ContentsPART IForewordvPreface vii1. RelationsandFunctions 11.1 Introduction 11.2 Types of Relations 21.3 Types of Functions 71.4 Composition of Functions and Invertible Function 121.5 Binary Operations 192. InverseTrigonometricFunctions 332.1 Introduction 332.2 BasicConcepts 332.3 Properties of Inverse Trigonometric Functions 423. Matrices 563.1 Introduction 563.2 Matrix 563.3 Types of Matrices 613.4 Operations on Matrices 653.5 Transpose of a Matrix 833.6 Symmetric and Skew Symmetric Matrices 853.7 Elementary Operation (Transformation) of a Matrix 903.8 Invertible Matrices 914. Determinants 1034.1 Introduction 1034.2 Determinant 1034.3 Properties of Determinants 1094.4 Area of a Triangle 1214.5 Minors and Cofactors 1234.6 Adjoint and Inverse of a Matrix 1264.7 Applications of Determinants and Matrices 1335. Continuity and Differentiability 1475.1 Introduction 1475.2 Continuity 1475.3 Differentiability 1615.4 Exponential and Logarithmic Functions 1705.5 Logarithmic Differentiation 1745.6 Derivatives of Functions in Parametric Forms 1795.7 Second Order Derivative 1815.8 Mean Value Theorem 1846. ApplicationofDerivatives 1946.1 Introduction 1946.2 Rate of Change of Quantities 1946.3 Increasing and Decreasing Functions 1996.4 Tangents and Normals 2066.5 Approximations 2136.6 Maxima and Minima 216Appendix1:ProofsinMathematics 247A.1.1Introduction 247A.1.2What is a Proof? 247Appendix2:MathematicalModelling 256A.2.1Introduction 256A.2.2Why Mathematical Modelling? 256A.2.3Principles of Mathematical Modelling 257Answers 268xivThereisnopermanentplaceintheworldforuglymathematics....Itmaybeveryhardtodefinemathematicalbeautybutthatisjustastrueofbeautyofanykind,wemaynotknowquitewhatwemeanbyabeautifulpoem,butthatdoesnotpreventusfromrecognisingonewhenwereadit.G.H.HARDY1.1IntroductionRecall that the notion of relations and functions, domain,co-domain and range have been introduced in Class XIalong with different types of specific real valued functionsandtheirgraphs.Theconceptofthetermrelationinmathematics has been drawn from the meaning of relationin English language, according to which two objects orquantities are related if there is a recognisable connectionor link between the two objects or quantities. LetA bethe set of students of Class XII of a school and B be theset of students of Class XI of the same school. Then someof the examples of relations from A to B are(i) {(a, b) A B: a is brother of b},(ii) {(a, b) A B: a is sister of b},(iii) {(a, b) A B: age of a is greater than age of b},(iv) {(a, b) A B: total marks obtained by a in the final examination is less thanthe total marks obtained by b in the final examination},(v) {(a, b) A B: a lives in the same locality as b}.However, abstracting fromthis, we define mathematically a relation R from A to B as an arbitrary subsetof A B.If (a,b) R, we say that a is related to b under the relation R and we write asaRb.Ingeneral,(a,b)R,wedonotbotherwhetherthereisarecognisableconnection or link between a and b. As seen in Class XI, functions are special kind ofrelations.In this chapter, we will study different types of relations and functions, compositionof functions, invertible functions and binary operations.Chapter 1RELATIONS AND FUNCTIONSLejeuneDirichlet (1805-1859)MATHEMATICS 21.2Types of RelationsIn this section, we would like to study different types of relations. We know that arelationinasetAisasubsetofAA.Thus,theemptysetandAAare twoextreme relations. For illustration, consider a relation R in the set A = {1, 2, 3, 4} given byR = {(a, b): a b = 10}. This is the empty set, as no pair (a, b) satisfies the conditiona b = 10. Similarly, R = {(a, b) : | a b | 0} is the whole set A A, as all pairs(a,b)in AAsatisfy|ab|0.Thesetwoextremeexamplesleadustothefollowing definitions.Definition1 ArelationRinaset Aiscalledemptyrelation,ifnoelementof Aisrelated to any element of A, i.e., R = A A.Definition 2A relation R in a set A is called universal relation, if each element of Ais related to every element of A, i.e., R = A A.Both the empty relation and the universal relation are some times called trivialrelations.Example 1 Let A be the set of all students of a boys school. Show that the relation Rin A given by R = {(a, b) : a is sister of b} is the empty relation and R = {(a, b) : thedifference between heights of a and b is less than 3 meters} is the universal relation.Solution Since the school is boys school, no student of the school can be sister of anystudent of the school. Hence, R = , showing that R is the empty relation. It is alsoobvious that the difference between heights of any two students of the school has to beless than 3 meters. This shows that R = A A is the universal relation.RemarkInClassXI,wehaveseentwowaysofrepresentingarelation,namelyroastermethodandsetbuildermethod.However,arelationRintheset{1,2,3,4}definedbyR={(a,b):b=a+1}isalsoexpressedasaRbifandonlyifb = a + 1 by many authors. We may also use this notation, as and when convenient.If (a, b) R, we say that a is related to b and we denote it as a R b.One of the most important relation, which plays a significant role in Mathematics,isanequivalencerelation.Tostudyequivalencerelation,wefirstconsiderthreetypes of relations, namely reflexive, symmetric and transitive.Definition 3 A relation R in a set A is called(i) reflexive, if (a, a) R, for every a A,(ii) symmetric, if (a1, a2) R implies that (a2, a1) R, for all a1, a2 A.(iii) transitive, if (a1, a2) R and (a2, a3) R implies that (a1, a3) R, for all a1, a2,a3 A.RELATIONS ANDFUNCTIONS 3Definition4ArelationRinaset AissaidtobeanequivalencerelationifRisreflexive, symmetric and transitive.Example 2 Let T be the set of all triangles in a plane with R a relation in T given byR = {(T1, T2) : T1 is congruent to T2}. Show that R is an equivalence relation.SolutionRisreflexive,sinceeverytriangleiscongruenttoitself.Further,(T1, T2) R T1 is congruent to T2 T2 is congruent to T1 (T2, T1) R. Hence,R is symmetric. Moreover, (T1, T2), (T2, T3) R T1 iscongruent to T2 and T2 iscongruent to T3 T1 is congruent to T3 (T1, T3) R. Therefore, R is an equivalencerelation.Example 3 Let L be the set of all lines in a plane and R be the relation in L defined asR={(L1,L2):L1isperpendiculartoL2}.ShowthatRissymmetricbutneitherreflexive nor transitive.Solution R is not reflexive, as a line L1 can not be perpendicular to itself, i.e., (L1, L1) R. R is symmetric as (L1, L2) R L1 is perpendicular to L2 L2 is perpendicular to L1 (L2, L1) R.R is not transitive. Indeed, if L1 is perpendicular to L2 andL2 is perpendicular to L3, then L1 can never be perpendicular toL3. In fact, L1 is parallel to L3, i.e., (L1, L2) R, (L2, L3) R but (L1, L3) R.Example 4 Show that the relation R in the set {1, 2, 3} given by R = {(1, 1), (2, 2),(3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor transitive.Solution R is reflexive, since (1, 1), (2, 2) and (3, 3) lie in R. Also, R is not symmetric,as (1, 2) R but (2, 1) R. Similarly, R is not transitive, as (1, 2) R and (2, 3) Rbut (1, 3) R.Example 5 Show that the relation R in the set Z of integers given byR = {(a, b) : 2 divides a b}is an equivalence relation.Solution R is reflexive, as 2 divides (a a) for all a Z. Further, if (a, b) R, then2 divides a b. Therefore, 2 divides b a. Hence, (b, a) R, which shows that R issymmetric. Similarly, if (a, b) R and (b, c) R, then a b and b c are divisible by2. Now, a c = (a b) + (b c) is even (Why?). So, (a c) is divisible by 2. Thisshows that R is transitive. Thus, R is an equivalence relation in Z.Fig 1.1MATHEMATICS 4In Example 5, note that all even integers are related to zero, as (0, 2), (0, 4)etc., lie in R and no odd integer is related to 0, as (0, 1), (0, 3) etc., do not lie in R.Similarly,alloddintegersarerelatedtooneandnoevenintegerisrelatedtoone.Therefore, the set E of all even integers and the set O of all odd integers are subsets ofZ satisfying following conditions:(i) All elements of E are related to each other and all elements of O are related toeachother.(ii) No element of E is related to any element of O and vice-versa.(iii) E and O are disjoint and Z = E O.The subset E is called the equivalence class containing zero and is denoted by[0]. Similarly, O is the equivalence class containing 1 and is denoted by [1]. Note that[0] [1], [0] = [2r] and [1] = [2r + 1], r Z. Infact, what we have seen above is trueforanarbitraryequivalencerelationRinasetX.Givenanarbitraryequivalencerelation R in an arbitrary set X, R divides X into mutually disjoint subsets Ai calledpartitions or subdivisionsof X satisfying:(i) all elements of Ai are related to each other, for all i.(ii) no element of Ai is related to any element of Aj, i j.(iii) Aj = X and Ai Aj = , i j.The subsets Ai are called equivalence classes. The interesting part of the situationis that we can go reverse also. For example, consider a subdivision of the set Z givenby three mutually disjoint subsets A1, A2 and A3 whose union is Z withA1 = {x Z : x is a multiple of 3} = {..., 6, 3, 0, 3, 6, ...}A2 = {x Z : x 1 is a multiple of 3} = {..., 5, 2, 1, 4, 7, ...}A3 = {x Z : x 2 is a multiple of 3} = {..., 4, 1, 2, 5, 8, ...}Define a relation R in Z given by R = {(a,b) : 3 divides a b}. Following thearguments similar to those used in Example 5, we can show that R is an equivalencerelation. Also, A1 coincides with the set of all integers in Z which are related to zero, A2coincides with the set of all integers which are related to 1 and A3 coincides with theset of all integers in Z which are related to 2. Thus, A1 = [0], A2 = [1] and A3 = [2].In fact, A1 = [3r], A2 = [3r + 1] and A3 = [3r + 2], for all r Z.Example6LetRbetherelationdefinedinthesetA={1,2,3,4,5,6,7}byR={(a, b):bothaandbareeitheroddoreven}.ShowthatRisanequivalencerelation. Further, show that all the elements of the subset {1, 3, 5, 7} are related to eachotherandalltheelementsofthesubset{2,4,6}arerelatedtoeachother,butnoelement of the subset {1, 3, 5, 7} is related to any element of the subset {2, 4, 6}.RELATIONS ANDFUNCTIONS 5SolutionGivenanyelementainA,bothaandamustbeeitheroddoreven,sothat(a,a)R.Further,(a,b)Rbothaandbmustbeeitheroddoreven (b, a) R. Similarly, (a, b) R and (b, c) R all elements a, b, c, must beeither even or odd simultaneously (a, c) R. Hence, R is an equivalence relation.Further, all the elements of {1, 3, 5, 7} are related to each other, as all the elementsof this subset are odd. Similarly, all the elements of the subset {2, 4, 6} are related toeach other, as all of them are even. Also, no element of the subset {1, 3, 5, 7} can berelated to any element of {2, 4, 6}, as elements of {1, 3, 5, 7} are odd, while elementsof {2, 4, 6} are even.EXERCISE 1.11. Determine whether each of the following relations are reflexive, symmetric andtransitive:(i) Relation R in the set A = {1, 2, 3, ..., 13, 14} defined asR = {(x, y) : 3x y = 0}(ii) Relation R in the set N of natural numbers defined asR = {(x, y) : y = x + 5 and x < 4}(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} asR = {(x, y) : y is divisible by x}(iv) Relation R in the set Z of all integers defined asR = {(x, y) : x y is an integer}(v) Relation R in the set A of human beings in a town at a particular time given by(a)R = {(x, y) : x and y work at the same place}(b)R = {(x, y) : x and y live in the same locality}(c)R = {(x, y) : x is exactly 7 cm taller than y}(d)R = {(x, y) : x is wife of y}(e)R = {(x, y) : x is father of y}2. Show that the relation R in the set R of real numbers, defined asR = {(a, b) : a b2} is neither reflexive nor symmetric nor transitive.3. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} asR = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.4. Show that the relation R in R defined as R = {(a, b) : a b}, is reflexive andtransitive but not symmetric.5. Check whether the relation R in R defined by R = {(a, b) : a b3} is reflexive,symmetric or transitive.MATHEMATICS 66. Show that the relation R inthe set {1, 2, 3} given by R = {(1, 2), (2, 1)} issymmetric but neither reflexive nor transitive.7. Show that the relation R in the set A of all the books in a library of a college,given by R = {(x, y) : x and y have same number of pages} is an equivalencerelation.8. Show that the relation R in the set A = {1, 2, 3, 4, 5} given byR={(a,b):|ab|iseven},isanequivalencerelation.Showthatalltheelements of {1, 3, 5} are related to each other and all the elements of {2, 4} arerelated to each other. Butno element of {1, 3, 5} is related to any element of {2, 4}.9. Show that each of the relation R in the set A = {x Z : 0 x 12}, given by(i) R = {(a, b) : |a b| is a multiple of 4}(ii) R = {(a, b) : a = b}is an equivalence relation. Find the set of all elements related to 1 in each case.10. Give an example of a relation. Which is(i) Symmetric but neither reflexive nor transitive.