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RD Sharma XII Vol 1 2020 Solutions for Class 12 Commerce Maths Chapter 7 Solution Of Simultaneous Linear Equations are provided here with simple step-by-step explanations. These solutions for Solution Of Simultaneous Linear Equations are extremely popular among class 12 Commerce students for Maths Solution Of Simultaneous Linear Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XII Vol 1 2020 Book of class 12 Commerce Maths 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 2020 Solutions. All RD Sharma XII Vol 1 2020 Solutions for class 12 Commerce Maths are prepared by experts and are 100% accurate.

Page No 7.14:

Question 1:

Solve the following system of equations by matrix method:
(i) 5x + 7y + 2 = 0
    4x + 6y + 3 = 0

(ii) 5x + 2y = 3
     3x + 2y = 5

(iii) 3x + 4y − 5 = 0
          x − y + 3 = 0

(iv) 3x + y = 19
         3x − y = 23

(v) 3x + 7y = 4
       x + 2y = −1

(vi) 3x + y = 23
      5x + 3y = 12

Answer:

(i) The given system of equations can be written in matrix form as folllows:5746 xy =-2-3AX=BHere,A=5746, X=xy and B=-2-3Now,  A  =5746                  =30-28                 = 2 ≠0         The given system has a unique solution given by X=A-1B.Let Cij be the cofactors of the elements aij in A=aij. Then,C11=-11+16=6 , C12=-11+24=-4C21=-12+17=-7, C22=-12+25       =5adj A=6-4-75T        =6-7-45A-1=1Aadj A⇒A-1=126-7-45X=A-1B   =126-7-45-2-3   =12-12+218-15⇒xy=92-72∴ x=92 and y=-72

ii The given system of equations can be written in matrix form as follows:5232 xy =35AX=BHere,A=5232, X=xy and B=35Now, A  =5232                  =10-6                  =4≠0The given system has a unique solution given by X=A-1B.Let Cij be the cofactors of the elements aij in A=aij. Then,C11=-11+12=2 ,C12=-11+23=-3C21=-12+12=-2, C22=-12+25 =5∴ adj A=2-3-25T         =2-2-35A-1=1 Aadj A              =142-2-35X=A-1B   =142-2-3535   =146-10-9+25xy=-44164∴ x=-1 and y=4

iii The given system of equations can be written in matrix form as follows:341-1 xy =5-3AX=BHere,A=341-1, X=xy and B=5-3Now,  A  =341-1                  =-3-4                 =-7≠0So, the given system has a unique solution given by X=A-1B.Let Cij be the cofactors of the  elements aij in A=aij. Then,C11=-11+1-1=-1, C12=-11+21=-1C21=-12+14=-4, C22=-12+23=3adj A=-1-1-43T=-1-4-13A-1=1Aadj A              =1-7-1-4-13X=A-1B   =1-7-1-4-135-3    =1-7-5+12-5-9⇒xy=7-7-14-7∴ x=-1 and y=2

iv The given system of equations can be written in matrix form as follows:313-1 xy =1923AX=BHere,A=313-1, X=xy and B=1923Now,  A  =313-1                  =-3-3                 =-6≠0So, the given system has a unique solution given by X=A-1B.Let Cij be the cofactors of the elements aij in A=aij. Then,C11=-11+1-1=-1,C12=-11+23 =-3C21=-12+11=-1, C22=-12+23=3adj A=-1-3-13T        =-1-1-33A-1=1Aadj A       =1-6-1-1-33X=A-1B  =1-6-1-1-331923  =1-6-19-23-57+69  =xy =-42-612-6∴ x=7 and y=-2

v The given system of equations can be written in matrix form as follows:3712 xy =4-1AX=BHere,A=3712, X=xy and B=4-1Now,  A  =3712                  =6-7                 =-1≠0So, the given system has a unique solution given by X=A-1B.Let Cij be the cofactors of the elements aij in A=aij. Then,C11=-11+12=2 ,C12=-11+21=-1C21=-12+17=-7, C22=-12+23       =3adj A=2-1-73T         =2-7-13A-1=1Aadj A             =1-12-7-13X=A-1B   =1-12-7-134-1    =1-18+7-4-3⇒xy=15-1-7-1∴ x=-15 and y=7

vi The given system of equations can be written in matrix form as follows:3153 xy =712AX=BHere,A=3153, X=xy and B=712Now,  A  =3153         =9-5        =4≠0So, the given system has a unique solution given by X=A-1B.Let Cij be the cofactors of the elements aij in A=aij. Then,C11=-11+13=3, C12=-11+25=-5C21=-12+11=-1, C22=-12+23=3adj A=3-5-13T         =3-1-53A-1=1 Aadj A             =143-1-53X=A-1B  =143-1-53712   =1421-12-35+36⇒xy=9414∴ x=94 and y=14

Page No 7.14:

Question 2:

Solve the following system of equations by matrix method:
(i) x + y − z = 3
2x + 3y + z = 10
3x − y − 7z = 1

(ii) x + y + z = 3
2x − y + z = − 1
2x + y − 3z = − 9

(iii) 6x − 12y + 25z = 4
4x + 15y − 20z = 3
2x + 18y + 15z = 10

(iv) 3x + 4y + 7z = 14
2x − y + 3z = 4
x + 2y − 3z = 0

(v)
2x-3y+3z=101x+1y+1z=103x-1y+2z=13

(vi) 5x + 3y + z = 16
2x + y + 3z = 19
x + 2y + 4z = 25

(vii) 3x + 4y + 2z = 8
2y − 3z = 3
x − 2y + 6z = −2

(viii) 2x + y + z = 2
x + 3y − z = 5
3x + y − 2z = 6

(ix) 2x + 6y = 2
3x − z = −8
2x − y + z = −3

(x) x − y + z = 2
2x − y = 0
2y − z = 1

(xi) 8x + 4y + 3z = 18
2x + y +z = 5
x + 2y + z = 5

(xii) x + y + z = 6
x + 2z = 7
3x + y + z = 12

(xiii) 2x+3y+10z=4, 4x-6y+5z=1, 6x+9y-20z=2;x, y, z≠0

(xiv) x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12

Answer:

(i)
Here,A=11-12313-1-7 A =11-12313-1-7      =1-21+1-1-14-3-1(-2-9)      =-20+17+11      =8Let Cij be the cofactors of the elements aij in Aaij. Then,C11=-11+131-1-7 =-20 ,            C12=-11+2213-7=17,                C13=-11+3233-1=-11C21=-12+11-1-1-7  =8  ,                C22=-12+2 1-13-7  =-4,           C23=-12+3113-1=4C31=-13+11-131 =4 ,                      C32=-13+21-121   =-3,          C33=-13+31123=1adj A=-2017-118-444-31T       = -208417-4-3-1141⇒A-1=1Aadj A=18-208417-4-3-1141X=A-1B⇒xyz=18-208417-4-3-11413101⇒xyz=18-60+80+451-40-3-33+40+1⇒xyz=182488⇒x=248, y=88 and z=88∴ x=3, y=1 and z=1

(ii)  Here,A=1112-1121-3 A =1112-1121-3     =13-1-1-6-2+1(2+2)     =2+8+4     =14Let Cij be the cofactors of the elements aij in Aaij. Then,C11=-11+1-111-3  =2 ,            C12=-11+2212-3 =8,                C13=-11+32-121=4C21=-12+1111-3 =4,                C22=-12+2 112-3  =-5,           C23=-12+31121=1C31=-13+111-11 =2 ,               C32=-13+21121   =1,                  C33=-13+3112-1=-3adj A=2844-5121-3T        = 2428-5141-3⇒A-1=1Aadj A=1142428-5141-3X=A-1B⇒xyz=1142428-5141-33-1-9⇒xyz=1146-4-1824+5-912-1+27⇒xyz=114-162038⇒x=-1614, y=2014and z=3814∴ x=-87, y= 107and z=197

(iii) Here,A=6-1225415-2021815 A =6-1225415-2021815      =6225+360+1260+40+25(72-30)     =3510+1200+1050     =5760Let Cij be the cofactors of the elements aij in Aaij. Then,C11=-11+115-201815  =585,             C12=-11+24-20215  =-100 ,              C13=-11+3415218=42C21=-12+1-12251815  = 630,            C22=-12+2 625215  =40,                       C23=-12+36-12218=-132C31=-13+1-122515-20 =-135,        C32=-13+26254-20   =220,              C33=-13+36-12415=138adj A=585-1004263040-132-135220138T        = 585630-135-1004022042-132138⇒A-1=1Aadj A             =15760585630-135-1004022042-132138X=A-1B⇒xyz=15760585630-135-1004022042-1321384310⇒xyz=157602340+1890-1350-400+120+2200168-396+1380⇒xyz=15760288019201152⇒x=28805760, y=19205760and z=11525760∴  x= 12, y= 13and z= 15

(iv) Here,A=3472-1321-3 A =3472-1321-3      =33-3-4-6-6+7(2+2)      =0+48+28      =76Let Cij be the cofactors of elements aij in Aaij. Then,C11=-11+1-131-3 = 0,           C12=-11+2232-3  =12,              C13=-11+32-121=4C21=-12+1471-3  = 19 ,           C22=-12+2 372-3  =-23 ,               C23=-12+33421=5C31=-13+147-13 =19,              C32=-13+23723   =5,                 C33=-13+3342-1=-11adj A=012419-235195-11T        = 0191912-23545-11⇒A-1=1Aadj A          =1760191912-23545-11X=A-1B⇒xyz=1760191912-23545-111440⇒xyz=1760+76+0168-92+056+20+0⇒xyz=176767676⇒x=7676, y=7676 and z=7676∴ x= 1, y= 1 and z=1


(v)
Let 1xbe a, 1ybe b and 1zbe c.Here,A=2-331113-12 A =2-331113-12      =22+1+32-3+3(-1-3)      =6-3-12      =-9Let Cij be the cofactors of the elements aij in Aaij. Then,C11=-11+111-12 =3,                  C12=-11+21132  =1 ,              C13=-11+3113-1=-4C21=-12+1-33-12  =3 ,               C22=-12+2 2332  =-5,               C23=-12+32-33-1=-7C31=-13+1-3311 =-6  ,           C32=-13+22311   =1,                C33=-13+32-311=5adj A=31-43-5-7-615T        = 33-61-51-4-75⇒A-1=1Aadj A            =1-933-61-51-4-75X=A-1B⇒abc=1-933-61-51-4-75101013⇒abc=1-930+30-7810-50+13-40-70+65⇒abc=1-9-18-27-45⇒ x=1a=-9-18, y=1b=-9-27and z=1c=-9-45∴ x=1a= 12, y=1b= 13and z=1c= 15

