RD Sharma XII Vol 1 2017 Solutions for Class 12 Science Math Chapter 7 Adjoint And Inverse Of Matrix are provided here with simple step-by-step explanations. These solutions for Adjoint And Inverse Of Matrix are extremely popular among class 12 Science students for Math Adjoint And Inverse Of Matrix Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XII Vol 1 2017 Book of class 12 Science Math Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XII Vol 1 2017 Solutions. All RD Sharma XII Vol 1 2017 Solutions for class 12 Science Math are prepared by experts and are 100% accurate.

Page No 7.22:

Question 1:

Find the adjoint of each of the following matrices:

(i) -3524

(ii) abcd

(iii) cos αsin αsin αcos α

(iv) 1tan α/2-tan α/21

Verify that (adj A) A = |A| I = A (adj A) for the above matrices.

Answer:


Given below are the squares matrices. Here, we will interchange the diagonal elements and change the signs of
the off-diagonal elements.

s.i A= -3524adjA=4-5-2-3(adjA)A=-2200-22A=-22AI=-2200-22A(adjA)=-2200-22(adjA)A=AI=A(adjA)Hence verified.ii B=abcdadjB=d-b-ca(adjB)B=ad-bc00-cb+adB=ad-bcBI=ad-bc00-cb+adB(adjB)=ad-bc00-cb+ad(adjB)B=BI=B(adjB)Hence verified.iii C=cosαsinαsinαcosαadjC=cosα-sinα-sinαcosα(adjC)C=cos2α-sin2α00cos2α-sin2αC=cos2α-sin2αCI=cos2α-sin2α00cos2α-sin2αC(adjC)=cos2α-sin2α00cos2α-sin2α(adjC)C=CI=C(adjC)Hence verified.iv D=1tanα2-tanα21adjD=1-tanα2tanα21(adjD)D=1+tan2α2001+tan2α2D=1+tan2α2DI=1+tan2α2001+tan2α2D(adjD)=1+tan2α2001+tan2α2(adjD)D=DI=D(adjD)Hence verified.

Page No 7.22:

Question 2:

Compute the adjoint of each of the following matrices:

(i) 122212221

(ii) 125231-111

(iii) 2-1342504-1

(iv) 20-1510113

Verify that (adj AA = |AI = A (adj A) for the above matrices.

Answer:

(i) A=122212221Now,C11=1221=-3, C12=-2221=2 and C13=2122=2C21=-2221=2, C22=1221=-3 and C23=-1222=2C31=2212=2, C32=-1222=2  and C33=1221=-3adjA=-3222-3222-3T=-3222-3222-3and (adjA)A=500050005Now, A=5AI=500050005and A(adjA)=500050005Thus, (adjA)A=AI= A(adjA)(ii) B=125231-111Now,C11=3111=2, C12=-21-11=-3 and C13=23-11=5C21=-2511=3, C22=15-11=6 and C23=-12-11=-3C31=2531=-13, C32=-1521=9 and C33=1223=-1adjB=2-3536-3-139-1T=23-13-3695-3-1(adjB)B=210002100021and B=21BI=210002100021and B(adjB)=210002100021Thus, (adjA)A=AI= A(adjA)(iii) C=2-1342504-1Now,C11=254-1=-22, C12=-450-1=4 and C13=4204=16C21=--134-1=11, C22=230-1=-2 and C23=-2-104=-8C31=-1325=-11, C32=-2345=2 and C33=2-142=8adjC=-2241611-2-8-1128T=-2211-114-2216-88(adjC)C=000000000and C=0CI=000000000and CadjC=000000000Thus, (adjA)A=AI= A(adjA)(iv) D=20-1510113Now,C11=1013=3, C12=-5013=-15 and C13=5111=4C21=-0-113=-1, C22=2-113=7 and C23=-2011=-2C31=0-110=1, C32=-2-150=-5 and C33=2051=2adjD=3-154-17-21-52T=3-11-157-54-22(adjD)D=200020002and D=2DI=200020002and D(adjD)=200020002Thus, (adjA)A=AI= A(adjA)

Page No 7.22:

Question 3:

For the matrix A=1-1123018210, show that A (adj A) = O.

Answer:

A=1-1123018210Now,C11=30210=30          C12=-201810=-20     C13=23182=-50C21=--11210=12   C22=111810=-8           C23=-1-1182=-20C31=-1130=-3       C32=-1120=2                C33=1-123=5adj A=30-20-5012-8-20-325T=3012-3-20-82-50-205A(adj A)=1-11230182103012-3-20-82-50-205=30+20-5012+18-20-3-2+560-60-024-24-0-6+6+0540-40-500216-16-200-54+4+50=000000000

Page No 7.22:

Question 4:

If A=-4-3-3101443, show that adj A = A.

Answer:

A=-4-3-3101443Now,C11=0143=-4, C12=-1143 =1 and C13=1044=4C21=--3-343=-3, C22=-4-343=0 and C23=--4-344=4C31=-3-301=-3, C32=--4-311=1 and C33=-4-310=3adj A=-414-304-313T=-4-3-3101443=AHence proved.



Page No 7.23:

Question 5:

If A=-1-2-221-22-21, show that adj A = 3AT.

Answer:

A=-1-2-221-22-21Now,C11=1-2-21=-3, C12=-2-221=-6 and C13=212-2=-6C21=--2-2-21=6, C22=-1-221=3 and C23=--1-22-2=-6C31=-2-21-2=6, C32=--1-22-2=-6 and C33=-1-221=3adj A=-3-6-663-66-63T=-366-63-6-6-63AT=-122-21-2-2-213AT=-366-63-6-6-633AT=adj A

Page No 7.23:

Question 6:

Find A (adj A) for the matrix A=1-2302-1-452.

Answer:

A=1-2302-1-452Now,C11=2-152=9, C12=-0-1-42=4 and C13=02-45=8C21=--2352=19, C22=13-42=14 and C23=-1-2-45=3C31=-232-1=-4, C32=-130-1=1 and C33=1-202=2adj A=94819143-412T=919-44141832A(adj A)=250002500025

Page No 7.23:

Question 7:

Find the inverse of each of the following matrices:

(i) cos θsin θ-sin θcos θ

(ii) 0110

(iii) abc1+bca

(iv) 25-31

Answer:

i A=cosθsinθ-sinθcosθA=cos2θ+sin2θ=10A is a singular matrix; therefore, it is invertible.Let Cij be a cofactor of aij in A.Now,C11=cosθ C12=sinθC21=-sinθC22=cosθadjA=cosθsinθ-sinθcosθT=cosθ-sinθsinθcosθA-1=1AadjA=cosθ-sinθsinθcosθii B=0110B=0-1=-10B is a singular matrix; therefore, it is invertible.Let Cij be a cofactor of bij in B.Now,C11=0 C12=-1C21=-1C22=0adjB=0-1-10T=0-1-10B-1=1BadjB=1-10-1-10=0110iii C=abc1+bcaC=1+bc-bc=10C is a singular matrix; therefore, it is invertible. Let Cij be a cofactor of cij in C.Now,C11=1+bca C12=-cC21=-bC22=aadjC=1+bca-c-baT=1+bca-b-caC-1=1CadjC=1+bca-b-caiv D=25-31D=2+15=170D is a singular matrix; therefore, it is invertible. Let Cij be a cofactor of dij in D.Now,C11=1 C12=3C21=-5C22=2adjD=13-52T=1-532D-1=1DadjD=1171-532

Page No 7.23:

Question 8:

Find the inverse of each of the following matrices.

(i) 123231312

(ii) 1251-1-123-1

(iii) 2-11-12-11-12

(iv) 20-1510013

(v) 01-14-343-34

(vi) 00-1345-2-4-7

(vii) 1000cos αsin α0sin α-cos α

Answer:

i A=123231312Now,C11=3112=5, C12=-2132=-1 and C13=2331=-7C21=-2312=-1, C22=1332=-7 and C23=-1231=5C31=2331= -7, C32=-1321=5 and C33=1223= -1adjA=5-1-7-1-75-75-1T=5-1-7-1-75-75-1and A=-18A-1=-1185-1-7-1-75-75-1ii B=1251-1-123-1Now,C11=-1-13-1=4, C12=-1-12-1=-1 and C13=1-123=5C21=-253-1=17, C22=152-1=-11 and C23=-1223=1C31=25-1-1= 3, C32=-151-1=6 and C33=121-1= -3adjB=4-1517-11136-3T=4173-1-11651-3and B=27B-1=1274173-1-11651-3iii C=2-11-12-11-12Now,C11=2-1-12=3, C12=--1-112=1 and C13=-121-1=-1C21=--11-12=1, C22=2112=3 and C23=-2-11-1=1C31=-112-1= -1, C32=-21-1-1=1 and C33=2-1-12= 3adjC=31-1131-113T=31-1131-113and C=4C-1=1431-1131-113iv D=20-1510013Now,C11=1013=3, C12=-5003=-15 and C13=5101=5C21=-0-113=-1, C22=2-103=6 and C23=-2001=-2C31=0-110=1, C32=-2-150=-5 and C33=2051=2adjD=3-155-16-21-52T=3-11-156-55-22and D=1D-1=3-11-156-55-22v  E=01-14-343-34Now,C11=-34-34=0, C12=-4434=-4 and C13=4-33-3=-3C21=-1-1-34=-1, C22=0-134=3 and C23=-013-3=3C31=1-1-34=1, C32=-0-144=-4 and C33=014-3=-4adjE=0-4-3-1331-4-4T=0-11-43-4-33-4and E=-1E-1=-10-11-43-4-33-4=01-14-343-34viF=00-1345-2-4-7Now,C11=45-4-7=-8, C12=-35-2-7=11 and C13=34-2-4=-4C21=-0-1-4-7=4, C22=0-1-2-7=-2 and C23=-00-2-4=0C31=0-145=4, C32=-0-135=-3 and C33=0034=0adjF=-811-44-204-30T=-84411-2-3-400and F=4F-1=14-84411-2-3-400vii G=1000cosαsinα0sinα-cosαNow,C11=cosαsinαsinα-cosα=-1, C12=-0sinα0-cosα=0 and C13=0cosα0sinα=0C21=-00sinα-cosα=0, C22=100-cosα=-cosα and C23=-100sinα=-sinαC31=00cosαsinα=0, C32=-100sinα=-sinα and C33=100cosα=cosαadjF=-1000-cosα-sinα0-sinαcosαT=-1000-cosα-sinα0-sinαcosαand F=-1F-1=-1-1000-cosα-sinα0-sinαcosα=1000cosαsinα0sinα-cosα

Page No 7.23:

Question 9:

Find the inverse of each of the following matrices and verify that A-1 A=I3.

(i) 133143134

(ii) 231341372

Answer:

i A=133143134Now,C11=4334=7, C12=-1314=-1 and C13=1413=-1C21=-3334=-3, C22=1314=1 and C23=-1313=0C31=3343=-3, C32=-1313=0 and C33=1314=1adjA=7-1-1-310-301T=7-3-3-110-101and A=1A-1=7-3-3-110-101Now, A-1A=7-3-3-110-101133143134=100010001=I3ii B=231341372Now,C11=4172=1, C12=-3132=-3 and C13=3437=9C21=-3172=1, C22=2132=1 and C23=-2337=-5C31=3141=-1, C32=-2131=1 and C33=2334=-1adjB=1-3911-5-11-1T=11-1-3119-5-1and  B=2B-1=1211-1-3119-5-1Now, B-1B=100010001=I3

Page No 7.23:

Question 10:

For the following pairs of matrices verity that AB-1=B-1 A-1:

(i) A=3275 and B 4632

(ii) A=2153 and B 4534

Answer:

i We have, A=3275 and B=4632AB=18224352AB=-10Since, AB0Hence, AB is invertible. Let Cij be the cofactor of ain in AB=aijC11=52 , C12=-43,  C21 =-22 and  C22=18adjAB=52-43-2218T=52-22-4318AB-1=-11052-22-4318     ...1Now, B=4632B=-10Since, B0Hence, B is invertible. Let Cij be the cofactor of ain in B=aijC11=2 , C12=-3,  C21 =-6 and  C22=4adjB=2-3-64T=2-6-34B-1=-1102-6-34A=1Since, A0Hence, A is invertible. Let Cij be the cofactor of ain in A=aijC11=5 , C12=-7,  C21 =-2 and  C22=3adjA=5-7-23T=5-2-73A-1=5-2-73Now, B-1A-1=-11052-22-4318      ...2From eq. 1 and 2, we haveAB-1=B-1A-1Hence verified.ii We have, A=2153 and B=4534AB=11142937Now,AB=1Since, AB0Hence, AB is invertible. Let Cij be the cofactor of ain in AB=aijC11=37 , C12=-29,  C21 =-14 and  C22=11adj(AB)=37-14-2911AB-1=37-14-2911            ...1B=1Since, B0Hence, B is invertible. Let Cij be the cofactor of ain in B=aijC11=4 , C12=-3,  C21 =-5 and  C22=4adjB=4-5-34B-1=4-5-34A=1Since, A0Hence, A is invertible. Let Cij be the cofactor of ain in A=aijC11=3 , C12=-5,  C21 =-1 and  C22=2adjA=3-1-52A-1=3-1-52Now, B-1A-1=37-14-2911         ...2From eq. 1 and 2, we haveAB-1=B-1A-1Hence verified.
(AB)1=B1 A1  (AB)1=B1 A1(AB)1=B1 A1

Page No 7.23:

Question 11:

Let A=3275 and B=6789. Find AB-1

Answer:

Given:A=3275B=6789AB=34398294Now,AB=-2Since, AB0Hence, AB is invertible. Let Cij be the cofactor of ain in AB=aijC11=94 , C12=-82,  C21 =-39 and  C22=34adj(AB)=94-82-3934T=94-39-8234AB-1=-1294-39-8234=-4739241-17

Page No 7.23:

Question 12:

Given A=2-3-47, compute A−1 and show that 2A-1=9I-A.

Answer:

We have, A=2-3-47Now, adj(A)=7342and A=2A-1=127342Now, 2A-1=9I-ALHS=2A-1=7342RHS=9I-A=91001-2-3-47=7342=LHSHence proved.
 

Page No 7.23:

Question 13:

If A=4521, then show that A-3I=2 I+3A-1.

Answer:

We have, A=4521Now,adj(A)=1-5-24and A=-6A-1=-161-5-24Now, A-3I=I+3A-1LHS= A-3I=4521-31001=152-2RHS=2I+3A-1=21001-3×161-5-24=20.52.51-1=152-2=LHSHence proved.

Page No 7.23:

Question 14:

Find the inverse of the matrix A=abc1+bca and show that aA-1=a2+bc+1 I-aA.

Answer:

We have, A=abc1+bcaSo, adj(A)=1+bca-b-caand A=1 A-1=1+bca-b-caNow, aA-1= a2+bc+1I-aALHS =aA-1=a1+bca-b-ca=1+bc-ba-caa2RHS =a2+bc+1I-aA=a2+bc+11001-a2baca1+bc=a2+bc+100a2+bc+1-a2baca1+bc= 1+bc-ba-caa2=LHSHence proved.

Page No 7.23:

Question 15:

Given A=504232121, B-1=133143134. Compute (AB)−1.

Answer:

We have,
A=504232121B-1=133143134We know (AB)-1=B-1A-1For matrix A, C11=3221=-1, C12=-2211=0 and C13=2312=1C21=-0421=8, C22=5411=1 and C23=-5012=-10C31=0432=-12, C32=-5422=-2 and C33=5023=15Now,adj (A)=-10181-10-12-215T=-18-1201-21-1015and A=-1A-1=--18-1201-21-1015=1-8120-12-110-15So, B-1A-1=1331431341-8120-12-110-15=-219-27-218-25-329-42

Page No 7.23:

Question 16:

Let F α=cos α-sin α0sin αcos α0001 and Gβ=cos β0sin β010-sin β0cos β

Show that
(i) F α-1=F -α
(ii) G β-1=G -β
(iii) F α G β-1=G -β F - α.

Answer:

  ​(i) F(α)=cosα-sinα0sinαcosα0001F(-α)=cos-α-sin-α0sin-αcos-α0001=cosαsinα0-sinαcosα0001Now,C11=cosα001=cosα, C12=-sinα001=-sinα and C13=sinαcosα00=0C21=--sinα001=sinα, C22=cosα001=cosα and C23=-cosα-sinα00=0C31=-sinα0cosα0=0, C32=-cosα0sinα0=0 and C33=cosα-sinαsinαcosα=1adj F(α)=cosα-sinα0sinαcosα0001T=cosαsinα0-sinαcosα0001F(α)=1Fα-1=cosαsinα0-sinαcosα0001             ...1Fα-1=F(-α)  (ii) G(β)=cosβ0sinβ010-sinβ0cosβG(-β)=cos-β0sin-β010-sin-β0cos-β=cosβ0-sinβ010sinβ0cosβNow, C11=100cosβ=cosβ, C12=-00-sinβcosβ=0 and C13=01-sinβ0=sinβC21=-0sinβ0cosβ=0, C22=cosβsinβ-sinβcosβ=1 and C23=-cosβ0-sinβ0=0C31=0sinβ10=-sinβ, C32=-cosβsinβ00=0 and C33=cosβ001=cosβadjG(β)=cosβ0sinβ010-sinβ0cosβT=cosβ0-sinβ010sinβ0cosβG(β)=1G(β)-1=cosβ0-sinβ010sinβ0cosβ            ...2        G(β)-1==G(-β)  (iii) F(α)=cosα-sinα0sinαcosα0001F(-α)=cos-α-sin-α0sin-αcos-α0001=cosαsinα0-sinαcosα0001        ... 3G(β)=cosβ0sinβ010-sinβ0cosβG(-β)=cos-β0sin-β010-sin-β0cos-β=cosβ0-sinβ010sinβ0cosβ        ... 4F(α)G(β)-1=G(β)-1F(α)-1                       =cosβ0-sinβ010sinβ0cosβcosαsinα0-sinαcosα0001     Using eqn 1 and 2                       =G(-β)F(-α)           Using eqn 3 and 4 

Page No 7.23:

Question 17:

If A=2312, verify that A2-4 A+I=O, where I=1001 and O=0000. Hence, find A−1.

