# RD Sharma Solutions for Class 12 Maths Exercise 8.1 Chapter 8 Solution Of Simultaneous Linear Equations

RD Sharma Solutions for Class 12 Maths Exercise 8.1 Chapter 8 Solution of Simultaneous Linear Equations is provided here. For students who wish to excel in their exams and build their confidence level, practising RD Sharma Class 12 textbook is essential. On regular practice, students can speed up the method of solving problems by using shortcut tips to secure high marks in their examination.

Students can download the pdf of RD Sharma Solutions for Class 12 Exercise 8.1 of Chapter 8 Solution of Simultaneous Linear Equations from the links provided below. This exercise mainly focused on the homogeneous and non-homogeneous system of linear equations. Let us have a look at some of the topics discussed in this exercise.

• Definition and meaning of consistent system
• Homogeneous and non-homogeneous systems
• Matrix method for the solution of a non-homogeneous system
• Solving the given system of linear equations when the coefficient matrix is non-singular
• Solving the given system of equations when the coefficient matrix is singular
• Solving a system of linear equations when the inverse of the coefficient matrix is obtained

## RD Sharma Solutions For Class 12 Chapter 8 – Solution of Simultaneous Linear Equations Exercise 8.1:-

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Exercise 8.2 Solutions

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Exercise 8.1 Page No: 8.14

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 â€“ 5 = 0

x â€“ y + 3 = 0

(iv) 3x + y = 19

3x â€“ y = 23

(v) 3x + 7y = 4

x + 2y = -1

(vi) 3x + y = 7

5x + 3y = 12

Solution:

(i) Given 5x + 7y + 2 = 0 and 4x + 6y + 3 = 0

Hence, x =Â 9/2 andÂ y =Â -7/2

(ii) Given 5x + 2y = 3

3x + 2y = 5

Hence, x =Â -1 andÂ y = 4

(iii) Given 3x + 4y â€“ 5 = 0

x â€“ y + 3 = 0

Hence, X = 1 Y = â€“ 2

(iv) Given 3x + y = 19

3x â€“ y = 23

(v) Given 3x + 7y = 4

x + 2y = -1

(vi) Given 3x + y = 7

5x + 3y = 12

2. Solve the following system of equations by matrix method:

(i) x + y â€“z = 3
2x + 3y + z = 10
3x â€“ y â€“ 7z = 1

(ii) x + y + z = 3

2x â€“ y + z = -1

2x + y â€“ 3z = -9

(iii) 6x â€“ 12y + 25z = 4

4x + 15y â€“ 20z = 3

2x + 18y + 15z = 10

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

2x â€“ y + 3z = 4

x + 2y â€“ 3z = 0

(v) (2/x) â€“ (3/y) + (3/z) = 10

(1/x) + (1/y) + (1/z) = 10

(3/x) â€“ (1/y) + (2/z) = 13

(vi) 5x + 3y + z = 16

2x + y + 3z = 19

x + 2y + 4z = 25

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

2y â€“ 3z = 3

x â€“ 2y + 6z = -2

(viii) 2x + y + z = 2

x + 3y – z = 5

3x + y â€“ 2z = 6

(ix) 2x + 6y = 2

3x â€“ z = -8

2x â€“ y + z = -3

(x) 2y â€“ z = 1

x â€“ y + z = 2

2x â€“ y = 0

(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) (2/x) + (3/y) + (10/z) = 4,

(4/x) â€“ (6/y) + (5/z) = 1,

(6/x) + (9/y) â€“ (20/z) = 2, x, y, z â‰  0

(xiv) x â€“ y + 2z = 7

3x + 4y â€“ 5z = -5

2x â€“ y + 3z = 12

Solution:

(i) Given x + y â€“z = 3

2x + 3y + z = 10

3x â€“ y â€“ 7z = 1

= (â€“ 20) â€“ 1(â€“ 17) â€“ 1(11)

= â€“ 20 + 17 + 11 = 8

So, the above system has a unique solution, given by

X = AÂ â€“ 1B

Cofactors of A are

C11Â = (â€“ 1)1 + 1Â â€“ 21 + 1 = â€“ 20

C21Â = (â€“ 1)2 + 1Â â€“ 7 â€“ 1 = 8

C31Â = (â€“ 1)3 + 1Â 1 + 3 = 4

C12Â = (â€“ 1)1 + 2Â â€“ 14 â€“ 3 = 17

C22Â = (â€“ 1)2 + 1Â â€“ 7 + 3 = â€“ 4

C32Â = (â€“ 1)3 + 1Â 1 + 2 = â€“ 3

C13Â = (â€“ 1)1 + 2Â â€“ 2 â€“ 9 = â€“ 11

C23Â = (â€“ 1)2 + 1Â â€“ 1 â€“ 3 = 4

C33Â = (â€“ 1)3 + 1Â 3 â€“ 2 = 1

(ii) Given x + y + z = 3

2x â€“ y + z = -1

2x + y â€“ 3z = -9

= (3 â€“ 1) â€“ 1(â€“ 6 â€“ 2) + 1(2 + 2)