(vi)

Here,A=531213124 A =531213124=54-6-38-3+1(4-1)=-10-15+3=-22Let Cij be the cofactors of the elements aij in Aaij. Then,C11=-11+11324 = -2,          C12=-11+22314  =-5 ,            C13=-11+32112=3C21=-12+13124  =-10 ,       C22=-12+2 5114  =19,             C23=-12+35312=-7C31=-13+13113 =8 ,              C32=-13+25123   =-13 ,          C33=-13+35321=-1adj A=-2-53-1019-78-13-1T        = -2-108-519-133-7-1⇒A-1=1Aadj A            =1-22-2-108-519-133-7-1X=A-1B⇒xyz=1-22-2-108-519-133-7-1161925⇒xyz=1-22-32-190+200-80+361-32548-133-25⇒xyz=1-22-22-44-110⇒x=-22-22, y=-44-22and z=-110-22∴  x= 1, y= 2 and z= 5

(vii)
Here,A=34202-31-26 A =34202-31-26       =312-6-40+3+2(0-2)       =18-12-4       =2Let Cij be the cofactors of the elements aij in Aaij. Then,C11=-11+12-3-26 = 6,             C12=-11+20-316   =-3 ,             C13=-11+3021-2=-2C21=-12+142-26  = -28 ,        C22=-12+2 3216  =16 ,                 C23=-12+3341-2 =10C31=-13+1422-3 =-16,           C32=-13+2320-3   =9,                  C33=-13+33402=6adj A=6-3-2-281610-1696T       = 6-28-16-3169-2106⇒A-1=1Aadj A=126-28-16-3169-2106X=A-1B⇒xyz=126-28-16-3169-210683-2⇒xyz=1248-84+32-24+48-18-16+30-12⇒xyz=12-462⇒x=-42, y=62and z=22∴ x=-2, y= 3 and z=1


(viiii)Here,A=21113-131-2 A =21113-131-2      =2-6+1-1-2+3+1(1-9)      =-10-1-8      =-19Let Cij be the cofactors of the elements aij in Aaij. Then,C11=-11+13-11-2 =-5,           C12=-11+21-13-2   =-1,            C13=-11+31331=-8C21=-12+1111-2  = 3,          C22=-12+2 2-13-2  =-7,              C23=-12+32131=1C31=-13+1113-1 =-4,           C32=-13+2211-1   =3,               C33=-13+32113=5adj A=-5-1-83-71-435T        = -53-4-1-73-815⇒A-1=1Aadj A=1-19-53-4-1-73-815X=A-1B⇒xyz=1-19-53-4-1-73-815256⇒xyz=1-19-10+15-24-2-35+18-16+5+30⇒xyz=1-19-191919⇒x=-19-19, y=19-19 and z=19-19∴ x= 1, y= 3 and z=-1


(ix) Here,A=26030-12-11 A =26030-12-11      =20-1-63+2+0(-3+0)      =-2-30       =-32Let Cij be the cofactors of the elements aij in Aaij. Then,C11=-11+10-1-11 =-1 ,          C12=-11+23-121  =-5 ,          C13=-11+3302-1=-3C21=-12+160-11  = -6 ,          C22=-12+2 2021  =2  ,               C23=-12+3262-1=14C31=-13+1600-1 =-6  ,          C32=-13+2203-1    =2  ,            C33=-13+32630=-18adj A=-1-5-3-6214-62-18T        = -1-6-6-522-314-18⇒A-1=1Aadj A           =1-32-1-6-6-522-314-18X=A-1B⇒xyz=1-32-1-6-6-522-314-182-8-3⇒xyz=1-32-2+48+18-10-16-6-6-112+54⇒xyz=1-3264-32-64⇒x=64-32, y=-32-32 and z=-64-32∴  x=-2, y= 1 and z= 2

(x) Here,A=1-112-1002-1 A =1-112-1002-1      =11-0+1-2-0+1(4-0)      =1-2+4      =3Let Cij be the cofactors of the elements aij in Aaij. Then,C11=-11+1-102-1 =1,           C12=-11+2200-1  =2,               C13=-11+32-102=4C21=-12+1-112-1  = 1,       C22=-12+2 110-1  =-1,           C23=-12+31-102=-2C31=-13+1-11-10 =1,            C32=-13+21120   =2,                 C33=-13+31-12-1=1adj A=1241-1-2121T        = 1112-124-21⇒A-1=1Aadj A           =111112-124-21X=A-1B⇒xyz=131112-124-21201⇒xyz=132+14+28+1⇒xyz=11369⇒x=33, y=63 and z=93∴ x= 1, y= 2 and z= 3

(xi) Here,A=843211121 A =843211121     =81-2-42-1+3(4-1)     =-8-4+9     =-3Let Cij be the cofactors of the elements aij in Aaij. Then,C11=-11+11121 = -1,         C12=-11+22111  =-1,           C13=-11+32112=3C21=-12+14321  =2,            C22=-12+2 8311  =5,               C23=-12+38412=-12C31=-13+14311 =1,            C32=-13+28321   =-2 ,            C33=-13+38421=0adj A=-1-1325-121-20T        = -121-15-23-120⇒A-1=1Aadj A           =1-3-121-15-23-120X=A-1B⇒xyz=1-3-121-15-23-1201855⇒xyz=1-3-18+10+5-18+25-1054-60⇒xyz=1-3-3-3-6⇒x=-3-3, y=-3-3and z=-6-3∴ x= 1, y= 1 and z= 2


(xii)Here,A=111102311 A =111102311      =10-2-11-6+1(1-0)      =-2+5+1      =4Let Cij be the cofactors of the elements aij in Aaij. Then,C11=-11+10211 = -2,          C12=-11+21231  =5,            C13=-11+31031 =1C21=-12+11111  =0,           C22=-12+2 1131  =-2,         C23=-12+31131=2C31=-13+11102 =2,           C32=-13+21112  =-1,           C33=-13+31110=-1adj A=-2510-222-1-1T        = -2025-2-112-1⇒A-1=1Aadj A           =14-2025-2-112-1X=A-1B⇒xyz=14-2025-2-112-16712⇒xyz=14-12+0+2430-14-126-14-12⇒xyz=14124-20⇒x=124, y=44and z=-204∴ x= 3, y= 1 and z= -5


(xiii) Let 1xbe a, 1ybe b and 1zbe c.Here,A=23104-6569-20 A =23104-6569-20      =2120-45-3-80-30+10(36+36)      =150+330+720      =1200Let Cij be the cofactors of the elements aij in Aaij. Then,C11=-11+1-659-20 =75,            C12=-11+2456-20  =110,             C13=-11+34-669=72C21=-12+13109-20  =150,             C22=-12+2 2106-20  =  -100,       C23=-12+32369=0C31=-13+1310-65 =75,                   C32=-13+221045   =30,               C33=-13+3234-6=-24adj A=7511072150-10007530-24T      = 7515075110-10030720-24⇒A-1=1Aadj A            =11200 7515075110-10030720-24X=A-1B⇒abc=11200 7515075110-10030720-24412⇒abc=11200300+150+150440-100+60288-48⇒abc=11200600400240⇒x=1a=1200600, y=1b=1200400 and z=1c=1200240∴ x= 2, y=3 and z= 5


(xiv) Here,A=1-1  23  4-52-1  3 A =1-1  23  4-52-1  3      =112-5+19+10+2(-3-8)      =7+19-22      =4Let Cij be the cofactors of elements aij in A=aij. Then,C11=-11+14-5-13=7,             C12=-11+23-523=-19,              C13=-11+3342-1=-11C21=-12+1-12-13=1 ,              C22=-12+2 1223=-1 ,                 C23=-12+31-12-1=-1C31=-13+1-124-5=-3,         C32=-13+2123-5=11,                  C33=-13+31-134=7adj A=7-19-111-1-1-3117T        =71-3-19-111-11-17⇒A-1=1Aadj A          =1471-3-19-111-11-17X=A-1B⇒xyz=1471-3-19-111-11-177-512⇒xyz=1449-5-36-133+5+132-77+5+84⇒xyz=148412⇒x=84, y=44 and z=124∴ x= 2, y= 1 and z=3.



Page No 7.15:

Question 3:

Show that each of the following systems of linear equations is consistent and also find their solutions:
(i) 6x + 4y = 2
9x + 6y = 3

(ii) 2x + 3y = 5
6x + 9y = 15

(iii) 5x + 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5

(iv) x − y + z = 3
2x + y − z = 2
−x −2y + 2z = 1

(v) x + y + z = 6
x + 2y + 3z = 14
x + 4y + 7z = 30

(vi) 2x + 2y − 2z = 1
4x + 4y − z = 2
6x + 6y + 2z = 3

Answer:

(i) Here,6x+4y=2       ...(1) 9x+6y=3       ...(2)AX=B Here,A=6496, X=xy and B=236496xy=23 A =6496       =36-36       =0So, A is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because adj AB≠0 or adj A=0.Let Cij be the co factors of the elements aij in Aaij. Then,C11=6, C12=-9, C21=-4, C22=6adj A=6-9-46T         = 6-4-96adj AB=  6-4-9623               =12-12-18+18               =00If A=0 and adj AB=0, then the system is consistent and has infinitely many solutions.Thus, AX=B has infinitely many solutions. Substituting y=k in the eq. (1), we get6x+4k=2⇒6x=2-4k⇒x=2-4k6⇒x=1-2k3∴ x=1-2k3 and y=kThese values of x and y satisfy the third equation.Thus, x=1-2k3 and y=k where k is a real number satisfy the given system of equations.


(ii) Here,2x+3y=5         ...(1) 6x+9y=15       ...(2) or , AX=B  where,A=2369, X=xy and B=515⇒2369xy=515∴ A  =2369            =18-18            =0So, A is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because adj AB≠0 or adj A=0.Let Cij be the co-factors of the elements aij in Aaij. Then,C11=9, C12=-6, C21=-3 and C22=2∴adj A=9-6-32T              = 2-3-69⇒adjAB=  9-3-62515                   =45-45-30+30                   =00IfA=0 and adjAB=0, then the system is consistent and has infinitely many solutions.Thus, AX=B has infinitely many solutions. Substituting y=k in eq. (1), we get2x+3k=5⇒2x=5-3k⇒x=5-3k2 and y=kThese values of x and y satisfy the third equation.Thus, x=5-3k2 and y=k where k is a real number satisfy the given system of equations.


