25 (ii)Removethe fractionalcoefficientsof the equation 3122 x- 2x + aX=o. (iii)Transform the equation 3x4 - 5x3 + x2 - X + 1 =0to onewith unity as itsleadingcoefficient. (iv)Diminish the rootsof the equationx3 + x- 2=0by2. (v)If a, /3"betherootsof theequation2x3 + 3x2 - X- 1findthe h .111 equation woserootsare-1--' -1/3'-1-. -a- -, (vi)If a, /3"betherootsof theequationx3 - 2x2+ X- 3=0,then form the equation whoserootsare a/3, /3",a. (vii)If a, /3"betherootsof theequationx3 - px2 + qx - r=0,findthe equation whoserootsare/3+", + a, a+ /3. (viii)Form the equation whose roots are reciprocal to those of x5 + 6x4 - 7x3 + 8x2 + 9x + 1 =O. SOLUTION: (i)Putting x=-y inthe givenequation. (_y)4 + 5{ _y)3- 6{ _y)2 + 8{ -y) - 9 =0 or,y4- 5y3 - 6y2- 8y - 9 =O. (ii)Putting x=y/a. ('!!.)3_ !('!!.)2 + =0 a2a3a 31222 or,y- 2Y a + aya=O. If wetakea=6(whichisthe l.c.m.of 2,3)thenthe fractionalcoeffi-cientsbecomeintegral. Thusy3- 3y2+ 24y=O. (iii)Dividingthegivenequation by3, x4- + !x2 - !x + !=0 3333. Putting x=y / a. -+ _+ =0 5111 or,y4_ _ ay3 + _a2y2_ _ ya3 + _a4 =0 3333 Choosinga= 3 the equation becomes y4_5y3 + 3y3 - 9y + 27=O. 1.26V.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) (iv)Applying successivesyntheticdivisionby2. 2101-2 10 8 The transformed equation isy3+ 6y2+ 13y + 8 =O. 1y-1 (v)Lety= -- or,a= --. 1-ay Sinceaisarootof the given equation,so 23 -+ - -a-1 =0. a3 a2 or,2 (y - 1) 3+ 3 (y - 1) 2_(y - 1) _1 =0 yyy or,2(y - 1)3+ 3(y - 1)2y - (y- 1)y2- y3=0 or,3y3- lly2 + 9y - 2 =O. (vi)Lety= 33 or,y=-- =- or,"(=-. "("(Y Also"(isarootof thegivenequation . .'."(3- 2"(2+ "(- 3 = 0 or, _ + _ 3 =0 yyY 961 or- - - + - - 1 =0 'y3y2Y or,y3- y2+ 6y - 9 =O. (vii)Lety=a+ or,Y = a+ + "(- "(= p- "(or,"(= p- y. Since "(isarootof thegiven equation, "(3- p"(2+ q"(_r=0 or, or,y3- 2py2+ (p2+q)y - (pq- r)=O. (viii)Putting x=in the givenequation,weget

CR.1:CLASSICALALGEBRA1.27 16789 -+---+-+-+1=0 y5y4y3y2Y or,y5+9y4+8y3 - 7y2+6y+1 =a whichistherequiredequation. Exercise 8(i)Diminish the rootsof theequation2x5 - x3 + lax - 8 =a by3. (ii)Find the equation whose roots are the roots of 3x4+7x3-15x2+x-2 =0, eachincreasedby5. (iii)If 01.,(3,"Ibetherootsof theequationx3 +2x2+3x+4=0,findthe equation whoserootsare1 +!., 1 +_(31, 1 +!.. 01."I (iv)If 01.,(3,"Ibe therootsof theequationx3 - 3x2 + 8x- 5 = 0,then form an equation whoserootsare201.+3,2(3 +3,2"1+3. [Ans.(i)2y5+30y4+179y3 +531y2 +793y+481= 0, (ii)3y4- 53y3 + 330y2- 824y + 618= 0, (iii)4y3 .:..9y2+ By- 2 = 0,(iv)y3- 15y2+ 95y- 217= 0.] Ch.211Linear Algebra 2.1Determinant 2.1.1Someproperties of .determinants (i)A determinant remainsunaltered by chl:l.ngingits rowsinto columns and itscolumnsinto rows. (ii)Theinterchangeof tworowsorcolumnsof adeterminantchangesthe sign of thedeterminantwithoutchangingitsnumerical value. (iii)If tworowsortwocolumns of a detecminantbe identical,then the value of the determinantiszero. (iv)If theelementsofadeterminanta berationalintegralfunctionsofx and tworowsor twocolumns become identical when x= a,then(x - a) isafactorof a. Ingeneral,if rrows(orrcolumns)becomeidenticalwhenx= a,then (x- a)r-lisafactorof a. (v)If everyelementof anyrow(orcolumn)of adeterminantbe multiplied byafactor,then the determinantismultipliedbythe samefactor. (vi)A determinant remains unchanged by adding (or subtracting) k times the elementsof anyrow(orcolumn)to(from)thecorrespondingelements of anyother row(orcolumn),wherek isanygivennumber. Ex.1(i)Iffl.beadeterminantoforder3andfl.1bethedeterminant obtained fromfl.byinterchanging firstrowand secondrow,whatisthe relation betweenfl.andfl.l? (ii)Justifythestatement'if tworowsofadeterminantareidenticalthen thedeterminantvanishes'. (iii)Find thevalueof adeterminantof order3 whentwocolumnsareiden-tical. (iv)Provethatif everyelementof athirdorderdeterminantbemultiplied bythe sameconstantp then thedeterminantismultipliedbyp3 .. SOLUTION: (i)Asweknow,interchangeof tworowsof adeterminantchangesthesign of the determinant,thereforerelation betweenfl.andfl.1isfl.= -fl.l. 2.2V.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) (ii)If weinterchange the two identical rows then the determinant .6.becomes ( -.6.). Hence.6.=-.6. or,2.6.=0 or,.6.=O. i.e.,if tworowsofadeterminantareidenticalthenthedeterminant vanishes. (iii)Sameas(ii). ala2a3 (iv)Letthethird orderdeterminantbe.6.= blb2b3 ClC2C3 palpa2pa3 Multiplying eachelementby p w ~getpblpb2pb3 PClPC2PC3 Takingcommon pfromallrowsweget palpa2pa3ala2a3 pblpb2pb3= p3bl b2b3= p3.6.. PClPC2PC3,ClC2C3 Hencethedeterminantismultipliedbyp3., 1ww2 Ex.2Simplifyww2 1wherew isan imaginary cube rootof unity. w2 1w 1ww2 1 + w + w2 ww2 SOLUTION:ww2 1=1 + w + w2 w2 1 w2 1w1 + w + w2 1w (Adding secondand third columnswith the firstcolumn) oww2 =,0w2 1=0[Since1 + w + w2 =0] w2 0w abc Ex.3ShowthatbCa= -(a3 + b3 + c3 - 3abc). cab abc SOLUTION:,bca cab cab baO+bc =aab- cbcca = a(bc - a2) - b(b2 - ac)+ c(ab - c2) = abc - a3 - b3 + abc + abc- c3 =-(a3 + b3 + c3 - 3abc) CH.2:LINEARALGEBRA Ex.4(i)Provethat(a- 1)isafactorof thedeterminant a+1 3 4 2 a+2 4 3 4 a+3 (ii)Showthat(x- a)2isafactorof the determinant xax axa aaa (iii)Showthat(3- ,isafactorof the determinant 111 A=a(3, a2 (32,2 (iv)Showthatx= a + b + c isasolution of the equation a - xbc bc - xa= O. cab - x (v)Provethatx = 1 isarootof theequation x + 233 3x + 45= O. 35x +4 SOLUTION: (i)Putting a= 1 inthe givendeterminantweget 223 334=0, 444 sincetwocolumnsareidentical. ...(a - 1)isafactorof thegivendeterminant. (ii)Putting x =ainthegivendeterminantweget aaa aaa=0, aaa sinceallthethreecolumnsareidentical. 2.3 Weknow,if rcolumnsbecomesidentical whenx = a then(x - a)r-lis afactorof A . ...(x- a)2isafactorof thegivendeterminant. 2.4V.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) 111001 (iii)=a(3"'I=a- "'I(3- "'I"'I a2 (32"'12a2 - "'12(32- "'12"'12 (Cif- Cl- f- C2- CS) o01 =(a - "'1)((3- "'I)11"'I a+ "'I(3+ "'I"'12 .'.((3- "'I)isafactorof the determinant. (iv)The givendeterminantis a-a;bCa+b+c-a;bc bc-a;a=a+b+c-a;c-a;a cab-a;a+b+c-a;ab-a; (cif- Cl+ C2+ cs) Putting a;= a + b + c weget obc o-a - ba= 0 oa-a - c . '.a;=a + b + c isasolution of the equation. (v)Putting a;=1 in the givendeterminantweget 333 355= 0, 355 sincesecondand third columnsareidentical. .'.(w- 1)isafactor,hence,IX=1 isarootof the givenequation. Exercise1Showthat(w- 1)2isafactorof thedeterminant a;1w 1a;1. 