Answer:

A=2312A2=71247andA2-4A+I= 71247-81248+1001A2-4A+I= 7-8+112-12+04-4+07-8+1=0000=OA2-4A+I=0A2-4A=-IA-1A2-4AA-1=-IA-1       Pre-multiplying both sides by A-1A-4I=-A-1A-1=4I-AA-1=4004-2312=2-3-12  



Page No 7.24:

Question 18:

Show that A=-8524 satisfies the equation A2+4A-42I=O. Hence, find A−1.

Answer:

A=-8524A2=74-20-826andA2+4A-42I=74-20-826+-3220816-420042A2+4A-42I=74-32-42-20+20-0-8+8-026+16-42=0000=ONow, A2+4A-42I=0A2+4A=42IA-1A2+4A-1A=42IA-1         Pre-multiplying both sides by A-1A+4I=42A-1A-1=142A+4IA-1=142-8524+4004=142-4528

Page No 7.24:

Question 19:

If A=31-12, show that A2-5A+7I=O.  Hence, find A−1.

Answer:

A=31-12A2=31-1231-12=9-13+2-3-2-1+4=85-53andA2-5A+7I=85-53-155-510+7007A2-5A+7I=8-15+75-5+0-5+5+03-10+7=0000=ONow, A2-5A+7I=0A2-5A=-7IA-1A2-5A-1A=-7A-1I    Pre-multiplying both sides by A-1A-5I=-7A-1A-1=-17A-5IA-1=-1731-12-5005=172-113

Page No 7.24:

Question 20:

If A=4325, find x and y such that A2=xA+yI=O. Hence, evaluate A−1.

Answer:

A=4325A2=22271831Now, A2-xA+yI=O22271831-4x3x2x5x+y00y=000022-4x-y27-3x18-2x31-5x-y=0000Thus, we have22-4x+y=0, 27-3x=0, 18-2x=0 and 31-5x+y=0-3x=-27x=9On putting x=9 in 22-4x+y=0, we get22-36+y=0-14=-yy=14Now,  A2-9A+14I=0A2-9A=-14IA-1A2-9AA-1=-14IA-1   Pre-multiplying both sides by A-1A-9I=-14A-1A-1=-114A-9IA-1=-1144325-9009=1145-3-24

Page No 7.24:

Question 21:

If A=3-24-2, find the value of λ so that A2=λA-2I. Hence, find A−1.

Answer:

A=3  -2  4  -2A2=1  -24  -4Given:  A2=λA-2I           ...11  -24  -4 =λ3   -24   -2-21     00     11  -24  -4=3λ   -2λ4λ   -2λ-2     00     2  1  -24  -4=3λ-2     -2λ4λ         -2λ-2On equating corresponding terms, we get -2λ=-2 λ=1 On substituting λ=1 in 1, we getA2=A-2I A2-A=-2IA-A2=2IA-1 A-A2 =A-1×2I               Pre-multiplying both sides with A-1I -A =2 A-12 A-1 =1     00    1- 3  -24  -2=1-3     0+20-4      1+2A-1 =12-2       2-4      3

Page No 7.24:

Question 22:

Show that A=53-1-2 satisfies the equation x2-3x-7=0. Thus, find A−1.

Answer:

A=  5       3-1  -2  A2=  22       9-3       1If I2 is the identity matrix of order 2, thenA2-3A-7I2= 22       9-3       1-3 5       3-1  -2-71    00    1                  A2-3A-7I2 = 22-15-7       9-9-0-3+3+0        1+6-7=0   00    0 =0A2-3A-7I2=0Thus, A satisfies x2-3x-7=0.Now,A2-3A-7I2=0A2-3A=7I2A-1A2-3A =A-1×7I2        Pre-multiplying both sides by A-1A-3I2=7A-1 5       3-1  -2-31    00    1=7A-1  A-1= 17 5-3         3-0-1-0     -2-3  = 17   2       3-1   -5

Page No 7.24:

Question 23:

Show that A=6576 satisfies the equation x2-12x+1=O. Thus, find A−1.

Answer:

A =6   5 7   6  A2 =71    60 84    71    If I2 is the identity matrix of order 2, thenA2 -12A+I2 =71    60 84    71 -126   5 7   6+1   0 0   1A2 -12A+I2=71-72+1      60-60+0 84-84+0      71-72+1A2 -12A+I2 =0Thus, A satisfies x2-12x+1=0.Now, A2 -12A+I2 =0I2 =12A-A2A-1I2 =A-112A-A2           Pre-multiplying both sides by A-1 A-1=12I2-AA-1 =12 1   0 0   1-6   5 7   6   A-1=12-6       0-5 0-7      12- 6     A-1=  6      -5 -7        6   

Page No 7.24:

Question 24:

For the matrix A=11112-32-13. Show that A-3-6A2+5A+11 I3=O. Hence, find A−1.

Answer:

A=1     1      11     2   -32   -1     3  A=1     1      11     2   -32   -1     3=1×3-1×9+1×-5=3-9-5=-11   Since, A0Hence, A-1 exists.Now,A2=1     1      11     2   -32   -1     31     1      11     2   -32   -1     3=1+1+2      1+2-1      1-3+31+2-6      1+4+3      1-6-92-1+6      2-2-3      2+3+9 =  4         2          1-3        8     -14  7      -3        14and A3 =A2.A=4     2      1-3     8   -147   -3     141     1      11     2   -32   -1     3 =  4+2+2            4+4-1             4-6+3-3+8-28     -3+16+14      -3-24-42  7-3+28         7-6-14            7+9+42=   8           7       1-23       27     -69  32     -13       58Now, A3 -6A2+5A+11I3=   8           7        1-23       27     -69  32     -13       58-6 4     2      1-3     8   -147   -3     14+5 1     1      11     2   -32   -1     3+11 1     0      00     1      00      0     1                                             =   8-24+5+11           7-12+5+0       1-6+5+0-23+18+5+0       27-48+10+11     -69+84-15+0  32-42+10+0     -13+18-5+0       58-84+15+11                                            = 0     0      00     0      00      0     0 =O (Null matrix)Again, A3 -6A2+5A+11I3 =OA-1×A3 -6A2+5A+11I3 =A-1×O             (Pre-multiplying both sides because A-1 exists) A2 -6A+5I3+11A-1 =0  4         2          1-3        8     -14  7      -3        14 -61     1      11     2   -32   -1     3  +51     0      00     1      00      0     1 =-11A-1  4-6+5         2-6+0          1-6+0-3-6+0        8-12+5     -14+18+0  7-12+0      -3+6+0        14-18+5 =-11A-1A-1 = - 111   3     -4     -5-9        1       4-5         3      1 

Page No 7.24:

Question 25:

Show that the matrix, A=10-2-2-12341 satisfies the equation, A3-A2-3A-I3=O. Hence, find A−1.

Answer:

We have, A=  1        0    -2-2   -1       2  3        4      1    A=  1        0    -2-2   -1        2  3        4       1 =1-9+0-2-8=-9+16 =7  Since, A0Hence, A-1 exists.Now,A2=    1        0    -2-2     -1       2  3        4        1   1        0    -2-2     -1       2  3         4       1=  1+0-6        0+0-8    -2+0-2-2+2+6     0+1+8       4-2+2  3-8+3        0-4+4      -6+8+1= -5     -8     -4  6        9         4-2       0         3A3=A2.A =-5     -8      -4   6        9         4-2       0          3   1        0    -2-2     -1       2   3        4      1    =-5+16-12       0+8-16          10-16-4   6-18+12        0-9+16       -12+18+4  -2+0+9         0+0+12             4+0+3  =  -1    - 8    -10    0        7       10    7       12         7 Now, A3-A2-3A-I3 =  -1    - 8    -10    0        7      10   7       12         7 --5     -8    -4  6         9       4-2        0        3-3  1        0    -2-2   -1       2   3        4      1 -  1        0       0  0        1       0  0        0       1                                         =  -1+5-3-1         -8+8+0+0      -10+4+6-00-6+6-0               7-9+3-1           10-4-6-0  7+2-9-0           12+0-12-0            7-3-3-1 =  0        0       0  0        0       0  0        0       0   =OHence proved.Now,  A3-A2-3A-I3=O (Null matrix)A-1A3-A2-3A-I3 =A-1O     (Pre-multiplying by A-1)A2-A1-3I3=A-1 -5     -8     -4  6        9         4-2       0         3- 1        0    -2-2   -1       2  3        4      1-3  1        0       0  0        1       0  0        0       1   =A-1 -5-1-3      -8-0-0      -4+2+0  6+2+0            9+1-3        4-2  -2-3-0         0-4-0     3-1-3 = -9        -8       -2    8           7          2 -5        -4       -1=A-1A-1 = -9        -8       -2    8           7          2 -5        -4       -1

Page No 7.24:

Question 26:

If A=2-11-12-11-12. Verify that A3-6A2+9A-4I=O and hence find A−1.

