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Matrices ( Fastracl« Revision ) > Arectangular array of ma numbers arranged in m rows and ncolumns |s called a matrix of order m xn. The numbers are called the elements or entries of the matrix. > The element of ithrow and/thcolumn is represented by ay > Two matrices are said to be equal, if the order of two matrices Is equal and corresponding elements are also equal. > A matrix having the same number of rows and columns Is called a square matrix. > A square matrix in which every non-diagonal element Is zero, is called a diagonal matrix. > A square matrix in which every non-diagonal element Is 0 and every diagonal element is 1, is called an identity or unit matrix. > A square matrix A=[ay] Is sald to be upper triangular matrix, iFay=OV />J. > A square matrix A=[ag] Is sald to be lower triangular matrix, if ay=0 ¥ i If Aand B are two matrices of same order, then their sum (A+) is a matrix of same order which Is obtained by adding the corresponding elements of A and 8. > IF A and B are two matrices of same order, then their difference (A- 8) Is equal to A+ (-8), i.e, sum of matrix A and the matrix (—&). > The product ABof two matrices Aand Bcan be determined only when number of columns in matrix A= number of rows in column B. > IF Aand Bare square matrices of the same order, say ‘n', then both the product AS and BAare defined and each is a square matrlx of order ‘n’. >In the matrix product A&, the matrix A Is called pre-multiplier (pre-Factor) and 8 Is called post-multiplier (Post-Factor). > Properties of Matrix Multiplication (|) AB 2 BA (il) A(BC) =(AB)C (ill) A(B 4 C)=AB + AC (iv) (A+ B)C=AC 4 BC (v) Al=A=IA » The matrix obtained by interchanging rows into columns and columns into rows of matrix Als called the transpose matrix of matrix A, denoted by A” or A’. > IF the order of matrix Ais m xn, then order of A’ =n xm > Some Results on Transpose of Matrices () (A= (ii) (A+ 8)’ =A" + BT (iii) (kA)! = kA (iv) (AB)" = BTA’. > A square matrix of order a xa Is sald to be orthogonal, if AA’ =1,=A'A. > Asquare matrix Als said to be symmetric matrix, if A” =A. > IF Aand B are symmetric matrices of the same order, then the product AB is symmetric, iff BA= AB. > A square matrix A Is sald to be skew-symmetric matrix, if Als—A. > All elements of main diagonal of a skew-symmetric matrix are zero. » Every square matrix can be expressed uniquely as a sum of asymmetric and a skew-symmetric matrices. If Als a square matrix, then A=P + Q; where, P= HAs A’) (where ks a constant) Is asymmetric matrix andQ=3(A-A’) Isaskew-symmetric matrix. > A square matrix A of order ms called invertible, if there exists a square matrix 8 of order m such that AB = BA=1, Also, Bls called the inverse matrix of Aand Is denoted by A’. > The inverse of every invertible matrix Is unique. » IF Aand Bare two invertible square matrices of same order then (Aa)"! = B14"! ren Practice Exercise 7) Multiple Choice Questions Q 1. The number of all possible matrices of order 2 x 3 with each entry 1 or 2 is: (CBSE 2021 Tarm-1) a.16 b.6 c. 64 d. 24 Q2. IfA= [a, is a square matrix of order 2 such that 1, wheni« j 2 = , then A* is: ay te wheniim'f then A° is: (CBSE SQP 2023-24, 21 Tarm-1, CBSE 2023) rel ool bal le] De Q3. Q4. Q5. Q6. Qa. Qa. Q10. Qu. A matrix A=[a,];,, is defined by: 2i+3j, ij The number of elements in A which are more than 5, is: (CBSE 2021 Term-1) a.3 b. es d.6 if 7 a then: ‘CBSE 2023) 2 I5lFlo mn: ( ) axsly=2 b.x=2.ys1 cxslys dx =3,y=2 2a+b a-2b 4 -3 if = , then the value of ae bes [as | enine value a+b-c+2d is: (CBSE SQP 2021 Term-1) a8 b.10 C4 d-8 iraslo 7 id kA banal ae the vali =lz _4| am lap 24° then the values of k,a and b respectively are: (CBSE SQP 2021 Term-1) a.-G,-12 -18 b.-6,-4,-9 c-6.4.9 d.-6, 12,18 b -b . The produa| ‘ [ | equal to: -b aj|b a (CBSE 2023) a? +b? ie) (a+b? O a b a a*+b? (a+b)? O a?+b? O a{? ie) “| ub? o ‘0 b fi o 0 IfA =| ft Jon = 2? then x equa: 2 1 11 (CBSE 2023) a. +1 b.