111 2.1.2Minor,cofactorand adjointof determinant Def.1Minor:The minor of anyelementaijin adeterminantisadeter-minantobtained byomitting fromtheith rowand jth column. Def.2Thecofactorofanyelementai,inadeterminantisa determinantobtained byomitting fromthe ith rowand jth column with a propersign.The sign of thecofactorof the elementai,is(-1)i+j. Note1If Mi,andAijbetheminorandcofactoroftheelementai,inthe determinantthenAij=(-1) i+ j Mij . CH.2:LINEARALGEBRA Ex.5(i)Find the cofactors of the elementsaandb in 3-4a 27b 5-92 2.5 (ii)Provethat the minor and co-factor of xin the followingdeterminant are the same x 'II Z x2 '112 z2 'liZ zxxy SOLUTION: (i)Cofactorof a=(_1)1+3 27 =-18 - 35=-53. 5-9 Cofactorof b = (_1)2+3 3-4 =-(-27+20) = 7. 5-9 22 (ii)Minor of x='IIz= xy3- xz3. zxxy '112z2 Cofactorof x =(_1)1+1= xy3- xz3. zxxy minor of x =cofactorof x. Ex.6(i)Prove,withoutexpanding,that 523 734= o. 945 (ii)Without expanding findthevalueof thedeterminant 712-3 914-1 813-2 (iii)Prove,withoutexpanding,that 1ab+ e 1be + a= O. 1ea + b (iv)Prove,withoutexpanding,that 1aa2 - be 1bb2 - ea= O. 1ee2 - ab 2.6V.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) (v)Prove,withoutexpanding,that 1abe1aa2 1bea=1bb2 1eab (vi)Prove,withoutexpanding,that ad3a- 4d be3b - 4e=O. e/3c - 4/ SOLUTION: (i) (ii) (iii) 523553 734 =774 ( O ~f- 02+ Oa) 945995 = 0,since firstandsecondcolumnsareidentical. 712-3712-3 [ R\+-R,-R,] 914-1 = 111 813-2111 R ~f- Ra- Rl = 0,sincesecondandthird rowsareidentical. 1ab+e 1ba+e 1ea+b 1aa+b+e = 1ba+b+e( O ~f- 02t' Oa) 1ea+b+e 1a1 =(a+b+e)1b1=0, 1e1 sincefirstand thirdcolumnsareidentical. 1aa2 - be1aa2 - be (iv)1bb2 - ae=0b - ab2 - ae- a2 + be (v) 1ee2 - ab0e - ae2 - ab- a2 + be ( R ~f- R2- Rl, R ~f- Ra- R1) b - a(b- a)(a + b + c) =e - a(e- a)(a + b + c) 1a+b+e =(b-a)(e-a)1a+b+e=0 (sincefirstand second rowsareidentical). 1abe1aa2 abc 1bea= -bbb2 abc 1eabaCee2 abc (multiplying first,secondandthird rows = CH.2:LINEARALGEBRA bya, b, c respectively) aa2 1 bb2 1(takingcommon abcfromthird columns) cc2 1 1aa2 1bb2 1Cc2 (interchanging columns) ad3a - 4d (vi)be3b- 4e c13c - 41 adO be0+-- C3 - 3C1 + 4C2] c10 =0 2.7 Exercise2 1w (i)Ww2 w2 1 Find the values of the followingdeterminantwithout expanding. w2 1w3 w2 530480438 1,(ii)w3 1w,(iii)480450396 ww2 w1438396362 [Ans.(i)0,(ii)3,(iii)0.] Exercise3(i)Find theminorand cofactor of xin 3-4-3 27x 5-92 (ii)Showthat(x+ a + b + c)isafactorof the determinant x+ abc bx+ ca cax+ b 111 (iii)Showthata - b and a - c are twofactorsofabc a2 b2 c2 [Ans.(i)-7,7.] Der.3Adjointoradjugatedeterminant: Theadjointoradjugateofadeterminantisthedeterminant whose elementsarethecofactorsof the corresponding elementsof Theorem1Jacobi'sTheorem If 0) adeterminantof ordernand be itsadjointthen = Q.1StateJacobi's theorem. 2.8U.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) Def.4Adeterminantfj.issaid tobe skew-symmetric if aij=-aji forall iand jinfj.. oa-b The determinant-a0cisaskew-symmetric. b-c0 Def.5Adeterminantfj.issaid tobe symmetric if aij=aji forall iand j in b.. axy The determinantxbzisasymmetric. yzc Ex.7Prove that the value of every skew-symmetric determinant of odd order iszero. SOLUTION:Letfj.beaskew-symmetricdeterminantofoddorder.Let fj.'bethedeterminantobtainedbychangingrowsoffj.intocolumns,then fj.=fj.'. Again,fj.'isalsoobtainedbymultiplyingeachrowof fj.by(-1).Hence fj.'=(-1)nfj. =-fj., sincenisodd . ...fj.=-fj. or,2b.= 0or,fj.= O. Thus everyskew-symmetricdeterminantof odd orderiszero. Exercise4Giveanexampleof athirdorderskew-symmetricdeterminant and findits value. 2.1.3Cramer'srule Ex.8(i)SolvebyCramers rule2x - Y = 1, x+ y=2. (ii)UsingCramers rule,solve,if possible,2x - 3y= 1, x+ y- 5 = O. SOLUTION: (i) -; or, wehavebyCramer'srule xy .,-------;-=.,..----'----;-1-1 21 21 12 xy1 333 or, 1 2-1 11 x=y=1. CH.2:LINEARALGEBRA (ii)Here~= I ~- ~I = 2 + 3 = 5 iO. wehavebyCramer's rule or, or, Exercise5SolvethefollowingequationsbyCramers rule (i)x+ y =5, 3x + 2y = 12. (ii)-x + y=1, 4x + 3y=3. (iii)x+ y= 0, 6x + 8y = O. [Ans.(i)x= 2, y= 3,(ii)x= 0, y= 1,(iii)x= 0, y= O.J 2.2Matrices Ex.1If A =[ ~ 1= ~l' then findthe valueof IAI2. 1-1-1 SOLUTION:IAI=2-2=2 + 2=4 . . '.IAI2=16. Theorem1LetAand Bbe twomatrices.ThenIABI=IAIIBI. 2 . ~ Def.1Singularmatrix:Amatrix Aissaid to be singular if IAIordet(A) isequalto zero. Def.2Non-singularmatrix:AmatrixAissaidtobe non-singularif IAI or det(A)isnotequal tozero. Q.1Definesingularand non-singularmatrices. Ex.2(i)Show that of the twomatrices[ ~;1and[ ~:lone issin. gular and the otherisnon-singular. 2.10V.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) (ii)Forwhat valuesof x,the matrix SOLUTION: (i)HereI ~~I =10- 4 =6 # 0 andI ~: 1= 20- 20=O. [3-2 x -2 2 4-x -4 ~1issingular? -1- x . '.the firstmatrixisnon-singularand secondmatrix issingular. 3 - x22 (ii)If thematrix issingularthen24 - x1=O. -2-4-1- x or,(3- x)[-(4 - x)(l +x)+4]- 2[-2(1 +x)+2] +2[-8 +2(4 - x)]=0 or,(3- x)(x2 - 3x) + 4x - 4x = or,(3- x)(x2 - 3x)=0 or,x= 3, x (3- x)= or,x=0,3,3. whenx=0,3,3then thematrix issingular. Ex.3(i)If A=[ _ ~~ ]andIbe the2 x2 unitmatrix,finaHIevalue of A- 21. (ii)Evaluate3A - 4BwhereA=[ ~~ 4~ ]and B = [ ~~!]. (iii) (iv) CH.2:LINEARALGEBRA2.11 SOLUTION: (i)WehaveA=[ _ ~~ ] . A- 2I =[_ ~~ ] - 2[ ~~ ] - [4 ~ ~1 - ~]- [ _ ~_ ~ ] . (ii)3A - 4B = 3[ ~- ~~ ] - 4[ ~~!] =[9 - 4-1218- 4] =[5-123 149] 15- 8321- 127 (iii)For equality of twomatrices [x-y y-t]=[y-ZX3+-yZ] z+tx+z2+t Comparing weget x- y= y- z,y- t= x- z,z+ t= 2 + t,x+ z= 3 + y or,x- 2y + z= 0,x- y- z+ t= 0,z= 2,x+ z= 3 + y or,x- 2y + 2 = 0,x- y= 1(sincez= 2). Solvingx- 2y = -2 andx- y= 1 weobtain x= 4, y= 3. Thuswegetx=4, y=3, z=2and t=l. (iv)Wehave 3[ ~~ ]=[_ ~2 ~ ]+ [ c! da 1b] =[4+a6+a+b] c + d -12d + 3 Comparing weget3a =4 + aor,2a =4 or,a=2 3b=6 + a + b or,a- 2b= -6 or,b = 4 3d = 2d + 3 or,d= 3 3c =c + d - 1 or,2c=d - 1 or,2c =2 or,c =1 Hencea= 2,b = 4,c = I,d = 3. Def.3Transposeof amatrix:The transpose of amatrix Aisthe matrix obtained fromAby changingits rowsinto columnsorcolumnsinto rows. Generally,it isdenotedbyA' orAT. Q.2Definethe transpose of amatrix. 2.12V.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) Ex.4 (i)If A=[ ! 2 3][-3 1 _ ],thenfindthe 5 7andB=5 -3 matrix CsuchthatA + B= C. Then verifythat(CTf = C. (ii) Find A,when[!5 3 2][ -1 3