(ii) Transitive but neither reflexive nor symmetric.(iii) Reflexive and symmetric but not transitive.(iv) Reflexive and transitive but not symmetric.(v) Symmetric and transitive but not reflexive.11. ShowthattherelationRinthesetAofpointsinaplanegivenbyR = {(P, Q) : distance of the point P from the origin is same as the distance of thepoint Q from the origin}, is an equivalence relation. Further, show that the set ofall points related to a point P (0, 0) is the circle passing through P with origin ascentre.12. Show that the relation R defined in the set A of all triangles as R = {(T1, T2) : T1is similar to T2}, is equivalence relation. Consider three right angle triangles T1with sides 3, 4, 5,T2 with sides 5, 12, 13 and T3 with sides 6, 8, 10. Whichtriangles among T1, T2 and T3 are related?13. Show that the relation R defined in the set A of all polygons as R = {(P1, P2) :P1 and P2 have same number of sides}, is an equivalence relation. What is theset of all elements in A related to the right angle triangle T with sides 3, 4 and 5?14. Let L be the set of all lines in XY plane and R be the relation in L defined asR = {(L1, L2) : L1 is parallel to L2}. Show that R is an equivalence relation. Findthe set of all lines related to the line y = 2x + 4.RELATIONS ANDFUNCTIONS 715. Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4,4),(1, 3), (3, 3), (3, 2)}. Choose the correct answer.(A) R is reflexive and symmetric but not transitive.(B) R is reflexive and transitive but not symmetric.(C) R is symmetric and transitive but not reflexive.(D) R is an equivalence relation.16. Let R be the relation in the set N given by R = {(a, b) : a = b 2, b > 6}. Choosethecorrectanswer.(A) (2, 4) R (B) (3, 8) R (C) (6, 8) R (D) (8, 7) R1.3Types of FunctionsThe notion of a function along with some special functions like identity function, constantfunction, polynomial function, rational function, modulus function, signum function etc.along with their graphs have been given in Class XI.Addition, subtraction, multiplication and division of two functions have also beenstudied. As the concept of function is of paramount importance in mathematics andamong otherdisciplines as well, we would like to extend our study about function fromwhere we finished earlier. In this section, we would like to study different types offunctions.Consider the functionsf1, f2, f3 and f4 given by the following diagrams.In Fig 1.2, we observe that the images of distinct elements of X1 under the functionf1 are distinct, but the image of two distinct elements 1 and 2 of X1 under f2 is same,namely b. Further, there are some elements like e and f in X2 which are not images ofany element of X1 under f1, while all elements of X3 are images of some elements of X1under f3. The above observations lead to the following definitions:Definition 5 A function f : X Y is defined to be one-one (or injective), if the imagesofdistinctelementsofXunderfaredistinct,i.e.,foreveryx1,x2X,f (x1)=f (x2)implies x1 = x2. Otherwise, f is called many-one.The function f1 and f4 in Fig 1.2 (i) and (iv) are one-one and the function f2 and f3in Fig 1.2 (ii) and (iii) are many-one.Definition 6 A function f : X Y is said to be onto (or surjective), if every elementof Y is the image of some element of X under f, i.e., for every y Y, there exists anelement x in X such that f (x) = y.The function f3 and f4 in Fig 1.2 (iii), (iv) are onto and the function f1 in Fig 1.2 (i) isnot onto as elements e, f in X2 are not the image of any element in X1 under f1.MATHEMATICS 8Remarkf : X Y is onto if and only if Range of f = Y.Definition 7 A function f : X Y is said to be one-one and onto (or bijective), if f isboth one-one and onto.The function f4 in Fig 1.2 (iv) is one-one and onto.Example 7 Let A be the set of all 50 students of Class X in a school. Let f : A N befunctiondefinedbyf (x)=rollnumberofthestudentx.Showthatfisone-onebut not onto.Solution No two different students of the class can have same roll number. Therefore,f must be one-one. We can assume without any loss of generality that roll numbers ofstudents are from 1 to 50. This implies that 51 in N is not roll number of any student ofthe class, so that 51 can not be image of any element of X under f. Hence,f is not onto.Example 8 Show that the function f : N N, given by f (x) = 2x, is one-one but notonto.Solution The function f is one-one, for f (x1) = f (x2) 2x1 = 2x2 x1 = x2. Further,f is not onto, as for 1 N, there does not exist any x in N such that f (x) = 2x = 1.Fig 1.2 (i) to (iv)RELATIONS ANDFUNCTIONS 9Example 9Prove that the function f : R R, given by f (x) = 2x, is one-one and onto.Solutionf is one-one, as f (x1) = f (x2) 2x1 = 2x2 x1 = x2. Also, given any realnumber y in R, there exists 2y in R such that f (2y) = 2 . (2y) = y. Hence, f is onto.Fig 1.3Example 10Show that the function f : N N, given by f (1) = f (2) = 1 and f (x) = x 1,for every x > 2, is onto but not one-one.Solutionf is not one-one, as f (1) = f (2) = 1. But f is onto, as given any y N, y 1,wecanchoosexasy+1suchthatf (y+1)=y+11=y. Alsofor1N,wehave f (1) = 1.Example 11Show that the function f : R R,defined as f (x) = x2, is neither one-one nor onto.SolutionSince f ( 1) = 1 = f (1), f is not one-one. Also, the element 2 in the co-domain R isnotimageofanyelementxinthedomainR(Why?). Therefore f is not onto.Example 12Show that f : N N, given by1, if is odd,( )1, if is evenx xf xx x+ = is both one-one and onto.Fig 1.4MATHEMATICS 10SolutionSuppose f (x1) = f (x2). Note that if x1 is odd and x2 is even, then we will havex1 + 1 = x2 1, i.e., x2 x1 = 2 which is impossible. Similarly, the possibility of x1 beingeven and x2 being odd can also be ruled out, using the similar argument. Therefore,bothx1andx2mustbeeitheroddoreven.Supposebothx1andx2areodd.Thenf (x1) = f (x2) x1 + 1 = x2 + 1 x1 = x2. Similarly, if both x1 and x2 are even, then alsof (x1) = f (x2) x1 1 = x2 1 x1 = x2. Thus, f is one-one. Also, any odd number2r + 1 in the co-domain N is the image of 2r + 2 in the domain N and any even number2r in the co-domain N is the image of 2r 1 in the domain N. Thus, f is onto.Example 13 Show that an onto function f : {1, 2, 3} {1, 2, 3} is always one-one.Solution Suppose f is not one-one. Then there exists two elements, say 1 and 2 in thedomain whose image in the co-domain is same. Also, the image of 3 under f can beonly one element. Therefore, the range set can have at the most two elements of theco-domain {1, 2, 3}, showing that fis not onto, a contradiction. Hence, f must be one-one.Example 14 Show that a one-one function f : {1, 2, 3} {1, 2, 3} must be onto.Solution Since f is one-one, three elements of {1, 2, 3} must be taken to 3 differentelements of the co-domain {1, 2, 3} under f. Hence, f has to be onto.Remark The results mentioned in Examples 13 and 14 are also true for an arbitraryfinite set X, i.e., a one-one function f : X X is necessarily onto and an onto mapf : X X is necessarily one-one, for every finite set X. In contrast to this, Examples 8and 10 show that for an infinite set, this may not be true. In fact, this is a characteristicdifference between a finite and an infinite set.EXERCISE 1.21. Show that the function f : R R defined by f (x) = 1x is one-one and onto,where R is the set of all non-zero real numbers. Is the result true, if the domainR is replaced by N with co-domain being same as R ?2. Check the injectivity and surjectivity of the following functions:(i) f : N N given by f (x) = x2(ii) f : Z Z given by f (x) = x2(iii) f : R R given by f (x) = x2(iv) f : N N given by f (x) = x3(v) f : Z Z given by f (x) = x33. Prove that the Greatest Integer Function f : R R, given by f (x) = [x], is neitherone-one nor onto, where [x] denotes the greatest integer less than or equal to x.RELATIONS ANDFUNCTIONS 114. Show that the Modulus Function f : R R, given by f (x) = | x |, is neither one-one nor onto, where | x | is x, if x is positive or 0 and | x | is x, if x is negative.5. Show that the Signum Function f : R R, given by1, if 0( ) 0, if 01, if 0xf x xx> = = = =8. Given a non empty set X, consider P(X) which is the set of all subsets of X.MATHEMATICS 30Define the relation R in P(X) as follows:For subsets A, B in P(X), ARB if and only if A B. Is R an equivalence relationon P(X)? Justify your answer.9. Given a non-empty set X, consider the binary operation : P(X) P(X) P(X)given by A B = A BA, B in P(X), where P(X) is the power set of X.Show that X is the identity element for this operation and X is the only invertibleelement in P(X) with respect to the operation .10. Find the number of all onto functions from the set {1, 2, 3, ... , n} to itself.11. Let S = {a, b, c} and T = {1, 2, 3}. Find F1 of the following functions F from Sto T, if it exists.(i) F = {(a, 3), (b, 2), (c, 1)} (ii) F = {(a, 2), (b, 1), (c, 1)}12. Consider the binary operations : R R R and o : R R R defined asa b = |a b| and a o b = a,a, b R. Show that is commutative but notassociative, o is associative but not commutative. Further, show thata, b, c R,a (b o c) = (a b) o (a b). [If it is so, we say that the operation distributesover the operation o]. Does o distribute over ? Justify your answer.13. Givenanon-emptysetX,let:P(X)P(X)P(X)bedefinedasA * B = (A B) (B A), A, B P(X). Show that the empty set is theidentity for the operation and all the elements A of P(X) are invertible withA1 = A. (Hint : (A ) ( A) = A and (A A) (A A) = A A = ).14. Define a binary operation on the set {0, 1, 2, 3, 4, 5} as,if66 if6a b a ba ba b a b+ + < = + + Show that zero is the identity for this operation and each element a of the set isinvertible with 6 a being the inverse of a.15. Let A = { 1, 0, 1, 2}, B = { 4, 2, 0, 2} and f, g : A B be functions definedbyf (x)=x2x,xAand 1( ) 2 1,2g x x = xA. Arefandgequal?Justifyyouranswer.(Hint:Onemaynotethattwofunctionsf: ABandg : A B such that f (a) = g(a) a A, are called equal functions).16. Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which arereflexive and symmetric but not transitive is(A) 1 (B) 2 (C) 3 (D) 417. Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is(A) 1 (B) 2 (C) 3 (D) 4RELATIONS ANDFUNCTIONS 3118. Let f : R R be the Signum Function defined as1, 0( ) 0, 01, 0xf x xx> = = 1Solution Wehave,1cos sintancos sina x b xb x a x + =1cos sincostancos sincosa x b xb xb x a xb x + = 1tantan1 tanaxbaxb + =1 1tan tan (tan )axb = 1tanaxb INVERSETRIGONOMETRICFUNCTIONS51Example 13 Solve tan1 2x + tan1 3x = 4Solution We have tan1 2x + tan1 3x = 4or12 3tan1 2 3x xx x + =4i.e.125tan1 6xx =4Therefore251 6xx =tan 14 =or 6x2 + 5x 1 = 0 i.e., (6x 1) (x + 1) = 0which gives x =16 or x = 1.Since x = 1 does not satisfy the equation, as the L.H.S. of the equation becomesnegative, 16x= is the only solution of the given equation.Miscellaneous Exercise on Chapter 2Find the value of the following:1.113cos cos6 2.17tan tan6 Provethat3.1 13 242sin tan5 7=4.1 1 18 3 77sin sin tan17 5 36+ =5.1 1 14 12 33cos cos cos5 13 65+ =6.1 1 112 3 56cos sin sin13 5 65+ =7.1 1 163 5 3tan sin cos16 13 5= +8.1 1 1 11 1 1 1tan tan tan tan5 7 3 8 4 + + + = 52 MATHEMATICSProvethat9.1 11 1tan cos2 1xxx = +, x [0, 1]10.11 sin 1 sincot2 1 sin 1 sinx x xx x + + = + , 0,4x 11.1 11 1 1tan cos4 2 1 1x xxx x + = + + , 112x [Hint: Put x = cos 2]12.1 19 9 1 9 2 2sin sin8 4 3 4 3 =Solve the following equations:13. 2tan1 (cos x)=tan1(2cosecx) 14.1 11 1tan tan ,( 0)1 2xx xx= >+15. sin (tan1 x), | x | < 1 is equal to(A)21xx (B)211 x (C)211 x +(D)21xx +16. sin1 (1 x) 2 sin1 x = 2, then x is equal to(A) 0, 12(B) 1, 12(C) 0 (D)1217.1 1tan tanx x yy x y + is equal to(A)2(B)3(C)4(D)34 INVERSETRIGONOMETRICFUNCTIONS53Summary