= 2 + 8 + 4

= 14

So, the above system has a unique solution, given by

X = AÂ â€“ 1B

Cofactors of A are

C11Â = (â€“ 1)1 + 1Â 3 â€“ 1 = 2

C21Â = (â€“ 1)2 + 1Â â€“ 3 â€“ 1 = 4

C31Â = (â€“ 1)3 + 1Â 1 + 1 = 2

C12Â = (â€“ 1)1 + 2Â â€“ 6 â€“ 2 = 8

C22Â = (â€“ 1)2 + 1Â â€“ 3 â€“ 2 = â€“ 5

C32Â = (â€“ 1)3 + 1Â 1 â€“ 2 = 1

C13Â = (â€“ 1)1 + 2Â 2 + 2 = 4

C23Â = (â€“ 1)2 + 1Â 1 â€“ 2 = 1

C33Â = (â€“ 1)3 + 1Â â€“ 1 â€“ 2 = â€“ 3

(iii) Given 6x â€“ 12y + 25z = 4

4x + 15y â€“ 20z = 3

2x + 18y + 15z = 10

= 6(225 + 360) + 12(60 + 40) + 25(72 â€“ 30)

= 3510 + 1200 + 1050

= 5760

So, the above system has a unique solution, given by

X = AÂ â€“ 1B

Cofactors of A are

C11Â = (â€“ 1)1 + 1Â (225 + 360) = 585

C21Â = (â€“ 1)2 + 1Â (â€“ 180 â€“ 450) = 630

C31Â = (â€“ 1)3 + 1Â (240 â€“ 375) = â€“ 135

C12Â = (â€“ 1)1 + 2Â (60 + 40) = â€“ 100

C22Â = (â€“ 1)2 + 1Â (90 â€“ 50) = 40

C32Â = (â€“ 1)3 + 1Â (â€“ 120 â€“ 100) = 220

C13Â = (â€“ 1)1 + 2Â (72 â€“ 30) = 42

C23Â = (â€“ 1)2 + 1(108 + 24) = â€“ 132

C33Â = (â€“ 1)3 + 1Â (90 + 48) = 138

(iv) Given 3x + 4y + 7z = 14

2x â€“ y + 3z = 4

x + 2y â€“ 3z = 0

= 3(3 â€“ 6) â€“ 4(â€“ 6 â€“ 3) + 7(4 + 1)

= â€“ 9 + 36 + 35

= 62

So, the above system has a unique solution, given by

X = AÂ â€“ 1B

Cofactors of A are

C11Â = (â€“ 1)1 + 1Â 3 â€“ 6 = â€“ 3

C21Â = (â€“ 1)2 + 1Â â€“ 12 â€“ 14 = 26

C31Â = (â€“ 1)3 + 112 + 7 = 19

C12Â = (â€“ 1)1 + 2Â â€“ 6 â€“ 3 = 9

C22Â = (â€“ 1)2 + 1Â â€“ 3 â€“ 7 = â€“ 10

C32Â = (â€“ 1)3 + 1Â 9 â€“ 14 = 5

C13Â = (â€“ 1)1 + 2Â 4 + 1 = 5

C23Â = (â€“ 1)2 + 1Â 6 â€“ 4 = â€“ 2

C33Â = (â€“ 1)3 + 1Â â€“ 3 â€“ 8 = â€“ 11

(v) Given (2/x) â€“ (3/y) + (3/z) = 10

(1/x) + (1/y) + (1/z) = 10

(3/x) â€“ (1/y) + (2/z) = 13

= 5(4 â€“ 6) â€“ 3(8 â€“ 3) + 1(4 â€“ 2)

= â€“ 10 â€“ 15 + 3

= â€“ 22

So, the above system has a unique solution, given by

X = AÂ â€“ 1B

Cofactors of A are

C11Â = (â€“ 1)1 + 1Â (4 â€“ 6) = â€“ 2

C21Â = (â€“ 1)2 + 1(12 â€“ 2) = â€“ 10

C31Â = (â€“ 1)3 + 1(9 â€“ 1) = 8

C12Â = (â€“ 1)1 + 2Â (8 â€“ 3) = â€“ 5

C22Â = (â€“ 1)2 + 1Â 20 â€“ 1 = 19

C32Â = (â€“ 1)3 + 1Â 15 â€“ 2 = â€“ 13

C13Â = (â€“ 1)1 + 2Â (4 â€“ 2) = 2

C23Â = (â€“ 1)2 + 1Â 10 â€“ 3 = â€“ 7

C33Â = (â€“ 1)3 + 1Â 5 â€“ 6 = â€“ 1

(vi) Given 5x + 3y + z = 16

2x + y + 3z = 19

x + 2y + 4z = 25

= 3(12 â€“ 6) â€“ 4(0 + 3) + 2(0 â€“ 2)

