# RD Sharma Solutions for Class 12 Maths Chapter 8 Solution of Simultaneous Linear Equations

## RD Sharma Solutions Class 12 Maths Chapter 8 – Free PDF Download Updated for (2021-22)

RD Sharma Solutions for Class 12 Maths Chapter 8 – Solution of Simultaneous Linear Equations provides accurate answers to all the questions of the chapter. Experts have designed the solutions in a systematic manner to help students grasp the concepts more effectively. By practising these solutions they are able to get their doubts cleared instantly. RD Sharma Solutions are essential reference books to score high in Mathematics board exams as well as in competitive exams.

RD Sharma Solutions for Class 12 provides answers that are easy to understand and remember. Further helps students to comprehend formulae and solving techniques. This chapter of RD Sharma Solutions for Class 12 mainly focuses on the homogeneous and non-homogeneous system of equations. Students can download the solutions in PDF format for effective 2021-22 exam preparation from the links given below.

Let us have a look at some of the important concepts that are discussed in the RD Sharma Solutions of this chapter.

• 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
• Applications of simultaneous linear equations
• Solution of homogeneous system of linear equations
• The determinant of the coefficient matrix is non-singular
• The determinant of the coefficient matrix is singular

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

### Access answers to Maths RD Sharma Solutions For Class 12 Chapter 8 – Solution of Simultaneous Linear Equations

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

Exercise 8.2 Page No: 8.20

Solve the following systems of homogeneous linear equations by matrix method:

1. 2x â€“ y + z = 0

3x + 2y â€“ z = 0

x + 4y + 3z = 0

Solution:

Given

2x â€“ y + z = 0

3x + 2y â€“ z = 0

X + 4y + 3z = 0

The system can be written as

A X = 0

Now, |A| = 2(6 + 4) + 1(9 + 1) + 1(12 â€“ 2)

|A| = 2(10) + 10 + 10

|A| = 40 â‰  0

Since, |A|â‰  0, hence x = y = z = 0 is the only solution of this homogeneous equation.

2. 2x â€“ y + 2z = 0
5x + 3y â€“ z = 0
X + 5y â€“ 5z = 0

Solution:

Given 2x â€“ y + 2z = 0

5x + 3y â€“ z = 0

X + 5y â€“ 5z = 0

A X = 0

Now, |A| = 2(â€“ 15 + 5) + 1(â€“ 25 + 1) + 2(25 â€“ 3)

|A| = â€“ 20 â€“ 24 + 44

|A| = 0

Hence, the system has infinite solutions

Let z = k

2x â€“ y = â€“ 2k

5x + 3y = k

3. 3x â€“ y + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z = 0

Solution:

Given 3x â€“ y + 2z = 0

4x + 3y + 3z = 0

5x + 7y + 4z = 0

A X = 0

Now, |A| = 3(12 â€“ 21) + 1(16 â€“ 15) + 2(28 â€“ 15)

|A| = â€“ 27 + 1 + 26

|A| = 0

Hence, the system has infinite solutions

Let z = k

3x â€“ y = â€“ 2k

4x + 3y = â€“ 3k

4. x + y â€“ 6z = 0
x â€“ y + 2z = 0
â€“ 3x + y + 2z = 0

Solution:

Given x + y â€“ 6z = 0

x â€“ y + 2z = 0

â€“ 3x + y + 2z = 0

A X = 0

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

|A| = â€“ 4 â€“ 8 + 12

|A| = 0

Hence, the system has infinite solutions

Let z = k

x + y = 6k

x â€“ y = â€“ 2k

### Also, access RD Sharma Solutions for Class 12 Maths Chapter 8 Solution of Simultaneous Linear Equations

Exercise 8.1 Solutions

## Frequently Asked Questions on RD Sharma Solutions for Class 12 Chapter 8

### Is RD Sharma Solutions for Class 12 Maths Chapter 8 available as a PDF at BYJUâ€™S?

Yes, BYJUâ€™S provides the RD Sharma Solutions for Class 12 Maths Chapter 8 as a PDF that can be downloaded by the students at no cost. RD Sharma Solutions are curated by subject experts at BYJUâ€™S according to the latest CBSE syllabus and marking schemes. Referring to these solutions while solving textbook problems help students to ace their board exams.

### How can students effectively use the RD Sharma Solutions Class 12 Maths Chapter 8?

Students can effectively utilise the RD Sharma Solutions Class 12 Maths while solving the exercise questions of any chapter. These solutions provide precise answers in a step-by-step manner following the latest CBSE guidelines. Further, these solutions are very handy during revisions and instant doubt clearance.

### Will the RD Sharma Class 12 Solutions Chapter 8 PDF help students in their exam preparations?

Yes, the RD Sharma Solutions for Class 12 Maths Chapter 8 PDF will definitely help students with their exam preparations. Referring to these solutions will strengthen conceptual knowledge and problem-solving skills. RD Sharma Solutions also lends out valuable tips and tactics for students to score high marks.

### Does RD Sharma Solutions for Class 12 Maths Chapter 8 provide precise answers to all the textbook questions?

Yes, RD Sharma Solutions for Class 12 Maths Chapter 8 provide precise answers to all the textbook questions. Students can make use of solutions PDF designed by the subject experts from the links given here. The solutions are completely based on the current CBSE syllabus 2021-22. The textbook problems are solved in a comprehensive manner as per the marks weightage in the exams.

### Where can I get exercise wise answers of RD Sharma Solutions for Class 12 Maths Chapter 8?

Students of Class 12 can access the exercise wise answers of RD Sharma Solutions Maths Chapter 8 in BYJUâ€™S. Those who aspires to score good marks in academics must follow the right study material to solve the textbook problems efficiently. High professional teachers designed the solutions for both chapter wise and exercise wise to help students in their learning process.