# 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 2023-24

RD Sharma Solutions for Class 12 Maths Chapter 8 – Solution of Simultaneous Linear Equations provide 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 provide answers that are easy to understand and remember. Further, they help students to comprehend formulae and solving techniques. This chapter of RD Sharma Solutions for Class 12 mainly focuses on the homogeneous and non-homogeneous systems of equations. Students can download the solutions in PDF format for effective exam preparation for 2023-24 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 a homogeneous system of linear equations
• The determinant of the coefficient matrix is non-singular
• The determinant of the coefficient matrix is singular

### 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 – 5 = 0

x – y + 3 = 0

(iv) 3x + y = 19

3x – 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 = 9/2 and y = -7/2

(ii) Given 5x + 2y = 3

3x + 2y = 5

Hence, x = -1 and y = 4

(iii) Given 3x + 4y – 5 = 0

x – y + 3 = 0

Hence, X = 1 Y = – 2

(iv) Given 3x + y = 19

3x – 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 –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) (2/x) – (3/y) + (3/z) = 10

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

(3/x) – (1/y) + (2/z) = 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) 2y – z = 1

x – y + z = 2

2x – 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) – (6/y) + (5/z) = 1,

(6/x) + (9/y) – (20/z) = 2, x, y, z ≠ 0

(xiv) x – y + 2z = 7

3x + 4y – 5z = -5

2x – y + 3z = 12

Solution:

(i) Given x + y –z = 3

2x + 3y + z = 10

3x – y – 7z = 1

= (– 20) – 1(– 17) – 1(11)

= – 20 + 17 + 11 = 8

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

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 – 21 + 1 = – 20

C21 = (– 1)2 + 1 – 7 – 1 = 8

C31 = (– 1)3 + 1 1 + 3 = 4

C12 = (– 1)1 + 2 – 14 – 3 = 17

C22 = (– 1)2 + 1 – 7 + 3 = – 4

C32 = (– 1)3 + 1 1 + 2 = – 3

C13 = (– 1)1 + 2 – 2 – 9 = – 11

C23 = (– 1)2 + 1 – 1 – 3 = 4

C33 = (– 1)3 + 1 3 – 2 = 1

(ii) Given x + y + z = 3

2x – y + z = -1

2x + y – 3z = -9

= (3 – 1) – 1(– 6 – 2) + 1(2 + 2)

= 2 + 8 + 4

= 14

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

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 3 – 1 = 2

C21 = (– 1)2 + 1 – 3 – 1 = 4

C31 = (– 1)3 + 1 1 + 1 = 2

C12 = (– 1)1 + 2 – 6 – 2 = 8

C22 = (– 1)2 + 1 – 3 – 2 = – 5

C32 = (– 1)3 + 1 1 – 2 = 1

C13 = (– 1)1 + 2 2 + 2 = 4

C23 = (– 1)2 + 1 1 – 2 = 1

C33 = (– 1)3 + 1 – 1 – 2 = – 3

(iii) Given 6x – 12y + 25z = 4

4x + 15y – 20z = 3

2x + 18y + 15z = 10

= 6(225 + 360) + 12(60 + 40) + 25(72 – 30)

= 3510 + 1200 + 1050

= 5760

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

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 (225 + 360) = 585

C21 = (– 1)2 + 1 (– 180 – 450) = 630

C31 = (– 1)3 + 1 (240 – 375) = – 135

C12 = (– 1)1 + 2 (60 + 40) = – 100

C22 = (– 1)2 + 1 (90 – 50) = 40

C32 = (– 1)3 + 1 (– 120 – 100) = 220

C13 = (– 1)1 + 2 (72 – 30) = 42

C23 = (– 1)2 + 1(108 + 24) = – 132

C33 = (– 1)3 + 1 (90 + 48) = 138

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

2x – y + 3z = 4

x + 2y – 3z = 0

= 3(3 – 6) – 4(– 6 – 3) + 7(4 + 1)

= – 9 + 36 + 35

= 62

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

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 3 – 6 = – 3

C21 = (– 1)2 + 1 – 12 – 14 = 26

C31 = (– 1)3 + 112 + 7 = 19

C12 = (– 1)1 + 2 – 6 – 3 = 9

C22 = (– 1)2 + 1 – 3 – 7 = – 10

C32 = (– 1)3 + 1 9 – 14 = 5

C13 = (– 1)1 + 2 4 + 1 = 5

C23 = (– 1)2 + 1 6 – 4 = – 2

C33 = (– 1)3 + 1 – 3 – 8 = – 11

(v) Given (2/x) – (3/y) + (3/z) = 10

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

(3/x) – (1/y) + (2/z) = 13

= 5(4 – 6) – 3(8 – 3) + 1(4 – 2)