(iii) Here,5x+3y+7z=4        ...(1) 3x+26y+2z=9       ...(2)7x+2y+10z=5       ...(3)or ,AX=B where, A=53732627210, X=xyz and B=49553732627210xyz=495 A =53732627210     =5260-4-330-14+7(6-182)     =1280-48-1232     =0So, A is singular. Thus,the given system of equations is either inconsistent or it is consistent with infinitely many solutions because adj AB≠0 or adj AB=0. Let Cij be the co-factors of the elements aij in Aaij. Then,C11=-11+1262210 =256 ,              C12=-11+232710 =-16,              C13=-11+332672=-176C21=-12+137210  =-16 ,             C22=-12+2 57710  =1,                C23=-12+35372 =11C31=-13+137262 =-176,             C32=-13+25732   =11,                   C33=-13+353326=121adj A=256-16-176-16111-17611121T        = 256-16-176-16111-17611121adj AB=256-16-176-16111-17611121495            =1024-144-880-64+9+55-704+99+605            =000if A=0 and adjAB=0, then the system is consistent and has infinitely many solutions.Thus, AX= B has infinitely many solutions. Substituting z=k in eq. (1) and eq. (2), we get5x+3y=4-7k and 3x+26y=9-2k53326xy=4-7k9-2kNow,A=53326   =130-9    =121≠ 0adj A=26-3-35⇒A-1=1Aadj A            =112126-3-35∴X=A-1B⇒xy=112126-3-354-7k9-2k⇒xy=1121104-182k-27+6k-12+21k+45-10k⇒xy=77-176k12133+11k121⇒x=117-16k121, y=113+k121 and z=k∴ x= 7-16k11, y= 3+k11and z=kThese values of x, y and z also satisfy the third equation.Thus, x= 7-16k11, y= 3+k11 and z=k where k is a real number satisfy the given system of equations.


(iv) Here,x-y+z=3           ...(1)2x+y-z=2          ...(2)-x-2y+2z=1          ...(3)or,AX=Bwhere, A=1-1121-1-1-22, X=xyz and B=3211-1121-1-1-22xyz=321 A =1-1121-1-1-22      =12-2+14-1+1(-4+1)      =0+3-3      =0So, A is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because adj AB≠0 or adj AB=0.Let Cij be the co-factors of the elements aij in Aaij. Then,C11=-11+11-1-22 =0,             C12=-11+22-1-12  =-3,               C13=-11+321-1-2=-3C21=-12+1-11-22  =0,               C22=-12+2 11-12  =3,                    C23=-12+31-1-1-2=3C31=-13+1-111-1 =0,             C32=-13+2112-1  =3 ,                     C33=-13+31-121=3adj A=0-3-3033033T        = 000-333-333adj AB= 000-333-333321             =0-9+6+3-9+6+3            =000If A=0 and adj AB=0, then the system is consistent and has infinitely many solutions.Thus, AX=B has infinitely many solutions. Substituting z=k in eq. 1 and eq. 2, we getx-y=3-k and 2x+y=2+k1-121xy=3-k2+kNow,A=1-121     =1+2=3 ≠ 0adj A=12-11⇒A-1=1Aadj A           =1311-21∴X=A-1B⇒xy=1311-213-k2+k⇒xy=133-k+2+k-6+2k+2+k⇒xy=533k-43∴ x=53, y=3k-43 and z=kThese values of x, y and z also satisfy the third equation.Thus, x= 53, y= 3k-43and z=k where k is a real number satisfy the given system of equations.


(v) Here,x+y+z=6        ...(1)x+2y+3z=14       ...(2)x+4y+7z=30          ...(3)or, AX=B where, A=111123147, X=xyz and B=61430111123147xyz=61430 A =111123147     =114-12-17-3+1(4-2)     =2-4+2     =0So, A is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because adj AB≠0 or adj A=0.Let Cij be the co-factors of the elements aij in Aaij. Then,C11=-11+12347 =2,           C12=-11+21317  =-4 ,              C13=-11+31214=2C21=-12+11147  =-3,        C22=-12+2 1117  =6,                  C23=-12+31114=-3C31=-13+11123 =1 ,            C32=-13+21113   =-2 ,              C33=-13+31112=1adj A=2-42-36-31-21T      = 2-31-46-22-31adj AB=  2-31-46-22-3161430             =12-42+30-24+84-6012-42+30            =000IfA=0 and adjAB=0, then the system is consistent and has infinitely many solutions.Thus, AX=B has infinitely many solutions. Substituting z=k in eq. (1) and eq. (2), we get x+y=6-k and x+2y=14-3k1112xy=6-k14+3kNow,A=1112    =2-1=1 ≠ 0adj A=2-1-11⇒A-1=1Aadj A            =112-1-11∴X=A-1B⇒xy=112-1-116-k14-3k⇒xy=1112-2k-14+3k-6+k+14-3k⇒xy=k-218-2k1∴ x=k-2, y=8-2k and z=kThese values of x, y and z also satisfy the third equation.Thus,  x=k-2, y=8-2k and z=k where k is a real number satisfy the given system of equations.


(vi) Here,2x+2y-2z=1     ...(1) 4x+4y-z=2      ...(2) 6x+6y+2z=3    ...(3) or, AX=Bwhere,A=22-244-1662, X=xyz and B=12322-244-1662xyz=123 A =22-244-1662      =28+6-28+6-2(24-24)      =28-28-0      =0So, A is singular. Thus, the system of equations is either inconsistent or it is consistent withinfinitely many solutions because adj AB≠0 or adj AB=0.Let Cij be the co-factors of the elements aij in Aaij. Then,C11=-11+14-162 =14,             C12=-11+24-162  =-14,              C13=-11+34466=0C21=-12+12-262  =-16,        C22=-12+2 2-262  =16,                C23=-12+32266=0C31=-13+12-24-1 =6,               C32=-13+22-24-1   =-6,               C33=-13+32244=0adj A=14-140-161606-60T        = 14-166-1416-6000adj AB=  14-166-1416-6000123            =14-32+18-14+32-180            =000IfA=0 and adjAB=0, then the system is consistent and has infinitely many solutions.Thus, AX=B has infinitely many solutions. Substituting y=k in eq. (1) and eq. (2), we get 2x-2z=1-2k and 4x-z=2-4k2-24-1xz=1-2k2-4kNow,A=2-24-1    =-2+8=6≠ 0adj A=-12-42⇒A-1=1Aadj A=16-12-42∴X=A-1B⇒xz=16-12-421-2k2-4k⇒xz=16-1+2k+4-8k-4+8k+4-8k⇒xz=3-6k60∴ x=1-2k2, y=k and z=0  These values of x, y and z satisfy the third equation.Thus, x=1-2k2, y=k and z=0 where k is real number satisfy the given system of equations.

Page No 7.15:

Question 4:

Show that each one of the following systems of linear equation is inconsistent:
(i) 2x + 5y = 7
6x + 15y = 13

(ii) 2x + 3y = 5
6x + 9y = 10

(iii) 4x − 2y = 3
6x − 3y = 5

(iv) 4x − 5y − 2z = 2
5x − 4y + 2z = −2
2x + 2y + 8z = −1

(v) 3x − y − 2z = 2
2y − z = −1
3x − 5y = 3

(vi) x + y − 2z = 5
x − 2y + z = −2
−2x + y + z = 4

Answer:


(i) The given system of equations can be expressed as follows:AX=B Here, A=25615, X=xy and B=713Now,  A =25615               =30-30               =0Let Cij be the cofactors of the elements aij in A =aij. Then,C11=-11+1 15=15,C12=-11+2 6 =-6C21=-12+1 5=-5,C22=-12+2 2=2adj A=15-6-52T        =15-5-62adj A B=15-5-62713                     =105-65-42+26                    =40-16≠ 0Hence, the given system of equations is inconsistent.


(ii) The given system of equations can be expressed as follows:AX=B Here, A=2369, X=xy and B=510Now,  A =2369               =18-18               =0Let Cij be the cofactors of the elements aij in A =aij. Then,C11=-11+1 9=9 ,C12=-11+2 6 =-6 C21=-12+1 3=-3, C22=-12+2 2=2adj A=9-6-32T       =9-3-62adj A B=9-3-62510                 =45-30-30+20                 =15-10≠ 0Hence, the given system of equations is inconsistent.


(iii) The given system of equations can be expressed as follows:AX=BHere, A=4-26-3, X=xy and B=35  A =4-26-3      =-12+12       =0Let Cij be the cofactors of the elements aij in A =aij. Then,C11=-11+1 -3=-3, C12=-11+2 6 =-6C21=-12+1 -2=2,    C22=-12+2 4=4adj A=-3-624T         =-32-64adj A B=-32-6435                =-9+10-18+20               =12≠ 0Hence, the given system of equations is inconsistent.


(iv) The given system of equations can be written as follows:AX=B Here, A=4-5-25-42228, X=xyz and B=2-2-1 A =4-5-25-42228     =4-32-4+540-4-2(10+8)     =-144+180-36     =0Let Cij be the cofactors of the elements aij in Aaij. Then,C11=-11+1-4228 =28,                      C12=-11+25228  =-36,                C13=-11+35-422=18C21=-12+1-5-228  =36 ,                C22=-12+2 4-228  =36  ,              C23=-12+34-522=-18C31=-13+1-5-2-42 =-18,            C32=-13+24-252   =-18,               C33=-13+34-55-4=9adj A=28-36183636-18-18-189T        = 2836-18-3636-1818-189adj AB= 2836-18-3636-1818-1892-2-1            =56-72+18-72-72+1836+36-9            =2-12663 ≠ 0Hence, the given system of equations is consistent.


(v) The given system of equations can be written as follows:AX=B Here, A=3-1-202-13-50, X=xyz and B=2-13 A =3-1-202-13-50     =30-5+10+3-2(0-6)     =-15+3+12     =0Let Cij be the cofactors of the elements aij in Aaij. Then,C11=-11+12-1-50 =-5,            C12=-11+20-130  =-3,              C13=-11+3023-5=-6C21=-12+1-1-2-50  =10,             C22=-12+2 3-230  =6,               C23=-12+33-13-5=12C31=-13+1-1-22-1 =5,               C32=-13+23-20-1   =3,               C33=-13+33-102=6adj A=-5-3-610612536T       = -5105-363-6126adj AB=-5105-363-61262-13=-10-10+15-6-6+9-12-12+18=-5-3-6 ≠ 0Hence, the given system of equations is consistent.