The domains and ranges (principal value branches) of inverse trigonometricfunctions are given in the following table:Functions Domain Range(PrincipalValueBranches)y = sin1 x [1, 1],2 2 y=cos1x [1, 1][0, ]y=cosec1x R (1,1) ,2 2 {0}y=sec1x R (1, 1) [0, ] { }2y = tan1 x R,2 2 y = cot1 x R (0, )

sin1xshouldnotbeconfusedwith(sin x)1.Infact(sinx)1= 1sin xandsimilarly for other trigonometric functions.

The value of an inverse trigonometric functions which lies in its principalvaluebranchiscalledtheprincipalvalueofthatinversetrigonometricfunctions.For suitable values of domain, we have

y = sin1 x x = sin y

x = sin y y = sin1 x

sin (sin1 x) = x

sin1 (sin x) = x

sin1 1x=cosec1 x

cos1 (x)=cos1 x

cos1 1x=sec1x

cot1(x)= cot1 x

tan1 1x=cot1x

sec1(x)=sec1 x 54 MATHEMATICS

sin1 (x) = sin1x

tan1(x)=tan1x

tan1 x + cot1 x = 2

cosec1(x)=cosec1x

sin1 x + cos1 x = 2

cosec1 x+sec1 x= 2

tan1x+tan1y=tan1 1x yxy+

2tan1x=tan1 221xx

tan1xtan1y=tan1 1x yxy+

2tan1 x = sin1 221xx += cos1 2211xx+HistoricalNoteThestudyoftrigonometrywasfirststartedinIndia.TheancientIndianMathematicians, Aryabhatta(476A.D.),Brahmagupta(598 A.D.),BhaskaraI(600 A.D.) and Bhaskara II (1114 A.D.) got important results of trigonometry. Allthis knowledge went from India to Arabia and then from there to Europe. TheGreekshadalsostartedthestudyoftrigonometrybuttheirapproachwassoclumsy that when the Indian approach became known, it was immediately adoptedthroughout the world.In India, the predecessor of the modern trigonometric functions, known asthe sine of an angle, and the introduction of the sine function represents one ofthemaincontributionofthesiddhantas(Sanskritastronomicalworks)tomathematics.Bhaskara I (about 600 A.D.) gave formulae to find the values of sine functionsforanglesmorethan90.AsixteenthcenturyMalayalamworkYuktibhasacontains a proof for the expansion of sin (A + B). Exact expression for sines orcosines of 18, 36, 54, 72, etc., were given by Bhaskara II.The symbols sin1 x, cos1 x, etc., for arc sin x, arc cos x, etc., were suggestedbytheastronomerSirJohnF.W.Hersehel(1813)ThenameofThales(about 600 B.C.) is invariably associated with height and distance problems. Heis credited with the determination of the height of a great pyramid in Egypt bymeasuring shadows of the pyramid and an auxiliary staff (or gnomon) of knownINVERSETRIGONOMETRICFUNCTIONS55height, and comparing the ratios:HShs= = tan (suns altitude)Thales is also said to have calculated the distance of a ship at sea throughthe proportionality of sides of similar triangles. Problems on height and distanceusing the similarity property are also found in ancient Indian works. 56 MATHEMATICS

Theessenceof Mathematicsliesinitsfreedom.CANTOR

3.1IntroductionThe knowledge of matrices is necessary in various branches of mathematics. Matricesare one of the most powerful tools in mathematics. This mathematical tool simplifiesour work to a great extent when compared with other straight forward methods. Theevolutionofconceptofmatricesistheresultofanattempttoobtaincompactandsimple methods of solving system of linear equations. Matrices are not only used as arepresentation of the coefficients in system of linear equations, but utility of matricesfar exceeds that use. Matrix notation and operations are used in electronic spreadsheetprograms for personal computer, which in turn is used in different areas of businessand science like budgeting, sales projection, cost estimation, analysing the results of anexperiment etc. Also, many physical operations such as magnification, rotation andreflection through a plane can be represented mathematically by matrices. Matricesare also used in cryptography. This mathematical tool is not only used in certain branchesof sciences, but also in genetics, economics, sociology, modern psychology and industrialmanagement.Inthischapter,weshallfinditinterestingtobecomeacquaintedwiththefundamentals of matrix and matrix algebra.3.2MatrixSuppose we wish to express the information that Radha has 15 notebooks. We mayexpress it as [15] with the understanding that the number inside [ ] is the number ofnotebooks that Radha has. Now, if we have to express that Radha has 15 notebooksand6pens. Wemayexpressitas[156]withtheunderstandingthatfirstnumberinside [ ] is the number of notebooks while the other one is the number of pens possessedby Radha. Let us now suppose that we wish to express the information of possessionChapter 3MATRICESMATRICES 57ofnotebooksandpensbyRadhaandhertwofriendsFauziaandSimranwhichis as follows:Radha has 15 notebooks and 6 pens,Fauzia has 10 notebooks and 2 pens,Simran has 13 notebooks and 5 pens.Now this could be arranged in the tabular form as follows:Notebooks PensRadha 15 6Fauzia 10 2Simran 13 5and this can be expressed asorRadha Fauzia SimranNotebooks 15 10 13Pens 6 2 5which can be expressed as:Inthefirstarrangementtheentriesinthefirstcolumnrepresentthenumberofnote books possessed by Radha, Fauzia and Simran, respectively and the entries in thesecond column represent the number of pens possessed by Radha, Fauzia and Simran,58 MATHEMATICSrespectively. Similarly, in the second arrangement, the entries in the first row representthe number of notebooks possessed by Radha, Fauzia and Simran, respectively. Theentries in the second row represent the number of pens possessed by Radha, FauziaandSimran,respectively. Anarrangementordisplayoftheabovekindiscalledamatrix. Formally, we define matrix as:Definition 1A matrix is an ordered rectangular array of numbers or functions. Thenumbers or functions are called the elements or the entries of the matrix.We denote matrices by capital letters. The following are some examples of matrices:5 2A 0 53 6 = , 12 32B 3.5 1 253 57i + = ,31 3Ccos tan sin 2x xx x x += + In the above examples, the horizontal lines of elements are said to constitute, rowsof the matrix and the vertical lines of elements are said to constitute, columns of thematrix. Thus A has 3 rows and 2 columns, B has 3 rows and 3 columns while C has 2rows and 3 columns.3.2.1OrderofamatrixA matrix having m rows and n columns is called a matrix of order m n or simply m nmatrix (read as an m by n matrix). So referring to the above examples of matrices, wehave A as 3 2 matrix, B as 3 3 matrix and C as 2 3 matrix. We observe that A has3 2 = 6 elements, B and C have 9 and 6 elements, respectively.In general, an m n matrix has the following rectangular array:or A = [aij]m n, 1 i m, 1 j n i, j NThus the ithrow consists of the elements ai1, ai2,ai3,..., ain, while the jth columnconsists of the elements a1j, a2j, a3j,..., amj,In general aij, is an element lying in the ith row and jth column. We can also callitasthe(i,j)thelementofA.Thenumberofelementsinanmnmatrixwillbeequal to mn.MATRICES 59$NoteIn this chapter1. We shall follow the notation, namely A = [aij]m n to indicate that A is a matrixof order m n.2. We shall consider only those matrices whose elements are real numbers orfunctions taking real values.We can also represent any point (x, y) in a plane by a matrix (column or row) asxy (or [x, y]). For example point P(0, 1) as a matrix representation may be given as0P1 = or [0 1].Observe that in this way we can also express the vertices of a closed rectilinearfigure in the form of a matrix. For example, consider a quadrilateral ABCD with verticesA (1, 0), B (3, 2), C (1, 3), D (1, 2).Now, quadrilateral ABCD in the matrix form, can be represented as2 4A B C D1 3 1 1X0 2 3 2 = or4 2A 1 0B 3 2YC 1 3D 1 2 = Thus, matrices can be used as representation of vertices of geometrical figures ina plane.Now, let us consider some examples.Example 1 Consider the following information regarding the number of men and womenworkers in three factories I, II and IIIMenworkers WomenworkersI 30 25II 25 31III 27 26Represent the above information in the form of a 3 2 matrix. What does the entryin the third row and second column represent?60 MATHEMATICSSolution The information is represented in the form of a 3 2 matrix as follows:30 25A 25 3127 26 = The entry in the third row and second column represents the number of womenworkers in factory III.Example 2 If a matrix has 8 elements, what are the possible orders it can have?Solution We know that if a matrix is of order m n, it has mn elements. Thus, to findall possible orders of a matrix with 8 elements, we will find all ordered pairs of naturalnumbers, whose product is 8.Thus, all possible ordered pairs are (1, 8), (8, 1), (4, 2), (2, 4)Hence, possible orders are 1 8, 8 1, 4 2, 2 4Example 3 Construct a 3 2 matrix whose elements are given by 1| 3 |2ija i j = .Solution In general a 3 2 matrix is given by 11 1221 2231 32Aa aa aa a = .Now1| 3 |2ija i j = , i = 1, 2, 3 andj = 1, 2.Therefore111|1 3 1| 12a = =121 5|1 3 2|2 2a = =211 1| 2 3 1|2 2a = =221| 2 3 2| 22a = =311| 3 3 1| 02a = =321 3| 3 3 2 |2 2a = =Hence the required matrix is given by 5121A 22302 = .MATRICES 613.3Types of Matrices In this section, we shall discuss different types of matrices.(i) Column matrixA matrix is said to be a column matrix if it has only one column.For example,03A 11/ 2 = is a column matrix of order 4 1.In general, A = [aij]m 1 is a column matrix of order m 1.(ii) Row matrixA matrix is said to be a row matrix if it has only one row.For example, 1 41B 5 2 32 = is a row matrix.In general, B = [bij]1 n is a row matrix of order 1 n.(iii) Square matrixA matrix in which the number of rows are equal to the number of columns, issaid to be a square matrix. Thus an m n matrix is said to be a square matrix ifm = n and is known as a square matrix of order n.For example 3 1 03A 3 2 124 3 1 = is a square matrix of order 3.In general, A = [aij]m m is a square matrix of order m.$NoteIf A = [aij] is a square matrix of order n, then elements (entries) a11, a22, ..., annare said to constitute the diagonal, of the matrix A. Thus, if 1 3 1A 2 4 13 5 6 = .Then the elements of the diagonal of A are 1, 4, 6.62 MATHEMATICS(iv) Diagonal matrixAsquarematrixB=[bij]mmissaidtobeadiagonalmatrixifallitsnondiagonal elements are zero, that is a matrix B = [bij]m m is said to be a diagonalmatrix if bij = 0, when i j.For example,A = [4], 1 0B0 2 = , 1.1 0 0C 0 2 00 0 3 = , are diagonal matricesof order 1, 2, 3, respectively.(v) Scalar matrixA diagonal matrix is said to be a scalar matrix if its diagonal elements are equal,that is, a square matrix B = [bij]n n is said to be a scalar matrix ifbij = 0,when i jbij=k,when i= j, for some constant k.For exampleA = [3], 1 0B0 1 = ,3 0 0C 0 3 00 0 3 = are scalar matrices of order 1, 2 and 3, respectively.(vi) IdentitymatrixA square matrix in which elements in the diagonal are all 1 and rest are all zerois called an identity matrix. In other words, the square matrix A = [aij]n n is anidentity matrix, if 1 if0 ifiji jai j= = .WedenotetheidentitymatrixofordernbyIn.Whenorderisclearfromthecontext, we simply write it as I.For example [1], 1 00 1 , 1 0 00 1 00 0 1 are identity matrices of order 1, 2 and 3,respectively.Observe that a scalar matrix is an identity matrix when k = 1. But every identitymatrix is clearly a scalar matrix.MATRICES 63(vii) Zero matrixA matrix is said to be zero matrix or null matrix if all its elements are zero.For example, [0], 0 00 0 , 0 0 00 0 0 , [0, 0] are all zero matrices. We denotezero matrix by O. Its order will be clear from the context.3.3.1EqualityofmatricesDefinition 2 Two matrices A = [aij] and B = [bij] are said to be equal if(i) they are of the same order(ii) each element of A is equal to the corresponding element of B, that is aij = bij forall i and j.For example, 2 3 2 3and0 1 0 1 are equal matrices but 3 2 2 3and0 1 0 1 arenot equal matrices. Symbolically, if two matrices A and B are equal, we write A = B.If 1.5 02 63 2x yz ab c = , then x = 1.5, y = 0, z = 2, a = 6, b = 3, c = 2Example 4 If3 4 2 7 0 6 3 26 1 0 6 3 2 23 21 0 2 4 21 0x z y ya cb b+ + = + + Find the values of a, b, c, x, y and z.Solution Asthegivenmatricesareequal,therefore,theircorrespondingelementsmust be equal. Comparing the corresponding elements, we getx + 3 = 0, z + 4 = 6, 2y 7 = 3y 2a 1 = 3, 0 = 2c + 2 b 3 = 2b + 4,Simplifying, we geta = 2, b = 7, c = 1, x = 3, y = 5, z = 2Example 5 Find the values of a, b, c, and d from the following equation:2 2 4 35 4 3 11 24a b a bc d c d+ = + 64 MATHEMATICSSolution By equality of two matrices, equating the corresponding elements, we get2a + b = 4 5c d = 11a 2b = 3 4c + 3d = 24Solving these equations, we geta = 1, b = 2, c = 3 and d = 4EXERCISE 3.11. In the matrix 2 5 1975A 35 2 12217 3 1 5 = , write:(i) The order of the matrix, (ii) The number of elements,(iii) Writetheelementsa13,a21,a33,a24,a23.2. If a matrix has 24 elements, what are the possible orders it can have? What, if ithas 13 elements?3. If a matrix has 18 elements, what are the possible orders it can have? What, if ithas 5 elements?4. Construct a 2 2 matrix, A = [aij], whose elements are given by:(i)2( )2iji ja+= (ii)ijiaj= (iii)2( 2 )2iji ja+=5. Construct a 3 4 matrix, whose elements are given by:(i)1| 3 |2ija i j = +(ii) 2ija i j = 6. Find the values of x, y and z from the following equations:(i)4 35 1 5y zx = (ii)2 6 25 5 8x yz xy+ = + (iii)957x y zx zy z+ + + = + 7. Find the value of a, b, c and d from the equation:2 1 52 3 0 13a b a ca b c d + = + MATRICES 658. A = [aij]m n\ is a square matrix, if(A) m < n (B) m > n (C) m = n (D) None of these9. Which of the given values of x and y make the following pair of matrices equal3 7 51 2 3xy x+ + , 0 28 4y (A)1, 73x y= =(B) Not possible to find(C) y = 7, 23x=(D)1 2,3 3x y = =10. The number of all possible matrices of order 3 3 with each entry 0 or 1 is:(A) 27 (B) 18 (C) 81 (D) 5123.4Operations on MatricesIn this section, we shall introduce certain operations on matrices, namely, addition ofmatrices, multiplication of a matrix by a scalar, difference and multiplication of matrices.3.4.1AdditionofmatricesSupposeFatimahastwofactoriesatplaces AandB.Eachfactoryproducessportshoes for boys and girls in three different price categories labelled 1, 2 and 3. Thequantities produced by each factory are represented as matrices given below:Suppose Fatima wants to know the total production of sport shoes in each pricecategory. Then the total productionIn category 1 : for boys (80 + 90), for girls (60 + 50)In category 2 : for boys (75 + 70), for girls (65 + 55)In category 3 : for boys (90 + 75), for girls (85 + 75)This can be represented in the matrix form as 80 90 60 5075 70 65 5590 75 85 75+ + + + + + .66 MATHEMATICSThis new matrix is the sum of the above two matrices. We observe that the sum oftwo matrices is a matrix obtained by adding the corresponding elements of the givenmatrices. Furthermore, the two matrices have to be of the same order.Thus, if11 12 1321 22 23Aa a aa a a = is a 2 3 matrix and 11 12 1321 22 23Bb b bb b b = is another23 matrix. Then, we define 11 11 12 12 13 1321 21 22 22 23 23A + Ba b a b a ba b a b a b+ + + = + + + .In general, if A = [aij] and B = [bij] are two matrices of the same order, say m n.Then, the sum of the two matrices A and B is defined as a matrix C = [cij]m n, wherecij = aij + bij, for all possible values of i and j.Example 6Given 3 1 1A2 3 0 = and 2 5 1B12 32 = , find A + BSince A, B are of the same order 2 3. Therefore, addition of A and B is definedand is given by2 3 1 5 1 1 2 3 1 5 0A+B1 12 2 3 3 0 0 62 2 + + + + = = + + $Note1. We emphasise that if A and B are not of the same order, then A + B is notdefined. For example if 2 3A1 0 = , 1 2 3B ,1 0 1 = then A + B is not defined.2. We may observe that addition of matrices is an example of binary operationon the set of matrices of the same order.3.4.2MultiplicationofamatrixbyascalarNow suppose that Fatima has doubled the production at a factory A in all categories(refer to 3.4.1).MATRICES 67Previously quantities (in standard units) produced by factory AwereRevised quantities produced by factory A are as given below:Boys Girls2 80 2 60 12 2 75 2 653 2 90 2 85 This can be represented in the matrix form as 160 120150 130180 170 . We observe thatthe new matrix is obtained by multiplying each element of the previous matrix by 2.Ingeneral,wemaydefinemultiplicationofamatrixbyascalarasfollows:ifA = [aij]m n is a matrix and k is a scalar, then kA is another matrix which is obtainedby multiplying each element of A by the scalar k.In other words, kA = k [aij]m n = [k (aij)]m n, that is, (i, j)th element of kA is kaijfor all possible values of i and j.For example, if A =3 1 1.55 7 32 0 5 , then3A =3 1 1.5 9 3 4.53 5 7 3 3 5 21 92 0 5 6 0 15 = NegativeofamatrixThenegativeofamatrixisdenotedby A. WedefineA = ( 1) A.68 MATHEMATICSFor example, let A =3 15 x , then A is given by A = ( 1)3 1 3 1A ( 1)5 5 x x = = Difference of matricesIf A = [aij], B = [bij] are two matrices of the same order,say m n, then difference A B is defined as a matrix D = [dij], where dij = aij bij,for all value of i and j. In other words, D = A B = A + (1) B, that is sum of the matrixA and the matrix B.Example 7 If 1 2 3 3 1 3A and B2 3 1 1 0 2 = = , then find 2A B.SolutionWehave2A B =1 2 3 3 1 322 3 1 1 0 2 =2 4 6 3 1 34 6 2 1 0 2 + =2 3 4 1 6 3 1 5 34 1 6 0 2 2 5 6 0 + = + + 3.4.3PropertiesofmatrixadditionThe addition of matrices satisfy the following properties:(i) Commutative LawIf A = [aij], B = [bij] are matrices of the same order, saym n, then A + B = B + A.Now A + B =[aij]+[bij]=[aij+bij]= [bij + aij] (addition of numbers is commutative)= ([bij] + [aij]) = B + A(ii) AssociativeLawForanythreematrices A=[aij],B=[bij],C=[cij]ofthesame order, say m n, (A + B) + C = A + (B + C).Now (A + B) + C =([aij]+[bij])+[cij]= [aij+bij]+[cij]=[(aij+bij)+cij]= [aij+(bij+cij)] (Why?)= [aij] + [(bij+ cij)] = [aij] + ([bij]+ [cij]) = A + (B + C)MATRICES 69(iii) ExistenceofadditiveidentityLetA=[aij]beanmnmatrixandO be an m n zero matrix, then A + O = O