Answer:

 A =   2     -1      1-1        2   -1   1      -1     2   A= 2      -1       1-1       2    -1   1     -1      2=2×4-1+1-2+1+11-2 =6-1-1=4  Since, A0 Hence, A-1 exists.Now,A2=   2     -1      1-1        2   -1   1      -1     2   2     -1      1-1        2   -1   1      -1     2 =   4+1+1     -2-2-1      2+1+2-2-2-1        1+4+1    -1-2-2   2+1+2      -1-2-2     1+1+4 =   6    -5        5-5       6      -5   5     -5       6  A3 =A2.A=   6    -5        5-5       6     -5   5      -5       6   2     -1      1-1        2   -1   1      -1     2   =   12+5+5    -6-10-5    6+5+10-10-6-5        5+12+5     -5-6-10   10+5+6      -5-10-6     5+5+12   =   22     -21      21-21        22   -21   21      -21     22   Now, A3-6A2+9A-4I=   22     -21      21-21        22   -21   21      -21     22-6   6    -5        5-5      6     -5   5      -5       6+9   2     -1      1-1        2   -1   1      -1     2-4 1     0      0 0     1       0  0     0      1                                          =  22-36+18-4    -21+30-9-0        21-30+9-0-21+30-9-0      22-36+18-4     -21+30-9-0  21-30+9-0      -21+30-9-0       22-36+18-4                                            =   0    0    0   0    0    0   0    0    0 =O   [Null matrix]Hence proved.Now, A3-6A2+9A-4I=OA-1×A3-6A2+9A-4I=A-1×O     A2-6A+9I-4A-1 =O4A-1 =A2-6A+9I4A-1 =   6    -5        5-5       6      -5   5     -5       6  -6   2     -1      1-1        2   -1   1      -1     2+9 1      0      0 0      1       0  0     0       14A-1 = 6-12+9      -5+6+0        5-6+0-5+6+0        6-12+9     -5+6+0  5-6+0        -5+6+0       6-12+9 4A-1 =   3     1    -1   1     3       1 -1     1      3A-1=14   3     1    -1   1     3       1 -1     1      3

Page No 7.24:

Question 27:

If A=19-8144471-84, prove that A-1=A3.

Answer:

A=19-8    1     4   4    4     7   1   -8    4 =-8 9     19     49   49      4 9    79   19   -8 9   49AT =-8 9      49        19   19       4 9    - 89   49      7 9      49 =19- 8        4        1    1        4      -8     4        7        4            ...(1)A=-8 9     19     49   49      4 9    79   19   -8 9   49 =19×9×9-8    1     4   4    4     7   1   -8    4    =19×9×9-8×72 -1×9+4×-36    =19×9×9×9×-64-1-16 =-9×819×9×9 =-1 If Cij  is a cofactor of aij such that A =aij, then we have C11=89          C 12 =-19      C 13 =-49C21  =-49       C22  =-4  9    C23 =-79 C31 =-19      C32=89       C33=-49Now,adj A =    89    -19     -49 -49    - 4 9    -79 -19       89    -49T=   89       -49      -19   -19      - 4 9       89 -49        -79      -49A-1=1Aadj A =-1×19   8      -4     -1-1      - 4      8  -4        -7    -4 =19- 8        4        1    1        4      -8     4        7        4=AT          [From (1)]A-1=AT

Page No 7.24:

Question 28:

If A=3-342-340-11, show that A-1=A3.

Answer:

We have, A=3  -3   42   -3   40  -1    1A2 =3    -3    42    -3    40   -1     13  -3   42   -3   40   -1    1 =9-6+0    -9+9-4     12-12+46-6+0    -6+9-4      8-12+40-2+0      0+3-1       0-4+1 =  3    -4       4   0    -1      0 -2    2    -3Now, A3=A2 ×A  =  3    -4       4   0    -1      0 -2     2    -33  -3    42   -3    40  -1     1 =  9-8    -9+12-4       12-16+4   0-2+0             0+3+0         -4 -6+4+0    6-6+3    -8+8-3=1    -1     0-2    3    -4  -2    3    -3Again, A3×A =1    -1     0-2    3    -4 -2     3    -33  -3   42   -3   40  -1    1=   3-2+0           -3+3+0              4-4+0-6+6         6-9+4     -8+12-4      0                  0                  1 =1       0       0   0       1       00       0       1 =I3    [Identity matrix of order 3]A3×A =I3  A3=A-1

Page No 7.24:

Question 29:

If A=-120-111010, show that A2=A-1.

Answer:

We have, A=-1    2   0-1    1   1   0    1   0Now,A2=-1    2   0-1    1   1   0    1   0-1    2   0-1    1   1   0    1   0=1-2+0    -2+2+0   0+2+01-1+0    -2+1+1   0+1+0 0-1+0     0+1+0    0+1+0=-1     0    2   0     0    1 -1    1    1and A2×A =-1     0    2   0     0    1 -1    1    1-1    2   0-1    1   1   0    1   0 =1+0+0    -2+0+2    00+0+0      0+0+1     0     0        -2+1+1     1 =  1     0    0   0    1    0   0    0    1 =I3                 [Identity matrix of order 3]A2×A =I3   A2=A-1  

Page No 7.24:

Question 30:

Solve the matrix equation 5411X=1-213, where X is a 2 × 2 matrix.

Answer:

Let: A =5  4 1  1    B =1  -21     3Now,A=5  4 1  1 =5-4=10  Hence, A is invertible. If Cij is cofactor of aij in A, then C11=1, C12 =-1, C21 =-4 and C22=5.adj A =   1    -1-4    -5T =       1    -4 -1      5A-1=1Aadj A =   1    -4 -1      5  Now, the given equation becomes AX=B.A-1AX = A-1×BA-1AX = A-1×BX = A-1×B          X =    1    -4 -1      5×1  -21     3 X=   1-4     -2-12-1+5      2+15X=-3    -14  4       17

Page No 7.24:

Question 31:

Find the matrix X satisfying the matrix equation

X53-1-2=14777.

Answer:

Let: A=   5     3-1   -2     A=  5       3-1   -2=-10+3=-70  Hence, A is invertible.If Cij is a cofactor of aij in A, then C11=-2, C12=1, C21=-3 and C22=5.Now,adj A=  -2   1-3   5T= -2   -3  1        5A-1=1Aadj A=-17 -2   -3  1        5    Let: B =14   77     7  B=14   77     7  =98-49=490  Hence, B is invertible.The given matrix equation becomes XA=B.XAA-1=BA-1     XAA-1=14   77     7  ×-17× -2   -3  1        5X=-17 -28+7    -42+35 -14+7     -21+35X=-17 -21   -7  -7     14X=  3       1  1   -2         

Page No 7.24:

Question 32:

Find the matrix X for which

3275 X -11-21=2-104

Answer:

Let A=  3    27    5, B=-1    1-2   1  and C=  2  -10     4Now,A=3    27    5 =15-14=1                   B=-1    1-2    1 =-1+2=1               Since, A0 and B0Hence,  A & B are invertible, so A-1 and B-1 exist. Cofactors of matrix A areA11=5          A12=-7        A21=-2      A22=3 Now,adj A=  5  -7-2    3T=  5  -2-7    3A-1=1Aadj A =  5  -2-7    3      Cofactors of matrix B areB11=1    B12=2     B21=-1   B22=-1Now,adj B=  1     2-1  -1T=  1   -1  2  -1B-1=1Badj B =  1     -1  2     -1        The given equation becomes AXB=CA-1AXBB-1=A-1CB-1IXI=A-1CB-1X =  5  -2-7    32  -10     4  1     -1  2     -1X=  5  -2-7    3  2-2     -2+1 0+8    0-4X=  5  -2-7    3  0     -1 8     -4X= -16   3   24    -5

Page No 7.24:

Question 33:

Find the matrix X satisfying the equation

2153 X 5332=1001.

Answer:

Let A=2   15   3 , B=  5  33   2 and I= 1   00   1A=2   15   3=6-5=1 Since, A0Thus, A is invertible.Also, B =5  33   2=10-9=1Thus, B is invertible.Cofactors of matrices A & B areA11=3         A12=-5      A21=-1       A22=2B11=2        B12=-3        B21=-3      B22=5Now,adj A=  3   -5-1     2T  =  3   -1-5     2     adj B =   2  -3 -3    5T=   2  -3 -3    5A-1=1Aadj A =  3   -1-5     2B-1=1Badj B=   2  -3 -3    5The given matrix equation becomes AXB=IA-1AXBB-1=IA-1B-1A-1AXBB-1=A-1B-1IXI=A-1B-1X=A-1B-1 X=  3   -1-5     2   2  -3 -3    5=   6+3       -9-5-10-6     15+10=  9      -14-16     25

Page No 7.24:

Question 34:

If A=122212221, find A-1 and prove that A2-4A-5I=O

Answer:

A=1   2   22   1   22   2   1          A=1    2    22    1    22    2    1=11-4-22-4+24-2=-3+4+4=5     Since, A0Hence, A is invertible.Now,A2=1   2   22   1   22   2   11   2   22   1   22   2   1=1+4+4    2+2+4    2+4+22+2+4    4+1+4   4+2+22+4+2    4+2+2   1+4+4=9   8  88   9   88   8  9Now, A2-4A-5I=9   8  88   9   88   8  9-41   2   22   1   22   2   1-51   0   00   1   00   0   1 = 9-4-5     8-8-0     8-8-08-8-0    9-4-5      8-8-08-8-0    8-8-0      9-4-5 = 0   0   00   0   00   0   0=O  A2-4A-5I=O        [Proved]Again, A2-4A-5I=OA-1A2-4A-5I=A-1O     [Pre-multiplying with A-1]A-1A2-4A-1A-5A-1=OA-4I=5A-15A-1=1   2   22   1   22   2   1-41   0   00   1   00   0   1=1-4   2-0   2-02-0   1-4   2-02-0   2-0   1-4=-3     2     2  2   -3     2  2      2  -3A-1=15-3     2     2  2   -3     2  2      2  -3



Page No 7.25:

Question 35:

Prove that A adj A=An.