= – 10 – 15 + 3

= – 22

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

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 (4 – 6) = – 2

C21 = (– 1)2 + 1(12 – 2) = – 10

C31 = (– 1)3 + 1(9 – 1) = 8

C12 = (– 1)1 + 2 (8 – 3) = – 5

C22 = (– 1)2 + 1 20 – 1 = 19

C32 = (– 1)3 + 1 15 – 2 = – 13

C13 = (– 1)1 + 2 (4 – 2) = 2

C23 = (– 1)2 + 1 10 – 3 = – 7

C33 = (– 1)3 + 1 5 – 6 = – 1

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

2x + y + 3z = 19

x + 2y + 4z = 25

= 3(12 – 6) – 4(0 + 3) + 2(0 – 2)

= 18 – 12 – 4

= 2

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

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 (12 – 6) = 6

C21 = (– 1)2 + 1(24 + 4) = – 28

C31 = (– 1)3 + 1(– 12 – 4) = – 16

C12 = (– 1)1 + 2 (0 + 3) = – 3

C22 = (– 1)2 + 1 18 – 2 = 16

C32 = (– 1)3 + 1 – 9 – 0 = 9

C13 = (– 1)1 + 2 (0 – 2) = – 2

C23 = (– 1)2 + 1 (– 6 – 4) = 10

C33 = (– 1)3 + 1 6 – 0 = 6

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

2y – 3z = 3

x – 2y + 6z = -2

= 2(– 6 + 1) – 1(– 2 + 3) + 1(1 – 9)

= – 10 – 1 – 8

= – 19

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

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 – 6 + 1 = – 5

C21 = (– 1)2 + 1(24 + 4) = – 28

C31 = (– 1)3 + 1 – 1 – 3 = – 4

C12 = (– 1)1 + 2 – 2 + 3 = – 1

C22 = (– 1)2 + 1 – 4 – 3 = – 7

C32 = (– 1)3 + 1 – 2 – 1 = 3

C13 = (– 1)1 + 21 – 9 = – 8

C23 = (– 1)2 + 12 – 3 = – 1

C33 = (– 1)3 + 1 6 – 1 = 5

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

x + 3y – z = 5

3x + y – 2z = 6

= 2(– 6 + 1) – 1(– 2 + 3) + 1(1 – 9)

= – 10 – 1 – 8

= – 19

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

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 – 6 + 1 = – 5

C21 = (– 1)2 + 1(24 + 4) = – 28

C31 = (– 1)3 + 1 – 1 – 3 = – 4

C12 = (– 1)1 + 2 – 2 + 3 = – 1

C22 = (– 1)2 + 1 – 4 – 3 = – 7

C32 = (– 1)3 + 1 – 2 – 1 = 3

C13 = (– 1)1 + 21 – 9 = – 8

C23 = (– 1)2 + 12 – 3 = – 1

C33 = (– 1)3 + 1 6 – 1 = 5

(ix) Given 2x + 6y = 2

3x – z = -8

2x – y + z = -3

= 2(0 – 1) – 6(3 + 2)

= – 2 – 30

= – 32

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

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 0 – 1 = – 1

C21 = (– 1)2 + 16 + 0 = – 6

C31 = (– 1)3 + 1 – 6 = – 6

C12 = (– 1)1 + 2 3 + 2 = 5

C22 = (– 1)2 + 1 2 – 0 = 2

C32 = (– 1)3 + 1 – 2 – 0 = 2

C13 = (– 1)1 + 2 – 3 – 0 = – 3

C23 = (– 1)2 + 1 – 2 – 12 = 14

C33 = (– 1)3 + 1 0 – 18 = – 18

(x) Given 2y – z = 1

x – y + z = 2

2x – y = 0

= 0 + 4 – 1

= 3

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

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 1 – 0 = 1

C21 = (– 1)2 + 11 – 2 = 1

C31 = (– 1)3 + 10 + 1 = 1

C12 = (– 1)1 + 2 – 2 – 0 = 2

C22 = (– 1)2 + 1 – 1 – 0 = – 1

C32 = (– 1)3 + 1 0 – 2 = 2

C13 = (– 1)1 + 2 4 – 0 = 4

C23 = (– 1)2 + 1 2 – 0 = – 2

C33 = (– 1)3 + 1 – 1 + 2 = 1

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

2x + y + z = 5

x + 2y + z = 5

= 8(– 1) – 4(1) + 3(3)