= 18 â€“ 12 â€“ 4

= 2

So, the above system has a unique solution, given by

X = AÂ â€“ 1B

Cofactors of A are

C11Â = (â€“ 1)1 + 1Â (12 â€“ 6) = 6

C21Â = (â€“ 1)2 + 1(24 + 4) = â€“ 28

C31Â = (â€“ 1)3 + 1(â€“ 12 â€“ 4) = â€“ 16

C12Â = (â€“ 1)1 + 2Â (0 + 3) = â€“ 3

C22Â = (â€“ 1)2 + 1Â 18 â€“ 2 = 16

C32Â = (â€“ 1)3 + 1Â â€“ 9 â€“ 0 = 9

C13Â = (â€“ 1)1 + 2Â (0 â€“ 2) = â€“ 2

C23Â = (â€“ 1)2 + 1Â (â€“ 6 â€“ 4) = 10

C33Â = (â€“ 1)3 + 1Â 6 â€“ 0 = 6

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

2y â€“ 3z = 3

x â€“ 2y + 6z = -2

= 2(â€“ 6 + 1) â€“ 1(â€“ 2 + 3) + 1(1 â€“ 9)

= â€“ 10 â€“ 1 â€“ 8

= â€“ 19

So, the above system has a unique solution, given by

X = AÂ â€“ 1B

Cofactors of A are

C11Â = (â€“ 1)1 + 1Â â€“ 6 + 1 = â€“ 5

C21Â = (â€“ 1)2 + 1(24 + 4) = â€“ 28

C31Â = (â€“ 1)3 + 1Â â€“ 1 â€“ 3 = â€“ 4

C12Â = (â€“ 1)1 + 2Â â€“ 2 + 3 = â€“ 1

C22Â = (â€“ 1)2 + 1Â â€“ 4 â€“ 3 = â€“ 7

C32Â = (â€“ 1)3 + 1Â â€“ 2 â€“ 1 = 3

C13Â = (â€“ 1)1 + 21 â€“ 9 = â€“ 8

C23Â = (â€“ 1)2 + 12 â€“ 3 = â€“ 1

C33Â = (â€“ 1)3 + 1Â 6 â€“ 1 = 5

(viii) Given 2x + y + z = 2

x + 3y – z = 5

3x + y â€“ 2z = 6

= 2(â€“ 6 + 1) â€“ 1(â€“ 2 + 3) + 1(1 â€“ 9)

= â€“ 10 â€“ 1 â€“ 8

= â€“ 19

So, the above system has a unique solution, given by

X = AÂ â€“ 1B

Cofactors of A are

C11Â = (â€“ 1)1 + 1Â â€“ 6 + 1 = â€“ 5

C21Â = (â€“ 1)2 + 1(24 + 4) = â€“ 28

C31Â = (â€“ 1)3 + 1Â â€“ 1 â€“ 3 = â€“ 4

C12Â = (â€“ 1)1 + 2Â â€“ 2 + 3 = â€“ 1

C22Â = (â€“ 1)2 + 1Â â€“ 4 â€“ 3 = â€“ 7

C32Â = (â€“ 1)3 + 1Â â€“ 2 â€“ 1 = 3

C13Â = (â€“ 1)1 + 21 â€“ 9 = â€“ 8

C23Â = (â€“ 1)2 + 12 â€“ 3 = â€“ 1

C33Â = (â€“ 1)3 + 1Â 6 â€“ 1 = 5

(ix) Given 2x + 6y = 2

3x â€“ z = -8

2x â€“ y + z = -3

= 2(0 â€“ 1) â€“ 6(3 + 2)