Answer:

Let A=ai j be a square matrix of order n×n.If Ci j is a cofactor of ai j in A, then adj A= Ci jT=Cj i. Also , it is a matrix of order n×n.Because A and adj A are matrices of order n×n, A×adj A exists and is of order n×n.A×adj Ai j =r=1nAi radj Ar j                            =r=1nai rCr j=A    if i=j0   otherwise   Thus, each diagonal element of A×adj A is A. Also, the non-diagonal elements are zero.A×adj A=   A     0      0              0 0      A     0           0 0      0      A    0          0 0      0     0   A A×adj A= A      0      0              0 0      A      0           0 0      0      A    0           0 0      0     0   A  = An  1      0     0              0 0      1     0           0 0      0     1    0           0 0      0     0   1  = An In= An A×adj A= An    Hence proved.

Page No 7.25:

Question 36:

If A-1=3-11-156-55-22 and B=12-2-1300-21, find AB-1.

Answer:

We know that (AB)1 = B1 A1.

B=12-2-1300-21B-1=1BAdj.BNow,B=12-2-1300-21     =13+0+12-4     =1Now, to find Adj. BB11=-11+13=3B12=-11+2-1=1B13=-11+32=2         B21=-12+12-4=2         B22=-12+21=1                       B23=-12+3-2=2B31=-13+16=6B32=-13+2-2=2B33=-13+33+2=5Therefore,Adj. B=326112225Thus,B-1=326112225.AB-1=B-1A-1          =3261122253-11-156-55-22          =9-30+30-3+12-123-10+123-15+10-1+6-41-5+46-30+25-2+12-102-10+10          =9-35-210102Hence, AB-1=9-35-210102

Page No 7.25:

Question 37:

If A=1-230-14-221, find AT-1.

Answer:

We know that (AT)1 = (A1)T.

A=1-230-14-221A-1=1AAdj.ANow,A=1-230-14-221     =1-1-8-2-8+3     =-9+10     =1Now, to find Adj. AA11=-11+1-9=-9A12=-11+28=-8A13=-11+3-2=-2         A21=-12+1-8=8         A22=-12+27=7                       A23=-12+3-2=2A31=-13+1-5=-5A32=-13+24=-4A33=-13+3-1=-1Therefore,Adj. A=-98-5-87-4-22-1Thus,A-1=-98-5-87-4-22-1.AT-1=A-1T         =-98-5-87-4-22-1T         =-9-8-2872-5-4-1Hence, AT-1=-9-8-2872-5-4-1.

Page No 7.25:

Question 38:

Find the adjoint of the matrix A=-1-2-221-22-21 and hence show that Aadj A=AI3.

Answer:

A=-1-2-221-22-21Now, to find Adj. AA11=-11+1-3=-3A12=-11+26=-6A13=-11+3-6=-6         A21=-12+1-6=6         A22=-12+23=3                       A23=-12+36=-6A31=-13+16=6A32=-13+26=-6A33=-13+33=3Therefore,Adj. A=-366-63-6-6-63Now,A=-1-2-221-22-21     =-11-4-2-2-4+24+2     =3+12+12     =27To show: Aadj A=AI3LHS=-1-2-221-22-21-366-63-6-6-63=3+12+12-6-6+12-6+12-6-6-6+1212+3+1212-6-6-6+12-612-6-612+12+3=270002700027=27 100010001  =AI3=RHSHence, Aadj A=AI3.

Page No 7.25:

Question 39:

If A=011101110, find A-1 and show that A-1=12A2-3I.

Answer:

A=011101110A-1=1AAdj. ANow,A=011101110     =-1-1+11     =2Now, to find Adj. AA11=-11+1-1=-1A12=-11+2-1=1A13=-11+31=1         A21=-12+1-1=1         A22=-12+2-1=-1                       A23=-12+3-1=1A31=-13+11=1A32=-13+2-1=1A33=-13+3-1=-1Therefore,Adj. A=-1111-1111-1Thus,A-1=12-1111-1111-1.Now,A2=011101110011101110    =0+1+10+0+10+1+00+0+11+0+11+0+00+1+01+0+01+1+0    =211121112Now, to show A-1=12A2-3IRHS=12A2-3I=12211121112-3100010001=12-1111-1111-1=A-1=LHSHence, A-1=12A2-3I.



Page No 7.34:

Question 1:

Find the inverse

714-3

Answer:

A=7     14   -3We knowA=IA7     14   -3=1   00   1 A1      17 4  - 3=17   0   0   1A              Applying R117R1 1           17 0    -257=17       0-47   1 A               Applying R2R2-4R11   17 0   1=17            0425  -725 A             Applying R2-725R21   0 0   1=325     125425   -725 A               Applying R1R1-17R2A-1=1253     14     -7

Page No 7.34:

Question 2:

Find the inverse

5221

Answer:

A=5   22   1We knowA=IA5   22   1=1   00   1 A5-4   2-2   2       1=1-0   0-2   0       1A              [Applying R1R1-2R2]1   02   1=  1    -2  0      1 A                         1     02-2   1=    1           -2 0-2       1+4 A               [Applying R2R2-2R1] 1   00   1     = 1    -2-2     5AA-1=  1    -2-2     5

Page No 7.34:

Question 3:

Find the inverse

16-35

Answer:

A=   1    6-3    5We knowA= IA   1    6-3    5= 1     0 0     1A     1          6-3+3    5+18=  1         0 0+3   1+0A                 [Applying R2R2+3R1] 1    6 0   23= 1     0 3    1 A 1    6-6 0     23= 1-18 23    0-623    3                1A                [Applying R1R1-623R2] 1     0 0     1= 523           -623 3 23            123A                      [Applying R2123R2]A-1= 523     -623   3 23     123=123 5   -6 3     1A-1=123 5   -6 3     1

Page No 7.34:

Question 4:

Find the inverse

2513

Answer:

A=2   51   3We knowA=I A2   51   3=1   00   1A2-1   5-3  1        3 =1-0   0-1    0       1A                [Applying R1R1-R2]1   21   3    =1   -10      1A   1        21-1   3-2 =  1          -10-1      1+1A             [Applying R2R2-R1]1   20   1 =   1   -1 -1     2A1   00   1 =   1+2   -1-4    -1          2A                      [Applying R1R1-2R2]1   00   1 =   3   -5-1      2AA-1=   3   -5-1      2

Page No 7.34:

Question 5:

Find the inverse

31027

Answer:

A=3  102   7We knowA=IA  3  102   7=1   00   1 A3-2    10-7   2         7 =1-0   0-1   0        1 A                     [Applying R1R1-R2]1   32   7 =1   -10      1A  1         32-2   7-6  =  1      -10-2   1+2                           [Applying R2R2-2R1]1    30    1 =  1      -1-2       3   A 1    3-30      1 =  1+6      -1-9   -2            3 A                     [Applying R1R1-3R2]1   00   1 =   7   -10-2      3AA-1=   7   -10-2      3

Page No 7.34:

Question 6:

Find the inverse

012123311

Answer:

A=012123311We knowA=IA  012123311=100010001 A012-212311=10001-1001 A                Applying R2R2-R3 -301-212311=10-101-1001 A                Applying R1R1-R3-301-212012=10-101-1100 A                Applying R3R3+R1-301034012=10-1-23-1100 A     Applying R23R2-2R1-3010340-2-2=10-1-23-13-31 A     Applying R3R3-R2-301034011=10-1-23-1-3232-12 A               Applying R3-12R3-3010-10011=10-14-31-3232-12 A               Applying R2R2-4R3-3010-10001=10-14-3152-3212 A               Applying R3R3+R2-3000-10001=-3232-324-3152-3212 A               Applying R1R1-R31000-10001=12-12124-3152-3212 A               Applying R1-13R1100010001=12-1212-43-152-3212 A               Applying R2-R2

Page No 7.34:

Question 7:

Find the inverse

20-1510013

Answer:

A=20-1510013We knowA=IA  20-1510013=100010001 A10-12510013=1200010001 A                Applying R112R1 10-120152013=1200-5210001 A                Applying R2R2-5R1  10-1201520012=1200-521052-1 1 A                Applying R3R3-R2 10-120152001=1200-52105-22 A              Applying R32R3100010001=3-11-156-55-22 A              Applying R1R1+12R3 and R2R2-52R3 A-1=3-11-156-55-22