-1 cl d.2 tae? + lg ot) oh, “4 -2 2f °"|3 1 x vt 15+ o-| 1 | su nat @4~36)¢ =D, then x = and a.3 b.—4 c.-6 d6 11 1)\[x 6 ff |0 1 1\|y/=/3}, then the value of 0 0 1i[z} {2 (2x +y =z) is: (CBSE 2023) al b.2 ea d5 [4 x ff} 2)[1 3 -3]|y|=0, then x + 3y -3zis: 3 z al b.3 a4 do Qi2. Q13. Q14. Q16. Q 16. qv. Q18, Q19. Q20. 1-3 If [> 7 | and A? -4A+10/=A, then k is equal to: a0 b.-4 c4and not 1 dlor4 If for a square matrix A, A*-A+1 =0, then A* equals: (CBSE 2023) aA b.A+l cl-A d.A-I If the matrix fF | is commutative with the ¢ aft f mati shen: aa=Ob=c cc=Od=a b.b=Oc=d d.d=Qa=b If a matrix A=[1 2 3], then the matrix AA’ (where A’ is the transpose of A) is: (CBSE2023) 100 bjO 2 0 00 4 a.14 12 3 c/2 3 1 6.(14) a 12 fA +; B Jonaat =3/, then: y =@. (CBSE SQP 2021 Term-1) alia?4py=9 b.l-a?-fy =0 c.3-u*-fy=0 d.3+u*+Py =0 If matrices A and B are of order 3xn and mx5 respectively, then the order of matrix C =5A+3B is: (CBSE SQP 2021 Term-1) a.3x5andmen b.3x5 c.3x3 d.5x5 Matrix A has m rows and (a + 5) columns, matrix B has m rows and (11-1) columns. If both AB and BA exists, then: a. ABand BA are square matrices b. AB and BA are of orders 8x8 and 3x13 respectively c. AB=BA d. None of the above If A is 3x 4 matrix and Bis a matrix such that A’ B and BA’ are both defined, then the order of matrix Bis: (CBSE 2023) adx4 b.3x3 4x4 d.4x3 o1f For the matrix X=|1 0 1], (X?-X)is: 110 (CBSE 2021 Term-1) a.2/ b.3I cl d.5) De ga. gaz. Q 43. gaa. 012 4/2 -1/2 1/2) ff A=|1 2 3;)andAt=|-4 3 |, 3al 5/2 -3/2 1/2] then the values of a and care: all b.1-1 ce d.-l1 If A and B are square matrices of the same order and AB =3/, then A~ is equal to: 1 a.36 b38 , Tiessy oe | d.=8 ° 3 If A and B are square matrices of the same order such that (A +B) (A -B) =A? -B?, then (ABA~*)? is equal to: a.6? b.1 c. A7E? d. A? If A? ~A +1 =0, then the inverse of A is: al-A b.A-I cA dA+l -{ Assertion & Reason type Questions correct. Q45. 0.46. Q47. Directions (Q. Nos. 45-56): /n the following questions, each question contains Assertion (A) and Reason (R). Each question has 4 choices (a), (b), (¢) and (d) out of which only one is The choices are: a. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A) b. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A) c Assertion (A) Is true but Reason (R) Is false d. Assertion (A) Is false and Reason (R) Is true Assertion (A): Matrix 3x 3,49 = — cannot be expressed as a sum of symmetric and skew-symmetric matrices. Reason (R): Matrix 3x 3a, = ! =r is neither i+2j symmetric nor skew-symmetric. ky izj 0; ia where, k is a scalar, is an identity matrix when kot, Reason (R): Every identity matrix is not a scalar matrix. Assertion (A): Scalar matrix A =toy1={ 300 Assertion (A):|0 4 0/is a diagonal matrix. 007 Reason (R): A= (a, ]is a square matrix such that a, =0, Vi j, then A is called diagonal matrix. Qa. gas. Q50. Q 61. Q 62. Q63. Q 54, 55. Q56. is a row 1x4 Assertion ): B=|-5 45 2 3| matrix. Reason (R): If B =[b; ), ,,, isa row matrix, then its order is 1x n. Assertion (A): If [ wy 4 | ( af then z+5 x+y} [0 4 x=2,y =2,z=-Sandw=4 Reason (R): Two matrices are equal, if their orders are same and their corresponding elements are equal. Assertion (A): The product of two diagonal matrices of order 3 x 3 is also a diagonal matrix. Reason (R): Matrix multiplication is always nan-commutative. Let A be a square matrix of order 3 satisfying AA =I. Assertion (A): A’ =A™*. Reason (R): (AB) = BA’. Assertion (A): Let A= [a, ]be an mx n matrix and O be an mx n zero matrix, thenA+O=0+A=A. In other words, O is the additive identity for matrix addition. Reason (R): Let A= la, Imxn be any matrix, then we have another matrix as -A=[-dy],,,, such that A+(-A)=(-A)+A=0. Then, -A is the additive inverse of A or negative of A. Assertion (A): For multiplication of two matrices A and B, the number of columns in A should be less than the number of rows in B. Reason (R): For getting the elements of the product matrix, we take rows of A and columns