NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 (Ex 3.2)

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NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 (Ex 3.2)

NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 help the students in scoring good marks in their examination from this particular chapter. The NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 are provided for the benefit of the students. Students tend to get confused while solving questions, and Extramarks provides the students with access to get the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 in one place. NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 provided by the Extramarks experts develop in students a basic understanding of the chapter and boost their confidence when they solve the questions. These solutions are provided in a step-by-step manner so that the students understand the concept and steps correctly.

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Access NCERT Solutions for Class 12 Maths Chapter 3 Matrice

Here is an example of NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2:

Question: The bookshop of a particular school has 10 dozen Chemistry books, 8 dozen Physics books, and 10 dozen Economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Answer: The bookshop has 10 dozen Chemistry books, 8 dozen Physics books, and 10 dozen Economics books.

The selling prices of

a Chemistry book is Rs 80,

a Physics book is Rs 60,

and an Economics book is Rs 40

The total amount of money that will be received from the sale of all three books can be represented in the form of a matrix as:

Thus, the amount received by the bookshop is Rs 20160 from the sale of these three books.

Topics Covered In Class 12 Maths Chapter 3 Exercise 3.2

NCERT solutions for Class 12 Maths Chapter 3 Exercise 3.2 include topics such as

  • Addition of matrices
  • Multiplication of matrices
  • Properties of matrix addition
  • Properties of scalar multiplication of a matrix
  • Properties of multiplication of matrices
  • Multiplication of a matrix by a scalar

The properties of multiplications included in this chapter are ranging from existence, associative, and distributive multiplicative identity. This is also provided and covered in the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2. The exercise has 22 questions that can be easily solved if the students grasp the concept firmly.

The NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 provided by Extramarks helps in redefining the students’ clarity as far as the exercises are concerned. The questions and topics covered in the NCERT Solutions help the students if they will be appearing in any kind of competitive examinations.

What is a Matrix?

As per the NCERT definition of a matrix, “A matrix is an ordered rectangular array of numbers or functions. The numbers or functions are called the elements or the entries of the matrix”.

What are the Operations of Matrices?

The chapter outline as provided by NCERT contains certain operations on matrices, namely, the addition of matrices, multiplication of a matrix by a scalar, difference, and multiplication of matrices. Combining any two or more matrices in a single matrix is also one of the operations.

NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2

While students start preparing for their examination, they can go for NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 provided by Extramarks’ subject-matter experts. This chapter includes a list of concepts which will make the students capable of solving questions when they appear in the examination by practising with the help of NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2. If students opt for getting the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2, then it will be much easier for them to score good marks. As mentioned above, students will have to solve more problems on their own to gain clarity of the concepts. It is easier said than done, but students become capable of solving questions with practice.

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The NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 are provided by expert academicians. The goal of students to crack the Class 12 exam is made easy with the NCERT solutions. This can come true only if they solve the questions in the easiest way possible. The guidance provided by the experts at Extramarks plays an important role to help them solve any confusion when the students are preparing for the board exams.

Extramarks provides easy-to-understand and comprehensible solutions for NCERT Exercise 3.2 Class 12. These answers may benefit in eliminating any uncertainties that students may have when attempting to answer the NCERT questions found in textbooks.

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The problems and questions from the exercise have been meticulously handled by the internal subject matter experts at Extramarks while adhering to the CBSE requirements and guidelines. Any student in Class 12 can score good marks on the final exam if they are thorough with the ideas from the Maths textbook and quite knowledgeable about the problems and questions from the exercises provided in it. Students may readily comprehend the type of questions that may be given in the exam from this chapter and learn the chapter or topic weightage in terms of overall grade by using NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2. Having said that, they can adequately study for the final exam.