Page No 7.34:

Question 8:

Find the inverse

231241372

Answer:

A=231241372We knowA=IA  231241372=100010001 A13212241372=1200010001 A                Applying R112R1 1321201005212=1200-110-3201 A                Applying R2R2-2R1 and R3R3-3R1  10120100012=2-320-1101-521 A                Applying R1R1-32R2 and  R3R3-52R2  1012010001=2-320-1102-52A              Applying R32R3100010001=11-1-1102-52A                    Applying R1R1-12R3 A-1=11-1-1102-52

Page No 7.34:

Question 9:

Find the inverse

3-342-340-11

Answer:

A=3-342-340-11We knowA=IA  3-342-340-11=100010001 A1-1432-340-11=1300010001 A                Applying R113R1 1-1430-1430-11=1300-2310001 A                 Applying R2R2-2R1 1-14301-430-11=130023-10001 A               Applying R2-R2 10001-4300-13=1-1023-1023-11 A                      Applying R1R1+R2 and R3R3+R210001-43001=1-1023-10-23-3 A                      Applying R3-3R3 100010001=1-10-23-4-23-3 A                      Applying R2R2+43R3  A-1=1-10-23-4-23-3

Page No 7.34:

Question 10:

Find the inverse

12023-11-13

Answer:

A=12023-11-13We knowA=IA  12023-11-13=100010001 A1200-1-10-33=100-210-101 A                 Applying R2R2-2R1 and R3R3-R1 1200110-33=1002-10-101 A                    Applying R2-R2 10-2011006=-3202-105-31 A                      Applying R1R1-2R2 and R3R3+3R210-2011001=-3202-1056-1216 A                      Applying R316R3 100010001=-4311376-12-1656-1216A                      Applying R1R1+2R3 and R2R2-R3  A-1=-4311376-12-1656-1216

Page No 7.34:

Question 11:

Find the inverse

2-13124311

Answer:

A=2-13124311We knowA=IA  2-13124311=100010001 A1-1232124311=1200010001A                 Applying R112R1  1-123205252052-72=1200-1210-3201 A                   Applying R2R2-R1 and R3R3-3R11-1232011052-72=1200-15250-3201 A                       Applying R225R2 10201100-6=25150-15250-1-11A                     Applying R1R1+12R2 and R3R3-52R2 102011001=25150-152501616-16 A                       Applying R3-16R2 100010001=115-21513-1130730161616-16 A                     Applying R2R2-R3 and R1R1-2R3  A-1=-130-24-1011-7-5-5-55

Page No 7.34:

Question 12:

Find the inverse

112311231

Answer:

A=112311231We knowA=IA  112311231=100010001 A1120-2-501-3=100-310-201A                 Applying R2R2-3R1 and R3R3-2R1112015201-3=10032-120-201 A                   Applying R2-12R2 10-12015200-112=-1212032-120-72121 A                         Applying R1R1-R2 and R3R3-R2 10-120152001=-1212032-120711-111-211 A                       Applying R3-211R3 100010001=-211511-111-111-311511711-111-211 A                     Applying R2R2-52R3 and R1R1+12R3  A-1=111-25-1-1-357-1-2 

Page No 7.34:

Question 13:

Find the inverse

2-144073-27

Answer:

A=2-144023-27We knowA=IA  2-144023-27=100010001 A1-1224023-27=1200010001 A                   Applying R112R1 1-12202-60-121=1200-210-3201A                 Applying R2R2-4R1 and R3R3-3R11-12201-30-121=1200-1120-3201 A                   Applying R212R2 101201-300-12=0140-1120-2141 A                         Applying R1R1+12R2 and R3R3+12R2 101201-3001=0140-11204-12-2 A                       Applying R3-2R3 100010001=-212111-1-64-12-2 A                     Applying R1R1-12R3 and R2R2+3R3  A-1=-212111-1-64-12-2  

Page No 7.34:

Question 14:

Find the inverse

30-1230041

Answer:

A=30-1230041We knowA=IA  30-1230041=100010001 A10-13230041=1300010001 A                   Applying R113R2 10-130323041=1300-2310001 A                 Applying R2R2-2R1 10-130129041=1300-29130001 A                   Applying R213R2 10-1301290019=1300-2913089-431 A              Applying R3R3-4R2 10-130129001=1300-291308-129 A                       Applying R39R3 100010001=3-43-23-28-129 A                      Applying R2R2-29R3 and R1R1+13R3  A-1=3-43-23-28-129 

Page No 7.34:

Question 15:

Find the inverse

13-2-30-1210

Answer:

A=13-2-30-1210We knowA=IA  13-2-30-1210=100010001 A13-209-70-54=100310-201 A                   Applying R2R2+3R1 and R3R3-2R1 13-201-790-54=10013190-201 A              Applying R219R2 10 1301-790019=0-13013190-13591 A                  Applying R1R1-3R2 and R3R3+5R2 101301-79001=0-13013190-359 A              Applying R39R3 100010001=1-2-3-247-359 A                      Applying R2R2+79R3 and R1R1-13R3  A-1=1-2-3-247-359 

Page No 7.34:

Question 16:

Find the inverse of each of the following matrices by using elementary row transformations:

-112123311

Answer:

Let A=-112123311.To find inverse, first write A=IA.i.e., -112123311=100010001A1-1-2123311=-100010001A      Applying R1-1R11-1-2035047=-100110301A      Applying R2R2-R1 and R3R3-3R11-1-20153047=-10013130301A      Applying R213R210-1301530013=-231301313053-431A      Applying R3R3-4R2 and R1R1+R210-130153001=-23130131305-43A      Applying R33R3100010001=1-11-87-55-43A      Applying R2R2-53R3 and R1R1+13R3Hence, A-1=1-11-87-55-43.



Page No 7.35:

Question 1:

Write the adjoint of the matrix A=-347-2.

Answer:

Let Cij be a cofactor of aij in A. Now,C11=-2C12=-7C21=-4C22=-3adj A=-2-7-4-3T=-2-4-7-3

Page No 7.35:

Question 2:

If A is a square matrix such that A (adj A)  5I, where I denotes the identity matrix of the same order. Then, find the value of |A|.

Answer:

We know
A(adj A)=AI
Here,
A(adj A)=5IA=5

Page No 7.35:

Question 3:

If A is a square matrix of order 3 such that |A| = 5, write the value of |adj A|.

Answer:

For any square matrix of order n,
adj A=An-1 
adj A=A2=52=25

Page No 7.35:

Question 4:

If A is a square matrix of order 3 such that |adj A| = 64, find |A|.

Answer:

For any square matrix of order n,
adj A=An-164=A2      adj A=64A=±8

Page No 7.35:

Question 5:

If A is a non-singular square matrix such that |A| = 10, find |A−1|.

Answer:

For any non-singular matrix A,
A-1=1AA-1=110       A=10

Page No 7.35:

Question 6:

If A, B, C are three non-null square matrices of the same order, write the condition on A such that AB = ACB = C.

Answer:

Consider AB=AC.
On multiplying both sides by A-1, we get

AA-1B=AA-1C
IB=IC    Because AA-1=I where I is the identity matrixB=C 

Therefore, the required condition is A must be invertible or A0.

Page No 7.35:

Question 7:

If A is a non-singular square matrix such that A-1=53-2-1, then find AT-1.

Answer:

For any invertible matrix A,
(AT)-1=(A-1)T

We haveA-1=53-2-1(AT)-1=5-23-1

Page No 7.35:

Question 8:

If adj A=234-1 and adj B=1-2-31, find adj AB.

Answer:

Given:

adj A=234-1adj B=1-2-31

For any two non-singular matrices,

adjAB=adj B×adj AadjAB=-65-2-10

Page No 7.35:

Question 9:

If A is symmetric matrix, write whether AT is symmetric or skew-symmetric.

Answer:

For any symmetric matrix, AT=A.

Hence, AT is also symmetric.

Page No 7.35:

Question 10:

If A is a square matrix of order 3 such that |A| = 2, then write the value of adj (adj A).

Answer:

For any square matrix A, we have

adj(adj A)=An-2Aadj(adj A)=2A        A=2 and n=3          

Page No 7.35:

Question 11:

If A is a square matrix of order 3 such that |A| = 3, then write the value of adj (adj A).

Answer:

For any square matrix A, we have

adj(adj A)=An-12adj(adj A)=34=81

Page No 7.35:

Question 12:

If A is a square matrix of order 3 such that adj (2A) = k adj (A), then write the value of k.

Answer:

For any matrtix A of order n, adj λA=λn-1adj A, where λ is a constant. Thus, for matrix A of order 3, we have adj (2A) = 23-1adj A adj (2A)= 22 adj Aadj (2A) =4 adj Akadj (A) =4 adj A        adj (2A)=k adj Ak=4

Page No 7.35:

Question 13:

If A is a square matrix, then write the matrix adj (AT) − (adj A)T.