+ A = A. In other words, O is theadditive identity for matrix addition.(iv) TheexistenceofadditiveinverseLetA=[aij]mnbeanymatrix,thenwehave another matrix as A = [ aij]m n such that A + ( A) = ( A) + A= O. So A is the additive inverse of A or negative of A.3.4.4PropertiesofscalarmultiplicationofamatrixIf A = [aij] and B = [bij] be two matrices of the same order, say m n, and k and l arescalars, then(i) k(A +B) = k A + kB, (ii) (k + l)A = k A + l A(ii) k(A+B)=k([aij]+[bij])=k[aij+bij]=[k(aij+bij)]=[(kaij)+(kbij)]=[kaij]+[kbij]=k[aij]+k[bij]=kA+kB(iii) ( k+l )A = (k + l) [aij]= [(k + l) aij] + [k aij] + [l aij] = k [aij] + l[aij] = k A + lAExample8If 8 0 2 2A 4 2 and B 4 23 6 5 1 = = ,thenfindthematrixX,suchthat2A + 3X = 5B.Solution We have 2A + 3X = 5Bor 2A + 3X 2A = 5B 2Aor 2A 2A + 3X = 5B 2A (Matrix addition is commutative)or O + 3X = 5B 2A ( 2A is the additive inverse of 2A)or 3X = 5B 2A (O is the additive identity)or X = 13 (5B 2A)or2 2 8 01X 5 4 2 2 4 235 1 3 6 = = 10 10 16 0120 10 8 4325 5 6 12 + 70 MATHEMATICS= 10 16 10 0120 8 10 4325 6 5 12 + + = 6 10112 14331 7 = 1023144331 73 3 Example 9 FindX and Y, if 5 2X Y0 9 + = and 3 6X Y0 1 = .SolutionWehave ( ) ( )5 2 3 6X Y X Y0 9 0 1 + + = + .or (X + X) + (Y Y) =8 80 8 8 82X0 8 = or X =8 8 4 4 10 8 0 4 2 = Also (X + Y) (X Y) =5 2 3 60 9 0 1 or (X X) + (Y + Y) =5 3 2 60 9 1 + 2 42Y0 10 = or Y =2 4 1 210 10 0 5 2 = Example 10 Find the values of x and y from the following equation:5 3 427 3 1 2xy + = 7 615 14 SolutionWehave 5 3 427 3 1 2xy + = 7 615 14 2 10 3 4 7 614 2 6 1 2 15 14xy + = MATRICES 71or2 3 10 414 1 2 6 2xy+ + + =7 615 14 2 3 6 7 615 2 4 15 14xy+ = or 2x + 3 = 7 and 2y 4 = 14 (Why?)or 2x = 7 3 and 2y = 18or x = 42and y = 182i.e. x = 2 and y = 9.Example11TwofarmersRamkishanandGurcharanSinghcultivatesonlythreevarietiesofricenamelyBasmati,PermalandNaura.Thesale(inRupees)ofthesevarieties of rice by both the farmers in the month of September and October are givenby the following matrices A and B.(i) FindthecombinedsalesinSeptemberandOctoberforeachfarmerineachvariety.(ii) Find the decrease in sales from September to October.(iii) If both farmers receive 2%profit on gross sales, compute the profit for eachfarmer and for each variety sold in October.Solution(i) Combined sales in September and October for each farmer in each variety isgiven by72 MATHEMATICS(ii) Change in sales from September to October is given by(iii) 2% of B = 2B100 = 0.02 B= 0.02 =Thus, in October Ramkishan receives Rs 100, Rs 200 and Rs 120 as profit in thesale ofeach variety of rice, respectively, and Grucharan Singh receives profit of Rs400, Rs 200 and Rs 200 in the sale of each variety of rice, respectively.3.4.5MultiplicationofmatricesSuppose Meera and Nadeem are two friends. Meera wants to buy 2 pens and 5 storybooks, while Nadeem needs 8 pens and 10 story books. They both go to a shop toenquire about the rates which are quoted as follows:Pen Rs 5 each, story book Rs 50 each.How much money does each need to spend? Clearly, Meera needs Rs (5 2 + 50 5)that is Rs 260, while Nadeem needs (8 5 + 50 10) Rs, that is Rs 540. In terms ofmatrix representation, we can write the above information as follows:Requirements Pricesperpiece(inRupees) Moneyneeded(inRupees)2 58 10 550 5 2 5 50 2608 5 10 50 540 + = + Suppose that they enquire about the rates from another shop, quoted as follows:pen Rs 4 each,story book Rs 40 each.Now,themoneyrequiredbyMeeraandNadeemtomakepurchaseswillberespectively Rs (4 2 + 40 5) = Rs 208 and Rs (8 4 + 10 40) = Rs 432MATRICES 73Again, the above information can be represented as follows:Requirements Pricesperpiece(inRupees) Moneyneeded(inRupees)2 58 10 440 4 2 40 5 2088 4 10 4 0 432 + = + Now, the information in both the cases can be combined and expressed in terms ofmatrices as follows:Requirements Pricesperpiece(inRupees) Moneyneeded(inRupees)2 58 10 5 450 40 5 2 5 50 4 2 40 58 5 10 5 0 8 4 10 4 0 + + + + = 260 208540 432 Theaboveisanexampleofmultiplicationofmatrices.Weobservethat,formultiplication of two matrices A and B, the number of columns in A should be equal tothe number of rows in B. Furthermore for getting the elements of the product matrix,we take rows of A and columns of B, multiply them element-wise and take the sum.Formally, we define multiplication of matrices as follows:The product of two matrices A and B is defined if the number of columns of A isequal to the number of rows of B. Let A = [aij] be an m n matrix and B = [bjk] be ann p matrix. Then the product of the matrices A and B is the matrix C of order m p.To get the (i, k)th element cik of the matrix C, we take the ith row of A and kth columnof B, multiply them elementwise and take the sum of all these products. In other words,if A = [aij]m n, B = [bjk]n p, then the ith row of A is [ai1 ai2 ... ain] and the kth column ofB is 12...kknkbbb , then cik =ai1 b1k +ai2 b2k + ai3 b3k + ... + ain bnk = 1nij jkja b=.The matrix C = [cik]m p is the product of A and B.For example, if 1 1 2C0 3 4 = and 2 71 D 15 4 = , then the product CD is defined74 MATHEMATICSand is given by 2 71 1 2CD 1 10 3 45 4 = . This is a 2 2matrix in which eachentry is the sum of the products across some row of C with the corresponding entriesdown some column of D. These four computations areThus 13 2CD17 13 = Example 12 Find AB, if 6 9 2 6 0A and B2 3 7 9 8 = = .Solution Thematrix Ahas2columnswhichisequaltothenumberofrowsofB.Hence AB is defined. Now6(2) 9(7) 6(6) 9(9) 6(0) 9(8)AB2(2) 3(7) 2(6) 3(9) 2(0) 3(8)+ + + = + + + =12 63 36 81 0 724 21 12 27 0 24+ + + + + + = 75 117 7225 39 24 MATRICES 75RemarkIf AB is defined, then BA need not be defined. In the above example, AB isdefined but BA is not defined because B has 3 column while A has only 2 (and not 3)rows. If A, B are, respectively m n, k l matrices, then both AB and BA are definedif and only if n = k and l = m. In particular, if both A and B are square matrices of thesame order, then both AB and BA are defined.Non-commutativityofmultiplicationofmatricesNow, we shall see by an example that even if AB and BA are both defined, it is notnecessary that AB = BA.Example13If 2 31 2 3A and B 4 54 2 52 1 = = ,thenfindAB,BA.ShowthatABBA.Solution Since A is a 2 3 matrix and B is 3 2 matrix. Hence AB and BA are bothdefined and are matrices of order 2 2 and 3 3, respectively. Note that2 31 2 3AB 4 54 2 52 1 = =2 8 6 3 10 3 0 48 8 10 12 10 5 10 3+ + = + + + + and2 3 2 12 4 6 6 151 2 3BA 4 5 4 20 8 10 12 254 2 52 1 2 4 4 2 6 5 + + = = + + + + 10 2 2116 2 372 2 11 = Clearly ABBAIn the above example both AB and BA are of different order and so AB BA. Butone may think that perhaps AB and BA could be the same if they were of the sameorder. But it is not so, here we give an example to show that even if AB and BA are ofsame order they may not be same.Example 14 If 1 0A0 1 = and 0 1B1 0 = , then 0 1AB1 0 = .and0 1BA1 0 = . Clearly AB BA.Thus matrix multiplication is not commutative.76 MATHEMATICS$NoteThis does not mean that AB BA for every pair of matrices A, B forwhich AB and BA, are defined. For instance,If1 0 3 0A , B0 2 0 4 = = , then AB = BA = 3 00 8 Observe that multiplication of diagonal matrices of same order will be commutative.ZeromatrixastheproductoftwononzeromatricesWe know that, for real numbers a, b if ab = 0, then either a = 0 or b = 0. This neednot be true for matrices, we will observe this through an example.Example 15 Find AB, if 0 1A0 2 = and 3 5B0 0 = .Solution Wehave 0 1 3 5 0 0AB0 2 0 0 0 0 = = .Thus, if the product of two matrices is a zero matrix, it is not necessary that one ofthe matrices is a zero matrix.3.4.6PropertiesofmultiplicationofmatricesThe multiplication of matrices possesses the following properties, which we state withoutproof.1. The associative law For any three matrices A, B and C. We have(AB) C = A (BC), whenever both sides of the equality are defined.2. The distributive law For three matrices A, B and C.(i) A (B+C) = AB + AC(ii) (A+B) C = AC + BC, whenever both sides of equality are defined.3. TheexistenceofmultiplicativeidentityForeverysquarematrixA,thereexist an identity matrix of same order such that IA = AI = A.Now, we shall verify these properties by examples.Example16If 1 1 1 1 31 2 3 4A 2 0 3 , B 0 2 and C2 0 2 13 1 2 1 4 = = = ,findA(BC), (AB)C and show that (AB)C = A(BC).MATRICES 77Solution Wehave 1 1 1 1 3 1 0 1 3 2 4 2 1AB 2 0 3 0 2 2 0 3 6 0 12 1 183 1 2 1 4 3 0 2 9 2 8 1 15 + + + = = + + + = + + (AB)(C)2 2 4 0 6 2 8 1 2 11 2 3 41 18 1 36 2 0 3 36 4 182 0 2 11 15 1 30 2 0 3 30 4 15+ + + = = + + + + + + =4 4 4 735 2 39 2231 2 27 11 Now BC =1 6 2 0 3 6 4 3 1 31 2 3 40 2 0 4 0 0 0 4 0 22 0 2 11 4 1 8 2 0 3 8 4 4+ + + = + + + + + + =7 2 3 14 0 4 27 2 11 8 Therefore A(BC)=7 2 3 1 1 1 12 0 3 4 0 4 23 1 2 7 2 11 8 =7 4 7 2 0 2 3 4 11 1 2 814 0 21 4 0 6 6 0 33 2 0 2421 4 14 6 0 4 9 4 22 3 2 16+ + + + + + + + + + + + + + + =4 4 4 735 2 39 2231 2 27 11 . Clearly, (AB) C = A(BC)78 MATHEMATICSExample17If 0 6 7 0 1 1 2A 6 0 8 , B 1 0 2 , C 27 8 0 1 2 0 3 = = = Calculate AC, BC and (A + B)C. Also, verify that (A + B)C = AC + BCSolution Now, 0 7 8A+B 5 0 108 6 0 = So (A + B) C =0 7 8 2 0 14 24 105 0 10 2 10 0 30 208 6 0 3 16 12 0 28 + = + + = + + Further AC =0 6 7 2 0 12 21 96 0 8 2 12 0 24 127 8 0 3 14 16 0 30 + = + + = + + and BC =0 1 1 2 0 2 3 11 0 2 2 2 0 6 81 2 0 3 2 4 0 2 + = + + = + So AC + BC =9 1 1012 