(vi) The given system of equations can be written as follows:AX=B Here, A=11-21-21-211, X=xyz and B=5-24 A =11-21-21-211     =1-2-1-11+2-2(1-4)     =-3-3+6     =0Let Cij be the cofactors of the elements aij in Aaij. Then,C11=-11+1-2111 =-3 ,             C12=-11+211-21   =-3,              C13=-11+31-2-21=-3C21=-12+11-211  =-3,            C22=-12+2 1-2-21  =-3 ,         C23=-12+311-21=-3C31=-13+11-2-21 =-3,         C32=-13+21-211   =-3,              C33=-13+3111-2=-3adj A=-3-3-3-3-3-3-3-3-3T        =-3-3-3-3-3-3-3-3-3adj AB=-3-3-3-3-3-3-3-3-35-24=-15+6-12-15+6-12-15+6-12=-21-21-21 ≠ 0Hence, the given system of equations is consistent.

Page No 7.15:

Question 5:

If A=1-102   340   12 and B=   2   2-4-4   2-4   2-1   5 are two square matrices, find AB and hence solve the system of linear equations:
x − y = 3, 2x + 3y + 4z = 17, y + 2z = 7

Answer:

Here,

A=1-10234012 and B=22-4-42-42-15Now,AB=1-10234012 22-4-42-42-15⇒AB=600060006⇒AB=6100010001⇒AB=6I3⇒16AB=I3⇒16BA=I3         ∵ AB=BA⇒A-1=16B⇒A-1=1622-4-42-42-15X=A-1BX=1622-4-42-42-153177xyz=166+34-28-12+34-286-17+35xyz=1612-624∴ x=2, y=-1 and z=4

Page No 7.15:

Question 6:

If A=2-3   53   2-41   1-2, find A−1 and hence solve the system of linear equations
2x − 3y + 5z = 11, 3x + 2y − 4z = −5, x + y + 2z = −3

Answer:

Here,A=2-3532-411-2 A =2-3532-411-2     =2-4+4+3-6+4+5(3-2)     =0-6+5     =-1Let Cij be the co factors of the elements aij in Aaij. Then,C11=-11+12-41-2 =0,                     C12=-11+23-41-2  =2,                C13=-11+33211=1C21=-12+1-351-2  =-1,               C22=-12+2 251-2  =-9,                  C23=-12+32-311=-5C31=-13+1-352-4 =2 ,                  C32=-13+2253-4   =23,                C33=-13+32-332=13adj A=021-1-9-522313T         = 0-122-9231-513⇒A-1=1Aadj A=1-10-122-9231-513The given system of equations can be written in matrix form as follows:2-3532-411-2xyz=11-5-3X=A-1B⇒xyz=1-10-122-9231-51311-5-3⇒xyz=1-10+5-622+45-6911+25-39⇒xyz=1-1-1-2-3⇒ x=-1-1, y=-2-1 and z=-3-1∴ x= 1, y=2 and z= 3



Page No 7.16:

Question 7:

Find A−1, if A=1   2   51-1-12   3-1. Hence solve the following system of linear equations:
x + 2y + 5z = 10, x − y − z = −2, 2x + 3y − z = −11

Answer:

Here,A=1251-1-123-1 A =1251-1-123-1     =11+3-2-1+2+5(3+2)     =4-2+25    =27Let Cij be the co factors of the elements aij in Aaij. Then,C11=-11+1-1-13-1 =4,               C12=-11+21-12-1  =-1,                 C13=-11+31-1-23=5C21=-12+1253-1  =17,               C22=-12+2 152-1  =-11,             C23=-12+31223=1C31=-13+125-1-1 =3,              C32=-13+2151-1   =6,                    C33=-13+3121-1=-3adj A=4-1517-11136-3T         = 4173-1-11651-3⇒A-1=1Aadj A=1274173-1-11651-3The given system of equations can be written in matrix form as follows:1251-1-123-1xyz=10-2-11X=A-1B⇒xyz=1274173-1-11651-310-2-11⇒xyz=12740-34-33-10+22-6650-2+33⇒xyz=127-27-5481∴ x=-2727 , y=-5427 and z=8127⇒x= -1, y= -2 and z=3

Page No 7.16:

Question 8:

(i) If A=1-202   130-21, find A−1. Using A−1, solve the system of linear equations
x − 2y = 10, 2x + y + 3z = 8, −2y + z = 7

(ii) A=3-422   351   01, find A−1 and hence solve the following system of equations:
3x − 4y + 2z = −1, 2x + 3y + 5z = 7, x + z = 2

(iii) A=1-202130-21 and B=72-6-21-3-42   5, find AB. Hence, solve the system of equations:
x − 2y = 10, 2x + y + 3z = 8 and −2y + z = 7

(iv) If A=120-2  -1-20-11, find A−1. Using A−1, solve the system of linear equations
x − 2y = 10, 2x − y − z = 8, −2y + z = 7

(v) Given A=22-4-42-42-1  5, B=1-10234012, find BA and use this to solve the system of equations
y + 2z = 7, x − y = 3, 2x + 3y + 4z = 17

(vi) If A=2311  22–3  1-1, find A–1 and hence solve the system of equations 2x + y – 3z = 13, 3x + 2y + z = 4, x + 2y – z = 8.
(vii) Use product 1-1202-33-24-20192-361-2 to solve the system of equations x + 3z = 9, −x + 2y − 2z = 4, 2x − 3y + 4z = −3.

Answer:

​(i) Here, A=1-202130-21 A=1 1+6+22-0+0-4-0     =7+4+0     =11Let Cij be the cofactors of the elements aij  in A=aij. Then,C11=-11+113-21=7, C12=-11+22301=-2, C13=-11+3 210-2=-4C21=-12+1-20-21=2, C22=-12+21001=1, C23=-12+31-20-2=2C31=-13+1-2013=-6, C32=-13+21023=-3, C33=-13+31-221=5∴adj A=7-2-4212-6-35T            = 72-6-21-3-425⇒A-1=1Aadj A=11172-6-21-3-425or,  AX=Bwhere,  A=1-202130-21,X=xyz and B=1087Now, ∴X=A-1B⇒X=11172-6-21-3-4251087⇒X=11170+16-42-20+8-21-40+16+35⇒xyz=11144-3311∴ x=4, y=-3 and z=1

(ii) Here, A=3-42235101 A=3 3-0+42-5+20-3      =9-12-6      =-9Let Cij be the cofactors of the elements aij  in A=aij. Then,C11=-11+13501=3, C12=-11+22511=3, C13=-11+3 2310=-3C21=-12+1-4201=4, C22=-12+23211=1, C23=-12+33-410=-4C31=-13+1-4235=-26, C32=-13+23225=-11, C33=-13+33-423=17adj A=33-341-4-26-1117T       = 34-2631-11-3-417⇒A-1=1Aadj A=1-934-2631-11-3-417AX=BHere, A=3-42235101, X=xyz and B=-172X=A-1B⇒X=1-934-2631-11-3-417-172⇒X=1-9-3+28-52-3+7-223-28+34⇒xyz=1-9-27-189∴ x=3, y=2 and z=-1

(iii) Here, A=1-202130-21 and B=72-6-21-3-425AB=1-202130-21 72-6-21-3-425⇒AB=7+4+02-2+0-6+6+014-2-124+1+6-12-3+150+4-40-2+20+6+5           =110001100011⇒AB=11100010001AB=11I3⇒111AB=I3⇒111BA=I3⇒A-1=111B⇒A-1=11172-6-21-3-425X=A-1BX=11172-6-21-3-4251087xyz=11170+16-42-20+9-21-40+16+35xyz=11144-3311∴ x=4, y=-3 and z=1

​(iv) Here, A=1 20-2-1-20-11A=1-1-2+22     =-3+4     =1Let Cij be the cofactors of the elements aij in A=aij. Then,C11=-11+1-1-2-1  1=-3, C12=-11+2-2-2  0  1=2, C13=-11+3 -2-1  0-1=2C21=-12+120-11=-2,   C22=-12+21001=1,       C23=-12+3120-1=1C31=-13+120-1-2=-4, C32=-13+210-2-2=2, C33=-13+312-2-1=3∴adj A=-322-211-423T            = -3-2-4212213⇒A-1=1Aadj A         =11-3-2-4212213         =-3-2-4212213We know that, AT-1=A-1T.Here, C=ATi.e.,C=1-202-1-10-21∴ C-1=-322-211-423or, CX=Bwhere, C=1-202-1-10-21, X=xyz and B=1087Now, ∴X=C-1B⇒X=-322-211-4231087⇒X=-30+16+14-20+8+7-40+16+21⇒xyz=0-5-3∴ x=0, y=-5 and z=-3.

(v) Here, A=22-4-42-42-15 and B=1-10234012BA=1-1023401222-4-42-42-15⇒BA=2+4+02-2+0-4+4+04-12+84+6-4-8-12+200-4+40+2-20-4+10        =600060006⇒BA=6100010001⇒BA=6I3⇒B16A=I3⇒B-1=16A⇒B-1=1622-4-42-42-15Now, BX=Cwhere, B=1-10234012, X=xyz and C=3177∴ X=B-1C⇒X=1622-4-42-42-153177⇒xyz=166+34-28-12+34-286-17+35⇒xyz=1612-624∴ x=2, y=-1 and z=4.

(vi) We have, A=2311  22–3  1-1.
∴ A=231122-31-1=2-2-2-3-1+6+11+6=-8-15+7=-16≠0
So, A is invertible.
Let Cij be the co-factors of the elements aij in A[aij]. Then,
C11=-11+1221-1=-2-2=-4C12=-11+212-3-1=-1-1+6=-5C13=-11+312-31=1+6=7
C21=-12+1311-1=3+1=4C22=-12+221-3-1=-2+3=1C23=-12+323-31=-12+9=-11
C31=-13+13122=6-2=4C32=-13+22112=-14-1=-3C33=-13+32312=4-3=1
∴Adj A=-4-5741-114-31T=-444-51-37-111⇒A-1=Adj AA=1-16-444-51-37-111
Now, the given system of equations is expressible as
21-332112-1xyz=1348
Or AT X = B, where X=xyz, B=1348
Now, AT=A=-16≠0
So, the given system of equations is consistent with a unique solution given by
X=AT-1B=A-1TB
xyz=-116-444-51-37-111T1348⇒xyz=-116-4-5741-114-311348⇒xyz=-116-52-20+5652+4-8852-12+8⇒xyz=-116-16-3248⇒xyz=12-3
Hence, x = 1, y = 2 and z = −3 is the required solution.