of B, multiply them elementwise and take the sum. A i wy: ft Axl? *} e-[t ° hi ssertion (A): =l1 4 =lq 4) then (A+B)? =A? +B? +2AB. Reason (R): For the matrices A and B given in Assertion (A), AB = BA. 1-2 2 Assertion (A): If Ass -2 1 24 then -2 -2 -1 A(AT ) = 1. Reason (R): For any square matrix A, (av y =A For any square matrix A with real number entries, consider the following statements: Assertion (A): A + A’ is a symmetric matrix. Reason (R): A A’ is a skew-symmetric matrix. Answers 1. (c) 2. (d) 3. (0) 4. (b) 5. (a) 6. (0) 7. (a) 8. (C) 9. (0) 10. (d) 11. () 12. (c) 13. () 14. (0) 15. (d) 16. (c) 17. (b) 18. (a) 19. (a) 20. (a) 21.(d) 22.0) oD). . ) 2.) 27.(a2) 28) 29. (a) 30. (0) 31.4) 32) 33@ 34. (bt) 35. (c) 36. (a) 37.) 38 ©) = 39.) 40. 41. (0) 42. (bo) 43. (@) 44. (a) 45. (d) 46. (c) 47. (a) 48. (a) 49. (a) 50. (c) 51. (b) 52. (b) 53. (d) 54. (a) 55. (b) 56. (b) el QD Case Study Based Questions * Solutions 2 1. Given data can be written in matrix form as below. Case Study L Pencil Eraser Sharpener Sale Price CostPrice A manufacturer produces three stationery products 3.50 3.25 pencil, eraser and sharpener which he sells in two Market | S500 8000 000 | 175 152 markets. Annual sale is mentioned below: WL .S600) 1.000 J2000) 555 2.00 O75 — . 3x1 . Syd Lat Aq 15,000 6000 8000 ~ {8.500 17.000 12,000 3.50 3.25 8 =| 175 |,C =| 150 2.00 075 GW. AB's ea 6000 uel sia intl Products (in numbers) #200 17.000 12.000 2.00 amet Pencil Eraser | Sharpener 4 [ae 3.50 +6000 x175 + 8000 2) = = x 15,000 6.000 8.000 9,500 x 3.50 + 17,000 x1.75+12,000 x2 Y 9,500 17,000 72,000 = 52,500 + 10,500 + 16.000 7 79,000 33250 + 29,7504 24,000 87,000, If the unit sale price of pencil, eraser and sharpener 18000 6000 aod 3.25 are %3.50, @ 1.75 and ® 2.00 respectively, and unit and Ac =| oon Wo00 vaono| 1.50 cost of the above three commodities are ? 3.25, % 1.50 " ‘ 0.75 and & 0.75 respectively. _J 15.000 3.25 + 6.000 «150+ 8,000 x 0.75 Based on the above information, solve the following 9,500 «3.25 + 17,000 «1.50 + 12,000 x0.75 questions: -| 48750+ 3500 -9000| -[ésa75| Q1. Total revenue of market X is: 30,875 + 25500 + 9,000 65.375 a. = 64,000 b.% 60,000 = c.% 79,000 d.& 81,000 Cn TiP Q 2. Total revenue of market Y is: Total revenue Is the sum of the product of each commodity a, 35,000 b.% 87,000 with corresponding unit sale price and total cost incurred c. ? 53,000 d. € 81,000 Js the sum of the product of each commodity with Q3. Cost incurred in market X is: corresponding unit cost price. a, 13,000 b.% 30,100 247400 d.% 63750 :. Total revenue of market X is 79,000. " : So, option (c) Is correct. 4. Profit in markets. and ¥ respectively are: 2. From the above data, total revenue of market Y is a.€15250 and¢ 21625 b.% 17,000 and ¥ 15,000 # 87,000. c.® 10,000 and & 20,000 d. & 51,000 and & 71,000 So, option (b) Is correct. Q 65. Gross profit in both market is: 3. From the above data, cost Incurred In market X Is a. € 23,000 b.& 32,000 € 63,750 c.% 36,875 d.€ 40,200 So, option (d) Is correct. De September sales (in @) [sooo 20,000 van Ramakishan {50,000 30,000 10,000 | Gurucharan October sales (in ®) 5,000 10,000 6,000 |Ramakishan = so000 10,000 a Based on the above information, solve the following questions: Q1. The total sales in September and October for each farmer in each variety can be represented as: a.A+B b.A-B cA>B dAB dA3 In A = 10,000 G-TiP Azz means, the element in matrix A represented by intersection of second row and third column. So, option (a) is correct. 3. The decrease in sales from September to October Is given byA-B. So, option (b) Is correct. 2 4 2H OB es we e028 5,000 10,000 6,000 20,000 10,000 10.000 Lai 200 fil uucianen -002| 400 200 200] Gurucharan .. Required profit of Ramakishan for each varlety sold In October are ¥ 100, ? 200 and 8 120. So, option (a) Is correct. 5. 