8 2030 2 28 + = Clearly, (A + B) C = AC + BCExample 18 If 1 2 3A 3 2 14 2 1 = , then show that A3 23A 40 I = OSolution Wehave 21 2 3 1 2 3 19 4 8A A.A 3 2 1 3 2 1 1 12 84 2 1 4 2 1 14 6 15 = = = MATRICES 79So A3= A A2 = 1 2 3 19 4 8 63 46 693 2 1 1 12 8 69 6 234 2 1 14 6 15 92 46 63 = NowA3 23A 40I = 63 46 69 1 2 3 1 0 069 6 23 23 3 2 1 40 0 1 092 46 63 4 2 1 0 0 1 =63 46 69 23 46 69 40 0 069 6 23 69 46 23 0 40 092 46 63 92 46 23 0 0 40 + + =63 23 40 46 46 0 69 69 069 69 0 6 46 40 23 23 092 92 0 46 46 0 63 23 40 + + + + + + + =0 0 00 0 0 O0 0 0 = Example 19 In a legislative assembly election, a political group hired a public relationsfirm to promote its candidate in three ways: telephone, house calls, and letters. Thecost per contact (in paise) is given in matrix A asA = 40 Telephone100 Housecall50 LetterCost per contact ThenumberofcontactsofeachtypemadeintwocitiesXandYisgivenbyTelephone Housecall Letter1000 500 5000 XBY 3000 1000 10, 000 = . Find the total amount spent by the group in the twocities X and Y.80 MATHEMATICSSolutionWehaveBA =40, 000 50, 000 250, 000 XY 120,000 +100,000 +500,000+ + =340, 000 XY 720,000 Sothetotalamountspentbythegroupinthetwocitiesis340,000paiseand720,000 paise, i.e., Rs 3400 and Rs 7200, respectively.EXERCISE 3.21. Let 2 4 1 3 2 5A , B , C3 2 2 5 3 4 = = = Find each of the following:(i) A + B (ii) A B (iii) 3A C(iv) AB (v) BA2. Compute the following:(i)a b a bb a b a + (ii)2 2 2 22 2 2 22 22 2a b b c ab bcac aba c a b + + + + + (iii)1 4 6 12 7 68 5 16 8 0 52 8 5 3 2 4 + (iv)2 2 2 22 2 2 2cos sin sin cossin cos cos sinx x x xx x x x + 3. Compute the indicated products.(i)a b a bb a b a (ii)123 [2 3 4] (iii)1 2 1 2 32 3 2 3 1 (iv)2 3 4 1 3 53 4 5 0 2 44 5 6 3 0 5 (v)2 11 0 13 21 2 11 1 (vi)2 33 1 31 01 0 23 1 MATRICES 814. If 1 2 3 3 1 2 4 1 2A 5 0 2 , B 4 2 5 and C 0 3 21 1 1 2 0 3 1 2 3 = = = ,thencompute(A+B) and (B C). Also, verify that A + (B C) = (A + B) C.5. If 2 5 2 31 13 3 5 51 2 4 1 2 4A and B3 3 3 5 5 57 2 7 6 223 3 5 5 5 = = , then compute 3A 5B.6. Simplify cos sin sin coscos + sinsin cos cos sin 7. Find X and Y, if(i)7 0 3 0X + Y and X Y2 5 0 3 = = (ii)2 3 2 22X + 3Y and 3X 2Y4 0 1 5 = + = 8. Find X, if Y = 3 21 4 and 2X + Y = 1 03 2 9. Find x and y, if 1 3 0 5 620 1 2 1 8yx + = 10. Solve the equation for x, y, z and t, if 1 1 3 52 3 30 2 4 6x zy t + = 11. If 2 1 103 1 5x y + = , find the values of x and y.12. Given 6 431 2 3x y x x yz w w z w+ = + + , find the values of x, y, z and w.82 MATHEMATICS13. If cos sin 0F ( ) sin cos 00 0 1x xx x x = , show that F(x) F(y) = F(x + y).14. Show that(i)5 1 2 1 2 1 5 16 7 3 4 3 4 6 7 (ii)1 2 3 1 1 0 1 1 0 1 2 30 1 0 0 1 1 0 1 1 0 1 01 1 0 2 3 4 2 3 4 1 1 0 15. Find A2 5A + 6I, if 2 0 1A 2 1 31 1 0 = 16. If 1 0 2A 0 2 12 0 3 = , prove that A3 6A2 + 7A + 2I = 017. If 3 2 1 0A and I=4 2 0 1 = , find k so that A2 = kA 2I18. If 0 tan2Atan 02 = and I is the identity matrix of order 2, show thatI + A = (I A) cos sinsin cos 19. A trust fund has Rs 30,000 that must be invested in two different types of bonds.The first bond pays 5% interest per year, and the second bond pays 7% interestper year. Using matrix multiplication, determine how to divide Rs 30,000 amongthe two types of bonds. If the trust fund must obtain an annual total interest of:(a) Rs 1800 (b) Rs 2000MATRICES 8320. Thebookshopofaparticularschoolhas10dozenchemistrybooks,8dozenphysics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60and Rs 40 each respectively. Find the total amount the bookshop will receivefrom selling all the books using matrix algebra.Assume X, Y, Z, W and P are matrices of order 2 n, 3 k, 2 p, n 3 and p k,respectively. Choose the correct answer in Exercises 21 and 22.21. The restriction on n, k and p so that PY + WY will be defined are:(A) k = 3, p = n (B) k is arbitrary, p = 2(C) p is arbitrary, k = 3 (D) k = 2, p = 322. If n = p, then the order of the matrix 7X 5Z is:(A) p 2 (B) 2 n (C) n 3 (D) p n3.5.Transpose of a MatrixIn this section, we shall learn about transpose of a matrix and special types of matricessuch as symmetric and skew symmetric matrices.Definition 3If A = [aij] be an m n matrix, then the matrix obtained by interchangingthe rows and columns of A is called the transpose of A. Transpose of the matrix A isdenoted by A or (AT). In other words, if A = [aij]m n, then A = [aji]n m. For example,if 2 33 2353 3 0A 3 1 , then A15 10 155 = = 3.5.1PropertiesoftransposeofthematricesWe now state the following properties of transpose of matrices without proof. Thesemay be verified by taking suitable examples.For any matrices A and B of suitable orders, we have(i) (A) = A, (ii) (kA)=kA(wherekisanyconstant)(iii) (A + B) = A + B (iv) (A B) = B AExample 20 If 2 1 2 3 3 2A and B1 2 44 2 0 = = , verify that(i) (A) = A, (ii) (A + B) = A + B,(iii) (kB) =kB, wherekisanyconstant.84 MATHEMATICSSolution(i) WehaveA =( )343 3 2 3 3 2A 3 2 A A4 2 0 4 2 02 0 = = = Thus (A)= A(ii) WehaveA = 3 3 2,4 2 0 B =2 1 2 5 3 1 4A B1 2 45 4 4 + = Therefore (A + B) =553 1 44 4 Now A =34 2 13 2 , B 1 2 ,2 0 2 4 = So A + B =553 1 44 4 Thus (A + B) = A + B(iii) WehavekB= k2 1 2 2 21 2 4 2 4k k kk k k = Then (kB)=2 2 12 1 2 B2 4 2 4k kk k k kk k = = Thus (kB)= kBMATRICES 85Example 21 If[ ]2A 4 , B 1 3 65 = = , verify that (AB) = BA.SolutionWehaveA =[ ]24 , B 1 3 65 = then AB = [ ]24 1 3 65 = 2 6 124 12 245 15 30 Now A = [2 4 5] , 1B 36 = BA= [ ]1 2 4 53 2 4 5 6 12 15 (AB)6 12 24 30 = = Clearly (AB)= BA3.6Symmetric and Skew Symmetric MatricesDefinition4 AsquarematrixA=[aij] issaidtobesymmetricifA= A,thatis,[aij] = [aji] for all possible values of i and j.For example 3 2 3A 2 1.5 13 1 1 = is a symmetric matrix as A = ADefinition5AsquarematrixA=[aij] issaidtobeskewsymmetricmatrixifA = A, that is aji = aij for all possible values of iand j. Now, if we put i = j, wehave aii = aii. Therefore 2aii = 0 or aii = 0 for all is.This means that all the diagonal elements of a skew symmetric matrix are zero.86 MATHEMATICSFor example, the matrix 0B 00e fe gf g = is a skew symmetric matrix as B= BNow,wearegoingtoprovesomeresultsofsymmetricandskew-symmetricmatrices.Theorem 1 For any square matrix A with real number entries, A + A is a symmetricmatrix and A A is a skew symmetric matrix.Proof Let B = A + A, thenB = (A + A)= A + (A) (as (A + B) = A + B)= A + A (as (A) = A)= A + A (as A + B = B + A)= BTherefore B = A + A is a symmetric matrixNow let C = AAC = (A A) = A (A) (Why?)= A A(Why?)= (A A) = CTherefore C = A A is a skew symmetric matrix.Theorem2Anysquarematrixcanbeexpressedasthesumofasymmetricandaskew symmetric matrix.Proof Let A be a square matrix, then we can write1 1A (A A ) (A A )2 2 = + + From the Theorem 1, we know that (A + A) is a symmetric matrix and (A A) isa skew symmetric matrix. Since for any matrix A, (kA) = kA, it follows that 1(A A )2 +issymmetricmatrixand 1(A A )2 isskewsymmetricmatrix.Thus,anysquarematrix can be expressed as the sum of a symmetric and a skew symmetric matrix.MATRICES 87Example 22 Express the matrix 2 2 4B 1 3 41 2 3 = as the sum of a symmetric and askew symmetric matrix.SolutionHereB =2 1 12 3 24 4 3 Let P =4 3 31 1(B + B ) 3 6 22 23 2 6 = = 3 322 233 1231 32 ,Now P=3 322 233 1231 32 = PThus P =1(B+ B )2 is a symmetric matrix.Also, let Q =1 502 20 1 51 1 1(B B ) 1 0 6 0 32 2 25 6 053 02 = = Then Q =1 502 310 3 Q253 02 = 88 MATHEMATICSThus Q =1(B B )2 is a skew symmetric matrix.Now3 3 1 52 02 2 2 22 2 43 1P + Q 3 1 0 3 1 3 4 B2 21 2 33 51 3 3 02 2 = + = = Thus, B is represented as the sum of a symmetric and a skew symmetric matrix.EXERCISE 3.31. Find the transpose of each of the following matrices:(i)5121 (ii)1 12 3 (iii)1 5 63 5 62 3 1 2. If 1 2 3 4 1 5A 5 7 9 and B 1 2 02 1 1 1 3 1 = = , then verify that(i) (A + B) = A + B, (ii) (A B) = A B3. If 3 41 2 1A 1 2 and B1 2 30 1 = = , then verify that(i) (A + B) = A + B (ii) (A B) = A B4. If 2 3 1 0A and B1 2 1 2 = = , then find (A + 2B)5. For the matrices A and B, verify that (AB) = BA, where(i) [ ]1A 4 , B 1 2 13 = = (ii) [ ]0A 1 , B 1 5 72 = = MATRICES 896. If (i)cos sinAsin cos = , then verify that A A = I(ii) If sin cosAcos sin = , then verify that A A = I7.(i) Show that the matrix 1 1 5A 1 2 15 1 3 = is a symmetric matrix.(ii)Show that the matrix 0 1 1A 1 0 11 1 0 = is a skew symmetric matrix.8. For the matrix 1 5A6 7 = , verify that(i) (A + A) is a symmetric matrix(ii) (A A) is a skew symmetric matrix9. Find ( )1A A2 +and ( )1A A2 , when 0A 00a ba cb c = 10. Express the following matrices as the sum of a symmetric and a skew symmetricmatrix:(i)3 51 1 (ii)6 2 22 3 12 1 3 (iii)3 3 12 2 14 5 2 (iv)1 51 2 90 MATHEMATICSChoose the correct answer in the Exercises 11 and 12.11. If A, B are symmetric matrices of same order, then AB BA is a(A) Skew symmetric matrix (B) Symmetric matrix(C) Zero matrix (D) Identity matrix12. If cos sinA ,sin cos = thenA + A = I, if the value of is(A)6(B)3(C) (D)323.7Elementary Operation (Transformation) of a MatrixThere are six operations (transformations) on a matrix, three of which are due to rowsandthreeduetocolumns,whichareknownaselementaryoperationsortransformations.(i) The interchange of any two rows or two columns. Symbolically the interchangeof ith and jth rows is denoted by Ri Rj and interchange of ith and jth column isdenoted by Ci Cj.For example, applying R1 R2 to 1 2 1A 1 3 15 6 7 = , we get 1 3 11 2 15 6 7 .