(vii) Suppose, A = 1-1202-33-24 B=-20192-361-2

A×B=1-1202-33-24-20192-361-2=-2-9+120-2+21+3-40+18-180+4-30-6+6-6-18+240-4+43+6-8=100010001

Since, A × B = I,
∴ B = A−1                                 .....(1)

Now, the given system of equations is

x + 3z = 9
−x + 2y − 2z = 4
2x − 3y + 4z = −3

This can also be represented as,

103-12-22-34xyz=94-3
Here, we can observe that 103-12-22-34=AT
So, ATxyz=94-3

Multiply the above expression by AT-1.

xyz=AT-194-3xyz=A-1T94-3     xyz=BT94-3                       Using (1)

xyz=-20192-361-2T94-3=-2960211-3-294-3=-18+36-180+8-39-12+6=053

Hence, x = 0, y = 5 and z = 3.

Page No 7.16:

Question 9:

The sum of three numbers is 2. If twice the second number is added to the sum of first and third, the sum is 1. By adding second and third number to five times the first number, we get 6. Find the three numbers by using matrices.

Answer:

Let the three numbers be x, y and z.

According to the question,x+y+2x+2y+z=15x+y+z=6The given system of equations can be written in matrix form as follows: 111121511xyz=216AX=BHere,A=  111121511, X=xyz and B=216A=1 2-1-11-5+11-10    =1+4-9    =-4Let Cij be the cofactors of t he elements aij  in A=aij. Then,C11=-11+12111=1, C12=-11+21151=4, C13=-11+3 1251=-9C21=-12+11111=0, C22=-12+21151=-4, C23=-12+31151=4C31=-13+11121=-1, C32=-13+21111=0, C33=-13+31112=1adjA=14-90-44-101T        = 10-14-40-941⇒A-1=1Aadj A=1-410-14-40-941X=A-1B⇒X=1-410-14-40-941216⇒X=1-42+0-68-4+0-18+4+6⇒xyz=1-4-44-8∴ x=1, y=-1 and z=2

Page No 7.16:

Question 10:

An amount of Rs 10,000 is put into three investments at the rate of 10, 12 and 15% per annum. The combined income is Rs 1310 and the combined income of first and  second investment is Rs 190 short of the income from the third. Find the investment in each using matrix method.

Answer:

Let x , y and z be the investments at the rates of interest of 10%, 12% and 15% per annum respectively.

Total investment = Rs 10,000
⇒x+y+z=10,000
Income from the first investment of Rs x = Rs10x100=Rs 0.1xIncome from the second investment of Rs x = Rs12y100=Rs 0.12y Income from the third investment of Rs x = Rs15z100=Rs 0.15z∴ Total annual income =Rs 0.1x+0.12y+0.15z⇒0.1x+0.12y+0.15z=1310      ∵ Total annual income = Rs 1310It is given that the combined income from the first two incomes is Rs 190 short of the income from the third.∴ 0.1x+0.12y=0.15z-190⇒-0.1x-0.12y+0.15z=190

Thus, we obtain the following system of simultaneous linear equations:x+y+z=100000.1x+0.12y+0.15z=1310-0.1x-0.12y+0.15z=190The given system of equation can be written in matrix form as follows: 1110.10.120.15-0.1-0.120.15xyz=100001310190AX=BHere,A= 1110.10.120.15-0.1-0.120.15,X=xyz and B=100001310190 A=1 0.15×0.12+0.15×0.12-10.15×0.1+0.15×0.1+1-0.1×0.12+0.12×0.1     =0.036-0.03+0      =0.006Let Cij be the cofactors of the elements aij  in A=aij. Then,C11=-11+10.120.15-0.120.15=0.036, C12=-11+20.10.15-0.10.15=-0.03, C13=-11+3 0.10.12-0.1-0.12=0C21=-12+111-0.120.15=-0.27, C22=-12+211-0.10.15=0.25, C23=-12+311-0.1-0.12=0.02C31=-13+1110.120.15=0.03, C32=-13+2110.10.15=-0.05, C33=-13+3110.10.12=0.02adj A=0.036-0.030-0.270.250.020.03-0.050.02T     = 0.036-0.270.03-0.030.25-0.0500.020.02⇒A-1=1Aadj A           =10.0060.036-0.270.03-0.030.25-0.0500.020.02X=A-1B⇒X=10.0060.036-0.270.03-0.030.25-0.0500.020.0210000 1310190⇒X=10.006360-353.7+5.7-300+327.5-9.50+26.2+3.8⇒xyz=10006121830∴ x=2000, y=3000 and z=5000Thus, the three investments are of Rs 2000, Rs 3000 and Rs 5000, respectively.

Page No 7.16:

Question 11:

A company produces three products every day. Their production on a certain day is 45 tons. It is found that the production of third product exceeds the production of first product by 8 tons while the total production of first and third product is twice the production of second product. Determine the production level of each product using matrix method.

Answer:

Let x, y and z be the production level of the first, second and third product, respectively.

According to the question,x+y+z=45         ...(1)-x+z=8              ...(2)x+z=2y             Since the production of first and third product is twice the production of second productx-2y+z=0         ...(3)The given system of equation can be written in matrix form as follows: 111-1011-21xyz=4580AX=BA= 111-1011-21 X=xyz B=4580Now, A=1 -0+2-1-1-1+12-0=2+2+2=6Let Cij be the cofactors of the elements aij  in A=aij. Then,C11=-11+101-21=2, C12=-11+2-1111=2, C13=-11+3 -101-2=2C21=-12+111-21=-3, C22=-12+21111=0, C23=-12+3111-2=3C31=-13+11101=1, C32=-13+211-11=-2, C33=-13+311-10=1adj A=222-3031-21T       = 2-3120-2231⇒A-1=1Aadj A=162-3120-2231X=A-1B⇒X=162-3120-22314580⇒X=1690-24+090+0+090+24+0⇒xyz=166690114 ∴ x=11, y=15 and z=19Thus, the production level of first, second and third product is 11, 15 and 19, respectively.

Page No 7.16:

Question 12:

The prices of three commodities P, Q and R are Rs x, y and z per unit respectively. A purchases 4 units of R and sells 3 units of P and 5 units of Q. B purchases 3 units of Q and sells 2 units of P and 1 unit of R. C purchases 1 unit of P and sells 4 units of Q and 6 units of R. In the process A, B and C earn Rs 6000, Rs 5000 and Rs 13000 respectively. If selling the units is positive earning and buying the units is negative earnings, find the price per unit of three commodities by using matrix method.

Answer:

The prices of three commodities P, Q and R are Rs x, Rs y and Rs z per unit, respectively.

According to the question,3x+5y-4z=60002x-3y+z=5000-x+4y+6z=13000The given system of equations can be written in matrix form as follows: 35-42-31-146xyz=6000500013000AX=BHere,A=35-42-31-146 X=xyz B=6000500013000Now, A=3 -18-4-512+1-48-3=-66-65-20=-151≠0So, A-1 exists.Let Cij be the cofactors of the elements aij  in A=aij. Then,C11=-11+1-3146=-22, C12=-11+221-16=-13, C13=-11+3 2-3-14=5C21=-12+15-446=-46, C22=-12+23-4-16=14, C23=-12+335-14=-17C31=-13+15-4-31=-7, C32=-13+23-421=-11, C33=-13+3352-3=-19adj A=-22-135-4614-17-7-11-19T         = -22-46-7-1314-115-17-19⇒A-1=1Aadj A=1-151-22-46-7-1314-115-17-19X=A-1B⇒X=1-151-22-46-7-1314-115-17-196000500013000⇒X=1-151-453000-151000-302000⇒xyz=300010002000∴ x=3000, y=1000 and z=2000
Thus, the prices of the three commodities P, Q and R are Rs 3000, Rs 1000 and Rs 2000 per unit, respectively.



Page No 7.17:

Question 13:

The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards.

Answer:

As per the information given in the question, the following equations hold true:

.

The above three equations can be represented in the form of a matrix as

Or AX = B, where,

Thus, A is non-singular. Therefore, its inverse exists.

Adj A is given by∴

Therefore, the number of awardees for Honesty, Cooperation and Supervision are 3, 4, and 5 respectively.

One more value which the management of the colony must include for awards may be Sincerity.

Page No 7.17:

Question 14:

A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs 6,000. Three times the award money for Hard work added to that given for honesty amounts to Rs 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.

Answer:

Let the award money given for Honesty, Regularity and Hard work be x, y and z respectively.

Since total cash award is Rs 6,000.

∴ x + y + z = Rs 6,000 ...(1)

Three times the award money for Hard work and Honesty is Rs 11,000.

∴ x + 3 z = Rs 11,000

⇒ x + 0.y + 3 z = Rs 11,000 ...(2)

Award money for Honesty and Hard work is double the one given for regularity.

∴ x + z = 2y

⇒ x − 2y + z = 0 ...(3)

The above system can be written in matrix form as,

Or AX = B, where

Thus, A is non-singular. Hence, it is invertible.

Adj A =

∴

Thus, award money given for Honesty, Regularity and Hard work are Rs 500, Rs 2000 and Rs 3500 respectively.

School can include sincerity for awards.

Page No 7.17:

Question 15:

Two institutions decided to award their employees for the three values of resourcefulness, competence and determination in the form of prices at the rate of Rs. x, y and z respectively per person. The first institution decided to award respectively 4, 3 and 2 employees with a total price money of Rs. 37000 and the second institution decided to award respectively 5, 3 and 4 employees with a total price money of Rs. 47000. If all the three prices per person together amount to Rs. 12000 then using matrix method find the value of x, y and z. What values are described in this equations?

Answer:

A.T.Q4x+3y+2z=370005x+3y+4z=47000x+y+z=12000We can expressed these equations as AX=B.Where A=432534111,X=xyz  and B=370004700012000A=43-4-35-4+25-3=-4-3+4=-3≠0So, A is non singular therefore inverse exists.A11=-1   A12=-1   A13=2A21=-1   A22=2   A23=-1A31=6   A32=-6   A33=-3adj A=-1-16-12-62-1-3A-1=1Aadj A=-13-1-16-12-62-1-3X=A-1B=-13-1-16-12-62-1-3 370004700012000    =-13-37000-47000+72000-37000+94000-7200074000-47000-36000=-13-12000-15000-9000=400050003000So, x=4000 , y= 5000  and z=3000.