1 3+ 5A =5 6 SOLUTION: (i)C=A + B = [123] + [-312] =[ -2 3575-3-18 35] 26 NowaT =[-:!] and ((/T)T=[-; 3 =C. 2 Hence(CT)T= C. (ii)The givenequation can be written as [-137][432][-5 0 ] 5A =563- -513=10 5 1[-505][-101] or,A= 5"1050=210. Ex.5(i)Find (A + B)', whereA= nand B=[!(ii)Find twonon-zeromatriceswhoseproductisazeromatrix.-(iii)Cite oneexampletoshowthatmatrix multiplicationis,in general,not commutative. (iv)GivenA = : and B=are twomatrices.360789 ineif they satisfythecommutativelawof multiplication. (v)Write[xyz] fhbf9c][X;]asasinglematrix. (vi)If A= showthatA2+ 1= O. (I and0 being identityand null matrices of order2). CH.2:LINEARALGEBRA ..[cos 0- sin 0 ] (VU)If A(O)=.00' then showthat smcos A(O)A()= A()A(O)= A(O + ). (viii)If A=[_ = andB= - ] showthatA2+ B2=(A + B)2 . SOLUTION: (i)A + B=+ [! = (A+B? =(A+B)(A+B) ==(ii)LetA::::and B=be twonon-zeromatrices. Bm AB ==isanull matrix. (iii)LetA=[; and B=be anytwomatrices.' NowAB =[;= and BA = 1[; = ...AB-#BA. Hencematrix multiplication is,in general,notcommutative. (iv)AB [ 4 0 1[0 1 500 680 P 0 4 o 1[00 1 05o=00 786o40122 .'.AB -#BA. 2.13 ThusthematricesAandBdoesnotsatisfythecommutativelawof matrix multiplication. 2.14U.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) (vl[ xyz1 ;1 =[ax + hy + gzhx + by + ! zgx + !y + cz1[ 1 = [ax2 +by2+cz2 + 2hxy +2gxz +2Jyz]. (vi)A=[- - ] A2== .'.A2+ 1=+ =(vii)A(8)= [C?S8-sin8]andA()= [C?S-sin] sm8cos 8smcos A(8)A() =[()- sin8]. [- sin] sm ()cos 8sm cos =[cos cos 8 - sin sin 8- sin cos 8 - cos sin 8] cos sin 8 +sin cos 8- sin sin 8 +cos cos 8 =[C?S(8+)-Sin(8+)] =A(8+). sm(8 +)cos(8 +) SimilarlA(A.)A(8)=[C?s( 8 +)- sin( 8 +)] y,'f'sm(8 +)cos(8 +) .'.A(8).A()=A().A(8)=A(8 + ). (viii)A2== B2== ...A2+ B2=+ = A + B= +=...(A + B)2== HenceA2 + B2=(A + B)2. CH.2:LINEARALGEBRA2.15 Exercise1(i)If nbe apositiveintegerandA=[_ l' then hhtA2=[cos 2()sin 2()1 sowta.2()2() -smcos (ii)If A= l' then find(A2- 3A - 131). [Ans.(ii)0] 2.2.1Symmetricandskew-symmetricmatrices Def.4Symmetricmatrix:AmatrixA=[Aij]issaidtobe symmetricif aij= ajiforall i,j. Alternative:Amatrix issaid tobe symmetric if AT =A. Def.5Skew-symmetricmatrix:AmatrixA=[Aij]issaidtobeskew-symmetricif aij=-aji forall i,j. Alternative:Amatrix issaid to be skew-symmetric if AT =-A. Q.3Definesymmetric and skew-symmetricmatricesand giveexamples. Ex.6(i)If A=(-l then showthatA2issymmetric. (ii)If Abe asquarematrix,provethatA + AT isasymmetricmatrix and A- AT isskew-symmetric. (iii)ShowthatforanymatrixA,AAT issymmetric. (iv)IfbothAandBare2x2realskewsymmetricmatrices,showthat AB=BA. ( 073:1)(v)ExpressthematrixA=asthesum of twomatricesof whichoneissymmetrical and the other isskew-symmetrical. SOLUTION: (i)A2=A.A =(-Js (-Js =. ....A 2 isasymmetric matrix. 2.16U.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) (ii)Wehave(A + AT)T =AT + (ATf =AT + A. Thus(A + AT)issymmetricmatrix. Again(A - AT)T =AT - (AT)T =AT - A=-,(A - AT) . . '.(A - AT)isskew-symmetricmatrix. (iii)LetAbe an mxnmatrix,thenAT isan nxmmatrix. .AATisanmxmmatrix. Now(AAT)T= (AT)T AT = AAT. HenceA4. Tisasymmetric matrix. (iv)LetA=( ~- ~ ) ,B=( ~- ~ ) . Then AB =, ( ~- ~)( ~- ~ )=( ~- ~ )and BA =( ~- ~ )( ~- ~ )= ( ~- ~ ) .AB =BA. (v)If Abe asquare matrix then~ ( A+ AT)issymmetric and~ ( A- AT)is skew-symmetric. 11 Then A=2(A + AT) + 2(A - AT) . ~-2) o2. -20 Ex.7If A, B, C are matrices of appropriate orders with AB =AC, then does it imply thatB=C?Givean examplein support of yourconclusion. CH.2:LINEARALGEBRA2.17 SOLUTION:LetA= B= and C= NowAB =andAC =ThusAB =AC but Bi=C. Hence,AB =AC doesnotimplyB=C. 2.2.2Adjointandinverseof amatrix Def.6Adjointofamatrix:If A=[aij]nxnbeasquarematrix,then thetranspose of thematrix[Aij]nxnwhoseelementsarethe cofactorsof the corresponding elementsinIAIiscalledtheadjoint oradjugatematrix of A. It isgenerally denotedbyadj.A. Def.7Inverseof amatrix:LetAbeasquarematrix of ordernxnand thereexistsanothersquarematrixBofthesameordersuchthatA.B= B.A =Iwhere Iis the unitmatrix of order nx n, then Biscalled the inverse of Aand isdenotedbyA -1. Q.4Defineadjointandinverseof amatrix. Ex.8Find theadjointof the matrix [ cosa -sina sina] cos a SOLUTION:Letan = cos a, a12= sin a, a21= - sin a, a22= cos a. The cofactorsof an, a12, a21, a22arerespectively An = (_1)1+1 cos a= cos a,A12= (-1)1+2(-sina) =sina, A21= (-1?+1(sina) = - sina,A22= (_1)2+2 cos a=cos a . . '.the adjointof Ais AdjA = (AnA21)= -sina). A12A22smacos a Ex.9(i)Show that the inverse of the matrixis equal to its trans-pose. (ii)Writedowntheinverseof the matrix A = !]. 2.18U.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) (iii)Find A,if A= l (iv)If A= then showthatA2- 4A + 31=O. (v)LetA, Bbe twomatrices such thatAB =0,where 0isthe null matrix. Doesit imply thatA= 0 orB= O?Givean example in support of your answer. (vi)If Abeanon-singularsquarematrixandB, Caresquarematricesof the same order suchthatAB = AC,showthatB= C . SOLUTION: (i)LetA=adj.A =andIAI= 0 - 1 = -1. A-I = = - =. .T[Ol]T[01] AgamA=10=10. HenceAT =A-I. (ii)HereA= ! l IAI= 8 - 3 = 5.adjA = [_!.A-I = adj A= ![4-1]. . .IAI5-32 (iii)LetB=[; IBI= 8 - 3 = 5.adjB =-!] . . '.B-1 == -!].' WehaveA [: = lor, AB = 1 or,A =B-1 =-!] CH.2:LINEARALGEBRA =~[ ~ ~~ ~ ]=[-!- ~l (iv)A2=[2 -1] [ 2 -1]=[5 -4]. -12-12-45 A2[_ 4t +_!I][2-1][ 1.0][ 0 =-45- 4-12+ 301=0~]. ...A2- 4A + 31=O. [ -11][-22] (v)LetA=-11andB=_ 22. AB =[=~~ ][ =~~ ]=[ ~~ ] . ThusAB =0 doesnotimplyA =0 orB=O. (vi)WehaveAB =AG.GiventhatAisnon-singular.SoA-I exists . ...A-I(AB) =A-1(AC) or,(A-1 A)B =(A-1 A)C or,I B= IC or,B= G. 2.2.3Orthogonalmatrix 2.19 Def.8Orthogonal matrix:A matrix A is said to be orthogonal if AAT =I, ATisthe transpose of A. Q.5Defineorthogonalmatrix. Ex.10(i)Showthatthe matrix[ ~- ~]isorthogonal. Y272. (ii)Provethatthe inverseof an orthogonal matrixisorthogonal. (iii)Showthattheproduct of twoorthogonal matricesisorthogonal. (iv)Showthattheunitmatricesareorthogonal. (v)Show that the value of the determinant of an orthogonal matrix is either 1 or -1. (vi)Showthatthe transpose of an orthogonal matrix isorthogonal. SOLUTION: (i)LetA =[~ Y2 HenceAisan orthogonal matrix. ~ ] 1. Y2 ~]=I. 2.20V.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) (ii)LetAbean orthogonalmatrix.Weknowforanorthogonalmatrix A, AT =A-1 and AT A=I. Now(A-1)(A-1)T=(AT)(AT)T =AT A=I. HenceA -1isorthogonal. (iii)LetAand Bbe twoorthogonal matrices.Then AT A=Iand BT B=I. Now(ABf(AB) = (BT AT)(AB) = BT(AT A)B = BTIB =BTB =I . . '.AB isan orthogonal matrix. (iv)SinceIT1= 1.1 = I, thereforeunitmatricesare orthogonal. (v)LetAbe an orthogonal matrix,then AT A=I . . '.IAT AI=IIIor,IATIIAI=III=1 or,IAI2= 1(sinceIATI= IAI) or,IAI=l. (vi)LetAbe an orthogonal matrix thenAAT =I. Now(AT)T AT = AAT = I . . '.ATisan orthogonalmatrix. Exercise2(i)Showthat thematrix[ ~~1isorthogonal. (ii)Showthat the matrix[ ~~ 11isorthogonal. .