(ii) Themultiplicationoftheelementsofanyroworcolumnbyanonzeronumber.Symbolically, the multiplication of each element of theith row by k,where k 0 is denoted by Ri k Ri.The corresponding columnoperation is denoted by Ci kCiFor example, applying 3 31C C7,to1 2 1B1 3 1 = , we get11 2711 37 (iii) Theadditiontotheelementsofanyroworcolumn,thecorrespondingelementsofanyotherroworcolumnmultipliedbyanynonzeronumber.Symbolically, the addition to the elements of ith row, the corresponding elementsof jth row multiplied by k is denoted by Ri Ri + kRj.MATRICES 91The corresponding column operation is denoted by Ci Ci + kCj.For example, applying R2 R2 2R1, to 1 2C2 1 = , we get 1 20 5 .3.8Invertible MatricesDefinition6IfAisasquarematrixoforderm,andifthereexistsanothersquarematrix B of the same order m, such that AB = BA = I, then B is called the inversematrix of A and it is denoted by A 1. In that case A is said to be invertible.For example, let A =2 31 2 and B = 2 31 2 be two matrices.Now AB =2 3 2 31 2 1 2 =4 3 6 6 1 0I2 2 3 4 0 1 + = = + Also BA =1 0I0 1 = . Thus B is the inverse of A, in otherwords B = A 1 and A is inverse of B, i.e., A = B1$Note1. A rectangular matrix does not possess inverse matrix, since for products BAand AB to be defined and to be equal, it is necessary that matrices A and Bshould be square matrices of the same order.2. If B is the inverse of A, then A is also the inverse of B.Theorem 3 (Uniqueness of inverse) Inverse of a square matrix, if it exists, is unique.ProofLet A=[aij]beasquarematrixoforderm.Ifpossible,letBandCbetwoinverses of A. We shall show that B = C.Since B is the inverse of AAB = BA = I ...(1)Since C is also the inverse of AAC = CA = I ...(2)Thus B = BI = B (AC) = (BA) C = IC = CTheorem 4 If A and B are invertible matrices of the same order, then (AB)1 = B1 A1.92 MATHEMATICSProof From the definition of inverse of a matrix, we have(AB)(AB)1= 1or A1(AB)(AB)1= A1I (Pre multiplying both sides by A1)or (A1A)B(AB)1= A1(SinceA1I=A1)or IB(AB)1= A1or B(AB)1= A1or B1B(AB)1= B1A1or I(AB)1= B1A1Hence (AB)1= B1A13.8.1InverseofamatrixbyelementaryoperationsLet X, A and B be matrices of, the same order such that X = AB. In order to apply asequence of elementary row operations on the matrix equation X = AB, we will applythese row operations simultaneously on X and on the first matrix A of the product ABon RHS.Similarly, in order to apply a sequence of elementary column operations on thematrix equation X = AB, we will apply, these operations simultaneously on X and on thesecond matrix B of the product AB on RHS.In view of the above discussion, we conclude that if A is a matrix such that A1exists,thentofind A1usingelementaryrowoperations,write A=IAandapplyasequence of row operation on A = IA till we get, I = BA. The matrix B will be theinverse of A. Similarly, if we wish to find A1 using column operations, then, writeA = AI and apply a sequence of column operations on A = AI till we get, I = AB.Remark In case, after applying one or more elementary row (column) operations onA = IA (A = AI), if we obtain all zeros in one or more rows of the matrix A on L.H.S.,then A1 does not exist.Example23Byusingelementaryoperations,findtheinverseofthematrix1 2A =2 1 .Solution In order to use elementary row operations we may write A = IA.or1 2 1 0 1 2 1 0A, then A2 1 0 1 0 5 2 1 = = (applying R2 R2 2R1)MATRICES 93or1 20 1 =1 0A2 15 5 (applying R2 15R2)or1 00 1 =1 25 5A2 15 5 (applying R1 R1 2R2)Thus A1=1 25 52 15 5 Alternatively, in order to use elementary column operations, we write A = AI, i.e.,1 22 1 =1 0A0 1 Applying C2 C2 2C1, we get1 02 5 =1 2A0 1 Now applying C2 21C5 , we have1 02 1 =215A105 Finally, applying C1 C1 2C2, we obtain1 00 1 =1 25 5A2 15 5 Hence A1=1 25 52 15 5 94 MATHEMATICSExample 24 Obtain the inverse of the following matrix using elementary operations0 1 2A 1 2 33 1 1 = .Solution Write A = I A, i.e., 0 1 21 2 33 1 1 = 1 0 00 1 0 A0 0 1 or1 2 30 1 23 1 1 = 0 1 01 0 0 A0 0 1 (applying R1 R2)or1 2 30 1 20 5 8 = 0 1 01 0 0 A0 3 1 (applying R3 R3 3R1)or1 0 10 1 20 5 8 = 2 1 01 0 0 A0 3 1 (applying R1 R1 2R2)or1 0 10 1 20 0 2 = 2 1 01 0 0 A5 3 1 (applying R3 R3 + 5R2)or1 0 10 1 20 0 1 = 2 1 01 0 0 A5 3 12 2 2 (applying R3 12 R3)or1 0 00 1 20 0 1 = 1 1 12 2 21 0 0 A5 3 12 2 2 (applying R1 R1 + R3)MATRICES 95or1 0 00 1 00 0 1 =1 1 12 2 24 3 1 A5 3 12 2 2 (applying R2 R2 2R3)Hence A1=1 1 12 2 24 3 15 3 12 2 2 Alternatively, write A = AI, i.e.,0 1 21 2 33 1 1 =1 0 0A 0 1 00 0 1 or1 0 22 1 31 3 1 =0 1 0A 1 0 00 0 1 (C1C2)or1 0 02 1 11 3 1 =0 1 0A 1 0 20 0 1 (C3C32C1)or1 0 02 1 01 3 2 =0 1 1A 1 0 20 0 1 (C3C3+C2)or1 0 02 1 01 3 1 =10 12A 1 0 110 02 (C3 12C3)96 MATHEMATICSor1 0 00 1 05 3 1 =12 12A 1 0 110 02 (C1C12C2)or1 0 00 1 00 3 1 =1 112 2A 4 0 15 102 2 (C1 C1 + 5C3)or1 0 00 1 00 0 1 =1 1 12 2 2A 4 3 15 3 12 2 2 (C2C2 3C3)Hence A1=1 1 12 2 24 3 15 3 12 2 2 Example 25 Find P1, if it exists, given 10 2P5 1 = .Solution We have P = I P, i.e., 10 2 1 0P5 1 0 1 = .or1155 1 =10P 100 1 (applying R1 110R1)MATRICES 97or1150 0 =1010P112 (applying R2 R2 + 5R1)Wehaveallzerosinthesecondrowofthelefthandsidematrixoftheaboveequation. Therefore, P1 does not exist.EXERCISE 3.4Using elementary transformations, find the inverse of each of the matrices, if it existsin Exercises 1 to 17.1.1 12 3 2.2 11 1 3.1 32 7 4.2 35 7 5.2 17 4 6.2 51 3 7.3 15 2 8.4 53 4 9.3 102 7 10.3 14 2 11.2 61 2 12.6 32 1 13.2 31 2 14.2 14 2 . 15.2 3 32 2 33 2 2 16.1 3 23 0 52 5 0 17.2 0 15 1 00 1 3 18. Matrices A and B will be inverse of each other only if(A) AB = BA (B) AB = BA = 0(C) AB = 0, BA = I (D) AB = BA = I98 MATHEMATICSMiscellaneous ExamplesExample 26 If cos sinAsin cos = , then prove that cos sinAsin cosnn nn n = , n N.Solution We shall prove the result by using principle of mathematical induction.Wehave P(n) : If cos sinAsin cos = , then cos sinAsin cosnn nn n = , n NP(1) : cos sinAsin cos = , so 1cos sinAsin cos = Therefore, the result is true for n = 1.Let the result be true for n = k. SoP(k) : cos sinAsin cos = , then cos sinAsin coskk kk k = Now, we prove that the result holds for n = k +1Now Ak + 1 =cos sin cos sinA Asin cos sin coskk kk k = =cos cos sin sin cos sin sin cossin cos cos sin sin sin cos cosk k k kk k k k + + + =cos ( ) sin( ) cos( 1) sin ( 1)sin( ) cos ( ) sin ( 1) cos ( 1)k k k kk k k k + + + + = + + + + Therefore, the result is true for n = k+ 1. Thus by principle of mathematical induction,we have cos sinAsin cosnn nn n = , holds for all natural numbers.Example 27 If A and B are symmetric matrices of the same order, then show that ABis symmetric if and only if A and B commute, that is AB = BA.Solution Since A and B are both symmetric matrices, therefore A = A and B = B.Let AB be symmetric, then (AB) = ABMATRICES 99But (AB)= BA= BA (Why?)Therefore BA = ABConversely, if AB = BA, then we shall show that AB is symmetric.Now (AB)= BA= B A (as A and B are symmetric)= ABHence AB is symmetric.Example28Let 2 1 5 2 2 5A , B , C3 4 7 4 3 8 = = = .FindamatrixDsuchthatCD AB = O.SolutionSince A,B,Careallsquarematricesoforder2,andCD ABiswelldefined, D must be a square matrix of order 2.Let D = a bc d . Then CD AB = 0 givesor2 5 2 1 5 23 8 3 4 7 4a bc d = Oor2 5 2 5 3 03 8 3 8 43 22a c b da c b d+ + + + =0 00 0 or2 5 3 2 53 8 43 3 8 22a c b da c b d+ + + + = 0 00 0 By equality of matrices, we get2a + 5c 3 = 0 ...(1)3a + 8c 43 = 0 ...(2)2b + 5d = 0 ...(3)and 3b + 8d 22 = 0 ...(4)Solving (1) and (2), we get a = 191, c = 77. Solving (3) and (4), we get b = 110,d = 44.Therefore D =191 11077 44a bc d = 100 MATHEMATICSMiscellaneous Exercise on Chapter 31. Let 0 1A0 0 = , show that (aI + bA)n = an I + nan 1 bA, where I is the identitymatrix of order 2 and n N.2. If 1 1 1A 1 1 11 1 1 = , prove that 1 1 11 1 11 1 13 3 3A 3 3 3 , .3 3 3n n nn n n nn n nn = N3. If 3 4 1 2 4A , then prove that A1 1 1 2nn nn n + = = , where n is any positiveinteger.4. If A and B are symmetric matrices, prove that AB BA is a skew symmetricmatrix.5. Show that the matrix BAB is symmetric or skew symmetric according as A issymmetric or skew symmetric.6. Find the values of x, y, z if the matrix 0 2Ay zx y zx y z = satisfy the equationAA= I.7. For what values of x :[ ]1 2 0 01 2 1 2 0 1 21 0 2 x = O?8. If 3 1A1 2 = , show that A2 5A + 7I = 0.9. Find x, if [ ]1 0 25 1 0 2 1 4 O2 0 3 1xx = MATRICES 10110. A manufacturer produces three products x, y, z which he sells in two markets.Annual sales are indicated below:Market ProductsI 10,000 2,000 18,000II 6,000 20,000 8,000(a) If unit sale prices of x, y and z are Rs 2.50, Rs 1.50 and Rs 1.00, respectively,find the total revenue in each market with the help of matrix algebra.(b) If the unit costs of the above three commodities are Rs 2.00, Rs 1.00 and 50paise respectively. Find the gross profit.11. Find the matrix X so that 1 2 3 7 8 9X4 5 6 2 4 6 = 12. If A and B are square matrices of the same order such that AB = BA, then proveby induction that ABn = BnA. Further, prove that (AB)n = AnBn for all n N.Choose the correct answer in the following questions:13. If A = is such that A = I, then(A) 1 + + = 0 (B) 1 + = 0(C) 1 = 0 (D) 1 + = 014. If the matrix A is both symmetric and skew symmetric, then(A) A is a diagonal matrix (B) A is a zero matrix(C) A is a square matrix (D) None of these15. If A is square matrix such that A2 = A, then (I + A) 7 A is equal to(A) A (B) I A (C) I (D) 3ASummary