2% of A=—2-xA=002 xA 100 10.000 20,000 30000 002 | aun 30,000 fal 200 400 600] Ramakishan -[con0 600 500 ». Required profit of Gurucharan for each variety sold in September are %1.000. % 600 and % 200. So, option (b) Is correct. Case Study 4 To promote the making of toilets for women, an organisation tried to generate awareness through (i) house calls, (ii) emails and (iii) announcements. 1 jaa The cost for each model per attempt is given below: (i) 250 (ii) 220 (iii) F 40 The number of attempts made in the villages X, Y and Z are given below: ji) Gi) X 400 300 100 Y 300 250 75 Z 500 400 150 Also, the chance of making of toilets corresponding to one attempt of given model is: (i) 2% (ii) 4% (iii) 20% Based on the above information, solve the following questions: Q1. The cost incurred by the organisation on village X is: a.%10,000 b.%15000 c%30,000 d.% 20,000 Q2. The cost incurred by the organisation on village Y is: a.% 25,000 b.€18.000 c.% 23.000 d.% 28,000 Qa. The cost incurred by the organisation on village Z Gurucharan is: a.% 19,000 b. % 39,000 c.% 45,000 d.% 50,000 Q 4. The total number of toilets that can be expected after the promotion in village X, is: a. 20 b. 30 c. 40 d.50 Q 5. The total number of toilets that can be expected after the production in village Z, is: a. 26 b. 36 c.46 d. 56 De «—————— Solutions 1. Let A. & B and & C be the cost incurred by the organisation for villages X, Yand Zrespectively. Then A.8, Cwill be given by the following matrix equation: 400 300 100][50)] [A 300 250 75 || 20|=|8 500 400 150}|40 E. [A 400 x50 + 300 x20 + 100 x 40 > B}=| 300 x50+250x20+75 x40 Cc 500 x50 + 400 x20 + 150 x 40 20.000 + 6.000 + 4.000 30.000 15,000 + 5.000 + 3.000 | =| 23,000 25,000 + 8,000 + 6,000 39,000 The cost Incurred by the organisation on village X Is % 30,000. So, option (c) Is correct. 2. From the above data, the cost Incurred by the organisation on village Y is = 23,000. So, option (c) Is correct. 3. From the above data, the cost Incurred by the organisation on village Z Is ¥ 39,000. So. option (b) is correct. 4. Total number of toilets that can be expected In each village Is given by the following matrix: 400 300 100} 2/100 yY)=|300 250 75 }) 4/100 Z 500 400 150}] 20/100 84+12+20 40 Y}=/ 6+10+15 |=] 31 x 10416 +30 56 The total number of tallets that can be expected after promotion in village X Is 40. So, option (c) Is correct. 5. The total number of toilets that can be expected after the production In village Z Is 56. So, option (d) Is correct. Case Study 5 Assume X,Y,Z,W and P are matrices of order 2xn,3xk, 2x p,nx3and p xk, respectively. Based on the above information, solve the following questions: (NCERT EXERCISE) Q 1. The restriction on n,k and p, so that define the order of PY + WY. Q2. If =p, then find the order of the matrix 7X -5Z. e—————_ Solutions 1. Given, order of the matrix P = pxk order of the matrix Y =3 xk and order of the matrix W =n x3 PY is defined when, Number of columns of matrix P=Number of rows of matrix Y = kea (1) Also, WY Is defined when, Number of columns of matrix W =Number of rows of matrix Y => 324 (True) Now, PY + WY is defined when both PY and WY have same order. Order of matrix PY = p x3 and order of matrix WY =n xk Here, restriction for PY + WY are p=nandk=3 2. Matrix X Is of the order 2 xn. Therefore. matrix 7X is also of the same order. Matrix Z is of the order 2 xp |.e.,2 xn (Since, n = p) Therefore, matrix 5Z Is also of the same order. Now, both the matrices 7X and 5Z are of the order 2xn. Thus, matrix 7X —5Z Is well-defined and Is of order 2xn. Case Study 6 Sanjeev, Amit and Nitika were given the task of creating a square matrix of order 3..X, Yand Zare the matrices created by Sanjecv, Amit and Nitika respectively, which is given below: fi. 2-1) [2 1-1] [3 1 0] x-[0 3 1|,Ysll 0 3 /Z=|-I 2: 1 [-1 0 2] 2] = [Oo 1 2 Based on the above information, solve the following questions: Q1.If a=5 and b=~-3, then find the value of (bX) + (@2)". Q 2. Find the value of (XY -¥Z). Qa. If a=-4 and b=-2, then find the value of (a-b) v2)". ° Salutions ° 10 -T 3-10 1. Here,X?