[cosa Showthat the matrIX. sma - sin al' ISorthogonal. cos a (iii) (iv) [ -1 Showthat the matrix~- ~ 2 ':-2] 12isorthogonal. 21 (v)Find k,such that[: ~ ~ ~ [Ans.(v)k=l.] - sinO0] cos 00isan orthogonal matrix. ok 2.2.4Rankof amatrix Def.9Rank of aMatrix:The rank of anon-null matrix Aisthe order of thelargestsquare sub-matrix in Awhosedeterminantdoesnotvanish. Ex.11(i)Find the rank of[!!~ ] . 4316 CH.2:LINEARALGEBRA (ti)Find the rank of the matrix :!). (iii)Find the rank of[J 2.21 (iv)Whatisthe rank of unitmatrix of ordertwo?Justify youranswer. (v)Find the value of xlor which the rank of the matrix !istwo . SOLUTION: [1'2 [ 2 -Iq [112+- R, - 3R,] (i)34-2 43160-3 o f- R3- (Rl + R2) NP 2 -I; ](112+- R, - -2 .0 018 Herealltheelementsof lowertriangularpartarezeroandthenumber of non-zerorowsis3 sothe rank of the givenmatrix is3. (ii)'" [ f- R2- 2Rl1 366000R3f- R3- (Rl + R2) Herealltheelementsof lowertriangularpartarezeroandthenumber of non-zerorowsis2 sothe rank of the givenmatrix is2. (iii)LetA[J 1 NowIAI=v'3=-V6 O. 0 Hencethe rank of the givenmatrix is2. (iv)Let 1=1be aunitmatrix of order two. 10 NowIII=01=1 O. Hencethe rank of aunitmatrix of order twois2. (v)If the rank of the givenmatrix is2 then its determinant value should be equal to zero.Thatis '-2.22V.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) 242 212=0 lOx or,2(x - 0)- 4(2x - 2)+ 2(0 - 1)=0 or,- 6x + 6 =0 or,x=1. Theorem2If AandBbetwomatricesthenrank(A + B):c::;rank(A)+ rank(B). Theorem 3If A and Bbe two matrices then rank(AB):c::;min{rank(A), rank(B)}. 2.2.5Solution of equationsby matrix inversemethod Ex.12(i)Writethe matrix equation forthe system of equations: x+ y + z=3, 2x - 3y + 4z =17, x+ 6y - z=-8. (ii)GiventhematricesA~[!- ~nX~[~1andC~[n write downthe three linear equationsgivenbyAX =C . SOLUTION: (i)The equations arewritten in matrix notation asAX =B, whereA=[ ~ - !! 1,X =[ :]andB=[1~]. 16-1z-8 (il):T f ~ n][ ~]~[~1 The three linear equationsarex+ 2y + 3z =6, 3x - 2y + z=2,4x + 2y + z=7. Ex.13(i)Solvebymatrix method 2x - y= 1, x+ y= 2. (ii)Solvebymatrix inversemethod3x - 2y =4,4x - 3y =5 . SOLUTION: (i)The givenequations canbe written as AX=B whereA =[ ~- ~1'X= [ :land B= [ ~l (1) CH.2:LINEARALGEBRA 12-11[1 IAI=11= 2 + 1 = 3andadj A=-1 .A-I = adj A= ![1 1]. . .IAI3-12 From(1),X=A-I B or,[:]=~[- ~~ ][~]=~[~] Hencex= 1, Y = 1. (ii)The givenequations can be written in theform AX=B whereA= [!= ~ ] ,X= [ :]and B= [ :]. IAI=! = ~= -9 + 8 = -1 and adjA = [ =!.~l .A-I = adjA= -1 [-32]= [3-2]. . .IAI--434-3 From(1),X=A-I B or,[:]= [!= ~ ][: ] = [ ~] Hencex="2,y =1. 2.3VectorSpace 2.23 (1) Def.1Vectorspace:LetVbeanon-empty setwhoseelementsarecalled vectorsand Fbe ascalar field.The setViscalledavectorspaceoralinear space overthe fieldFif the followingaxiomsaresatisfied: (i)a + (3EVforall a, (3EV. (ii)a + :a =(3 + a forall a, (3E V (iii)a + ((3 + 1) =(a + (3)+ 1 forall a, (3, 1 E V (iv)There existsaunique vector (j EVsuch that a+ (j = a = (j + a forall a E V (v)Foranyvector a E Vthere existsauniquevector-a E V such that a + (-a) =(j (vi)aa EVforall a EFand a EV (vii)a(a + (3)=aa + b(3forall a EFand a,(3 E V (viii)(a + b)a =aa + ba,forall a, bE Fand a EV (ix)(ab)a =a(ba), forall a,b EFand a EV (x)la =a,1 E Fand a E V. 2.24V.G.MATHEMATICS(SHORTQUb:::HIONSANDANSWERS) The set{(x, y, Z): x, y, zER}isavector spaceoverthe fieldR. Def.2Subspace:Anon-empty subset Wof avectorspace Vovera fieldF iscalledavectorsubspaceoralinearsubspaceorsimplyasubspaceof Vif WisavectorspaceoverFw.r.t.addition and scalarmultiplication. Theorem_1AWof Visasubspace of Viff (i)a, j3EW=} a + j3EEWand (ii)aEW, eEF=} cal EW. Ex.1(i)LetV={(x, Y/'z): x, y, zER}andif W={(x, y, z): x- 3y + 4z=O}then prove Wisasubspace of V. I (ii)LetV={(x,y,z): i,y,z ER}andif W={(x,y,z):x2+ y2=z2} then provethat Wisnotasubspace of V . SOLUTION: (i)Leta=(Xl, yl, zI)and 7J=(X2' Y2,Z2)be twovectorsof W. Then Xl- 3Yl+ 4Zl=0andX2- 3Y2+ 4Z2=o. Addingtheseequations, (Xl+ X2)- 3(yl+Y2)+4(Zl+ Z2)=O. Thisa+7J=(Xl+ X2,Yl+Y2,Zl+ Z2)EV. Again ea =(exl' eYl, eZl)EWas eXl- 3eYl+4ezl=0[by(1)]. HenceWisasubspace of V. (1) (ii)Let a=(-3,4,5) and 7J=(3,0,3) be two vectors ofW as (_3)2+42 =52 and 32 + 02 = 32. But a+ 7J=(-3,4,5) +(3,0,3)=(0,4,8) Was02 +42 # 82. HenceWisnotasubspaceof V. 2.3.1Linearlydependenceandindependenceof vectors Def.3Linearlydependent:Aset of vectors{Xl, X2,... , xn} of Enissaid tobelinearlydependentif thereexistsasetof scalarsei, i=1,2, ... ,n not all of whichare zero,such that where0isthe null vectorin En. CH.2:LINEARALGEBRA2.25 Def.4Linearlyindependent:Asetofvectors{Xl, X2,... ,xn}ofEnis said tobe linearlyindependentif theonlysetof Ciforwhich CIXI+ C2X2+ ... + CnXn=0 holds,be Ci= 0, i= 1,2, ... , n. Q.1Definelinearlydependentand independentvectors. albl CJ Formula 2LetLl=a2b2C2. a3b3 C3 If Ll=0then thevectors(al,bl,cl),(a2,b2,c2)and(a3,b3,c3)arelinearly dependentand if Ll=f.0then theyarelinearlyindependent. Ex.2(i)Showthatthevectors(1,0,0)'(0,1,1),(1,1,1)in therealvector space ll3arelinearlydependent. (ii)Determineksothatthevectors(1, -1, 2), (0, k, 3)and(-1,2,3)are linearly dependent. (iii)Showthatthevectors(2,1,0), (1, 1,0), (4,2,0)ofR3arenotlinearly independent . SOLUTION: (i)The value of the determinantformby the givenvectorsis 1 0 11= 111 11 11 =1-1 =O. The determinant value iszero,therefore the given vectors are linearly dependent. (ii)Sincethe vectorsare linearly dependent 1-12 k3= -123 10 or,0k3= 0[ C ~+- G2+ Gl,G3 +- C3 - 2G1] -115 k3 or,15=0or,5k - 3 =0or,k=3/5. 210 (iii)Here110=(sinceall the elementsof third column are zero). 420 Hencethe vectorsarenotlinearlyindependent. 2.26V.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) Exercise1(i)Arethe vectors(1, -2, 0), (3,0, -2) and (0, -1, -5) linearly independentin thereal vectorspaceR3? (ii)Show that the vectors (2,4,0),(0,1,0) and (2,6,2)are linearly independent in the realvectorspaceR3. [Ans.(i)Yes.] Ex.3(i)Expressthevector(7,11)asalinearcombinationof thevectors (2,3)and(3,5). (ii)Express the vector(15,18)asalinearcombination of the vectors(3,4) and(3,3) . SOLUTION: (i)Let(7,11)= cl(2, 3)+ c2(3, 5)= (2Cl+ 3C2,3Cl+ 5C2) Comparing both sidesweget2Cl+ 3C2= 7 and3Cl+ 5C2= 11. SolvingwegetCl=2, C2=1.Hencethelinear combination is(7,11)= 2(2,3) + (3,5). (ii)Let(15,18)= Cl (3,4) + c2(3, 3)= (3Cl+ 3C2,4Cl+ 3C2) Comparing both sidesweget3Cl+ 3C2= 15and 4Cl+ 3C2= 18. Solving wegetc!=3, C2=2.Hence the linear combination is(15,18)= 3(3,4) + 2(3,3). 