A matrix is an ordered rectangular array of numbers or functions.

A matrix having m rows and n columns is called a matrix of order m n.

[aij]m 1 is a column matrix.

[aij]1 n is a row matrix.

An m n matrix is a square matrix if m = n.

A = [aij]m m is a diagonal matrix if aij = 0, when i j.102 MATHEMATICS

A=[aij]nnisascalarmatrixifaij=0,whenij,aij=k,(kissomeconstant), when i = j.

A = [aij]n n isan identity matrix, if aij = 1, when i = j, aij = 0, when i j.

A zero matrix has all its elements as zero.

A=[aij] =[bij]=Bif(i) AandBareofsameorder,(ii)aij=bijforallpossible values of i and j.

kA=k[aij]mn=[k(aij)]mn

A = (1)A

A B = A + (1) B

A + B = B + A

(A + B) + C = A + (B + C), where A, B and C are of same order.

k(A + B) = kA + kB, where A and B are of same order, k is constant.

(k + l ) A = kA + lA, where k and l are constant.

If A = [aij]m n and B = [bjk]n p, then AB = C = [cik]m p, where 1nik ij jkjc a b==

(i) A(BC) = (AB)C,(ii) A(B + C) = AB + AC,(iii) (A + B)C = AC + BC

If A = [aij]m n, then A or AT = [aji]n m

(i) (A) = A,(ii) (kA) = kA,(iii) (A + B) = A + B,(iv)(AB) = BA

A is a symmetric matrix if A = A.

A is a skew symmetric matrix if A = A.

Anysquarematrixcanberepresentedasthesumofasymmetricandaskew symmetric matrix.

Elementary operations of a matrix are as follows:(i) Ri Rj or Ci Cj(ii) RikRi orCikCi(iii) RiRi + kRj orCiCi + kCj

If AandBaretwosquarematricessuchthat AB=BA=I,thenBistheinverse matrix of A and is denoted by A1 and A is the inverse of B.