=|2 3 OlandZ7=/1 2 1 112 o12 Now, (bx)74(aZ)" =bx™+0z7 10 -1 3-10 =(-3)}2 3 O]+5/1 2 1 112 o12 -30 3] /15 -5 QO] fl2 -5 3 =|-6 -9 Of+/5 10 S]=/-1 1 5 3 -3 -6} |O 5 i] [3a 2 4 e—————_ Solutions pe 1. In 2019, dealer A sold 120 Hatchback, 50 Sedan and 10 SUV dealer 8 sold 100 Hatchback, 30 Sedan and 5 SUV and dealer C sald 90 Hatchback. 40 Sedan and 2 SUV. . Required matrlx, say P, is given by Hatchback Sedan SUV A 120 50 «10 P=B8 100 30 5 Cc 90 40 2 In 2020, dealer A sald 300 Hatchback, 150 Sedan, 20 SUV dealer 8 sold 200 Hatchback, 50 Sedan, 6 SUV and dealer C sold 100 Hatchback, 60 Sedan, 5 SUV. Required matrix, say Q, is given by Hatchback Sedan SUV A 300 150 «20 | Q=8 200 50 6 C 100 €0 5 2. Total number of cars sold In two given years, by each dealer. Is given by Hatchback Sedan SUV A| 120+300 50+150 10+20} P+Q=6/100+200 30+50 5+6 C{ 90+100 40+60 2+5 Hatchback Sedan SUV Af 420 200 30 oB 300 80 611 Cc 190 oo 7 Or The Increase In sales from 2019 to 2020 Is given by Hatchback Sedan SUV A/300-120 150-50 20-10 Q-P=8/200-100 50-30 3] C{ 100-90 GO-40 5-2 Hatchback Sedan SUV A 180 100 10 =B 100 20 1 ej 10 20 3 3, The amount of profit In 2020 received by each dealer is glven by the matrix Hatchback Sedan SUV A 300 150 20 50,000) 8 200 50 6 100000 G 100 60 5 }{| 2.00000} A{1.50.00.000 + 1.50.00.000 + 40,00,000 w B] 1.00,00,000 +. 50.00.000 + 12,00,000 C| 50.00.000 + 60.00.000 + 10,00.000 A/[3.40.00.000 » B| 1.62,00,000 C | 1,.20,00.000 ay Very Short Answer type Questions Q1. Ifa matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements? (NCERT EXERCISE) Q2. How many number of matrices are passible of order 3x 3.with each entry O or 1? (NCERT EXERCISE) Q3. Construct a 3x 2 matrix whose elements are given byay = ; V3. (NCERT EXERCISE) Q4, Find the values of x, y and z fram the: x+ysz| [9 x+y [=|5 (NCERTEXERCISE) Yrz z 2+/ 4 1+/ 2 06. IfA=|"- andB=|". ~ |,thenfindA+ 8. 3° A 2i3 4 2 13) 2 0 3 Q6. If A=|}O 5 7] and B=|3 10 5j, then find 68 9 5 7 0 (A-28). Q7. Find the value of x—y, if alo el*[t alla a} (NCERT EXERCISE; CBSE 2019) 5 0] 14] 3 Q8. If sa and > } then find the 25 matrix A. (CBSE 2019) 21 Q9. If X4+¥ [i 2 “1 2 | 2X -Y ly | then find the value of X. 010, ral ‘ 32 25 4 i 3 and B , then find BA. 1. ifa-| + —Janda=|? *),then show that AB Qu. al 4 [an =|4. 2 }rthen show that is a zero matrix, Q 12. Find the value of the matrices 2 : l 10 j x . —1 2 1 =—— 4 im an Oa -3 Q13. If the matrix A =| 2 0 -1| Is skew-symmetric, |b 1 0 find the values of a and b, (CBSE 2018) y Short Answer Type-! Questions [2 +2 «2 2 Qt fas pats ae and Be ou See) [sin® x cos” x cos’ x sin” x then find (A+ 8). (NCERT EXERCISE) ita=[? “Janae=|* °), then sind (44 8)and gz =l3 2/8" =|_> 5)» then find (A+ B)ani (A-B). Qa. xere|s gfenex-¥ =[9 “| then find the 09 QO -1 values of X and Y. 2 0 f Q4. IfA=|2 1 3|,thenfind (A? -5A). 1-10 Q5. Find a matrix A such that 2A-—38+5C =O, where B -2 2 0 dc 20 -2 =] 4 4lMdCal7 4 gf (cosezor9) 0 -1 3.5 Q6. Ka-{y 3 |one a-[) o}ten find the values of AB and BA. (NCERT EXERCISE) Q7. Show that all the diagonal elements of a skew- symmetric matrix are zero. (CBSE 2017) 1 -2 3 4 27 . Q8. Wael, 2 s]e4 [5 5 i] then fina AB. Qo. rae 8 of then prove that (A’) =A. del 8 baa 7 j ; then fi Q10. If lig 2 and B=} = 7}, then find 210 (Ay. 2 -3 oi Qu. GivenA= 47 |, compute A~ with the help of 2A-! = 9/-A. Y Short Answer Type-ll Questions guy t” 2 6 a = S+z oy 5 2 a} then find the values of x, yandz. (NCERT EXERCISE) Q 2. Simplify: cos§ -sin@ . sin® cos0 cosO] | +sin0 i sinOd cosd -cos0 sind (NCERT EXERCISE) Qa. Q4. Q5. Q6. Q7. Qa. Qa. Q10. Qu. Q12. Q13. Q14. Q16. Q16. 1 -1 3 IfA= 14 , then prove that A~ =4A. Li 5 IfA=|/3 -2 1), then show that 421 A} -23A-40/ =0. (CBSE 2023) 40 125 If A= and B=/2 1),then find AB and 346 1.5 BA ls AB= BA? cos x If F(x)=|sinx cosx 0|,then prove that (0) 0 1 F(x) Fy) =F (x+y). IfA |; p is such that A? =/, then prove that yY ~a -sinx 0 (NCERT EXERCISE) 1-a? —py=0. Aid iekeotxi et 21a 8? ‘ind matrix X so that 45 6fl2 4 6 (NCERT EXERCISE) (CBSE 2017) 2 -1 5 2 25 Let a|3 a} e-[) neces sp tter find a matrix Dsuch thatCD-AB=0. (CBSE 2017) Find matrix A such that: 2 -1 -1 -8 2. 