Page No 7.17:

Question 16:

Two factories decided to award their employees for three values of (a) adaptable tonew techniques, (b) careful and alert in difficult situations and (c) keeping clam in tense situations, at the rate of ₹ x, ₹ y and ₹ z per person respectively. The first factory decided to honour respectively 2, 4 and 3 employees with a total prize money of ₹ 29000. The second factory decided to honour respectively 5, 2 and 3 employees with the prize money of ₹ 30500. If the three prizes per person together cost ₹ 9500, then
i) represent the above situation by matrix equation and form linear equation using matrix multiplication.
ii) Solve these equation by matrix method.
iii) Which values are reflected in the questions?

Answer:

According to question,2x+3y+4z=29000     .....15x+2y+3z=30500     .....2x+y+z=9500     .....3From 1, 2 and 3 we get the matrix equation,234523111xyz=29000305009500⇒-2-102-10111xyz=-900020009500     R1→R1-4R3  and  R2→R2-3R3⇒-2-10-210111xyz=-9000-20009500       R2→-R2⇒-400-210301xyz=-11000-200011500       R1→R1+R2  and  R3→R3-R2⇒100-210301xyz=2750-200011500           R1→-14R1⇒100010001xyz=275035003250       R2→R2+2R1  and  R3→R3-3R1∴x= 2750, y= 3500 and z= 3250

Page No 7.17:

Question 17:

Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award â‚¹x each, â‚¹y each and â‚¹z each for the three respective values to 3, 2 and 1 students respectively with a total award money of â‚¹1,600. School B wants to spend â‚¹2,300 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is â‚¹900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.

Answer:

Let the award money given for sincerity, truthfulness and helpfulness be â‚¹x, â‚¹y and â‚¹z respectively.

Since, the total cash award is â‚¹900.
∴ x + y + z = 900                    ....(1)

Award money given by school A is â‚¹1,600.
∴ 3x + 2y + z = 1600              ....(2)

Award money given by school B is â‚¹2,300.
∴ 4x + y + 3z = 2300              ....(3)

The above system of equations can be written in matrix form CX = D as
111321413xyz=90016002300Where, C=111321413, X=xyz and D=90016002300Now,C=111321413     =16-1-19-4+1(3-8)     =5-5-5     =-5Let Cij be the cofactors of elements cij in C=cij. Then,C11=-11+12113=5,                   C12=-11+23143=-5,                  C13=-11+33241=-5C21=-12+11113=-2 ,              C22=-12+2 1143=-1 ,                 C23=-12+31141=3C31=-13+11121=-1,               C32=-13+21131=2,                      C33=-13+31132=-1adj C=  5-5-5-2-1  3-1  2-1T        =  5-2-1-5-1  2-5  3-1⇒C-1=1Cadj C          =1-5  5-2-1-5-1  2-5  3-1X=C-1D⇒xyz=-15  5-2-1-5-1  2-5  3-190016002300⇒xyz=-154500-3200-2300-4500-1600+4600-4500+4800-2300⇒xyz=-15-1000-1500-2000⇒x=-1000-5, y=-1500-5 and z=-2000-5∴ x= 200, y= 300 and z=400.

Hence, the award money for each value of sincerity, truthfulness and helpfulness is â‚¹200, â‚¹300 and â‚¹400.

One more value which should be considered for award hardwork.

Page No 7.17:

Question 18:

Two schools P and Q want to award their selected students on the values of Discipline, Politeness and Punctuality. The school P wants to award â‚¹x each, â‚¹y each and â‚¹z each the three respectively values to its 3, 2 and 1 students with a total award money of â‚¹1,000. School Q wants to spend â‚¹1,500 to award its 4, 1 and 3 students on the respective values (by giving the same award money for three values as before). If the total amount of awards for one prize on each value is â‚¹600, using matrices, find the award money for each value. Apart from the above three values, suggest one more value for awards.

Answer:

​Let the award money given for Discipline, Politeness and Punctuality be â‚¹x, â‚¹y and â‚¹z respectively.

Since, the total cash award is â‚¹600.
∴ x + y + z = 600                    ....(1)

Award money given by school P is â‚¹1,000.
∴ 3x + 2y + z = 1000              ....(2)

Award money given by school Q is â‚¹1,500.
∴ 4x + y + 3z = 1500              ....(3)

The above system of equations can be written in matrix form AX = B as
111321413xyz=60010001500Where, A=111321413, X=xyz and B=60010001500Now,A=111321413     =16-1-19-4+1(3-8)     =5-5-5     =-5Let Cij be the cofactors of elements aij in A=aij. Then,C11=-11+12113=5,                   C12=-11+23143=-5,                  C13=-11+33241=-5C21=-12+11113=-2 ,              C22=-12+2 1143=-1 ,                 C23=-12+31141=3C31=-13+11121=-1,               C32=-13+21131=2,                      C33=-13+31132=-1adj A=  5-5-5-2-1  3-1  2-1T        =  5-2-1-5-1  2-5  3-1⇒A-1=1Aadj A          =1-5  5-2-1-5-1  2-5  3-1X=A-1B⇒xyz=-15  5-2-1-5-1  2-5  3-160010001500⇒xyz=-153000-2000-1500-3000-1000+3000-3000+3000-1500⇒xyz=-15-500-1000-1500⇒x=-500-5, y=-1000-5 and z=-1500-5∴ x= 100, y= 200 and z=300.

Hence, the award money for each value of Discipline, Politeness and Punctuality is â‚¹100, â‚¹200 and â‚¹300.

​One more value which should be considered for award is Honesty.

Page No 7.17:

Question 19:

Two schools P and Q want to award their selected students on the values of Tolerance, Kindness and Leadership. The school P wants to award â‚¹x each, â‚¹y each and â‚¹z each for the three respective values to 3, 2 and 1 students respectively with a total award money of â‚¹2,200. School Q wants to spend â‚¹3,100 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as school P). If the total amount of award for one prize on each values is â‚¹1,200, using matrices, find the award money for each value.
Apart from these three values, suggest one more value which should be considered for award.
 

Answer:

​​Let the award money given for Tolerance, Kindness and Leadership be â‚¹x, â‚¹y and â‚¹z respectively.

Since, the total cash award is â‚¹1,200.
∴ x + y + z = 1200                    ....(1)

Award money given by school P is â‚¹2,200.
∴ 3x + 2y + z = 2200                ....(2)

Award money given by school Q is â‚¹3,100.
∴ 4x + y + 3z = 3100                ....(3)

The above system of equations can be written in matrix form AX = B as
111321413xyz=120022003100Where, A=111321413, X=xyz and B=120022003100Now,A=111321413     =16-1-19-4+1(3-8)     =5-5-5     =-5Let Cij be the cofactors of elements aij in A=aij. Then,C11=-11+12113=5,                   C12=-11+23143=-5,                  C13=-11+33241=-5C21=-12+11113=-2 ,              C22=-12+2 1143=-1 ,                 C23=-12+31141=3C31=-13+11121=-1,               C32=-13+21131=2,                      C33=-13+31132=-1adj A=  5-5-5-2-1  3-1  2-1T        =  5-2-1-5-1  2-5  3-1⇒A-1=1Aadj A          =1-5  5-2-1-5-1  2-5  3-1X=A-1B⇒xyz=-15  5-2-1-5-1  2-5  3-1120022003100⇒xyz=-156000-4400-3100-6000-2200+6200-6000+6600-3100⇒xyz=-15-1500-2000-2500⇒x=-1500-5, y=-2000-5 and z=-2500-5∴ x= 300, y= 400 and z=500.

Hence, the award money for each value of Tolerance, Kindness and Leadership is â‚¹300, â‚¹400 and â‚¹500.

​One more value which should be considered for award is Honesty.



Page No 7.18:

Question 20:

A total amount of â‚¹7000 is deposited in three different saving bank accounts with annual interest rates 5%, 8% and 812% respectively. The total annual interest from these three accounts is â‚¹550. Equal amounts have been deposited in the 5% and 8% saving accounts. Find the amount deposited in each of the three accounts, with the help of matrices.

Answer:

​​​Let the amount deposited in each of the three accounts be â‚¹x, â‚¹x and â‚¹y respectively.

Since, the total amount deposited is â‚¹7,000.
∴ x + x + y = 7000
⇒ 2x + y = 7000                      ....(1)

Total annual Interest is â‚¹550.
∴ 5100x+8100x+17200y=550
⇒26x+17y=110000                ....(2)

The above system of equations can be written in matrix form AX = B as
212617xy=7000110000where, A=212617, X=xy and B=7000110000Now,A=212617     =34-26     =8Let Cij be the cofactors of elements aij in A=aij. Then,C11=-11+117=17,               C12=-11+226=-26C21=-12+11=-1 ,              C22=-12+2 2=2adj A=17-26-12T        =17-1-262⇒A-1=1Aadj A          =1817-1-262X=A-1B⇒xy=1817-1-2627000110000⇒xy=18119000-110000-182000+220000⇒xy=18900038000⇒x=90008 and y=380008∴ x= 1125 and y=4750.

Hence, the amount deposited in each of the three accounts is â‚¹1125, â‚¹1125 and â‚¹4750.

Page No 7.18:

Question 21:

A shopkeeper has 3 varieties of pens 'A', 'B' and 'C'. Meenu purchased 1 pen of each variety for a total of Rs 21. Jeevan purchased 4 pens of 'A' variety 3 pens of 'B' variety and 2 pens of 'C' variety for Rs 60. While Shikha purchased 6 pens of 'A' variety, 2 pens of 'B' variety and 3 pens of 'C' variety for Rs 70. Using matrix method, find cost of each variety of pen.