2.3.2Basisof avectorspace Def.5Linearspan:LetVbeavectorspaceoverthefieldFandWbe any non-empty sub-set of V.Then the linear span L(W) of Wisthe set of all linearcombination of finitesetsof elements of W. L(W) issaid to be generated or spanned bythe setWand Wissaid to be the setof generatorsof L(W). Def.6Basis:LetVbeavectorspaceoverthefieldFand Wbeasub-set of Vsuch that (i)Wisasetof linearlyindependentvectorsin Vand (ii)WgeneratesV, then Wiscalledabasisset orsimplybasisof V. Forexample,{(1, 0, 0), (0, 1,0), (0,0, 1)}isthe standard basis of R3. Def.7The number of vectors in any basis of a finite dimensional vector space Viscalledthe dimension of the vectorspaceand itisdenoted by dimV. Note1The basis of a vector space isnot unique but the dimension is unique. CH.2:LINEARALGEBRA '2.27 :' , Theorem 2If Wbe aproper sub-space of afinitedimensional vector-space V,then Wisfinitedimensional and dim W:Sdim V. Theorem 3If WIandW2betwosub-spacesof afinitedimensionalvector space V,then dim(WI U W2)=dim WI + dim W2- dim(WI n'W2). Ex.4Given thatS= {(x, y, z)E R3: 3x - 4y + z= o}isasub-spacein R3. Find abasisand the dimension of S . SOLUTION:Leta= (x, y, z)be avectorof S.Then 3x - 4y + z= 0,i.e., z=-3x +4y. Now,a=(x, y, -3x + 4y)=x(l, 0,-3)+ y(O, 1,4).Thusthevectorais alinearcombinationof thevectors(1,0, -3)and(0,1,4).Therefore,W= {(I, 0,-3), (0, 1, 4)}generatesS. Also,the vectors(1,0, -3) and(0,1,4)arelinearindependent. Hence Wisabasis of S.The number of vectors of Wistwo,the dimension of Wis2. \ 2.3.3Eigenvalues Def.8Eigenvalues:Theeigenvaluesof.asquarematrixAaretheroots of the equationIA - All=0,i.e.,thevaluesof thevariableA. This equation isknownascharacteristicequation. Ex.5(i)Find the eigenvaluesof the matrix[!~ ] . (il)Find the eigen values of thematrix[ ~~~ ]. SOLUTION: (i)LetA=[!~J. Now,A- AI =[!~ ] _A [ ~~ ]=[1 ~A ThenIA - All=11 ~A5 ~AI =A2- 6A- 7. The eigenvaluesaregivenbyA2- 6A- 7= or,(A+ 1) (A- 7)=or,A = -1, 7 ...the eigenvaluesare-1,7. 2.28B.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) (ii)LetA =303 Now,A - >'1=>. 30300 1 ]. 303 - >. 3 - >.03 ThenIA - >'11=o 3 3 - >.0=(3- >.)3- 3.3(3 - >.). o3 - >. The eigen valuesaregivenby(3- >.)3- 3.3(3 - >.)=0 or,(3- >'){(3- >.)2- 9}=0 or,(3- >')(9- 6>'+ >.2- 9)=0 or,>'(3- >.)(>.- 6)=0 or,>.=0,3,6 . . '.the eigenvalues are0,3,6. Exercise2(i)Find the eigen valuesof thematrix[! (ii)Find the eigen valuesof the matrix111 [Ans.(i)4, -1, (ii)1,1,3.] Theorem 4Caley-Hamiltontheorem:Everysquarematrixsatisfiesits owncharacteristic equation. Somepropertiesof eigen values: (i)The eigen valuesof areal symmetric matrix are real. (ii)Theeigenvaluesofarealskew-symmetricmatrixarezeroorpurely imaginary. (iii)The eigen valuesof an orthogonalmatrix areof unitmodulus. Ch.3\\Abstract Algebra 3.1SetTheory Def.1Set:Acollection of welldefinedobjectsarecalled set. Upper caselettersareusedtodenotea set. Someexamplesof setare (i)A={I, 3, 5, 7,... ,99}, (ii)B={I, 4, 9,16,25, ... }, (iii)B= {x : xisareal number satisfyingx(x2 - 4)= O}, (iv)Collection of allchairsand tablesof aclassroom, (v)Collection of allIndian cricketers, (vi)Allbooksof classX, (vii)Allstudents of B.Sc.class. The followingcollecticmsarenotthe examples of set: (i)Alltall men of aclass, (ii)Allgood students of acollege. Def.2Null set or empty set:Asethaving no element iscalled null setor empty set. It isgenerallydenotedbythe symbolcp. The followingarethe examplesof null sets. (i)A= {x : xisrealnumbersatisfyingx2 + 4 =O}, (ii)Collection of Indian Presidents of agemorethan 150years. Def.3Singletonsetorunitset:Asethavingonly oneelementiscalled aunitsetorasingleton. Def.4Finite setandinfiniteset:Asetiscalledfiniteif itbe emptyor containsafinitenumberof elements,otherwiseasetiscalledinfinite. The setcontaining allevenintegersbetween2 and 10including2 and 10is afiniteset,whilethe setof all odd integersisan infiniteset. 3.2V.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) Note1The notation xEA isused to denote the statement that the element xbelongstothesetA,whilethenotationxrt.Adenotesthattheelementx doesnotbelong tothe setA. Def.5Sub-setandsuperset:If everyelementofasetAbealsoan element of another setB, then Aiscalled asub-set of Band Biscalled super setof A.Symbolically,AcBorB::::>A. Forexample,if A= {I, 3, 4, 5}and B= {I, 2, 3,4,5, 7},thenAcB. If Bcontainssomeelementswhicharenotthe elementsof A,thenAisa proper sub-set of B. Note2Todistinguish thesetwocases,weusethe symbols~or:;2to denote asub-setand useCor::::>todenoteaproper sub-set. Def.6Disjointsets:TwosetsAandBarecalleddisjointif theyhaveno elementin common. Forexample,A={I, 2, 3}andB={1O, 11, 12}aredisjoint. Def.7Equal oridentical sets:TwosetsAand Baresaid tobe equal or identical, i.e.,A= B, if they have the same elements.In this case,all elements of Aarealsotheelementsof B. Forexample,A={I, 2, 3}andB=={3, 1, 2}are equal. Some properties on sub-setarelistedbelow. (i)If Abe asub-setof Band Bbeasub-setof C,thenAisasub-setof C,Le.,if A~Band B~C,thenA~C. (ii)If A~Band B~A,thenA=B. (iii)The null set=4> (vii)A n AI=4> (viii)If A~B,then An B=A. 3.1.3Cartesianproductof sets Def.9Cartesian product:The cartesian productbetweentwosetsAand BisdefinedasAxB={(x, y): xEAandyEB}. Forexample,letA= {1,2}andB= {a,b}then AxB={(I, a), (1, b), (2, a), (2, b)}. Q.2Definecartesian productof setsandgiveoneexample of it. Ex.3(i)A= {1,2},B= {3, 4, 5}andC= {4, 5, 6}then showthatAx (B n C)=(AxB) n (AxC). (xisthecartesian product). (ii)If A={I, 2, 3}andB={3,4},thenfindAU B,An B,(AxB)and (BxA). (iii)Forthe setsA= {a,b,c}andB= {1,2}verifythat Ax B=1=Bx A. (iv)If A= {a, b}, B= {I, 2}and C= {2, 3},find(AxB) U (AxC). CH.3:ABS'fRACTALGELRA SOLUTION: (i)Bn C= {4, 5},Ax(B n C)= {(I, 4), (1,5), (2,4), (2, 5n AxB={(I, 3), (1,4), (1, 5), (2,3), (2,4), (2, 5n AxC={(1,4),(1,5),(1,6),(2,4),(2,5),(2,6n . . '.(AxB) n (AxC)={(I, 4), (1,5), (2,4), (2, 5n. HenceAx(B n C) =(AxB) n (AxC). (ii)A={I, 2, 3}andB={3,4}. AUB = {1,2,3,4},AnB = {3} AxB= {(I, 3), (1,4), (2,3), (2,4), (3, 3), (3, 4n and BxA= {(3,1),(3,2),(3,3),(4,1),(4,2),(4,3n. (iii)AxB={(a, 1), (a,2), (b,1), (b,2),(e,1), (e,2n and BxA ={(I, a),(1, b),(1, e),(2, a), (2, b),(2, en . . '.AxBofBxA. (iv)AxB={(a, 1), (a, 2), (b,1), (b,2n and AxC ={(a, 2), (a,3), (b,2),(b,3n. (AxB) u (AxC)={(a, 1), (a,2), (a,3), (b,1), (b,2), (b,3n. 3.1 Exercise2(i)Find the cartesian product A x Bof the sets Aand Bwhere A= {1,3,5}andB= {2,4}. (ii)If A= {I, 2}, B= {2, 3}, C={3,4}.FindAx(B U C). [Ans.