Inverse of a square matrix, if it exists, is unique. AllMathematicaltruthsarerelativeandconditional. C.P.STEINMETZ 4.1IntroductionInthepreviouschapter,wehavestudiedaboutmatricesand algebra of matrices. We have also learnt that a systemofalgebraicequationscanbeexpressedintheformofmatrices. This means, a system of linear equations likea1x + b1y = c1a2x + b2y = c2canberepresentedas 1 1 12 2 2a b c xa b c y = .Now,thissystemofequationshasauniquesolutionornot,isdeterminedbythenumbera1b2a2 b1.(Recallthatif1 12 2a ba b or,a1b2a2 b10,thenthesystemoflinearequations has a unique solution). The number a1 b2 a2 b1which determines uniqueness of solution is associated with the matrix 1 12 2Aa ba b = and is called the determinant of A or det A. Determinants have wide applications inEngineering, Science, Economics, Social Science, etc.In this chapter, we shall study determinants up to order three only with real entries.Also, we will study various properties of determinants, minors, cofactors and applicationsof determinants in finding the area of a triangle, adjoint and inverse of a square matrix,consistencyandinconsistencyofsystemoflinearequationsandsolutionoflinearequations in two or three variables using inverse of a matrix.4.2DeterminantToeverysquarematrixA=[aij]ofordern,wecanassociateanumber(realorcomplex) called determinant of the square matrix A, where aij = (i, j)th element of A.Chapter 4DETERMINANTSP.S. Laplace(1749-1827)104 MATHEMATICSThismaybethoughtofasafunctionwhichassociateseachsquarematrixwithaunique number (real or complex). If M is the set of square matrices, K is the set ofnumbers (real or complex) and f : M K is defined by f (A) = k, where A M andk K, then f (A) is called the determinant of A. It is also denoted by | A| or det A or .If A = a bc d , then determinant of A is written as | A| = a bc d = det (A)Remarks(i) For matrix A, | A| is read as determinant of A and not modulus of A.(ii) Only square matrices have determinants.4.2.1DeterminantofamatrixoforderoneLet A = [a ] be the matrix of order 1, then determinant of A is defined to be equal to a4.2.2DeterminantofamatrixofordertwoLet A = 11 1221 22a aa a be a matrix of order 2 2,then the determinant of A is defined as:det (A) = |A|===a11a22a21a12Example 1Evaluate 2 41 2.Solution We have 2 41 2 = 2(2) 4(1) = 4 + 4 = 8.Example 2Evaluate 1 1x xx x+SolutionWehave1 1x xx x+ = x (x) (x + 1) (x 1)= x2 (x2 1) = x2 x2 + 1 = 14.2.3Determinantofamatrixoforder33Determinant of a matrix of order three can be determined by expressing it in terms ofsecondorderdeterminants.Thisisknownasexpansionofadeterminantalongarow(oracolumn).TherearesixwaysofexpandingadeterminantoforderDETERMINANTS1053 corresponding to each of three rows (R1, R2 and R3) and three columns (C1, C2 andC3) giving the same value as shown below.Consider the determinant of square matrix A = [aij]3 3i.e., | A | =21 22 2331 32 33a a aa a a11 12 13a a aExpansionalongfirstRow(R1)Step 1 Multiply first element a11 of R1 by (1)(1 + 1) [(1)sum of suffixes in a11] and with thesecond order determinant obtained by deleting the elements of first row (R1) and firstcolumn (C1) of | A | as a11 lies in R1 and C1,i.e., (1)1+1a11 22 2332 33a aa aStep 2 Multiply 2nd element a12 of R1 by (1)1 + 2 [(1)sum of suffixes in a12] and the secondorder determinant obtained by deleting elements of first row (R1) and 2nd column (C2)of | A | as a12 lies in R1 and C2,i.e., (1)1+2a12 21 2331 33a aa aStep 3 Multiply third element a13 of R1 by (1)1 + 3 [(1)sum of suffixes in a13] and the secondorder determinant obtained by deleting elements of first row (R1) and third column (C3)of | A | as a13 lies in R1 and C3,i.e., (1)1+3a13 21 2231 32a aa aStep 4 Now the expansion of determinant of A, that is, | A | written as sum of all threeterms obtained in steps 1, 2 and 3 above is given bydet A = |A| = (1)1 + 1a11 22 23 21 23 1 21232 33 31 33(1)a a a aaa a a a++ + 21 22 1 31331 32(1)a aaa a+or |A| = a11(a22a33a32 a23)a12(a21a33a31 a23)+a13(a21a32a31a22)106 MATHEMATICS= a11a22a33a11a32 a23a12a21a33+a12a31 a23+a13a21a32a13a31a22...(1)$NoteWe shall apply all four steps together.Expansionalongsecondrow(R2)| A | =11 12 1331 32 33a a aa a a21 22 23a a aExpanding along R2, we get| A | =12 13 11 13 2 1 2 221 2232 33 31 33(1) (1)a a a aa aa a a a+ ++11 12 2 32331 32(1)a aaa a++= a21(a12a33a32 a13)+a22(a11a33a31 a13)a23(a11a32a31a12)| A | = a21a12a33+a21a32 a13+a22a11a33a22a31 a13a23a11a32+a23a31a12= a11a22a33a11a23a32a12a21a33+a12a23a31+a13a21a32a13a31a22...(2)ExpansionalongfirstColumn(C1)| A | =12 1322 2332 33112131aaaa aa aa aBy expanding along C1, we get| A | =22 23 12 13 1 1 2 111 2132 33 32 33(1) ( 1)a a a aa aa a a a+ ++ + 12 13 3 13122 23(1)a aaa a+= a11(a22a33a23a32)a21(a12a33a13a32)+a31(a12a23a13a22)DETERMINANTS107| A | = a11a22a33a11a23a32a21a12a33+a21a13a32+a31a12a23a31a13a22= a11a22a33a11a23a32a12a21a33+a12a23a31+a13a21a32a13a31a22...(3)Clearly, values of | A| in (1), (2) and (3) are equal. It is left as an exercise to thereader to verify that the values of |A| by expanding along R3, C2 and C3 are equal to thevalue of | A| obtained in (1), (2) or (3).Hence, expanding a determinant along any row or column gives same value.Remarks(i) For easier calculations, we shall expand the determinant along that row or columnwhich contains maximum number of zeros.(ii) While expanding, instead of multiplying by (1)i + j, we can multiply by +1 or 1according as (i + j) is even or odd.(iii) Let A = 2 24 0 and B = 1 12 0 . Then, it is easy to verify that A = 2B. Also| A| = 0 8 = 8 and | B| = 0 2 = 2.Observe that, | A| = 4( 2) = 22| B| or| A| = 2n| B|, where n = 2 is the order ofsquare matrices A and B.In general, if A = kB where A and B are square matrices of order n, then | A| = kn| B |, where n = 1, 2, 3Example 3 Evaluate the determinant = 1 2 41 3 04 1 0.Solution Note that in the third column, two entries are zero. So expanding along thirdcolumn (C3), we get =1 3 1 2 1 24 0 04 1 4 1 1 3+= 4 (1 12) 0 + 0= 52Example 4 Evaluate = 0 sin cossin 0 sincos sin 0 .108 MATHEMATICSSolution Expanding along R1, we get =0 sin sin sin sin 00 sin cossin 0 cos 0 cos sin = 0 sin (0 sin cos ) cos (sin sin 0)= sin sin cos cos sin sin = 0Example 5 Find values of x for which 3 3 21 4 1xx= .Solution We have 3 3 21 4 1xx=i.e. 3 x2 = 3 8i.e. x2= 8Hence x = 2 2 EXERCISE 4.1Evaluate the determinants in Exercises 1 and 2.1.2 45 12. (i)cos sinsin cos (ii)2 1 11 1x x xx x++ +3. If A = 1 24 2 , then show that | 2A | = 4 | A |4. If A = 1 0 10 1 20 0 4 , then show that | 3 A | = 27 | A |5. Evaluate the determinants(i)3 1 20 0 13 5 0(ii)3 4 51 1 22 3 1DETERMINANTS109(iii)0 1 21 0 32 3 0(iv)2 1 20 2 13 5 06. If A = 1 1 22 1 35 4 9 , find | A |7. Find values of x, if(i)2 4 2 45 1 6xx= (ii)2 3 34 5 2 5xx=8. If 2 6 218 18 6xx = , then x is equal to(A) 6 (B) 6 (C) 6 (D) 04.3 Properties of DeterminantsIn the previous section, we have learnt how to expand the determinants. In this section,we will study some properties of determinants which simplifies its evaluation by obtainingmaximumnumberofzerosinaroworacolumn.Thesepropertiesaretruefordeterminants of any order. However, we shall restrict ourselves upto determinants oforder 3 only.Property 1 The value of the determinant remains unchanged if its rows and columnsare interchanged.Verification Let = 1 2 31 2 31 2 3a a ab b bc c cExpanding along first row, we get =2 3 1 3 1 21 2 32 3 1 3 1 2b b b b b ba a ac c c c c c += a1(b2 c3b3c2)a2(b1c3b3c1)+a3(b1c2b2c1)By interchanging the rows and columns of , we get the determinant1 =1 1 12 2 23 3 3a b ca b ca b c110 MATHEMATICSExpanding 1 along first column, we get1 = a1(b2c3c2b3)a2(b1c3b3c1)+a3(b1c2b2c1)Hence = 1RemarkItfollowsfromabovepropertythatifAisasquarematrix,thendet (A) = det (A), where A = transpose of A.$NoteIf Ri = ith row andCi =ith column, then for interchange of row andcolumns, we will symbolically write Ci RiLet us verify the above property by example.Example 6 Verify Property 1 for = 2 3 56 0 41 5 7Solution Expanding the determinant along first row, we have =0 4 6 4 6 02 (3) 55 7 1 7 1 5+= 2 (0 20) + 3 ( 42 4) + 5 (30 0)= 40 138 + 150 = 28By interchanging rows and columns, we get1 =2 6 13 0 55 4 7 (Expanding along first column)=0 5 6 1 6 12 (3) 54 7 4 7 0 5+= 2 (0 20) + 3 ( 42 4) + 5 (30 0)= 40 138 + 150 = 28Clearly = 1Hence, Property 1 is verified.Property 2 If any two rows (or columns) of a determinant are interchanged, then signof determinant changes.Verification Let = 1 2 31 2 31 2 3a a ab b bc c cDETERMINANTS111Expanding along first row, we get=a1(b2c3b3c2) a2 (b1c3b3c1)+a3(b1c2b2c1)Interchanging first and third rows, the new determinant obtained is given by1 =1 2 31 2 31 2 3c c cb b ba a aExpanding along third row, we get1 = a1(c2b3b2c3)a2(c1b3c3b1)+a3(b2c1b1c2)= [a1(b2c3b3c2)a2(b1c3b3c1)+a3(b1c2b2c1)]Clearly1 = Similarly, we can verify the result by interchanging any two columns.$NoteWe can denote the interchange of rows by Ri Rj and interchange ofcolumns by Ci Cj.Example 7 Verify Property 2 for = 2 3 56 0 41 5 7.Solution = 2 3 56 0 41 5 7 = 28 (See Example 6)Interchanging rows R2 and R3 i.e., R2 R3, we have1 =2 3 51 5 76 0 4Expanding the determinant 1 along first row, we have1 =5 7 1 7 1 52 (3) 50 4 6 4 6 0+= 2 (20 0) + 3 (4 + 42) + 5 (0 30)= 40 + 138 150 = 28112 MATHEMATICSClearly 1 = Hence, Property 2 is verified.Property 3 If any two rows (or columns) of a determinant are identical (all correspondingelements are same), then value of determinant is zero.Proof If we interchange the identical rows (or columns) of the determinant , then does not change. However, by Property 2, it follows that has changed its signTherefore = or = 0Let us verify the above property by an example.Example 8 Evaluate = 3 2 32 2 33 2 3Solution Expanding along first row, we get = 3 (6 6) 2 (6 9) + 3 (4 6) = 0 2 (3) + 3 (2) = 6 6 = 0Here R1 and R3 are identical.Property 4 If each element of a row (or a column) of a determinant is multiplied by aconstant k, then its value gets multiplied by k.Verification Let = 1 1 12 2 23 3 3a b ca b ca b cand 1 be the determinant obtained by multiplying the elements of the first row by k.Then1 =1 1 12 2 23 3 3k a k b k ca b ca b cExpanding along first row, we get1 = ka1 (b2c3b3c2)kb1(a2c3c2a3)+kc1(a2b3b2 a3)= k[a1(b2c3b3c2)b1(a2c3c2a3)+c1(a2b3b2a3)]= kDETERMINANTS113Hence1 1 12 2 23 3 3k a k b k ca b ca b c = k 1 1 12 2 23 3 3a b ca b ca b cRemarks(i) By this property, we can take out any common factor from any one row or anyone column of a given determinant.(ii) If corresponding elements of any two rows (or columns) of a determinant areproportional (in the same ratio), then its value is zero. For example =1 2 31 2 31 2 3a a ab b bk a k a k a = 0 (rows R1 and R2 are proportional)Example 9Evaluate 102 18 361 3 417 3 6Solution Note that 6(17) 6(3) 6(6) 17 3 6 102 18 361 3 4 1 3 4 6 1 3 4 017 3 6 17 3 6 17 3 6= = =(Using Properties 3 and 4)Property 5 If some or all elements of a row or column of a determinant are expressedas sum of two (or more) terms, then the determinant can be expressed as sum of two(or more) determinants.For example,1 1 2 2 3 31 2 31 2 3a a ab b bc c c+ + + = 1 2 3 1 2 31 2 3 1 2 31 2 3 1 2 3a a ab b b b b bc c c c c c +Verification L.H.S. = 1 1 2 2 3 31 2 31 2 3a a ab b bc c c+ + + 114 MATHEMATICSExpanding the determinants along the first row, we get = (a1+1)(b2c3 c2b3)(a2+2)(b1c3b3c1)+(a3+3)(b1c2b2c1)= a1(b2c3c2b3)a2(b1c3b3c1)+a3(b1c2b2c1)+1(b2c3c2b3) 2(b1c3b3c1)+3(b1c2b2c1)(by rearranging terms) =1 2 3 1 2 31 2 3 1 2 31 2 3 1 2 3a a ab b b b b bc c c c c c + = R.H.S.Similarly, we may verify Property 5 for other rows or columns.Example 10Show that 2 2 2 0a b ca x b y c zx y z+ + + =Solution We have2 2 2a b ca x b y c zx y z+ + += 2 2 2a b c a b ca b c x y zx y z x y z+(by Property 5)= 0 + 0 = 0 (Using Property 3 and Property 4)Property 6 If, to each element of any row or column of a determinant, the equimultiplesof corresponding elements of other row (or column) are added, then value of determinantremains the same, i.e., the value of determinant remain same if we apply the operationRiRi+kRjorCiCi+k Cj.VerificationLet =1 2 31 2 31 2 3a a ab b bc c c and1 = 1 1 2 2 3 31 2 31 2 3a k c a k c a k cb b bc c c+ + +,where 1 is obtained by the operation R1 R1 + kR3 .Here, we have multiplied the elements of the third row (R3) by a constant k andadded them to the corresponding elements of the first row (R1).Symbolically, we write this operation as R1 R1 + k R3.DETERMINANTS115Now, again1 =1 2 3 1 2 31 2 3 1 2 31 2 3 1 2 3a a a k c k c k cb b b b b bc c c c c c+ (Using Property 5)= + 0 (since R1 and R3 are proportional)Hence = 1Remarks(i) If1isthedeterminantobtainedbyapplyingRikRiorCikCitothedeterminant , then 1 = k.(ii) If more than one operation like Ri Ri + kRj is done in one step, care should betaken to see that a row that is affected in one operation should not be used inanother operation. A similar remark applies to column operations.Example11Provethat 32 3 2 4 3 23 6 3 10 6 3a a b a b ca a b a b c aa a b a b c+ + ++ + + =+ + +.SolutionApplyingoperationsR2R22R1andR3R33R1tothegivendeterminant , we have =0 20 3 7 3a a b a b ca a ba a b+ + +++Now applying R3 R3 3R2 , we get =0 20 0a a b a b ca a ba+ + ++Expanding alongC1, we obtain =20a a baa+ + 0 + 0= a (a2 0)= a (a2) = a3116 MATHEMATICSExample 12 Without expanding, prove that =01 1 1x y y z z xz x y+ + +=Solution Applying R1 R1 + R2 to , we get =1 1 1x y z x y z x y zz x y+ + + + + +Since the elements of R1 and R3 are proportional, = 0.Example13Evaluate =111a bcb cac abSolution Applying R2 R2 R1 and R3 R3 R1, we get =10 ( )0 ( )a bcb a c a bc a b a c Taking factors (b a) and (c a) common from R2 and R3, respectively, we get =1( ) ( ) 0 1 0 1 a bcb a c a cb = (b a) (c a) [( b + c)] (Expanding along first column)= (ab) (bc) (ca)Example14Provethat4b c a ab c a b abcc c a b++ =+Solution Let = b c a ab c a bc c a b+++DETERMINANTS117Applying R1 R1 R2 R3 to , we get =0 2 2 c bb c a bc c a b++Expanding along R1, we obtain =0 (2 )c a b b bcc a b c a b++ + (2 )b c abc c++= 2 c (a b + b2 bc) 2 b (b c c2 ac)= 2 a b c + 2 cb2 2 bc2 2 b2c + 2 bc2 + 2 abc= 4abcExample 15 If x, y, z are different and 2 32 32 311 01x x xy y yz z z+= + =+, thenshow that 1 + xyz = 0SolutionWehave =2 32 32 3111x x xy y yz z z+++ =2 2 32 2 32 2 3111x x x x xy y y y yz z z z z+(Using Property 5)=2 22 2 22 21 1( 1) 1 11 1x x x xy y xyz y yz z z z + (Using C3C2 and then C1 C2)=22211 (1 )1x xy y xyzz z+118 MATHEMATICS = ( )22 22 211 00x xxyz y x y xz x z x+ (UsingR2R2R1andR3R3R1)Taking o