Answer:

As there are 3 varieties of pen A, B and CMeenu purchased 1 pen of each variety which costs her Rs 21Therefore, A+B+C=21Similarly, For Jeevan4A+3B+2C=60For Shikha 6A+2B+3C=70111432623ABC=216070where P=111432623, Q=216070P=19-4-112-12+18-18=-5≠0∴P-1 existsX=P-1QC11=5        C12=0          C13=-10C21=-1    C22=-3     C23=4C31=-1     C32=2         C33=-1adj P=   5   0-10-1-3    4-1   2-1T=5-1-10-3   2-10   4-1P-1=1-5     5-1-1     0-3   2-10   4-1X=P-1Q

1-55-1-10-32-104-1216070=1-5105-60-700-180+140-210+240-70=1-5-25-40-40∴X=588Therefore, cost of A variety of pens =Rs 5Cost of B variety of pens =Rs 8Cost of C variety of pens =Rs 8



Page No 7.20:

Question 1:

2x − y + z = 0
3x + 2y − z = 0
x + 4y + 3z = 0

Answer:


(i) The given system of homogeneous equations can be written in matrix form as follows:2-1132-1143 xyz=000or, AX=Owhere, A=2-1132-1143, X= xyzand O=000 A=2-1132-1143    =26+4+19+1+1(12-2)     =40∴ A ≠0So, the given system has only trivial solution, which is given below:x=y=z=0

Page No 7.20:

Question 2:

2x − y + 2z = 0
5x + 3y − z = 0
x + 5y − 5z = 0

Answer:

Here,
2x − y + 2z = 0                 ...(1)
5x + 3y − z = 0                 ...(2)
x + 5y − 5z = 0                 ...(3)

 The given system of homogeneous equations can be written in matrix form as follows:2-1253-115-5 xyz=000AX=OHere,A=2-1253-115-5, X= xyzand O=000Now, A=2-1253-115-5             =2-15+5+1-25+1+2(25-3)             =-20-24+44             =0∴ A ≠ 0So, the given system of homogeneous equations has non-trivial solution. Substituting z=k in eq. (1) and eq. (2), we get2x-y=-2k and 5x+3y=kAX=BHere,A=2-153, X=xyand B=-2kk⇒2-153xy=-2kkA=2-153              =3×2+1×5              =11So, A-1exists.We have adj A=31-52A-1=1Aadj A⇒A-1=11131-52X=A-1B⇒xy=11131-52-2kk            =111-6k+k10k+2kThus, x=-5k11,y=12k11and z=k where k is any real number satisfy the given system of equations.

Page No 7.20:

Question 3:

3x − y + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z = 0

Answer:

Here,
3x − y + 2z = 0                  ...(1)
4x + 3y + 3z = 0                ...(2)
5x + 7y + 4z = 0                ...(3)


The given system of homogeneous equations can be written in matrix form as follows:3-12433574 xyz=000AX=OHere,A=3-12433574, X= xyzand O=000Now, A=3-12433574=312-21+116-15+2(28-15)=-27+1+26=0∴ A ≠0So, the given system of homogeneous equations has non-trivial solution. Substituting z=k in eq. (1) & eq. (2), we get 3x-y=-2k and 4x+3y=-3kAX=BHere, A=3-143, X=xyand B=-2k-3k⇒3-143xy=-2k-3kA=3-143=3×3+4×1=13So, A-1exists.We have adj A=31-43A-1=1Aadj A⇒A-1=11331-43X=A-1B⇒xy=11331-43-2k-3k            =113-6k-3k8k-9kThus, x=-9k13,y=-k13and z=k where k is any real number  satisfy the given system of equations.

Page No 7.20:

Question 4:

x + y − 6z = 0
x − y + 2z = 0
−3x + y + 2z = 0

Answer:

Here,
x + y − 6z = 0                  ...(1)
x − y + 2z = 0                  ...(2)
−3x + y + 2z = 0              ...(3)

The given system of homogeneous equations can be written in matrix form as follows:11-61-12-312 xyz=000AX=OHere,A=11-61-12-312, X= xyzand O=000Now, A=11-61-12-312     =1-2-2-12+6-6(1-3)     =-4-8+12     =0∴  A = 0So, the given system of homogeneous equations has non-trivial solution. Substituting  z=k in eq. (1) and eq. (2), we getx+y=6k and x-y=-2kAX=BHere,A=111-1, X=xyand B=6k-2k⇒111-1xy=6k-2kNow, A=111-1    =1×-1-1×1     =-2So, A-1exists.We have adj A=-1-1-11A-1=1Aadj A⇒A-1=1-2-1-1-11X=A-1B⇒xy=1-2-1-1-116k-2k             =1-2-6k+2k-6k-2kThus, x=2k,y =4k and z=k where k is any real number satisfy the given system of equations.

Page No 7.20:

Question 5:

x + y + z = 0
x − y − 5z = 0
x + 2y + 4z = 0

Answer:

Here,
x + y + z = 0             ...(1)
x − y − 5z = 0           ...(2)
x + 2y + 4z = 0         ...(3)

The given system of homogeneous equations can be written in matrix form as follows:1111-1-5124 xyz=000AX=OHere,A=1111-1-5124, X= xyzand O=000Now, A=1111-1-5124              =1-4+10-14+5+1(2+1)              =6-9+3              =0∴  A ≠0So, the given system of homogeneous equations has non-trivial solution. Substituting z=k in eq. (1) and eq, (2), we get x+y=-k and x-y=5kAX=BHere, A=111-1, X=xyand B=-k5k⇒ 111-1xy=-k5kA=111-1              =1×-1-1×1             =-2So, A-1exists.We have adj A=-1-1-11A-1=1Aadj A⇒A-1=1-2-1-1-11X=A-1B⇒xy=1-2-1-1-11-k5k            =1-2k-5kk+5kThus, x=2k, y=-3k and z=k where k is any real number satisfy the given system of equations.

Page No 7.20:

Question 6:

x + y − z = 0
x − 2y + z = 0
3x + 6y − 5z = 0

Answer:

x + y − z = 0                ...(1)
x − 2y + z = 0              ...(2)
3x + 6y − 5z = 0          ...(3)

The given system of homogeneous equations can be written in matrix form as follows:11-11-2136-5xyz=000AX=OHere, A=11-11-2136-5, X=xyzand O=000Now, A =11-11-2136-5                =110-6-1-5-3-16+6                =4+8-12                =0So, the given system of homogeneous equations has non-trivial solution. Substituting z=k in eq. (1) & eq. (2), we get x+y=k and x−2y=-k⇒111-2xy=k-kAX=BHere,A=111-2, X=xy and B=k-k A =111-2=-3So, A-1 exists.adj A=-2-1-11A-1=1 A adj A⇒A-1=1-3-2-1-11X=A-1B⇒xy=1-3-2-1-11k-k⇒xy=1-3-2k+k-k-kThus, x= k3 , y= 2k3and z=k where k is any real number satisfy the given system of equations.



Page No 7.21:

Question 7:

3x + y − 2z = 0
x + y + z = 0
x − 2y + z = 0

Answer:

The given system of homogeneous equations can be written in matrix form as follows:31-21111-21xyz=000or,  AX=Owhere, A=31-21111-21, X=xyzand O=000Now, A =31-21111-21               =31+2-11-1-2-2-1               =9-0+6               =15≠0So, the given system has only trivial solution, which is given below:x=y=z=0

Page No 7.21:

Question 8:

2x + 3y − z = 0
x − y − 2z = 0
3x + y + 3z = 0

Answer:

The given system of homogeneous equations can be written in matrix form as follows:23-11-1-2313xyz=000AX=OHere, A=23-11-1-2313, X=xyzand O=000Now,  A =23-11-1-2313               =2-3+2-33+6-11+3               =-2-27-4               =-33≠0So, the given system of homogeneous equations has only trivial solution, which is given below:x=y=z=0

Page No 7.21:

Question 1:

The system of equation x + y + z = 2, 3x − y + 2z = 6 and 3x + y + z = −18 has
(a) a unique solution
(b) no solution
(c) an infinite number of solutions
(d) zero solution as the only solution

Answer:

(a) a unique solution

The given system of equations can be written in matrix form as follows:1113-12311xyz=26-18AX=B  Here, A=1113-12311, X=xyz and B=26-18 A=1 -1-2-13-6+13+3=-3+3+6=6≠0So, the given system of equations has a unique solution.

Page No 7.21:

Question 2:

The number of solutions of the system of equations
2x + y − z = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
is
(a) 3
(b) 2
(c) 1
(d) 0

Answer:


(d) 0The given system of equations can be written in matrix form as follows:21-11-3214-3 xyz=715AX=B Here, A=21-11-3214-3, X=xyz and B=715Now,A=2 9-8-1-3-2-14+3    =2+5-7    =0Let Cij be the cofactors of the elements aij  in A=aij. Then,C11=-11+1-324-3=1, C12=-11+2121-3=5, C13=-11+3 1-314=7C21=-12+11-14-3=-1, C22=-12+22-11-3=-5, C23=-12+32114=-7C31=-13+11-1-32=-1, C32=-13+22-112=-5, C33=-13+3211-3=-7adj A=157-1-5-7-1-5-7T         =1-1-15-5-57-7-7⇒adj A B=1-1-15-5-57-7-7715                       =7-1-535-5-2549-7-35                       =157 ≠0So, the given system of equations has no solution.

Page No 7.21:

Question 3:

Let X=x1x2x3, A=1-122   013   21 and B=314. If AX = B, then X is equal to
(a) 123

(b) -1-2-3

(c) -1-2-3

(d) -1   2   3

(e) 021

Answer:

(a) 123


Here, A=1-12201321, X=x1x2x3 and B=314           Given A=1 0-2+12-3+24-0     =-2-1+8     =5Let Cij be the cofactors of the elements aij  in A=aij. Then,C11=-11+10121=-2, C12=-11+22131=1, C13=-11+3 2032=4C21=-12+1-1221=5, C22=-12+21231=-5, C23=-12+31-132=-5C31=-13+1-1201=-1, C32=-13+21221=3, C33=-13+31-120=2adj A=-2145-5-5-132T         =-25-11-534-52⇒A-1=1Aadj A         =15-25-11-534-52∴X=A-1B⇒X=15-25-11-534-52314⇒X=15-6+5-43-5+1212-5+8⇒x1x2x3=15-51015⇒x1x2x3=-123

Page No 7.21:

Question 4:

The number of solutions of the system of equations:
2x + y − z = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
(a) 3
(b) 2
(c) 1
(d) 0

Answer:


(d) 0The given system of equations can be written in matrix form as follows: 21-11-3214-3xyz=715AX=BHere, A= 21-11-3214-3, X=xyz and B=715Now,A=2 9-8-1-3-2-14+3    =2+5-7    =0Let Cij be the cofactors of the elements aij  in A=aij. Then,C11=-11+1-324-3=1, C12=-11+2121-3=5, C13=-11+3 1-314=7C21=-12+11-14-3=-1, C22=-12+22-11-3=-5, C23=-12+32114=-7C31=-13+11-1-32=5, C32=-13+22-112=-5, C33=-13+3211-3=-7adj A=157-1-5-75-5-7T= 1-155-5-57-7-7⇒adj AB=1-155-5-57-7-7715=7-1+2535-5-2549-7-35=3256≠ 0The given system of equations is inconsistent. Thus, it has no solution.