(i){(I, 2), (1, 4 ~ . ( 3 ,2), (3, 4), (5,2), (5, 4n, (ii){(I, 2), (1,3), (1,4), (2, 2),(2,3), (2,4) }.] Def.10Powerset:The collectionof allsub-setsof asetAincludingA,is calledpowersetof Aanditis,generally,denotedbyP(A). Ex.4(i)Writedownthesub-setsof the set{x, y, z}. (ii)Write down all the subsets of the set S={p, q, r},what isthe power set of the setS? SOLUTION: (i)Thesubsetsof{:r,y,z}are1>,{x},{y},{z},{x,y},{y,z},{z,x}arid {x,y,z}. (ii)The subsetsof Sare1>,{p},{q},{r},{p,q},{q,r},{r,p} and{p,q,r}. Thepowersetof Sis {1>,{p}, {q}, {r }, {p, q}, {q, r}, {r, p},{p, q, r} }. Note3If afinitesethasnelements,thenitspowersethas2n elements. 3.8V.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) Theorem 1DeMorgan'slaws ForanytwosetsA, B (i)(A U BY =A' n B' (ii)(A n B)' =A' U B', where'A"denotesthe complementof the setA. Proof: (i)Letxbe anyelementof (A U B)' xE(AUB)' x(AUB) xAandxB xEA' andxEB' xE(A' n B'). Hence(A U BY =A' n B'. (ii)Proof issimilarto(i). Ex.5(i)By Venndiagram provethatA- (B U C)=(A - B) n (A - C). (ii)Showby Venndiagram thatAn (B n C)=(A n B) n C . SOLUTION: (i)In 'figure(a),the verticallinesrepresentthe set(B U C)and horizontal linesrepresentthesetA.Theregionshadedonlybyhorizontallines representthe setA - (B U C). In figure(b),thehorizontallinesrepresentthe set(A - B) and vertical linesrepresentthe set(A - C).The double shaded region represents the set(A - B) n (A - C). Both the regionsare identical,henceA- (B U C)=(A - B) n (A - C). A (a) Figure:3.1.2 (b) (ii)In figure(a),theverticallinesrepresentsthe setAandhorizontallines representsthesetBn C.Thedoubleshadedregionrepresentstheset An (B n C). CH.3:ABSTRACT_\LGEBRA3.9 In figure(b),the vertical linesrepresentsthe set(AnB) and horizontal linesrepresentthesetC.Thedoubleshadedregionrepresentstheset (A n B) n C. Both the regionsareidentical,henceAn (B n C)=(A n B) n C. B 3.2Relation A C B of Figure:3.1.3 Def.1ArelationbetweentwosetsAandBisasub-setofAx Bandis denotedbyR.Thus RcA xB. ThenotationxRyisusedtorepresent(x, y)ER.xRyisreadas'xis relatedtoy'. If A=B, then wesaythatRisarelationin the setA. Def.2Forthe setAtherelation R ={(x,y): xEA,y EA,x =y}iscalled theidentity relation inAandisdenotedbyfA. For example,if A ={a, b, e}thenfA={(a, a), (b,b), (e, en 3.2.1Typesof relations Def.3Reflexiverelation:LetAbe asetand Rbe the relation definedin it.Rissaidto be reflexive,if (a, a)ERforalla EA i.e.,everyelementof Aisrelatedtoitself. A relation R in asetA isnot reflexive if there be at least one element aE A suchthat(a, a)rtR. Def.4Symmetricrelation:LetAbeasetandRbe therelationdefined init.Rissaid tobe symmetric,if (a, b)ER ::::}(b,a)ERi.e.,aRb::::}bRaforall(a, b)ER. ArelationRinasetAisnotsymmetricif(a, b)ERdoesnotimply (b,a)ER. 3.10U.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) Def.5Anti-symmetricrelation:LetAbeasetandRbetherelation defined in it.Rissaid to be anti-symmetric,if (a,b)ERand(b,a)ER=?a=b. A relation Rin Aisnot anti-symmetric, if there exist elements a, b EA, ai-b such that(a, b)ERand (b, a)E R. Def.6Transitiverelation:LetAbeasetandRbetherelationdefined in it.Rissaid to be transitive,if (a,b)ERand(b,c)ER=?(a,c)ER,a,b,c EA. ArelationRisnottransitiveif(a, b)ERand(b, c)ERdonotimply (a,c)ER. Def.7Equivalencerelation:ArelationR,definedinAissaidtobean equivalence relation if and only if (i)Risreflexive,i.e.,aRa forall aEA. (ii)Rissymmetric,i.e.,aRb=?bRa. (iii)Ristransitive,i.e.,aRb andbRc=?aRc. Ex.1If RbearelationinthesetofnaturalnumbersNdefinedbythe expression' (x- y)isdivisibleby 5',i.e., R={(x, y): x, yEN, (x- y)isdivisibleby5}, provethatRisan equivalencerelation . SOLUTION: (i)LetxEN.Then x- x=0whichisdivisibleby5. Therefore,xRx forallxEN. HenceRisreflexive. (ii)LetxRy hold.i.e.,x- yisdivisible by5 =?-(y - x)isdivisibleby 5 =?(y- x)isdivisible by5 =?yRx. HenceRissymmetric. (iii)LetxRy andyRz,i.e.,(x- y)and(y- z)aredivisibleby 5 ::}[(x- y)+ (y- z)]isdivisible:by5 =?(x- z)isdivisibleby5 =?xRz. HenceRistransitive. ThusRisan equivalencerelation. CH.3:ABSTRACTALGEBRA3.11 Ex.2LetAbethesetofintegers.ShowthattherelationRdefinedby 'a~b',thatis, R={(a, b): a,b EA,a ~b}, isnotan equivalencerelation . SOLUTION: (i)LetaEA.Then a~aholds. i.e.,aRa holds.HenceRisreflexive. (ii)LetaRb holds,i.e.,a~b. This doesnotimplyb ~a. HenceRisnotsymmetric. ThusRisnotan equivalence relation. 3.3Mapping Def.1Mapping:LetAandBbetwonon-emptysets.If thereexists acorrespondence,denotedbyf,whichassociatestoeachelementxof Aa uniqueelementyof B,thenwesaythatfisamappingof AintoB.The mapping of AintoBwillbe denotedbyf: A-+B. The setAandBarecalledrespectivelydomainandcodomain(orrange)of the mapping f. The elementyisdenotedbyf(x)and calledthefimage of xor the value of the functionforx. Def.2Range:Letf: A-+Bbeamapping.Therangeof fconsistsof thoseelements in Bwhich appear astheimageof atleastone elementof A. Def.3One-to-one mapping:Let f: A-+ Bbe amapping.If forXli=X2 =}f(XI)i=j(X2)thenthemappingfiscalledone-to-oneorone-oneor1-1 mapping. e.g.,the mappingf: R-+Rwheref(x)=xisone-to-onemapping. Def.4Ontomapping:Letf: A-+Bbeamapping.If f(A)= Bthen the mapping fiscalledontomapping. e.g.,the mappingf: R-+Rwheref(x)= xi'lontomapping. Def.5Identity mapping:If each element of aset ismapped on itself then itiscalledtheidentitymapping.The mappingf (x)=xforallxEAisan identitymapping. 3.12U.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) Def.6Many-to-onemapping:If twoormoreelementsof Acorrespond to the same elements of Bthrough the mapping fthen it is, called many-to-one