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NCERT Solutions for Class 12 Maths Chapter 3 Other Exercises

Chapter 3 Matrix Exercises
Exercise 3.1
10 Questions & Solutions (5 Short Answers, 5 Long Answers)
Exercise 3.3
12 Questions & Solutions (4 Short Answers, 8 Long Answers)
Exercise 3.4
18 Questions & Solutions (18 Short Answers)

Q.1 In the matrix A= [2 43 2], B=[   1 3−2 5], C=[−2 5   3 4]Find each of the following:(i) A+B (ii) A−B (iii) 3A−C (iv) AB(v) BA

Ans

In the matrix A=[2 43 2], B=[   1 3−2 5], C=[−2 5   3 4](i) A+B=[2 43 2]+[   1 3−2 5]       =[2+1 4+33+(−2) 2+5]       =[3 71 7](ii) A−B=[2 43 2]−[   1 3−2 5]       =[2 −1 4−33−(−2) 2−5]       =[1    15−3](iii)3A−C=3[2 43 2]−[−2 5   3 4]  =[6 129   6]−[−2 5   3 4]  =[6−(−2) 12 −59−3 6 −4]  =[8 76 2](iv)        AB=[2 43 2][   1 3−2 5]  =[2−8 6+203−4 9+10]  =[−6 26−1 19](v)        BA=[   1 3−2 5][2 43 2] =[   2+9     4+6−4+15 −8+10] =[11 1011   2]

Q.2

Compute the following:(i)    [  a b–b a]+[a bb a](ii) [a2+b2 b2+c2a2+c2 a2+b2](iii) [–1 4–68 5 162 8   5]+[12 7 68 0 53 2 4] (iv) [cos2x sin2xsin2x cos2x]+[sin2x cos2xcos2x sin2x]

Ans

(i)    [  a b−b a]+[a bb a]=[  a+a b+b− b+b a+a]        =[2a 2b0 2a](ii)[a2+b2 b2+c2a2+c2 a2+b2]+[   2ab 2bc−2ac      −2ab]     =[a2+b2+2ab b2+c2+2bca2+c2−2ac a2+b2−2ab]     =[(a+b)2 (b+c)2(a−c)2 (a−b)2](iii)[−1 4− 68 5 162 8   5]+[12 7 68 0 53 2 4]     =[−1+12 4+7        − 6+68+8 5+0 16+52+3 8+2   5+4]     =[11 11 016   5 21  5 10  9](iv) [cos2x sin2xsin2x cos2x]+[sin2x cos2xcos2x sin2x]     =[cos2x+sin2x sin2x+cos2xsin2x+cos2x cos2x+sin2x]     =[1 11 1]

Q.3

Compute the indicated products:(i)    [  a b−b a] [a–bb   a]      (ii) [123] [234](iii) [1 −22    3] [1 2 32 3 1](iv) [2 3 43 4 54 5 6] [1 –3 50    2 43    0 5]

( v )  [    2 1    3 2 −1 1 ][    1 0 1 −1 2 1 ] ( vi )[    3 −1 3 −1    0 2 ][ 2 −3 1    0 3    1 ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaamaabmaabaacbeGaa8NDaaGaayjkaiaawMcaaiaaykW7caaMc8+aamWaaqaabeqaaiaaykW7caaMc8UaaGPaVlaa=jdacaWLjaGaa8xmaaqaaiaaykW7caaMc8UaaGPaVlaa=ndacaWLjaGaa8NmaaqaaiabgkHiTiaa=fdacaWLjaGaa8xmaaaacaGLBbGaayzxaaWaamWaaqaabeqaaiaaykW7caaMc8UaaGPaVlaa=fdacaWLjaGaa8hmaiaaxMaacaWFXaaabaGaeyOeI0Iaa8xmaiaaxMaacaWFYaGaaCzcaiaa=fdaaaGaay5waiaaw2faaaqaamaabmaabaGaa8NDaiaa=LgaaiaawIcacaGLPaaadaWadaabaeqabaGaaGPaVlaaykW7caaMc8Uaa83maiaaxMaacqGHsislcaWFXaGaaCzcaiaa=ndaaeaacqGHsislcaWFXaGaaCzcaiaaykW7caaMc8UaaGPaVlaa=bdacaWLjaGaa8NmaaaacaGLBbGaayzxaaWaamWaaqaabeqaaiaa=jdacaWLjaGaeyOeI0Iaa83maaqaaiaa=fdacaWLjaGaaGPaVlaaykW7caWFWaaabaGaa83maiaaxMaacaaMc8UaaGPaVlaa=fdaaaGaay5waiaaw2faaaaaaa@858D@