Page No 7.21:

Question 5:

The system of linear equations:
x + y + z = 2
2x + y − z = 3
3x + 2y + kz = 4 has a unique solution if
(a) k ≠ 0
(b) −1 < k < 1
(c) −2 < k < 2
(d) k = 0

Answer:

(a) k ≠ 0

For a unique solution, A ≠ 0.The given system of equations can be written in matrix form as follows:11121-132k ≠0⇒1k+2-12k+3+14-3≠0⇒k+2-2k-3+1≠ 0⇒k≠0So, the given system of equations has a unique solution if k is not equal to 0.

Page No 7.21:

Question 6:

Consider the system of equations:
a1x + b1y + c1z = 0
a2x + b2y + c2z = 0
a3x + b3y + c3z = 0,
if a1b1c1a2b2c2a3b3c3 = 0, then the system has
(a) more than two solutions
(b) one trivial and one non-trivial solutions
(c) no solution
(d) only trivial solution (0, 0, 0)

Answer:

(a) more than two solutions Here,A=0 and B=0          GivenIf  A =0 and adj AB=0, then the system is consistent and has infinitely many solutions.Clearly, it has more than two solutions.



Page No 7.22:

Question 7:

Let a, b, c be positive real numbers. The following system of equations in x, y and z
x2a2+y2b2-z2c2=1, x2a2-y2b2+z2c2=1, -x2a2+y2b2+z2c2=1 has

(a) no solution
(b) unique solution
(c) infinitely many solutions
(d) finitely many solutions

Answer:

(b) unique solutionThe given system of equations can be written in matrix form as follows:1a21b2-1c21a2-1b21c2-1a21b21c2xyz=111Here,A=1a21b2-1c21a2-1b21c2-1a21b21c2, X=xyz and B=111Now,A=1a21b2-1c21a2-1b21c2-1a21b21c2=1a2b2c211-11-11-111=1a2b2c2×1-1-1-11+1-11-1=1a2b2c2×-2-2=-4a2b2c2⇒A≠ 0 So, the given system of equations has a unique solution.

Page No 7.22:

Question 8:

For the system of equations:
x + 2y + 3z = 1
2x + y + 3z = 2
5x + 5y + 9z = 4
(a) there is only one solution
(b) there exists infinitely many solution
(c) there is no solution
(d) none of these

Answer:

(a) there is only one solutionThe given system of equations can be written in matrix form as follows:123213559xyz=124Here,A=123213559, X=xyz and B=124Now,A=123213559    =19-15-218-15+310-5    =-6-6+15    =3≠0⇒A≠ 0 So, the given system of equations has a unique solution.

Page No 7.22:

Question 9:

The existence of the unique solution of the system of equations:
x + y + z = λ
5x − y + µz = 10
2x + 3y − z = 6
depends on
(a) µ only
(b) λ only
(c) λ and µ both
(d) neither λ nor µ

Answer:

(a) µ only
For a unique solution, A ≠ 0⇒1115-1μ23-1≠ 0⇒11-3μ-1-5-2μ+115+2≠ 0⇒1-3μ+5+2μ+17≠0⇒-μ+23≠ 0⇒μ ≠23So, existence of a unique solution depends only on μ.

Page No 7.22:

Question 10:

The system of equations:
x + y + z = 5
x + 2y + 3z = 9
x + 3y + λz = µ
has a unique solution, if
(a) λ = 5, µ = 13
(b) λ ≠ 5
(c) λ = 5, µ ≠ 13
(d) µ ≠ 13

Answer:

(b) λ ≠ 5For a unique solution, A ≠ 0.⇒11112313λ≠0⇒12λ-9-1λ-3+13-2≠ 0⇒2λ-9-λ+3+1 ≠ 0⇒λ-5≠ 0⇒λ ≠ 5

Page No 7.22:

Question 1:

If the system of equations x + ay = 0, az + y = 0, ax + z = 0 has infinitely many solutions then a = ___________________.

Answer:


The given system of homogeneous equations x + ay = 0, az + y = 0, ax + z = 0 has infinitely many solutions.

∴1a001aa01=0

⇒11-0-a0-a2+00-a=0

⇒1+a3=0

⇒1+a1-a+a2=0

⇒a+1=0               a2-a+1>0 ∀ a∈R

⇒a=-1

Thus, the value of a is −1.

If the system of equations x + ay = 0, az + y = 0, ax + z = 0 has infinitely many solutions then a = __−1__.
 

Page No 7.22:

Question 2:

If the system of equations x + y + z = 6, x + 2y + 3z = 10, x + 2y + λz = 12 is inconsistent then λ = __________________. 

Answer:


The system of equations x + y + z = 6, x + 2y + 3z = 10, x + 2y + λz = 12 is inconsistent.

∴∆=11112312λ=0

⇒12λ-6-1λ-3+12-2=0

⇒2λ-6-λ+3=0

⇒λ-3=0

⇒λ=3

Also, for λ = 3,

∆y=16111031123≠0 and ∆z=11612101212≠0

Thus, the value of λ is 3.

If the system of equations x + y + z = 6, x + 2y + 3z = 10, x + 2y + λz = 12 is inconsistent then λ = ___3___. 

Page No 7.22:

Question 3:

The number of solutions of the system of the system of equations  x + 2y + z = 3,  2x + 3y + z = 3,  3x + 5y + 2z = 1 is _____________.

Answer:


The given system of equations is x + 2y + z = 3, 2x + 3y + z = 3 and 3x + 5y + 2z = 1.

∆=121231352=16-5-24-3+14-3=1-2+1=0

∆x=321331152=36-5-26-1+115-3=3-10+12=5≠0

Here, ∆=0 and at least one of ∆x,∆y and ∆z is not zero, so the given system of equations has no solution.

Thus, the given system of equations has no solution.

The number of solutions of the system of the system of equations x + 2y + z = 3, 2x + 3y + z = 3, 3x + 5y + 2z = 1 is ___0___.

Page No 7.22:

Question 4:

If the system of equations  2x - y - z = 12,  x - 2y + z = -4,  x + y + λz = 4 has no solution, then λ = ________________.

Answer:


The given system of equations 2x − y − z = 12, x − 2y + z = −4 and x + y + λz = 4 has no solution.

∴∆=2-1-11-2111λ=0

⇒2-2λ-1--1λ-1-11+2=0

⇒-4λ-2+λ-1-3=0

⇒-3λ-6=0

⇒-3λ=6

⇒λ=-2

Thus, the value of λ = −2.

If the system of equations 2x − y − z = 12, x − 2y + z = −4, x + y + λz = 4 has no solution, then λ = ___−2___.

Page No 7.22:

Question 5:

If the system of equations  x - ky - z = 0,  kx - y - z = 0,  x + y - z = 0 has a non-zero solution then the values of k are __________________.

Answer:


The system of homogeneous equations x − ky − z = 0, kx − y − z = 0 and x + y − z = 0 has a non-zero solution or an infinite many solutions.

∴∆=1-k-1k-1-111-1=0

⇒11+1+k-k+1-1k+1=0

⇒2-k2+k-k-1=0

⇒k2=1

⇒k=±1

Thus, the values of k are −1 and 1.

If the system of equations x − ky − z = 0, kx − y − z = 0, x + y − z = 0 has a non-zero solution then the values of k are __−1 and 1__.
 

Page No 7.22:

Question 6:

The real value of λ for which the system of equations  λx + y + z = 0,  -x + λy + z = 0, - x - y + λz = 0 has a non-zero solution, is _____________.

Answer:


The system of homogeneous equations λx + y + z = 0, −x + λy + z = 0 and −x − y + λz = 0 has a non-zero solution or an infinite many solutions.

∴∆=λ11-1λ1-1-1λ=0

⇒λλ2+1-1-λ+1+11+λ=0

⇒λ3+λ+λ-1+1+λ=0

⇒λ3+3λ=0

⇒λλ2+3=0

⇒λ=0          (λ2 + 3 ≠ 0 for any real value of λ)

Thus, the real value of λ for which the given system of homogeneous equations has a non-zero solution is 0.

The real value of λ for which the system of equations λx + y + z = 0, −x + λy + z = 0, −x − y + λz = 0 has a non-zero solution, is __0__.

Page No 7.22:

Question 7:

The set of values of k for which the system of equations  x + y + z = 2, 2x + y - z = 3, 3x +2 y + kz = 4 has a unique solution, is _________.

Answer:


The system of equations x + y + z = 2, 2x + y − z = 3 and 3x + 2y + kz = 4 has a unique solution.

∴∆=11121-132k≠0

⇒1k+2-12k+3+14-3≠0

⇒k+2-2k-3+1≠0

⇒k≠0

Thus, the set of values of k for which the given system of equations has a unique solution is R − {0}.

The set of values of k for which the system of equations x + y + z = 2, 2x + y − z = 3, 3x + 2y + kz = 4  has a unique solution, is __R − {0}__.



Page No 7.23:

Question 1:

If 100010001xyz=   1-1   0, find x, y and z.

Answer:

Here,100010001xyz=1-10 ⇒xyz=1-10∴ x=1, y=-1 and z=0

Page No 7.23:

Question 2:

If 1   0   00-1   00   0-1xyz=101, find x, y and z.

Answer:

Here,1000-1000-1xyz=101 ⇒x-y-z=101∴ x=1, y=0 and z=-1

Page No 7.23:

Question 3:

If 1000y0001   x-1   z=101, find x, y and z.

Answer:

Here,1000y0001x-1z=101 ⇒x-yz=101∴ x=1, y=0 and z=1

Page No 7.23:

Question 4:

Solve the following for x and y:
3-49  2xy=10  2

Answer:

Here,3-492xy=102⇒3x-4y9x-2y=102⇒3x-4y=10           ...(1)      9x+2y=2             ... (2)Solving both the equations, we getx=1421  =23Substituting the value of x in eq. (1), we get  3×23-4y=10⇒2-4y=10⇒4y=-8⇒y=-2∴ x=23 and y=-2

Page No 7.23:

Question 5:

If 100001010xyz=   2-1   3, find x, y, z.

Answer:

Here, 100001010xyz=2-13 ⇒xzy=2-13∴ x=2, y=3 and z=-1

Page No 7.23:

Question 6:

If A=2443, X=n1, B= 811 and AX = B, then find n.

Answer:

Here,2443n1=811⇒2n+44n+3=811⇒2n+4=8 ⇒2n=4⇒n=2



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