= â€“ 2 â€“ 30

= â€“ 32

So, the above system has a unique solution, given by

X = AÂ â€“ 1B

Cofactors of A are

C11Â = (â€“ 1)1 + 1Â 0 â€“ 1 = â€“ 1

C21Â = (â€“ 1)2 + 16 + 0 = â€“ 6

C31Â = (â€“ 1)3 + 1Â â€“ 6 = â€“ 6

C12Â = (â€“ 1)1 + 2Â 3 + 2 = 5

C22Â = (â€“ 1)2 + 1Â 2 â€“ 0 = 2

C32Â = (â€“ 1)3 + 1Â â€“ 2 â€“ 0 = 2

C13Â = (â€“ 1)1 + 2Â â€“ 3 â€“ 0 = â€“ 3

C23Â = (â€“ 1)2 + 1Â â€“ 2 â€“ 12 = 14

C33Â = (â€“ 1)3 + 1Â 0 â€“ 18 = â€“ 18

(x) Given 2y â€“ z = 1

x â€“ y + z = 2

2x â€“ y = 0

= 0 + 4 â€“ 1

= 3

So, the above system has a unique solution, given by

X = AÂ â€“ 1B

Cofactors of A are

C11Â = (â€“ 1)1 + 1Â 1 â€“ 0 = 1

C21Â = (â€“ 1)2 + 11 â€“ 2 = 1

C31Â = (â€“ 1)3 + 10 + 1 = 1

C12Â = (â€“ 1)1 + 2Â â€“ 2 â€“ 0 = 2

C22Â = (â€“ 1)2 + 1Â â€“ 1 â€“ 0 = â€“ 1

C32Â = (â€“ 1)3 + 1Â 0 â€“ 2 = 2

C13Â = (â€“ 1)1 + 2Â 4 â€“ 0 = 4

C23Â = (â€“ 1)2 + 1Â 2 â€“ 0 = â€“ 2

C33Â = (â€“ 1)3 + 1Â â€“ 1 + 2 = 1

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

2x + y + z = 5

x + 2y + z = 5

= 8(â€“ 1) â€“ 4(1) + 3(3)

= â€“ 8 â€“ 4 + 9

= â€“ 3

So, the above system has a unique solution, given by

X = AÂ â€“ 1B

Cofactors of A are

C11Â = (â€“ 1)1 + 1Â 1 â€“ 2 = â€“ 1

C21Â = (â€“ 1)2 + 1Â 4 â€“ 6 = 2

C31Â = (â€“ 1)3 + 1Â 4 â€“ 3 = 1

C12Â = (â€“ 1)1 + 2Â 2 â€“ 1 = â€“ 1

C22Â = (â€“ 1)2 + 1Â 8 â€“ 3 = 5

C32Â = (â€“ 1)3 + 1Â 8 â€“ 6 = â€“ 2

C13Â = (â€“ 1)1 + 2Â 4 â€“ 1 = 3

C23Â = (â€“ 1)2 + 1Â 16 â€“ 4 = â€“ 12

C33Â = (â€“ 1)3 + 1Â 8 â€“ 8 = 0

(xii) Given x + y + z = 6

x + 2z = 7

3x + y + z = 12

= 1(â€“ 2) â€“ 1(1 â€“ 6) + 1(1)

= â€“ 2 + 5 + 1

= 4

So, the above system has a unique solution, given by

X = AÂ â€“ 1B

Cofactors of A are

C11Â = (â€“ 1)1 + 1Â 0 â€“ 2 = â€“ 2

C21Â = (â€“ 1)2 + 1Â 1 â€“ 1 = 0

C31Â = (â€“ 1)3 + 1Â 2 â€“ 0 = 2

C12Â = (â€“ 1)1 + 2Â 1 â€“ 6 = 5

C22Â = (â€“ 1)2 + 1Â 1 â€“ 3 = â€“ 2

C32Â = (â€“ 1)3 + 1Â 2 â€“ 1 = â€“ 1

C13Â = (â€“ 1)1 + 2Â 1 â€“ 0 = 1

C23Â = (â€“ 1)2 + 1Â 1 â€“ 3 = 2

C33Â = (â€“ 1)3 + 1Â 0 â€“ 1 = â€“ 1

(xiii) Given (2/x) + (3/y) + (10/z) = 4,

(4/x) â€“ (6/y) + (5/z) = 1,

(6/x) + (9/y) â€“ (20/z) = 2, x, y, z â‰  0

AX = B

Now,

|A| = 2(75) â€“ 3(â€“ 110) + 10(72)

= 150 + 330 + 720

= 1200

So, the above system has a unique solution, given by

X = AÂ â€“ 1B

Cofactors of A are

C11Â = (â€“ 1)1 + 1Â 120 â€“ 45 = 75

C21Â = (â€“ 1)2 + 1Â â€“ 60 â€“ 90 = 150

C31Â = (â€“ 1)3 + 1Â 15 + 60 = 75

C12Â = (â€“ 1)1 + 2Â â€“ 80 â€“ 30 = 110

C22Â = (â€“ 1)2 + 1Â â€“ 40 â€“ 60 = â€“ 100

C32Â = (â€“ 1)3 + 1Â 10 â€“ 40 = 30

C13Â = (â€“ 1)1 + 2Â 36 + 36 = 72

C23Â = (â€“ 1)2 + 1Â 18 â€“ 18 = 0

C33Â = (â€“ 1)3 + 1Â â€“ 12 â€“ 12 = â€“ 24

(xiv) Given x â€“ y + 2z = 7

3x + 4y â€“ 5z = -5

2x â€“ y + 3z = 12

A X = B

Now,

|A| = 1(12 â€“ 5) + 1(9 + 10) + 2(â€“ 3 â€“ 8)