e.g.,the mappingf: R--+Rwheref(x)=x2 ismany-to-onemapping. De(-a).b + a.b=0(byrightdistributivelaw) .'.(-a).b istheadditive inverse of a.b. Hence(-a).b =-(a.b),forall a,b EF. (iii)Letebe the multiplicative identity of F. Then a.a-1 =ewherea-IE Fisthe inverse of a. Alsowehave(a-I )-1.a-1 =e. ...(a-I )-1.a-1 = a.a-1 or,(a -1) -1=a(byrightcancellation law). Ex.10(i)In afieldF,provethata2 = b2 =>either Po= b ora= -b. (ii)ProvethataringRiscommutativeif (a+ b)2' =a2 + 2ab + b2 forall a,bE R (iii)If a, bEF(afield)anda=I0thenshowthatthereexistsaunique elementxsuchthata.x =b. 3.28V.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) SOLUTION: (i)Givena2 =b2. Adding(_b2)on both sidesweget, a2 + (-b2) =b2 + (_b2)(_b2 istheinverseof b2) =} a2 + (_b2)=0 =} a.a - b.b=0 =} a.a + a.b - a.b- b.b + 0 =0 =} a.(a + b)- (a+ b).b=0(bydistributive) =} a.(a + b)- b.(a + b)= 0(byadditive commutative) =} (a+ b).(a - b)=0(bydistributive) Since fieldiswithout zero divisor,therefore,either a - b = 0or a + b = 0, i.e.,either a=b ora==-b. (ii)Wehave(a+b)2= a2 +2ab+b2 or,(a + b).(a + b)=a.a + 2a.b + b.b or,a.(a + b)+ b.(a + b)=a.a + 2a.b + b.b (bydistributive law) or,a.a + a.b + b.a + b.b=a.a + 2a.b + b.b (bydistributive law) or,a.b + b.a + b.b=2a.b + b.b(byleftcancellation law) or,a.b + b.a=a.b + a.b(byrightcancellation law) or,b.a=a.b(byleftcancellation law). HenceRiscommutative. (iii)Sincea =f.O....a-I existsand a-I EF. Leftmultiplyingax = b bya -1weget a-l.ax =a-l.b or,e.x =a-l.b(leftidentity) or,x=a-l.b. Ex.11(i)If in aring R,a2 =a forall aER,prove that a+b =0 =} a=b foralla, b ER. (ii)If Dbe an integral domain andxbe an elementof Dsuch thatx2 =x, then provethatx= 0 or1. (iii)If a, bE F,Fisafieldand b =f.0,then provethata=1,when(ab)2= ab2 + bab - b2. (iv)If a, b, e,darethe membersof a~ e l dthen showthat~=~ifad =be . SOLUTION: (i)LetaER.Byadditive closure a+a =(a+a)2=(a+a)(a+a) CH.3:ABSTRACTALGEBRA =(a2 + a2)(a2 + a2)(bydistributive law) =(a+ a)(a + a)(sincea2 =a) or,(a+ a)+ 0 = (a+ a)+ (a+ a)(byadditive identity) or,0 =a + a(byleftcancellation) i.e.,a + a =O. Now,wehavea+ b =O. .a + a = a + b.. By leftcancellation,a=b. (ii)Letusconsidertherelation x(x - 1)=x2 - x(bydistributive law) =0(sincex2 = x) SinceDisan integral domain,soitiswithout zerodivisor. Thus,x=0orl. (iii)Given(ab)2=ab2 + bab- b2 '* (ab)(ab)=(ab)b+ bab- bb '* (aba)b=(ab)b + bab- bb '* aba=(ab)+ ba- b (by rightcancellation) '* aab= 2ab - b (asab= baforafield) '* aab=(2a- l)b '* aa=(2a- 1)(byrightcancellation) ,*aa-a-a+1=O '* (a- l)(a - 1)=0'* a-I =0 ,*a=l. (iv)Wehave~=:. '* ab-1 =ed-1 bd '* b-1a =ed-1 '* b(b-1a)=bed-1 '* a =bed-1 '* ad =bed-1d '* ad =bc. Againad =be'* add-1 =bed-1 '* a =bed-1 '* b-1a =b-1bed-1 '* b-1a =ed-1 '* ab-1 =ed-1 ae '* b ="d' Ex.12Provethateveryfieldiswithout zerodivisor. 3.29 3.30U.G.MATHEMATICS( ~ H O R TQUESTIONSANDANSWERS) SOLUTION:LetFbe afieldand a, bE F. Weshall provethat,if a.b= 0 then either a= 0 orb = 0 forall a, bE F. Suppose a.b =O.Leta=0,soa-Iexists. Now,a-l(a.b)=a-I.O =>(a-Ia).b = 0=>b = O. Similarly,if b =0 then a=O. Thus Fhas no zero-divisor. Note1Afieldisan integral domain. Ch.4IIGeometry ". TwoDimensions 4.1Transformationof Axes Changeof originwithoutchangeof directionof axes(translation): If (x, y)and(x', y')betheoldandnewcoordinatesrespectivelyof apointP thenx=x' + a,y =y'+ (3where(a,(3)istheneworiginw.r.t.totheold coordinates system. Rotation of rectangular axes without changing the origin (Rotation): If (x, y)and(x', y')bethecoordinatesof thesamepointPinoldandnew coordinatessystemthenx=x' cos 0 - y' sin 0,y=x' sin 0 + y' cos 0,where0 isthe angleof rotation. Equationofthecurveafterrotationatanangle0:If theaxesare rotatedatanangle0thenthecurveax2 + 2hxy + by2+ 2gx+ 2fy + c=0 becomesin the formax,2+ by,2+ 2g'x' + 2f'y' + c'=0,where0isgivenby 2h tan 20=--b' a-Combination of translation and rotation:If the origin is shifted to (a, (3) andtheaxesarerotatedthroughanangle0intheanticlockwisedirection (positive)thenthechangesthecoordinates(x, y)of apointParegivenby x=a+ x'cosO- y'sinO,y=(3+ x' sinO+ y'cosO,where(x',y')arethe coordinatesof Pin newsystem of axes. Ex.1Find the equation of the line ~+ ~=2,when the origin istransformed to thepoint(a, b). SOLUTION:Putting x=x' + aandy=y' + b. The transformed equation is ~ + a~ + b~~~~ -a- + -b- = 2 or,~+ b+ 2 = 2 or,~+ b= O. Ex.2(i)Find the transformed fromof the equation x2 - y2=4when the axesare turned through an angleof 45keepingthe origin fixed. 4.2V.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) (ii)Find the translation which transforms the equation x2 + y2- 2x + 14y + 20= 0intoX,2+ y,2= 30. (iii)Findtheanglethroughwhichtheaxesaretoberotated,sothatthe equation 17x2 + 18xy -7y2 =1 may be reduced to the formAX2 + By2= 1. (iv)Thecoordinatesaxesarerotatedthroughanangle60.If thetrans-formed coordinates of a point are (2V3, -6), find its original coordinates. (v)Showthattheequation4xy - 3x2 =1istransformedtox,2- 4y,2=1 by rotating theaxesthrough an angletan-1 2. (vi)Towhat point the origin isto be transferred to getrid of the firstdegree termsfromtheequation 8x2 + 10xy + 3y2+ 26x + 16y + 21=O? (vii)Transformtheequation3(12x- 5y + 39)2- 2(5x + 12y- 26)2=169, takingthelines12x- 5y + 39= 0and5x + 12y- 26= 0asthenew axesof xandyrespectively. (viii)Whatwillbetheformof theequationx2 - y2=4,if thecoordinates 7r axesare rotatedthrough an angle- 2"' (ix)Find the angle of rotation of the axes forwhich the equation x2 - y2=a2 willreducetoxy =c2.Determine c2. (x)Find the equations of the following when ax+by+c =0 and bx-ay+d = o areconsidered asaxesof xandyrespectively. (ax + by + c)2= a2 + b2 and(ax + by + c)(bx- ay + d)= a2 + b2. (xi)Showthatif theoriginistransferredto(0,1)andtheaxesarerotated through 45,theequation5x2 - 2xy + 5y2+ 2x - 10y- 7= 0referred x,2y,2 tonewaxesbecomes3+ ""2=1. (xii)Pisthepoint(2,4).If theaxesareturnedthroughanangleof30 keeping the position of the origin unchanged.Find the newcoordinates of P . SOLUTION: (i)For rotation x=x' cosO-y' sinO=x' cos 45-y' sin 45=x' /V2-y' /V2 andy= x' sinO - y' cos 0= x' /V2 + y' /V2. The transformed equation is (X'- y')2 _(X' + y')2= 4 V2vI2 or,-4x'y' =8or,:r'y' + 2 =O. CH.4:GEOMETRY4.3 (ii)Given equation isx2 + y2- 2x +14y + 20= 0 or,(x _1)2 +(y +7)2= 30. Substitutex-I = x'andy+7 = y'wegetx,2+y,2= 30 . . '.thetranslationisx'= x-I, y'= y +7. (iii)The angleof rotation e isgivenby 2h tan2e =--b wherea= 17,b = -7,2h = 18. a-.1831-13 tan 2e=17 _(-7)=4"or,e ="2tan4"' (iv)Let(x, y)and(x', y')be respectivelytheoldandnewcoordinatesafter transformation . . '.x= x' cose - y' sine andy= x' sine +y' cose. Herex' =2V3, y'=-6 ande =60. Then . '.the original coordinatesare(4V3, 0). (v)Lete = tan-12 or,tane = 2. ,.n2{)1 Slnu= J5' cos u= J5' ,,.x'2y' Then x= xcos e - y sm e =- - -J5J5' 2x'y' y= x' sin 0 +y' cos 0 =J5 +J5' Substituting the valuesof xandyto the equation 4xy - 3x2 =1 weget ~ ( x '- 2y'). (2x' + y')- ~ ( x '- 2y')2=1 J5J55 or,4(2x,2- 2y,2- 3x'y') - 3(x,2- 4x'y' + 4y'2)=5 or,5x,2- 20y,2=5 or,x,2- 4y,2=1. (vi)Putting x= x' +0'andy= y' +(3.Then 8(x' +0')2+ lO(x' +O')(y' + (3)+3(y' +(3)2 +26(x' +0') + 16(y' + (3) +21 = 0 4.4V.