Answer:

 In a non-singular matrix, adj AT =adj AT.adj AT  -adj AT=Null matrix

Page No 7.35:

Question 14:

Let A be a 3 × 3 square matrix, such that A (adj A) = 2 I, where I is the identity matrix. Write the value of |adj A|.

Answer:

We know that for a matix A of order n, A.adj A=AIn, where I is the identity matrix.Given: A.adj A=2IAI=2IA=2Now,  adj A=An-1adj A=22=4

Page No 7.35:

Question 15:

If A is a non-singular symmetric matrix, write whether A−1 is symmetric or skew-symmetric.

Answer:

Let A be an invertible symmetric  matrix.Then, A0 and AT=ANow, A-1T=AT-1    A-1T =A-1       [AT=A]Thus, A-1 is symmetric matrix. 

Page No 7.35:

Question 16:

If A=cos θsin θ-sin θcos θ and A adj A=k00k, then find the value of k.

Answer:

A= cos θ      sin θ-sin θ    cos θ A=cos θ      sin θ-sin θ    cos θ=cos2θ+ sin2 θ =10Thus, A-1 exists.Now, A-1 =adj AA= cos θ     -sin θsin θ    cos θA-1 =adj AA A-1 =A adj A A A-1=k  00  k     1  00  1=k  00  k            [  AA-1 =I]  k=1

Page No 7.35:

Question 17:

If A is an invertible matrix such that |A−1| = 2, find the value of |A|.

Answer:

We know, A-1=1A2=1A      A-1=2A=12

Page No 7.35:

Question 18:

If A is a square matrix such that A adj A=500050005, then write the value of |adj A|.

Answer:

Given:Aadj A=5  0  00  5  00  0  5    AIn=51  0  00  1  00  0  1A=5Now, adj A  =An-1 =53-1=25    

Page No 7.35:

Question 19:

If A=235-2 be such that A-1=k A, then find the value of k.

Answer:

A=2    35  -2A =2    35  -2 =-14-15=-19 The value is non-zero, so A-1 exists.By definition, we have A-1 A=I                      [I is the identity matrix]kA.A =I                  [Substituting A-1=kA]k2    35  -22    35  -2=1    00    1k4+15    6-610-10  15+4=1    00    1k19    00    19=1    00    1k=119

Page No 7.35:

Question 20:

Let A be a square matrix such that A2-A+I=O, then write A-1 interms of A.

Answer:

Given: A2-A+I =OA-1A2-A+I  =A-1O                         (Pre-multiplying both sides because A-1 exists)A-1A2-A-1A+A-1I  =O               (A-1O =O)A-I+A-1=O                                    (A-1I =A-1)A-1=I-A



Page No 7.36:

Question 21:

If Cij is the cofactor of the element aij of the matrix A=2-3560415-7, then write the value of a32C32.

Answer:

In the given matrix A=2-3560415-7,
C32 = (−1)3 + 2 (8 − 30) = 22

Therefore, a32C32 = 5 × 22 = 110.

Hence, the value of a32C32 is 110.

Page No 7.36:

Question 22:

Find the inverse of the matrix 3-2-75.

Answer:

A=3-2-75=10A is a non-singular matrix. Therefore, it is invertible.Let Cij be a cofactor of aij in A. The cofactors of element A are given by C11=5C12=7C21=2C22=3 A-1 =1A5723T=5273

Page No 7.36:

Question 23:

Find the inverse of the matrix co θsin θ-sin θcos θ.

Answer:

A=cosθsinθ-sinθcosθ A=cos2θ+sin2θ=10A is a singular matrix. Therefore, it is invertible.Let Cij be a cofactor of aij in A.The cofactors of element A are given byC11=cosθ C12=sinθC21=-sinθC22=cosθNow,adj A=cosθsinθ-sinθcosθT=cosθ-sinθsinθcosθ A-1=1Aadj A=cosθ-sinθsinθcosθ

Page No 7.36:

Question 24:

If A=1-320, write adj A.

Answer:

A=1-320=60A is a non-singular matrix. Therefore, it is invertible.Let Cij be a cofactor of aij in A.The cofactors of element A are given byC11=0C12=-2C21=3C22=1 adj A=0-231T=03-21

Page No 7.36:

Question 25:

If A=abcd, B=1001, find adj (AB).

Answer:

A×B=abcdA×B is a non-singular matrix. Therefore, it is invertible.Let Cij be a cofactor of aij in A.The cofactors of element A are given byC11=dC12=-cC21=-bC22=a adj A=d-c-baT=d-b-ca

Page No 7.36:

Question 26:

If A=312-3, then find |adj A|.

Answer:

A=312-3=-11 adj A=An-1=-112-1=-11

Page No 7.36:

Question 27:

If A=235-2, write A-1 in terms of A.

Answer:

A=235-2=-190A is a non-singular matrix. Therefore, it is invertible.Let Cij be a cofactor of aij in A.The cofactors of element A are given byC11=-2C12=-5C21=-3C22=2adj A=-2-5-32T=-2-3-52 A-1=1Aadj A=2/193/195/19-2/19

A-1=119A

Page No 7.36:

Question 28:

Write A-1 for A=2513

Answer:

A=2513=10Let Cij be the cofactor of aij in A.The cofactors of element A are given byC11=3C12=-1C21=-5C22=2adj A=3-1-52T=3-5-12A=6-5=1 A-1=1Aadj A=3-5-12

Page No 7.36:

Question 1:

If A is an invertible matrix, then which of the following is not true
(a) A2-1=A-12
(b) A-1=A-1
(c) AT-1=A-1T
(d) A0

Answer:

(a) A2-1=A-12

We have,  A-1=A-1, AT-1=A-1T and A0 all are the properties of the inverse of a matrix A



Page No 7.37:

Question 2:

If A is an invertible matrix of order 3, then which of the following is not true
(a) adj A=A2
(b) A-1-1=A
(c) If BA=CA, than BC, where B and C are square matrices of order 3
(d) AB-1=B-1 A-1, where Bbij3×3 and B0

Answer:

(c) If BA=CA, then BC where B and C are square matrices of order 3.

If A is an invertible matrix, then A-1 exists.

Now,
BA=CA
On multiplying both sides by A-1, we get
BAA-1=CAA-1

BI=CI     AA-1=I where I is the identity matrixB=C 

Therefore, the statement ​given in (c) is not true.

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Question 3:

If A=3424, B=-2-20-1, then A+B-1 =

(a) is a skew-symmetric matrix
(b) A−1 + B−1
(c) does not exist
(d) none of these

Answer:

(d) none of these

We have A+B=1223A+B=-10Thus, A+B-1 exists.Now,A+BT=1223Here,A+BT- A+BHence, it is not a skew symmetric matrix.We also know that A-1+B-1 is not the same as A+B-1.

Page No 7.37:

Question 4:

If S=abcd, then adj A is

(a) -d-b-ca

(b) d-b-ca

(c) dbca

(d) dcba

Answer:

(b) d-b-ca

Adjoint of a square matrix of order 2 is obtained by interchanging the diagonal elements and changing the signs of off-diagonal elements.

Here,

A=a  bc  dadj A=  d    -b-c      a

Page No 7.37:

Question 5:

If A is a singular matrix, then adj A is
(a) non-singular
(b) singular
(c) symmetric
(d) not defined

Answer:

(b) singular

A is singular, so A =0.By definition, we haveAadj A =O        A adj A=OA adj A =0                  adj A =0                   Hence, adj A is singular.

Page No 7.37:

Question 6:

If A, B are two n × n non-singular matrices, then
(a) AB is non-singular
(b) AB is singular
(c) AB-1A-1 B-1
(d) (AB)−1 does not exist

Answer:

(a) AB is non-singular

A and B are non-singular matrices of order n× n.A0  and   B0    ...1A and B are of the same order, so AB is defined and is of the same order. Thus,AB=AB     AB0         Using 1      Thus, AB is non-singular.

Page No 7.37:

Question 7:

If A=a000a000a, then the value of |adj A| is

(a) a27
(b) a9
(c) a6
(d) a2

Answer:

(c) a6

A=a  0  00  a  00   0 aA=a  0  00  a  00   0 a  =a30and   n=3Thus, we haveadj A =An-1=a32=a6

Page No 7.37:

Question 8:

If A=12-1-1122-11, then ded (adj (adj A)) is

(a) 144
(b) 143
(c) 142
(d) 14

Answer:

(a) 144

Given:A=  1     2   -1-1   1     2 2   -1    1A=1      2   -1-1    1     2 2    -1    1=11+2-2-1-4-11-2=3+10+1=14We have adjadj A=An-12adjadj A=1422=144

Page No 7.37:

Question 9:

If B is a non-singular matrix and A is a square matrix, then det (B−1 AB) is equal to
(a) Det (A−1)
(b) Det (B−1)
(c) Det (A)
(d) Det (B)

Answer:

(c) Det (A)

B is non-singular. This implies that B0, that B is invertible and that B-1 exists.Here, B is invertible.B-1=B-1=1BB-1AB=B-1ABB-1AB=B-1AB                           B-1AB =1BAB B-1AB =A

Page No 7.37:

Question 10:

For any 2 × 2 matrix, if A adj A=100010, then |A| is equal to
(a) 20
(c) 100
(d) 10
(d) 0

Answer:

 (c) 10Aadj A=10   00   10By definition, we have Aadj A=AI=adj AA           (Where I is the identity matrix)AI = Aadj AAI =101   00   1A=10

Page No 7.37:

Question 11:

If A5 = O such that AnI for 1 n4, then I-A-1 equals
(a) A4
(b) A3
(c) I + A
(d) none of these

Answer:

d  none of the theseI-A5=I-AI+A+A2+A3+A4Now,A5=0I=I-AI+A+A2+A3+A4II-A=I+A+A2+A3+A4I-A-1=I+A+A2+A3+A4

Page No 7.37:

Question 12:

If A satisfies the equation x3-5x2+4x+λ=0 then A−1 exists if
(a) λ1
(b) λ2
(c) λ-1
(d) λ0

Answer:

(d) λ0A satisfies x3-5x2+4x+λ =0.A3-5A2+4A=-λAssuming A-1 exists, we get  A-1A3-5A2+4A =-λA-1A2-5A+4=-A-1λ A-1=-A2-5A+4λThus, A-1 exists if λ0.