= 7 + 19 â€“ 22

= 4

So, the above system has a unique solution, given by

X = AÂ â€“ 1B

Cofactors of A are

C11Â = (â€“ 1)1 + 1Â 12 â€“ 5 = 7

C21Â = (â€“ 1)2 + 1Â â€“ 3 + 2 = 1

C31Â = (â€“ 1)3 + 1Â 5 â€“ 8 = â€“ 3

C12Â = (â€“ 1)1 + 2Â 9 + 10 = â€“ 19

C22Â = (â€“ 1)2 + 1Â 3 â€“ 4 = â€“ 1

C32Â = (â€“ 1)3 + 1Â â€“ 5 â€“ 6 = 11

C13Â = (â€“ 1)1 + 2Â â€“ 3 â€“ 8 = â€“ 11

C23Â = (â€“ 1)2 + 1Â â€“ 1 + 2 = â€“ 1

C33Â = (â€“ 1)3 + 1Â 4 + 3 = 7

3. Show that each one 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

(v) x + y + z = 6

x + 2y + 3z = 14

x + 4y + 7z = 30

(vi) 2x + 2y â€“ 2z = 1

4x + 4y â€“ z = 2

6x + 6y + 2z = 3

Solution:

(i) Given 6x + 4y = 2

9x + 6y = 3

|A| = 36 â€“ 36 = 0

So, A is singular, Now X will be consistence if (Adj A) x B = 0

C11Â = (â€“ 1)1 + 1Â 6 = 6

C12Â = (â€“ 1)1 + 2Â 9 = â€“ 9

C21Â = (â€“ 1)2 + 1Â 4 = â€“ 4

C22Â = (â€“ 1)2 + 2Â 6 = 6

(ii) Given 2x + 3y = 5

6x + 9y = 15

|A| = 18 â€“ 18 = 0

So, A is singular,

Now X will be consistence if (Adj A) x B = 0

C11Â = (â€“ 1)1 + 1Â 9 = 9

C12Â = (â€“ 1)1 + 2Â 6 = â€“ 6

C21Â = (â€“ 1)2 + 1Â 3 = â€“ 3

C22Â = (â€“ 1)2 + 2Â 2 = 2

(iii) Given 5x + 3y + 7z = 4

3x + 26y + 2z = 9

7x + 2y + 10z = 5

|A| = 5(260 â€“ 4) â€“ 3(30 â€“ 14) + 7(6 â€“ 182)

= 5(256) â€“ 3(16) + 7(176)

|A| = 0

So, A is singular. Thus, the given system is either inconsistent or it is consistent with infinitely many solution according to as:

(Adj A) x Bâ‰ 0 or (Adj A) x B = 0

Cofactors of A are

C11Â = (â€“ 1)1 + 1Â 260 â€“ 4 = 256

C21Â = (â€“ 1)2 + 1Â 30 â€“ 14 = â€“ 16

C31Â = (â€“ 1)3 + 1Â 6 â€“ 182 = â€“ 176

C12Â = (â€“ 1)1 + 2Â 30 â€“ 14 = â€“ 16

C22Â = (â€“ 1)2 + 1Â 50 â€“ 49 = 1

C32Â = (â€“ 1)3 + 1Â 10 â€“ 21 = 11

C13Â = (â€“ 1)1 + 2Â 6 â€“ 182 = â€“ 176

C23Â = (â€“ 1)2 + 1Â 10 â€“ 21 = 11

C33Â = (â€“ 1)3 + 1Â 130 â€“ 9 = 121

Now, AX = B has infinite many solution

Let z = k

Then, 5x + 3y = 4 â€“ 7k

3x + 26y = 9 â€“ 2k

ThisÂ can be written as

(v) Given x + y + z = 6

x + 2y + 3z = 14

x + 4y + 7z = 30

|A| = 1(2) â€“ 1(4) + 1(2)

= 2 â€“ 4 + 2

|A| = 0

So, A is singular. Thus, the given system is either inconsistent or it is consistent with infinitely many solution according to as:

(Adj A) x Bâ‰ 0 or (Adj A) x B = 0

Cofactors of A are

C11Â = (â€“ 1)1 + 1Â 14 â€“ 12 = 2

C21Â = (â€“ 1)2 + 1Â 7 â€“ 4 = â€“ 3

C31Â = (â€“ 1)3 + 1Â 3 â€“ 2 = 1

C12Â = (â€“ 1)1 + 2Â 7 â€“ 3 = â€“ 4

C22Â = (â€“ 1)2 + 1Â 7 â€“ 1 = 6

C32Â = (â€“ 1)3 + 1Â 3 â€“ 1 = 2

C13Â = (â€“ 1)1 + 2Â 4 â€“ 2 = 2

C23Â = (â€“ 1)2 + 1Â 4 â€“ 1 = â€“ 3

C33Â = (â€“ 1)3 + 1Â 2 â€“ 1 = 1

Now, AX = B has infinite many solution

Let z = k

Then, x + y = 6 â€“ k

x + 2y = 14 â€“ 3k

This can be written as:

(vi) Given x + y + z = 6

x + 2y + 3z = 14

x + 4y + 7z = 30

This can be written as

|A| = 2(14) â€“ 2(14) â€“ 2(0)

|A| = 0

So, A is singular. Thus, the given system is either inconsistent or it is consistent

with infinitely many solution according to as:

(Adj A) x Bâ‰ 0 or (Adj A) x B = 0

Cofactors of A are:

C11Â = (â€“ 1)1 + 18 + 6 = 14

C21Â = (â€“ 1)2 + 1Â 4 + 12 = â€“ 16

C31Â = (â€“ 1)3 + 1Â â€“ 2 + 8 = 6

C12Â = (â€“ 1)1 + 2Â 8 + 6 = â€“ 14

C22Â = (â€“ 1)2 + 1Â 4 + 12 = 16

C32Â = (â€“ 1)3 + 1Â â€“ 2 + 8 = â€“ 6

C13Â = (â€“ 1)1 + 2Â 24 â€“ 24 = 0

C23Â = (â€“ 1)2 + 1Â 12 â€“ 12 = 0

C33Â = (â€“ 1)3 + 1Â 8 â€“ 8 = 0

4. Show that each one of the following systems of linear equations is consistent:

(i) 2x + 5y = 7

6x + 15y = 13

(ii) 2x + 3y = 5

6x + 9y = 10

(iii) 4x â€“ 2y = 3

6x â€“ 3y = 5

(iv) 4x â€“ 5y â€“ 2z = 2

5x â€“ 4y + 2z = -2

2x + 2y + 8z = -1

(v) 3x â€“ y â€“ 2z = 2

2y â€“ z = -1

3x â€“ 5y = 3

(vi) x + y â€“ 2z = 5

x â€“ 2y + z = -2

-2x + y + z = 4

Solution:

(i) Given 2x + 5y = 7

6x + 15y = 13

|A| = 30 â€“ 30 = 0

So, A is singular,

Now X will be consistence if (Adj A) x B = 0

C11Â = (â€“ 1)1 + 1Â 15 = 15

C12Â = (â€“ 1)1 + 2Â 6 = â€“ 6

C21Â = (â€“ 1)2 + 1Â 5 = â€“ 5

C22Â = (â€“ 1)2 + 2Â 2 = 2

(ii) Given 2x + 3y = 5

6x + 9y = 10

|A| = 18 â€“ 18 = 0

So, A is singular,

Now X will be consistence if (Adj A) x B = 0

C11Â = (â€“ 1)1 + 1Â 9 = 9

C12Â = (â€“ 1)1 + 2Â 6 = â€“ 6

C21Â = (â€“ 1)2 + 1Â 3 = â€“ 3

C22Â = (â€“ 1)2 + 2Â 2 = 2

(iii) Given 4x â€“ 2y = 3

6x â€“ 3y = 5

|A| = â€“ 12 + 12 = 0

So, A is singular,

Now X will be consistence if (Adj A) x B = 0

C11Â = (â€“ 1)1 + 1Â â€“ 3 = â€“ 3

C12Â = (â€“ 1)1 + 2Â 6 = â€“ 6

C21Â = (â€“ 1)2 + 1Â â€“ 2 = 2

C22Â = (â€“ 1)2 + 2Â 4 = 4

(iv) Given 4x â€“ 5y â€“ 2z = 2

5x â€“ 4y + 2z = -2

2x + 2y + 8z = -1

|A| = 4(â€“ 36) + 5(36) â€“ 2(18)

|A| = 0

Cofactors of A are:

C11Â = (â€“ 1)1 + 1Â â€“ 32 â€“ 4 = â€“ 36

C21Â = (â€“ 1)2 + 1Â â€“ 40 + 4 = â€“ 36

C31Â = (â€“ 1)3 + 1Â â€“ 10 â€“ 8 = â€“ 18

C12Â = (â€“ 1)1 + 2Â 40 â€“ 4 = â€“ 36

C22Â = (â€“ 1)2 + 1Â 32 + 4 = 36

C32Â = (â€“ 1)3 + 1Â 8 + 10 = â€“ 18

C13Â = (â€“ 1)1 + 2Â 10 + 8 = 18

C23Â = (â€“ 1)2 + 1Â 8 + 10 = â€“ 18

C33Â = (â€“ 1)3 + 1Â â€“ 16 + 25 = 9

(v) Given 3x â€“ y â€“ 2z = 2

2y â€“ z = -1

3x â€“ 5y = 3

|A| = 3(â€“ 5) + 1(3) â€“ 2(â€“ 6)