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) ( vii) or,8x/2 + 10x'y' + 3y/2 + x'(16a + 10/3 + 26)+ y'(10a + 6/3 + 16) + 8a2 + 10a/3 + 3/32+ 26a + 16/3 + 21=O. Toremovethe firstdegreeterms,putting 16a + 10/3 + 26=0 and lOa + 6/3+ 16=O. Solving wegeta= -1, /3= -1. . '.the required pointis(-1, -1). L 12x - 5y + 3912x - 5y + 39d = J144 + 2513 _5x + 12y - 26_5x + 12y - 26 Y- - . J25 +14413 The givenequation reduces to 3(13X)2- 2(13y)2=169 or,3X2 - 2y2 = 1,whereX= 0and Y= 0arenewaxes. (viii)If (Xl,yl)bethenewcoordinatesofapoint(x, y)thenx=Xlcos e-yl sine,y=Xlsine + yl cose. Heretheangleof rotatione =-11'/2. Thenx= Xlcos( -11' /2)- yl sin( -11' /2)=yland y=Xlsin( -11' /2) + yl cos( -11' /2)=_Xl. The equation reducestoy/2- x/2=4. (ix)Putting x= Xlcos e - yl eand y= Xlsin e + yl cos e,weget(Xlcos e -yl sine)2 - (Xlsine + yl cose)2=a2 or,X/2( cos2 e - sin2 e) + y/2(sin2 e- cos2 e) + xlyl (-4 sin e cos e)=a2. Toreduce to the formxy = c2, putting cos2 e - sin2 e= 0 or,tan2 e=1 or,tane =1 or,()=11'/4,whichistheangleof rotation. The equation reducesto=r=4x'y'(sin1l'/4.cos1l'/4)=a2 II112II2/ or,=r=4xy. J2' J2 =aor,xy=a2. Obviouslyc2 =a2/2. (x)Let>X= ax + by + cand Y= bx- ay + d. Y The given equations reduce to (a2 + b2)X2=a2 + b2 and (a2 + b2)XY = a2 + b2,or,X2 =1andXY=1whereX=0andY=0arethe equations of the newaxes. CH.4:GEOMETRY4.5 (xi)If theoriginistransferredto(a,j3)i.e.,(0,1)andtheaxesarerotated through 45then x=a+ x' cos 0 - y' sin 0=0 + x' cos 45- y' sin 45 = v'2- v'2andy= 13+ x' sin 0 + y' cos 0 = 1 + v'2 + v'2 Putting thesevaluestothe givenequation weobtain 5x'y' or,2"(x'- y')2- 2() - (x'- y')(x' + y')+ 5 + + 5 +_(X'2 + 2x'y' + y'2)- 12=0 2 X,2y,2 or4x,2+ 6y,2= 12or- + - = 1 ,,32. (xii)Let(x', y')bethe newcoordinates of the point(x, y). Then x=x' cos 0 - y' sinO,y =x' sinO + y' cos O. Herex= 2,y= 4 and 0 = 30. .,V3,1,I,V3 2 =x. 2"" - y. 2"and 4 =x. 2"+ y. 2"" or,V3x'- y'=4andx' + V3y'=S. x'y'1 or==--,S + 4V3-4 + SV33 + 1 4.6U.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) ,8 + 4V3In,-4 + 8V3In orx== 2 + y3y==-1 + 2y 3 ,4'4. . '.the newcoordinatesare(2+ V3, -1 + 2V3). Exercise1(i)Findtheangleofrotationofthecoordinatesaxesabout originwhichwilltransform the equationx2 - y2= 4toxV = 2. (ii)Find thecoordinatesof the point(-2,4)referredtonewaxesobtained by rotating the old axes through an angle of 45in the positive direction. (iii)Find thetransformedequation of theliney=V3x,whentheaxesare rotated through an anglei in thepositive sense. (iv)Find aneworiginwithoutrotatingtheaxesin orderthattheequation y2+ 2y - 8x + 25= 0will turn outtobe y2=4ax.Determine a. [Ans.(i)37rj4,(ii)(V2, 3V2),(iii)y'= 0,(iv)(3,-1),a= 2.] Ex.3(i)If theexpressionax + bychangestoa' x' + b' y'byarotationof the rectangularaxesaboutthe origin,provethata2 + b2 =a,2+ b'2. (ii)By shifting the origin to the point (a, /3)without changing the directions of axes,each of the equation x - y + 3 = 0 and 2x - y + 1 = 0 isreduced tothe formax' + by'= O.Find aand /3 . SOLUTION: (i)Letthe transformation be x= x' cos B - y'sinB, y=x' sinB + y' cos B. Then ax + by= a( x' cos 8 - y' sin 8) + b( x' sin 8 + y' cos 8) =x' (a cos 8 + b sin B)+ y' ( -a sin B + b cos B) =a'x' + b'y' wherea'=acosB + bsinBand b'=-asinB + bcosB. Now,a,2+ b,2=(a cos B + bsinB)2 + (-asinB + bcosB)2 =a2 (cos2 B + sin2 B)+ b2 (cos2 B + sin2 B)+ 0 =a2 + b2 (ii)Putting x= x' + a,y= y' + /3.Then the given equation becomes (x' + a) - (y'+ /3)+ 3 =0and2(x' + a) - (y' + /3)+ 1 =0 or,x' - y' + (a - /3+ 3)=0and 2x' - y' + (2a - /3+ 1)=O. Accordingtotheproblem a- /3+ 3 =0and2a - /3+ 1 =O. Solving wegeta=2, /3= 5. eR.4:GEOMETRY4.7 Ex.4Show that the equation of the circlex2 +y2=a2 isinvariantunder the rotation of axes . SOLUTION:Putting x=x' cos 0 - y' sinO,y=x' sinO +y' cos O. Then x2 +y2=a2 becomes (x' cos 0 - y' sinO)2+(x' sinO + y' cos 0)2=a2 or,x'2(cos2 0 + sin2 0)+y'2(sin2 0 +cos2 0) +2x'y'( - sinO. cos 0 +cos O.sinO)=a2 or,x,2+y,2=a2. Hencethe givenequation isinv>triantw.r.t.rotation of axes. 4.2Pairof StraightLines Someusefulformulaeon pair of straightlines: (i)A second degree homogeneous equation of the formax2 + 2hxy + by2= 0 representsapair of straightlinesthrough the originiff h2 ~abo (ii)The anglebetween the linesax2 +2hxy +by2= 0are 0= tan-1 (2):: ~ab). (iii)Condition of coincidence:If 0 = 0then thelinesarecoincidentand henceh2 =abo (iv)Conditionofperpendicularity:If 0=rr/2thena+ b=0i.e., coefficientof x2+coefficientof y2= O. (v)The equation of bisectorsof the anglesbetween the linesax2 +2hxy + by2=0are (vi)The equation ax2 +2hxy + by2+ 2gx +2fy +c = 0representsapair of straightlinesif ah9 hbf=O. 9fc Ex.1Find the equation of the pair of straightlines joining theorigin to the points of intersection of thelinex+ 2y= 5andthe parabola y2= 8x. SOLUTION:Sincethepairof linespassingthroughtheoriginandthein-tersectionofx+ 2y=5andy2= 8x,soitisahomogeneousequationof .2x+ 2y xandy.Tomakehomogeneouswewntey=8x =1.8x =--5-.8x or, 5y2=8x2 + 16xy,which isthe requiredpair of the straightlines. 4.8V.G.MATHEMATICS(SHORTQUESTIONSANDANSWERS) Ex.2(i)Find the value of kfor which the equation x2_y2_2x+4y+k =0 mayrepresentapair of lines. (ii)Forwhatvalueof A,doesthe equationxy + 5x + AY + 15=0represent apair of straightlines? (iii)Show that 6x2 - 5xy - 6y2 + 14x + 5y + 4 =0 represents a pair of straight lineswhichareperpendicular toeach other. (iv)Show that the equation 4x2 + 12xy + 9y2 -16x - 24y + 16= 0 represents apair of coincidentstraightlines. (v)Show that the equation x2 + 2V3xy + 3y2 - 3x - 3V3y - 4 =0 represents apair of parallel lines,byfactorisingthelefthand side. (vi)Testwhether the equationx2 + 6xy + 9y2- 5x -15y + 6 =0represents apair of parallel straightlines. (vii)ForwhatvalueofA,3x2 + AXY- 5y2+ 2x + 2y=0willbeapairof straightlines? (viii)Showthat the equation x2 + 6xy + 9y2 + 4x + 12y - 5 =0 representpair of parallel linesand findthe distancebetween them. (ix)Findtheanglebetweenthepairofstraightlinesrepresentedbythe equation 3x2 - 10xy + 3y2= O. (x)Prove that the equation y3- x 3+3xy(y-x) =0 represent"s three straight lines. (xi)Find the equation of thelineswhich passthrough the origin and whose distance from(h, k)are equal to d. SOLUTION: (i)Comparing the given equation with ax2 + 2hxy + by2 + 29X + 21Y + c =0 wehavea= l,h = O,b= -1,9 = 1,1 = 2,c = k. If the curverepresentsapair of straightlinesthen ah 9 101 hb I 0-12=0 9I c12k or,1( -k - 4)+ 1(0 + 1)= 0 or,-k - 3 = 0or,k=-3. (ii)Herea= 0, h= 1/2, b = 0,9 = 5/2, 1= A/2andc = 15. If thegivencurverepresentsapair of straightlinesthen o1/25/2 1/20A/2=0 5/2A/215 eH.4:GEOMET'lY or_!(15_5),)+ ~ ( ~ - O )=Oor5),=15or).=3. '22424'44' (iii)Herea=6, h=-5/2, b =-6, 9=7, f=5/2 and c =4. ah96-5/27 Now,hbf=-5/2-65/2= O. 9fc75/24 Hencethe givencurverepres