Ans

(i)    [  ab−ba][a−bb   a]=[   a2+b2−ab+ab−ba+ab   b2+a2]      =[   a2+b2−ab+ab−ba+ab   b2+a2](ii)[123][234]=[2 3 44 6 86 9 12](iii)[1−22   3][1 2 32 3 1]=[1−42−63−22+64+96+3]     =[−3 −41   8 139](iv)[2 3 43 4 54 5 6][1−3 50    2 43    0 5]=[2+0+12−6+6+010+12+203+0+15−9+8+015+16+254+0+18−12+10+020+20+30]          =[1404218−15622−270](v)      [   21   32−11][   101−121]=[2−10+22+13−20+43+2−1−10+2−1+1]=[123145−220](vi) [   3−13−1   02][2−31  03  1]=[   6−1+9−9+0+3−2+0+6   3+0+2]=[14−645]

Q.4

If A= [12 −350    21      −1  1], B=[3 −124    252    03] and C=[4 120 321 −23],then compute (A+B) and (B−C).  Also, verify thatA+(B−C) = (A+B)−C.

Ans

A+B=[12 −350    21     −1  1]+[3−124    252    03]  =[1+32+(−1)−3+25+40+22+51+2    −1+0  1+3]=[4    1 −19    2    73 −1   4]B−C=[3 −124    252    03]−[4 120 321 −23]  =[3 −4−1−12−24 −0    2−35−22−1   0−(−2)3−3]  =[−1−20  4−13  1   20]A+(B−C)=[12−350   21     −1  1]+([3−124   252   03]−[4120321−23])     =[12−350   21     −1  1]+([3−4  −1−12−24−0     2−35−22−1   0+33−3])     =[12−350   21     −1  1]+[−1−20  4−13  1   30]     =[1−12−2       −3+05+40−1   2+31+1        −1+3  1+0]     =[00       −39    −1   52        2   1](A+B)−C=([12−350   21     −1  1]+[3−124   252   03])−[4120321−23]     =[1+32−1      −3+25+40+22+51+2       −1+0  1+3]−[4120321−23]     =[41     −19273       −14]−[4120321−23]     =[4−41−1     −1−29−0          2−3       7−23−1       −1+2      4−3]  =[00     −39     −1 52         1 1]Hence,  A+(B−C)=(A+B)−C.

Q.5

If  A=[23153132343 73223] & B=[25351152545756525], then compute 3A−5B.

Ans

Given, A=[23 1 5313 23 4373 2 23] and B=[25351152545756525]3A−5B=3[23 1 5313 23 4373 2 23]−5[25 35 115 25 4575 65 25]         =[3×233×13×533×133×233×433×733×23×23]−[5×255×355×15×155×255×455×755×655×25]       =[2 3 51 2 47 6 2]−[2 3 51 2 47 6 2]      =[2−23−35−51−12−24−47−76−62−2]      =[0 0 00 0 00 0 0]

Q.6

Simplify cosθ [cosθ sinθ−sinθ cosθ]+sinθ [sinθ    −cosθcosθ sinθ].

Ans

cosθ [cosθ sinθ−sinθ cosθ]+sinθ[sinθ     −cosθcosθ sinθ]=[  cos2θ     cosθ sinθ−sinθ cosθ cos2θ]+[sin2θ      −sinθ cosθsinθ cosθ sin2θ]=[  cos2θ+sin2θ    cosθ sinθ−sinθ cosθ−sinθ cosθ+sinθ cosθ cos2θ+ sin2θ]=[ 1 00 1].

Q.7

Find X and Y, if(i) X+Y=[7 02 5]  and  X−Y=[3 00 3](ii) 2X+3Y=[2 34 0]  and  3X+2Y=[   2 −2−1    5]

Ans

(i)       X+Y=[7 02 5]  ...(i)  and  X−Y=[3 00 3] ...(ii)  Adding both equations, we getX+Y+X−Y=[7 02 5]+[3 00 3]        2X=[7+30+02+05+3]     =[10028]X=12[10028]     =[5014]Putting value of X in equatin(i),  we get[5014]+Y=[7025] Y=[7025] −[5014]    =[7−50−02−15−4]    = [2011]Thus, X=[5014] and Y=[2011].(ii) 2X+3Y=[2340] ...(i) and  3X+2Y=[   2−2−1   5] ...(ii) Multiplying equation(i) by 2 and equation(ii) by 3, we get2(2X+3Y)=2[2340]⇒  4X+6Y=[4680]...(iii)3(3X+2Y)=3[   2−2−1   5]⇒ 9X+6Y=[   6−6−3   15]...(iv)Subtracting equation (iii) from equation(iv), we get    5X=[   6−6−3   15]−[4680]       X=15[   2−12−11   15] =[   25−125−115   3]Putting value of X in equation (i), we have2[   25−125−115   3]+3Y=[2340]   3Y=[2340]−[   45−245−225   6]        =[2−453−(−245)4−(−225)0−6] Y=13[65395425−6]=[25135145−2]Thus, X=[   25−125−115   3]and  Y=[25135145−2].