Page No 7.37:

Question 13:

If for the matrix A, A3 = I, then A−1 =
(a) A2
(b) A3
(c) A
(d) none of these

Answer:

a  A2 Given:A3=I     A3A-1=IA-1   Multiplying both sides by A-1A2=A-1

Page No 7.37:

Question 14:

If A and B are square matrices such that B = − A−1 BA, then (A + B)2 =
(a) O
(b) A2 + B2
(c) A2 + 2AB + B2
(d) A + B

Answer:

(b) A2+B2

B=-A-1BAAB=-AA-1BAAB=-BA                  ...1                AA-1=INow, A+B2=A+BA+B A+B2=A2+AB+BA+B2 A+B2=A2-BA+BA+B2       Using 1 A+B2=A2+B2



Page No 7.38:

Question 15:

If A=200020002, then A5=

(a) 5A
(b) 10A
(c) 16A
(d) 32A

Answer:

(c) 16A

 A=200020002A=2100010001A=2IA5=2I5A5=16×2IA5=16200020002A5=16A

Page No 7.38:

Question 16:

For non-singular square matrix A, B and C of the same order AB-1 C=
(a) A-1 B C-1
(b) C-1 B-1 A-1
(c) CBA-1
(d) C-1 B A-1

Answer:

Disclaimer: In Quesion, We are to find the inverse of AB-1 C. The inverse is missing in the question.

(d) C-1 B A-1

We have,
AB-1 C-1=C-1B-1-1A-1                     =C-1BA-1

Page No 7.38:

Question 17:

The matrix 5103-2-46-1-2b is a singular matrix, if the value of b is
(a) − 3
(b) 3
(c) 0
(d) non-existent

Answer:

(d) non-existent
For any singular matrix, the value of the determinant is 0.
Here,

A=5103-2-46-1-2bA=5(-4b+12)-10(-2b+6)+3(4-4)=0-20b+60+20b-12=0

Hence, b is non-existent.

Page No 7.38:

Question 18:

If d is the determinant of a square matrix A of order n, then the determinant of its adjoint is
(a) dn
(b) dn−1
(c) dn+1
(d) d

Answer:

(b) dn−1

We know,
adjA=An-1
adjA=dn-1

Page No 7.38:

Question 19:

If A is a matrix of order 3 and |A| = 8, then |adj A| =
(a) 1
(b) 2
(c) 23
(d) 26

Answer:

(d) 26

adjA=An-1        =82        =26

Page No 7.38:

Question 20:

If A2-A+I=0, then the inverse of A is
(a) A2
(b) A + I
(c) IA
(d) AI

Answer:

(c) IA

Given: A2-A+I =OA-1A2-A+I  =A-1O                              multiplying both sides by A-1A-1A2-A-1 A+A-1I  =O                A-1O =OA-I+A-1=O                                          A-1I =A-1A-1=I-A

Page No 7.38:

Question 21:

If A and B are invertible matrices, which of the following statement is not correct.
(a) adj A=AA-1
(b) det A-1=det A-1
(c) A+B-1=A-1+B-1
(d) AB-1=B-1 A-1

Answer:

(c) A+B-1=A-1+B-1

We have, adj A=AA-1, det A-1=det A-1 and AB-1=B-1 A-1 all are the properites of inverse of a matrix.

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Question 22:

If A is a square matrix such that A2 = I, then A1 is equal to
(a) A + I
(b) A
(c) 0
(d) 2A

Answer:

(b) A

Given:A2=I
On multiplying both sides by A-1 , we get

A-1A2=A-1IA=A-1IA=A-1

Page No 7.38:

Question 23:

Let A=123-5 and B=1002 and X be a matrix such that A = BX, then X is equal to

(a) 12243-5

(b) 12-2435

(c) 243-5

(d) none of these.

Answer:

(a) 12243-5

A=BXB-1A=B-1BXB-1A=IXX=B-1A        ...1Now, B=1002adjB=2001B=2 B-1=1BadjB=122001On putting the value of B-1in eq. 1, we getX=1220011   23-5X=122   43-5

Page No 7.38:

Question 24:

If A=235-2 be such that A-1=kA, then k equals
(a) 19
(b) 1/19
(c) − 19
(d) − 1/19

Answer:

(b) 1/19

adjA=-2-3-5   2A=-19A-1=1AadjAA-1=-119-2-3-5   2Now,   A-1=kA-119-2-3-5   2=kA1192   35-2=kA119A=kAk=119

Page No 7.38:

Question 25:

If A=1311221-2x2y is orthogonal, then x + y =
(a) 3
(b) 0
(c) − 3
(d) 1

Answer:

We have, A=1311221-2x2yAT=1312x1122-2yNow,  ATA=Ix2+52x+3xy-23+2x62yxy-62yy2+8=900090009The corresponding elements of two equal matrices are not equal.Thus, the matrix A is not orhtogonal.

Page No 7.38:

Question 26:

If A=101001ab2, then aI+bA+2 A2 equals
(a) A
(b) − A
(c) ab A
(d) none of these

Answer:

(d) none of these

A=101001ab2A2=1+ab3ab23a2ba+b+4Now,aI+bA+2A2=a000a000a+b0b00babb22b+2+2a2b62a2b46a6b2a+2b+8                    =3a+b+22bb+62aa+2bb+4ab+6ab2+6b3a+4b+8

Page No 7.38:

Question 27:

If 1-tan θtan θ1 1tan θ-tan θ1-1=a-bba, then

(a) a=1, b=1
(b) a=cos 2 θ, b=sin 2 θ
(c) a=sin 2 θ, b=cos 2 θ
(d) none of these

Answer:

(b) a=cos 2 θ, b=sin 2 θ

1tanθ-tanθ1-1=1sec2θ1-tanθtanθ1Given: 1-tanθtanθ11tanθ-tanθ1-1=a-bba1-tanθtanθ11sec2θ1-tanθtanθ1=a-bba1sec2θ1-tanθtanθ11-tanθtanθ1=a-bba1-tan2θsec2θ-2tanθsec2θ2tanθsec2θ1-tan2θsec2θ=a-bbaOn comparing, we geta=1-tan2θsec2θ and b=2tanθsec2θa=cos2θ-sin2θ and b=2sinθcosθa=cos2θ and b=sin2θ



Page No 7.39:

Question 28:

If a matrix A is such that 3 A3+2 A2+5 A+I=0, then A-1 equal to
(a) -3 A2+2 A+5
(b) 3 A2+2 A+5
(c) 3 A2-2 A-5
(d) none of these

Answer:

(d) none of these

3A3+2A2+5A+I=03A3+2A2+5A=-IA-1(3A3+2A2+5A)=-IA-13A2+2A+5I=-A-1A-1=-3A2-2A-5I

Page No 7.39:

Question 29:

If A is an invertible matrix, then det (A1) is equal to
(a) det A
(b) 1det A
(c) 1
(d) none of these

Answer:

(b) 1det A

We know that for any invertible matrix A, A-1 = 1A.

Page No 7.39:

Question 30:

If A=2-13-2, then An=

(a) A=1001, if n is an even natural number

(b) A=1001, if n is an odd natural number

(c) A=-1001, if nN

(d) none of these

Answer:

Disclaimer: In all option, the power of A (i.e. n is missing)

(a) An=1001, if n is an even natural number

A=2-13-2A2=1001A×A=IA-1=A

Generally,

An=(AA-1)n/2 when n is even.An=A(AA-1)n/2=A when n is odd.Thus, An=I when n is even.

Page No 7.39:

Question 31:

If x, y, z are non-zero real numbers, then the inverse of the matrix A=x000y000z, is
(a) x-1000y-1000z-1

(b) xyz x-1000y-1000z-1

(c) 1xyzx000y000z

(d) 1xyz 100010001

Answer:

(a) x-1000y-1000z-1


A=IAx000y000z=100010001A100010001=x-1000y-1000z-1 A          Applying R1=1xR1, R2=1yR2 and R3=1zR3A-1=x-1000y-1000z-1



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