0 Az|/1 -2 “344302 9 22), 02 (CBSE 2017) Find the value of x from the following: 1 0 2)[x [x-5-1]]/0 2 4]|4]=0. 2 0 3i{1 Using an example, prove that (A+B) =A'+B’, where A and B are matrices of same order. fi si If As| OS" im and A+-A’=/, then find the [sinw cosa value of o.. (CERT EXERCISE) f2 1 -1 2 IfA= and B= , then prove that [3 7 305 (ABy = BA’. f=-2 IfA=| 4|,B=[1 3 -6),then verify that: [5 (ABy =BA'. (NCERT EXERCISE) If Aand B are symmetric matrices, then prove that AB-BA is askew-symmetric matrix. De 6. = TiP IfA is a matrix and kis a scalar, then KA is another matrix which is obtained by multiplying each elements of A by a Scalar k. 4 2 13 2 0 2 3A-28=3x/0 5 7/-2x1/3 10 5 6689 5 7 0 i & 3g 4 0 6 =|0 15 21}/-|6 20 10 18 24 27] [10 14 QO 12-4 6-0 39-6] [8 6 33 =| 0-6 15-20 21-10/=|/-6 -5 11 18-10 24-14 2-0 8 10 27 7. We have, Wt 13 ie) 5 6 al hl 5| F | f2 6] fy 0) j5 6 = \o mde aI i al 2+y 5 5 6 ~ [ 1 wee "ty | TRICK Two matrices A and B are said to be equal, if: * order of Aand B is same. * corresponding elements of A and 8 are equal. uw On comparing both sides, we get 2+y=5>y=5-2=3 and 2x+2=8>2x=8-2=6 => So, X-y=3-3=0 8. We have, Adding the given matrix equations. (eov)oQx-n=| HL 4 2} {2 - xi ]= fh] wal) 2G 5] TRICK Product of two matrices is possible when number of columns of first matrix is equal to the number of rows of second matrix. {249 4+6] [0 10 =| aang -A+10} |1 2 i ae.|t * _f b+ (A714 (-).1 =) V4) Ye £1411 (A141 _[ 1-1 1-1]_f0 0 = load -141)° [0 0 Clearly, AB Is a zero matrix. COMMON ERROR Ritson | Mosily students commit error while multiplying matrices. 2 1 2.)3 2 | ; 1 ov heyy 2nd TiP Product of two matrices is possible when number of columns of first matrix is equal to the number of rows of second matrix. Hence proved. 3A-a-|+ a fene8-[ s| TiP Addition of two matrices is possible when its order are same. > 3JA= a0 +B= a0 + o 11 11 25 93 a1 samp sae] af} | ‘TiP Two matrices can be added, if they are of the same order. 2x141x(-1) 2x04+1x2 2x14+1x1 3xl42x(-1) 3x0+2x2 3x14+2x1 (A) x14 1x(A) AxO41x2 -1x14 1x1], 5 ul 2-1 0+2 241 i2a =/3-2 0+4 34+2]/=| 1 45 -1-1 0+2 -1+1 220 Oa -4 13. Given that, matrix A=}2 O -1|Is skew-symmetric b1 60 matrix. ATs-A Qo -3)' [Oa -3 = 2 0 -l| s-/2 0 -1 b 10 6160 TRICK Two matrices A and B are said to be equal, if: + order of A and B Is same. * corresponding elements of A and B are same Le, ays by V /and j. De Oo 2 6b] fO «3 => a O lJ=/-2 0 1 3-10 ob -1 0 On comparing corresponding elements. we get -a=2 => a=-2 and -d=-b => b=3 Short Answer Type-! Questions cos? x il ies el sin?x cos? x L A+B= cos? x sin? x _[sin? x +cas? x sin? x + cos? x les x+cos?x sin? x +cos? ‘| i qT “hy 24 13 ai 7 2 aves (3 7] and A-H= = & - ia 3 2) [25 1 [5 2 3. Adding the given matrix equations, we have (xaye(x-yy-[2 Io ‘] (- sin? 0+cos? 0 =9] " wo q N uw + 4 a 1 un iy a -5 -l -d =|-l -7 -10 “Hf 2 220 5. Given that B= 31 4 2 0 -2 and ae [ 716 ] TRICK Here, A must be taken of the order 2 x 3, because Band C are of the same order 2 x 3 and the sum of two or more matrices is possible only when they are af the same order. a 9}*{o a (TIP ] Two matrices can be added/subtracted, if they are of the def We have, 2A -36+5C =O jfe © <]_4f-2 2 9],,[2 0 2 = *lae fl la 1 4f*°-l7 1 6 _{9 0 0 “(og 0 0 TRICK Here, O i.e, zero matrix must be taken of the order 2 x 3, because two matrices are said to be equal, if they are ofthe same order. F b | Let A= same order. > 2X= o-anounoam Again, X4+Ve > Ye 201 4 Wehave,A=|2 1 4 1-10 it) 1 2 fz 01 Now, A*=A-Az=]2 3)x2 1 3 1-1 of [1 -10 4+0+1 0+0-1 2+0+0 =/44+2+3 O+1-3 24+34+0 2-240 O-1-O 1-3+0 5 -1 2 2065 -1 -2 -1 2 2 0 1) -2 5]|-5]/2 1 3 -1 -2 1-10 . A?