Q.8

Find X, if Y= [3 24 0] and 2X+ Y = [  2 −2−1    5].

Ans

Given, Y = [3 21 4] and 2X + Y=[  1 0−3 2]...(i)Putting value of Y in equation (i), we get2X+[3214]=[  1 0−3 2]                     2X=[  1 0−3 2]−[3 21 4]        =[   1−30−2−3−12−4]                        X =12[−2 −2−4 −2]       =[−1 −1−2  −1]

Q.9

If  x  and y, if 2 [1 30 x]+[y 01 2]=[5 61 8].

Ans

Given:     2[130x]+[y 01 2]=[5618]⇒[2 602 x]+[y 01 2]=[5 61 8]⇒     [2+y   6+00+12x+2]=[5618]⇒          2+y=5  and  2x+2=8⇒       y=3  and x=8−22=3Thus, x =3 and y=3.

Q.10

Solve thee quation for x,  y,  z and t, if2[xzyt]+3[1−10   2]=3[3546].

Ans

Given:  2[x zy t]+3[1 −10    2]=3[3 54 6]⇒          [2x2 z2y 2t]+[3−30    6]=[9 1 512 1 8]⇒                 [2x+32z−32y+02t+6]=[9 1 512 1 8]⇒2x+3=9,  2y=12,  2z−3=15  and  2t+6=18⇒x=3,y=6,  z=9​ and t=6.

Q.11

If  x [23] + y[−1   1]=[10  5], find the value of x and y.

Ans

Goven:x23+y[−1   1]=[10  5]⇒   [2x3x]+[−y   y]=[10  5]⇒          [2x−y3x+y]=[10  5]⇒2x−y=10,  3x+y=5⇒x=3  and  y=−4

Q.12

Given 3[xyzw]=[x6−12w]+[4   x+yz+w3], f in d thevalue of x, y, z and w.

Ans

Given :      3[xyzw]=[x6−12w]+[4   x+yz+w3]⇒[3x3y3z3w]=[x+4      6+x+y−1+z+w2w+3]⇒  3x=x+4⇒x=2  3y=6+x+y⇒y=4 3w=2w+3⇒w=3 and   3z=−1+z+w⇒z=1Therefore, x=2, y=4, z=1 and  w=3.

Q.13

If F(x) =cosx −sinx 0sinx cosx 00    0 1,  show that F(x) F(y)=F(x+y).

Ans

Given: F(x)=cosx− sinx 0sinx cosx 00    0 1      F(y)=cosy    − siny   0siny cosy 00    0 1L.H.S.:F(x).F(y)=cosx− sinx 0sinx cosx 00    0 1cosy       − siny   0siny cosy 00    0 1=[cosx cosy−sinx siny− siny cosx−sinx cosy0sinx cosy + cosx siny−sinx siny+cosx cosy00   01]=[cos(x+y)− sin(x+y)0sin(x+y) cos(x+y)00 01]=F(x+y)=R.H.S.

Q.14

Show that (i) [5−16   7][2134]≠[2134][5−16   7](ii) [123010110][−1100    −11234]≠[−1100   −11234][123010110]

Ans

 (i)  L.H.S.:[5−16   7][2134]=[10−35−412+216+28]        =[  7   13334]R.H.S.:[2134][5−16   7]=[10+6−2+715+24−3+28]        =[1653925]Therefore,[5−16   7][2134]≠[2134][5−16   7](ii)[123010110][−1100    −11234]≠[−1100   −11234][123010110]L.H.S:[123010110][−1100    −11234]=[−1+0+61−2+90+2+12   0+0+00−1+0 0+1+0−1+0+01−1+0  0+1+0]     =[   5814   0−11−10  1]R.H.S:[−1100   −11234][123010110]=[−1+0+0−2+1+0−3+0+00+0+10−1+10+0+02+0+44+3+46+0+0]     =[−1−1−3   1   0  0  6 11  6]∴   L.H.S.≠R.H.S