|A| = 0

Cofactors of A are

C11Â = (â€“ 1)1 + 1Â 0 â€“ 5 = â€“ 5

C21Â = (â€“ 1)2 + 1Â 0 â€“ 10 = 10

C31Â = (â€“ 1)3 + 1Â 1 + 4 = 5

C12Â = (â€“ 1)1 + 2Â 0 + 3 = â€“ 3

C22Â = (â€“ 1)2 + 1Â 0 + 6 = 6

C32Â = (â€“ 1)3 + 1Â â€“ 3 â€“ 0 = 3

C13Â = (â€“ 1)1 + 2Â 0 â€“ 6 = â€“ 6

C23Â = (â€“ 1)2 + 1Â â€“ 15 + 3 = 12

C33Â = (â€“ 1)3 + 1Â 6 â€“ 0 = 6

(vi) Given x + y â€“ 2z = 5

x â€“ 2y + z = -2

-2x + y + z = 4

|A| = 1(â€“ 3) â€“ 1(3) â€“ 2(â€“ 3) = â€“ 3 â€“ 3 + 6

|A| = 0

Cofactors of A are:

C11Â = (â€“ 1)1 + 1Â â€“ 2 â€“ 1 = â€“ 3

C21Â = (â€“ 1)2 + 1Â 1 + 2 = â€“ 3

C31Â = (â€“ 1)3 + 1Â 1 â€“ 4 = â€“ 3

C12Â = (â€“ 1)1 + 2Â 1 + 2 = â€“ 3

C22Â = (â€“ 1)2 + 1Â 1 â€“ 4 = â€“ 3

C32Â = (â€“ 1)3 + 1Â 1 + 2 = â€“ 3

C13Â = (â€“ 1)1 + 2Â 1 â€“ 4 = â€“ 3

C23Â = (â€“ 1)2 + 1Â 1 + 2 = â€“ 3

C33Â = (â€“ 1)3 + 1Â â€“ 2 â€“ 1 = â€“ 3

x â€“ y = 3, 2x + 3y + 4z = 17, y + 2z = 7

Solution:

2x â€“ 3y + 5z = 11, 3x + 2y â€“ 4z = -5, x + y â€“ 2z = -3.

Solution:

|A| = 2(0) + 3(â€“ 2) + 5(1)

= â€“ 1

Now, the cofactors of A

C11Â = (â€“ 1)1 + 1Â â€“ 4 + 4 = 0

C21Â = (â€“ 1)2 + 1Â 6 â€“ 5 = â€“ 1

C31Â = (â€“ 1)3 + 1Â 12 â€“ 10 = 2

C12Â = (â€“ 1)1 + 2Â â€“ 6 + 4 = 2

C22Â = (â€“ 1)2 + 1Â â€“ 4 â€“ 5 = â€“ 9

C32Â = (â€“ 1)3 + 1Â â€“ 8 â€“ 15 = 23

C13Â = (â€“ 1)1 + 2Â 3 â€“ 2 = 1

C23Â = (â€“ 1)2 + 1Â 2 + 3 = â€“ 5

C33Â = (â€“ 1)3 + 1Â 4 + 9 = 13

x + 2y + 5z = 10, x â€“ y â€“ z = -2, 2x + 3y â€“ z = -11.

Solution:

Given

|A| = 1(1 + 3) + 2(â€“ 1 + 2) + 5(3 + 2)

= 4 + 2 + 25

= 27

Now, the cofactors of A

C11Â = (â€“ 1)1 + 1Â 1 + 3 = 4

C21Â = (â€“ 1)2 + 1Â â€“ 2 â€“ 15 = 17

C31Â = (â€“ 1)3 + 1Â â€“ 2 + 5 = 3

C12Â = (â€“ 1)1 + 2Â â€“ 1 + 2 = â€“ 1

C22Â = (â€“ 1)2 + 1Â â€“ 1 â€“ 10 = â€“ 11

C32Â = (â€“ 1)3 + 1Â â€“ 1 â€“ 5 = 6

C13Â = (â€“ 1)1 + 2Â 3 + 2 = 5

C23Â = (â€“ 1)2 + 1Â 3 â€“ 4 = 1

C33Â = (â€“ 1)3 + 1Â â€“ 1 â€“ 2 = â€“ 3

Solution:

Given

|A| = 1(1 + 6) + 2(2 â€“ 0) + 0

= 7 + 4

= 11

Now, the cofactors of A

C11Â = (â€“ 1)1 + 1Â 1 + 6 = 7

C21Â = (â€“ 1)2 + 1Â â€“ 2 â€“ 0 = 2

C31Â = (â€“ 1)3 + 1Â â€“ 6 â€“ 0 = â€“ 6

C12Â = (â€“ 1)1 + 2Â 2 â€“ 0 = â€“ 2

C22Â = (â€“ 1)2 + 1Â 1 â€“ 0 = 1

C32Â = (â€“ 1)3 + 1Â 3 â€“ 0 = â€“ 3

C13Â = (â€“ 1)1 + 2Â â€“ 4 â€“ 0 = â€“ 4

C23Â = (â€“ 1)2 + 1Â â€“ 2 â€“ 0 = 2

C33Â = (â€“ 1)3 + 1Â 1 + 4 = 5

Solution:

Given

|A| = 3(3 â€“ 0) + 4(2 â€“ 5) + 2(0 â€“ 3)

= 9 â€“ 12 â€“ 6

= â€“ 9

Now, the cofactors of A

C11Â = (â€“ 1)1 + 1Â 3 â€“ 0 = 3

C21Â = (â€“ 1)2 + 1Â â€“ 4 â€“ 0 = 4

C31Â = (â€“ 1)3 + 1Â â€“ 20 â€“ 6 = â€“ 26

C12Â = (â€“ 1)1 + 2Â 2 â€“ 5 = 3

C22Â = (â€“ 1)2 + 1Â 3 â€“ 2 = 1

C32Â = (â€“ 1)3 + 1Â 15 â€“ 4 = â€“ 11

C13Â = (â€“ 1)1 + 2Â 0 â€“ 3 = â€“ 3

C23Â = (â€“ 1)2 + 1Â 0 + 4 = â€“ 4

C33Â = (â€“ 1)3 + 1Â 9 + 8 = 17

Solution:

Solution:

Given

|A| = 1(â€“ 1 â€“ 1) â€“ 2(â€“ 2 â€“ 0) + 0

= â€“ 2 + 4

= 2

Now, the cofactors of A

C11Â = (â€“ 1)1 + 1Â â€“ 1 â€“ 1 = â€“ 2

C21Â = (â€“ 1)2 + 1Â 2 â€“ 0 = 2

C31Â = (â€“ 1)3 + 1Â â€“ 2 â€“ 0 = â€“ 2

C12Â = (â€“ 1)1 + 2Â 2 â€“ 0 = â€“ 2

C22Â = (â€“ 1)2 + 1Â 1 â€“ 0 = 1

C32Â = (â€“ 1)3 + 1Â â€“ 1 â€“ 0 = 1

C13Â = (â€“ 1)1 + 2Â â€“ 2 â€“ 0 = â€“ 2

C23Â = (â€“ 1)2 + 1Â â€“ 1 â€“ 0 = 1

C33Â = (â€“ 1)3 + 1Â â€“ 1 + 4 = 3

Solution:

Given

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.

Solution:

Let the numbers are x, y, z

x + y + z = 2
â€¦â€¦ (i)

Also, 2y + (x + z) + 1

x + 2y + z = 1 â€¦â€¦ (ii)

Again,

x + z + 5(x) = 6

5x + y + z = 6 â€¦â€¦ (iii)

A X = B

|A| = 1(1) â€“ 1(â€“ 4) + 1(â€“ 9)

= 1 + 4 â€“ 9

= â€“ 4

Hence, the unique solution given by x = AÂ â€“ 1B

C11 =Â (â€“ 1)1 + 1Â (2 â€“ 1) = 1

C12Â = (â€“ 1)1 + 2Â (1 â€“ 5) = 4

C13Â = (â€“ 1)1 + 3Â (1 â€“ 10) = â€“ 9

C21Â = (â€“ 1)2 + 1Â (1 â€“ 1) = 0

C22Â = (â€“ 1)2 + 2Â (1 â€“ 5) = â€“ 4

C23Â = (â€“ 1)2 + 3Â (1 â€“ 5) = 4

C31Â = (â€“ 1)3 + 1Â (1 â€“ 2) = â€“ 1

C32Â = (â€“ 1)3 + 2Â (1 â€“ 1) = 0

C33Â = (â€“ 1)3 + 3Â (2 â€“ 1) = 1

10. An amount of â‚¹10,000 is put into three investments at the rate of 10, 12 and 15% per annum. The combined incomes are â‚¹1310 and the combined income of first and second investment is â‚¹ 190 short of the income from the third. Find the investment in each using matrix method.

Solution:

Let the numbers are x, y, and z

x + y + z = 10,000 â€¦â€¦ (i)

Also,

0.1x + 0.12y + 0.15z = 1310 â€¦â€¦ (ii)

Again,

0.1x + 0.12y â€“ 0.15z = â€“ 190 â€¦â€¦ (iii)

A X = B

|A| = 1(â€“ 0.036) â€“ 1(â€“ 0.03) + 1(0)

= â€“ 0.006

Hence, the unique solution given by x = AÂ â€“ 1B

C11 =Â â€“ 0.036

C12Â = 0.27

C13Â = 0

C21Â = 0.27

C22Â = â€“ 0.25

C23Â = â€“ 0.02

C31Â = 0.03

C32Â = â€“ 0.05

C33Â = 0.02