-5A= oun aw =(20 25 2c) 76 -6 0 },/19 9 -10 {2d 2e 2f|°[-9 -3 -12]"|35 5 30 ao0 [5 i) _ _, [20+16 2b-6 2c-10] fo 0 0 |2d+26 20+2 araal"|a Q ‘| On comparing the correspanding elements, we get 20+16=0 > ose eB 2b-6=0 > b=S=3 2c-10=0 > c= 2d+2650 > ds—«=-13 2e+250 > eaSe-t and 2 +18=O = fe——n-9 So, matrix A @ BG Aas ’ "Id e f{"[-13 -1 -9 « aae[2 [3 5]_fo+o o+07_fo o “to 2j{o o}"lo+0 o+0}"lo o and sel? 2][2 -]_fo+o -3+197_fo 7 “lo al{o 2}"[o+0 o+0 }"Jo o De G-TiP (22 Fe hE Z| 0 -2-2 242 4 4 - and sana TH | (2) a if [4 4] From eqs. (1) and (2). we get AR =4A Hence proved. 4. We know that. 12 371 2 3 A?=A-Az|3 -2 1|[3 -2 1 42 ¥ 421 146412 2-44+6 34+2+3 3-64+4 64442 9-241 [44+6+4 8-442 12+2+1 9 4 8 7 2 B 4 6 15 fl 2 3/719 4 8 sA-A?=/3 -2 1])1 12 8 4 2 1[14 6 15 19+2+42 4424418 8+164+45 =| 57-2414 12-24+6 24-16+15 7642414 164+24+6 32416415 > ws 7 63 46 69 =|69 -6 23 [92 46 63 Now, A? -23A - 401 [ea 46 69 123 loo =|69 -6 23/-23}3 -2 1]-40/0 1 0 92 46 63 4 21 oa1 [63 46 69 -23 -46 -69 =/69 -6 23/+/-69 46 -23 92 46 63 -92 -46 -23 “40 0 0 +/ 0 -40 0 o Oo -40 [63-23-40 46-46+0 69-69+0 69-69+0 -6+46-40 23-2340 [92-92+0 46-4640 63-23-40 fo ao =!0 O O|=0 Hence proved. Some students make square and cube the elements in A and obtain the answer. Give ample practice on problems based on multiplication of two matrices. AB fl 2 ‘| ; oo [44445 oa aal"|se = [124846 044430)" [26 34 Pale 15 and BA [4+0 8+0 20+0 2+3 4+4 1046 [1+15 2420 5+30 f4 8 20 =|5 8 16 [16 22 35 Here, AB #BA COMMONJER RO R Some students commit an error while calculating. cosx -sinx O 6 Glven, F(x) =|siInx cosx O (1) ie) QO 1 cosy -siny O F(y)=|siny cosy O (Replace x byy) Oo 0 1 cosx -sinx O|fcosy -siny QO) F(x)F(y)=| sinx cosx Ol|siny casy O Q 0 1 a is) 1 cos x cosy -sinxsiny +O =|sin x cosy+cos xsiny +0 0+0+0 -cosxsiny-sinxcosy+O 0+0+0 -sinxsiny+cosxcosy+O O+0+0 0+0+0 04+0+1 cosxcosy-sinxsiny -cosxsiny-sinxcosy O =|sinxcosy+cosxsiny -sinxsiny+cosxcosy O ie) i°) 1 [nea e sin(A+8)=sinA-cos B+ cosA-sinB ¢ cos (A+B)=cosA-cosB=sinA-sinB cos(x+y) -sin(x+y) O s|sin(x+y) cos(x+y) O oO Q 1 Replace x by(x + y)in eq. (1), we get cos(x+y) -sin(x+y) O F(x+y)=| sin(x+y) cos(x+y) O Q Q 1 F (x)F (y)=F (x+y) Hence proved. De 1 Given, A~|" ‘| 1 [re J If two matrix are equivalent, then its crept elements are equal. weaaz[™ Pile P - - y -ally -« _| a? +By oB-ap ay -ay By +a? (Pim o | Q By +a] Given: A al u7+By 0 -(; "] O py+a?] [o 1 On comparing corresponding elements, we get a*+By=1 > l-a*-By =0 Hence proved. 8. Given, 1 22 -7 -8 -9 The thelz «the oo Let X= ab C diye TR!ICK The product of two matrices are possible when number of columns of first matrix is equal to the number of rows of second matrix. As, RHS have a matrix of order 2 «3. So, LHS must contain the matrix of order 2 x 3. For this X should be a matrix of order 2 x 2. ie,2x[22)x3=2x3 Fram eq. (1), we get gd b|[1l 2 3 -7 -8 -9) [? alla 5 sf[2 4 6! _.fa+4b 20+5b 30+6b -7 -B -9 her 2c+5d xaeul"| 2 4 é| On comparing both sides, we get a+4bs-7 => a=-7-4b .(2) 3q+6b=-9 = a+2b=-3 .(3) c+4d=2 = c=2-4d (4) and 3c+6d=6 => c+2d=2 .(5) Fram eqs. (2) and (3), we get -7-4b+2b=-3 = 2b=-4 = b=-2 Fram eq. (2), a=-7-4(-2) =-7+8=1 > a=] From eqs. (4) and (5). we get 2-4d+2d=2 => d=0 From eq. (5), 2 -T Bi 2 9. Ghen.A~] a }e-[7 ‘ 25 and c-| | 38 TRICK The order of matrix D must be 2 x 2, as product of two matrices is possible when number of columns of first matrix is equal to the number of rows of second matrix. After that the addition or subtraction of two matrices is possible onlywhen the order of both matrices is same. Let. matrix o-|? ‘| cd We have, CD- AB=0 2 5]fo b] [2 -1f5 2] so o *13 alle alla 4|l7 4]"|o o (Here, Ols zero matrix) 2a+5c 2b+5i 10-7 4-4 oo *|30+8¢ 3b+8d| {15+2a 6+16|"|0 0 5 20+5c 2b+5 3 oO] jo 0 3a48c 3b+8d} {43 22]~|0 0 leona 2b+45d rls a| 3q+8c-43 3b+8d-22 aoa On comparing both sides, we get 2a+5c-3=0 = 204+5c=3 -(l) 2b+5d=0 => 2b=-5d ..(2) 3o0+8c-43=0 = 30+8c=43 ...