Q.15

Find   A 2 −5A+6I, if A=[ 2 0 1 2 1 3 1     −1 0 ]. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaaieqacaWFgbGaa8xAaiaa=5gacaWFKbGaaGPaVlaaykW7caWFbbWaaWbaaSqabeaacaWFYaaaaOGaeyOeI0Iaa8xnaiaa=feacaWFRaGaa8Nnaiaa=LeacaWFSaGaa8hiaiaa=LgacaWFMbGaa8hiaiaa=bcacaWFbbGaa8xpamaadmaaeaqabeaacaWFYaGaaCzcaiaa=bdacaWLjaGaa8xmaaqaaiaa=jdacaWLjaGaa8xmaiaaxMaacaWFZaaabaGaa8xmaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaeyOeI0Iaa8xmaiaaxMaacaWFWaaaaiaawUfacaGLDbaacaGGUaaaaa@608E@

Ans

  A 2 −5A+6I, A=[ 2 0 1 2 1 3 1     -1 0 ]      A 2 =A×A =[ 2 0 1 2 1 3 1     −1 0 ][ 2 0 1 2 1 3 1     −1 0 ] =[ 4+0+1 0+0−1 2+0+0 4+2+3 0+1−3 2+3+0 2−2+0    0−1+0 1−3+0 ] =[ 5 −1 2 9 −2 5 0 −1−2 ] Now,   A 2 −5A+6I=[ 5 −1 2 9 −2 5 0 −1−2 ]−5[ 2 0 1 2 1 3 1     −1 0 ]+6[ 1 0 0 0 1 0 0 0 1 ]        =[ 5 −1 2 9 −2 5 0 −1−2 ]−[ 10 0 5 10 5 15 5     −5 0 ]+[ 6 0 0 0 6 0 0 0 6 ]        =[ 5−10+6 −1 2−5 9−10 −2−5+6 5−15 0−5 −1+5 −2+6 ]        =[ 1 −1 −3 −1 −1  −10 −5   4  4 ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@AE37@

Q.16

If A = 1 0 2 0 2 12     0 3 ,   prove that  A3−6A2+7A+2I =0.

Ans

A =1 0 20 2 12     0 3,      A2 =A×A= 1 0 20 2 12     0 3 1 0 20 2 12    0 3= [1+0+4  0+0+02+0+60+0+2   0+4+00+2+32+0+6       0+0+04+0+9]=[5082458013]6A2 = 6[5082458013]  =[30   04812243016   078]A3=A2.A      =[5082458013][102021203]      =[5+0+160+0+010+0+242+0+100+8+04+4+158+0+260+0+016+0+39]      =[210341282334055]7A=7[102021203]          =[7014014714021]2I=2[100010001]      =[200020002]L.H.S:    A3−6A2+7A+2I      =[210341282334055]−6[5082458013]+[7014014714021]+[200020002]      =[210341282334055]−[3004812243048078]+[7014014714021]+[200020002]      =[21−30+7+20−0+0+034−48+14+012−12+0+08−24+14+223−30+7+034−48+14+00+0+0+055−78+21+2]      =[000000000]=0

Q.17

If A=[3−24−2],  and I=(1001),f indk so that A2= kA−2I.

Ans

∵ A=[3−24−2]  A2=[3−24−2][3−24−2]        =[9−8−6+412−8−8+4]        =[1−24−4]Now,  A2=kA−2I[1−24−4]=k[3−24−2]−2[1001][1−24−4]=[3k−2k4k−2k]−[2002][1−24−4]=[3k−2−2k4k   −2k−2]⇒       4=4k⇒       k=1Thus, the value of k is 1.