(3) and 3b4+8d-22=0 =» 3b18d=22 (4) From eqs. (2) and (4), we get 3 (- 2) +8d =22 = -15d + 16d = 44 = d=44 From eq, (2), we get 2b=-5 x44 > b=a-5x22 > b=-110 From egs. (1) and (3). we get 3(=*)+ 8c = 43 => 9-18c+16c =86 > ca7/ Put the value of c in eq. (1), we get 20+5x77 =3 = 20 =3-385 = -382 = a=-191 So, matrix D (7 ul 77 44 De -1 3][2 3 and Rus =Ara'=| > all; ;| _f-2+3 -3+21] [1 18 {445 6435] [9 41 Hence, LHS =RHS Hence proved. [COMMON] ERR()R Students commit error when multiplying the matrices together. -2 -2 -6 12 15. AB=| 4 |{1 3 -6)=| 4 12 -24 Give ample practice in problem based on multiplication [ of two matrices. 5 5 15 -30 TiP 2-6 2] = (AB)'=| 4 12 -24 5 15 -30 2 4 = «5 =|-6 12 15 (I) 12 -24 -30 1 Now, B'=| 3 JandA'=[-2 4 5} -6 1 2 4 = «5 =>8A'=| 3 |[-2 4 5)=|/-6 12) 15 a2) -6 12 -24 -30 From eqs. (1) and (2), we get (AB)! =B'A’ Hence proved. . Given, Aand B are symmetric matrices. A'= AandB'=8 AB)-(BA) [+ (X-Y)'=x'-Y] =B'A'-A'B' (: (xy) BA-AB (- B=BA' =-(AB-BA) .. AS-BAls askew-symmetric matrix. Hence proved. . Since, A and B are both symmetric matrices, therefore A’= A and8’=B Let AB be symmetric, then (AB)'= AB => B'A'= AB . BA= AB (. A’= AandB’= 8) Conversely, if AB = BA, then we shall show that ABIs symmetric. Now, (AB)'=B'A'= GAs AB (.:AandBare symmetric) Hence, ABIs symmetric. Hence proved. 18. (i) Let A be a symmetric matrix. then A'sA (@:AB)' = (5 (48) = (AB)’(B’)'=(B'A') B= BAB (- (AB)'=B'A' and A’ = A) Hence,g'ABis asymmetric matrix. Hence proved. (i) Let A be a skew-symmetric matrix, then A'=-A Now. (8' (AB))'= (AB) (6’)’ B'A')B=6'(-A)B [> A’=-A) B'AB = - (B'AB) Hence, 9'AB Is a skew-symmetric matrix. Hence proved. 18. Let Abe a square matrix, then Aad(ALA)+R(A-A)=P +O where, P =3(4+A')and o=3(4 -A’) Now, P* ={ha +A Peds A’) bs (kA) =kAY =3a4 (ay) [.(A+6)= AB] =3(4'+A) ((a'y=4) =A +A’) (From the commutative law of addition of matrices) =P So. P Is a symmetric matrix. Again o={3(A-A } =3(A-a'y=sta'-(A'Y) ib Tapa. ep =} (A'-A)=-2(A-A)=-0 Sa, Qis a skew-symmetric matrix. Hence, the square matrix A can be expressed as the sum of a symmetric matrix 3(A+A’) and a skew-symmetric matrix (A —A'). Hence proved. Long Answer Type Questions 15 TiP Adequate practice is required in problems based on multiplication of two matrices. 5 2 a(2 3 2aAAs woaal? als a] 4-3 646 1 @ “|-2-2 -3+4]"|-4 1 -B8 -12 sal 2 ~Y] vO id Ts an 0 | 1 12 -B -12 70 2 = A’-4A+ T= ah ha wala “| 1-847 12-1240 "(-444+0 1-847 0 ao 0 af Oo Hence proved. De > ? = 4A-7I Now, A? =A-A? =A (4A-71)= 44? -7Al =4(4A-7I)-7A (-Al=A) =1GA-281-7A=9A -281 Again. AP = A.A? =(9A-281)-(44-71) =36A?-63A-112A +196! = 36 (4A-7/)-175A+ 1961 =-31A-561 [2 3 1d 3/2 2|-ss|¢ | _[-1B -93 ~{ 31-118 ERROR Most students find difficulty in finding the correct value of A’. Oo -tanZ 2. Given, A=] 2 tan— ans ie} Let tanSe=t. z We know that, 2tan= sina = 2 1+ tar? S 2 => sina = 2h 7 +t ]-tan? S and cosas 2 14 tan? 1-0? > cosa =——— |+t4 O -t af J] and oe B/E =e a1 £0 -t 1 T © Oo -t] Now, LHS =/+A= | low, + Py ack a| ] -t] t 1] and as = ('-a)[ ee | sina cosa filet? __2t a[) ier ne -t aff] 2e 1-¢? ie he TRICKS 2 2 Ix(I-t Di. xt x -2t efylat l+t l+t let l+t 1? at -2t) 1-t x +1x tx +1x, let let? let l4t -t tae 2t? __2t tt?) -| Ge he Ge Ge t(l-t?) 2t 2t? 1-t? “Ve ne oe ae 1-t?4+2t? -2t+t-t? =| et 1+t? —t+t?42t 2t7+1-17 let? Tet it -t (+t?) iz ire Te? i(+t?) lat? 1+t? let? is = Jus td cosa -sing Hence,/+ A=(I- al | Hence proved. sina cosa We shall prove it by using principle of mathematical Induction. _— Here, P(n):If A= ee ee -siné cos6} 6 Sl Then, A.a| 8 snp ineN -sinné cosné cos@€ siné Now, P(l):A= i (0 -sin® cosé A coso sin@] “[-sine case} Therefore, the result Is true forn=1. Let the result be true forn =k. cos@ sing Pk}: A= -sind cosd 4 _[ cosko sink® Then 7 -( Sie cos ké Now, we shall prove that the result holds forn =k +1 also. Now, Att'= A: AK = fcos® sin®]{coske sinke [-siné cosé ‘ea aul cos 8-cosk@—sin @:sinkO | cbt cost dh sink6-cos @+cosk0-sin@ ae era © sin(A+B)=sinA-cosB+cosA-sinB © cos (A+ B)=cosA-cos B-sinA:sin8 TiP Use the correct formula in the right place. -[Srioce ae -sin (k0+6) cos(0+k0) cos(k+1)@ sin(k+1)@ a tend cos(k-+1)6| Therefore, the result Is true forn =K+1also. De