Q.18

If  A=[0−tanα2tanα20]and I is the identity matrix of order 2,show that I+A=(I−A)[cosα− sinαsinα cosα]

Ans

I+ A=[1001]+[0−tanα2tanα20]=[1−tanα2tanα21]I− A=[1001]−[0−tanα2tanα20]=[1tanα2−tanα21]R.H.S.=(I−A)[cosα−sinαsinα  cosα]  =[1tanα2−tanα21][cosα−sinαsinα  cosα]  =[1sinα2cosα2−sinα2cosα21][cosα−sinαsinα  cosα]  =[cosα+sinα.sinα2cosα2−sinα+cosαsinα2cosα2−cosα.sinα2cosα2+sinαsinα.sinα2cosα2+cosα]  =[cosα.cosα2+sinα.sinα2cosα2−(sinα.cosα2−cosαsinα2)cosα2sinαcosα2−cosα.sinα2cosα2cosαcosα2+sinα.sinα2cosα2]  =[cos(α−α2)cosα2−sin(α−α2)cosα2sin(α−α2)cosα2cos(α−α2)cosα2]  =[cos(α2)cosα2−sin(α2)cosα2sin(α2)cosα2cos(α2)cosα2]  =[1−tanα2tanα21]=L.H.S.

Q.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% interest per year. Using matrix multiplication, determine how to divide ₹ 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: (a) Rs 1800 (b) Rs 2000.

Ans

      Total money to invest=   ₹ 30,000Money invested in Ist type of bond=  ₹xMoney invested in 2nd type of bond= ₹30,000-xRate of interest on first type of bond=5%Rate of interest on secon type of bond=7%i                              Total obtained interest=1800ii                              Total obtained interest=2000           IIIx30,000-x5%7%=18002000i 5%  of x+7%30,000-x=1800        5100x+710030,000-x=1800                       5x+210000-7x=180000                      210000-180000=2x      30000=2x       x=300002 =15000To get annual interest of 1800, he should invest ₹ 15,000 in I type of bond and ₹ 15,000 in II type of bond.ii 5%  of x+7%30,000-x=2000        5100x+710030,000-x=2000                       5x+210000-7x=200000                      210000-200000=2x      10000=2x       x =100002 = 5000To get annual interest of ₹ 2000 he should invest ₹ 5000 & ₹ 25000 respectively in I and II type of bond.

Q.20 The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹80, ₹60 and ₹40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Ans

Number of chemistry books on bookshop= 10 dozen= 10×12= 120Number of physics books on bookshop= 8 dozen= 8×12=  96Number of economics books on bookshop= 10 dozen= 10×12= 120Selling price of one chemistry book = ₹ 80Selling price of one physics book = ₹ 60Selling price of one economics book = ₹ 40The total amount the bookshop will receive= 12096120806040= 9600+5760+4800= 20160Thus, the total amount the bookshop will receive from selling  all the books is ₹ 20, 160.

Q.21 Assume X, Y, Z, W and P are matrices of order 2 x n, 3 x k, 2 x p, n x 3 and p x k respectively. Choose the correct answer in Exercises 21 and 22.

The restriction of 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 = 3

Ans

Matrix Order
X 2 x n
Y 3 x k
Z 2 x p
W n x 3

PY + WY=Order of P× Order of Y+ Order of W× Order of Y

= ( p×k )×( 3×k )+( n×3 )×( 3×k )

= ( p×k )+( n×k )

Multiplication of PY is possible if k=3 and sum of PY and WY is possible if p = n.

Thus, option ( A ) is correct.

Q.22 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.

If n = p, then the order of the matrix 7X – 5Z is:

(A) P × 2

(B) 2 × n

(C) n × 3

(D) p × n

Ans

The order of matrix X is 2×nThe order of matrix Z is 2×p7X−5Z is defined when X and Z an of the same order.⇒n=p(Given)Thus the order of 7X−5Z is 2×n.Thus, option (B) is correct answer.

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FAQs (Frequently Asked Questions)

  • 3.1 Introduction
  • 3.2 Matrix
  • 3.2.1 Order of a matrix
  • 3.3 Types of Matrices
  • 3.3.1 Equality of Matrices
  • 3.4 Operations on Matrices
  • 3.4.1 Addition of Matrices
  • 3.4.2 Multiplication of a Matrix by a scalar
  • 3.4.3 Properties of matrix addition
  • 3.4.4 Properties of scalar multiplication of a Matrix
  • 3.4.5 Multiplication of Matrices
  • 3.4.6 Properties of multiplication of Matrices
  • 3.5. Transpose of a Matrix
  • 3.5.1 Properties of the transpose of the Matrices
  • 3.6 Symmetric and Skew Symmetric Matrices
  • 3.7 Elementary Operation (Transformation) of a Matrix
  • 3.8 Invertible Matrices
  • 3.8.1 Inverse of a Matrix by Elementary Operations

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