= – 8 – 4 + 9

= – 3

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

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 1 – 2 = – 1

C21 = (– 1)2 + 1 4 – 6 = 2

C31 = (– 1)3 + 1 4 – 3 = 1

C12 = (– 1)1 + 2 2 – 1 = – 1

C22 = (– 1)2 + 1 8 – 3 = 5

C32 = (– 1)3 + 1 8 – 6 = – 2

C13 = (– 1)1 + 2 4 – 1 = 3

C23 = (– 1)2 + 1 16 – 4 = – 12

C33 = (– 1)3 + 1 8 – 8 = 0

(xii) Given x + y + z = 6

x + 2z = 7

3x + y + z = 12

= 1(– 2) – 1(1 – 6) + 1(1)

= – 2 + 5 + 1

= 4

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

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 0 – 2 = – 2

C21 = (– 1)2 + 1 1 – 1 = 0

C31 = (– 1)3 + 1 2 – 0 = 2

C12 = (– 1)1 + 2 1 – 6 = 5

C22 = (– 1)2 + 1 1 – 3 = – 2

C32 = (– 1)3 + 1 2 – 1 = – 1

C13 = (– 1)1 + 2 1 – 0 = 1

C23 = (– 1)2 + 1 1 – 3 = 2

C33 = (– 1)3 + 1 0 – 1 = – 1

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

(4/x) – (6/y) + (5/z) = 1,

(6/x) + (9/y) – (20/z) = 2, x, y, z ≠ 0

AX = B

Now,

|A| = 2(75) – 3(– 110) + 10(72)

= 150 + 330 + 720

= 1200

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

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 120 – 45 = 75

C21 = (– 1)2 + 1 – 60 – 90 = 150

C31 = (– 1)3 + 1 15 + 60 = 75

C12 = (– 1)1 + 2 – 80 – 30 = 110

C22 = (– 1)2 + 1 – 40 – 60 = – 100

C32 = (– 1)3 + 1 10 – 40 = 30

C13 = (– 1)1 + 2 36 + 36 = 72

C23 = (– 1)2 + 1 18 – 18 = 0

C33 = (– 1)3 + 1 – 12 – 12 = – 24

(xiv) Given x – y + 2z = 7

3x + 4y – 5z = -5

2x – y + 3z = 12

A X = B

Now,

|A| = 1(12 – 5) + 1(9 + 10) + 2(– 3 – 8)

= 7 + 19 – 22

= 4

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

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 12 – 5 = 7

C21 = (– 1)2 + 1 – 3 + 2 = 1

C31 = (– 1)3 + 1 5 – 8 = – 3

C12 = (– 1)1 + 2 9 + 10 = – 19

C22 = (– 1)2 + 1 3 – 4 = – 1

C32 = (– 1)3 + 1 – 5 – 6 = 11

C13 = (– 1)1 + 2 – 3 – 8 = – 11

C23 = (– 1)2 + 1 – 1 + 2 = – 1

C33 = (– 1)3 + 1 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 – 2z = 1

4x + 4y – z = 2

6x + 6y + 2z = 3

Solution:

(i) Given 6x + 4y = 2

9x + 6y = 3

|A| = 36 – 36 = 0

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

C11 = (– 1)1 + 1 6 = 6

C12 = (– 1)1 + 2 9 = – 9

C21 = (– 1)2 + 1 4 = – 4

C22 = (– 1)2 + 2 6 = 6

(ii) Given 2x + 3y = 5

6x + 9y = 15

|A| = 18 – 18 = 0

So, A is singular,

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

C11 = (– 1)1 + 1 9 = 9

C12 = (– 1)1 + 2 6 = – 6

C21 = (– 1)2 + 1 3 = – 3

C22 = (– 1)2 + 2 2 = 2

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

3x + 26y + 2z = 9

7x + 2y + 10z = 5

|A| = 5(260 – 4) – 3(30 – 14) + 7(6 – 182)

= 5(256) – 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:

Cofactors of A are

C11 = (– 1)1 + 1 260 – 4 = 256

C21 = (– 1)2 + 1 30 – 14 = – 16

C31 = (– 1)3 + 1 6 – 182 = – 176

C12 = (– 1)1 + 2 30 – 14 = – 16

C22 = (– 1)2 + 1 50 – 49 = 1

C32 = (– 1)3 + 1 10 – 21 = 11

C13 = (– 1)1 + 2 6 – 182 = – 176

C23 = (– 1)2 + 1 10 – 21 = 11

C33 = (– 1)3 + 1 130 – 9 = 121

Now, AX = B has infinite many solution

Let z = k

Then, 5x + 3y = 4 – 7k

3x + 26y = 9 – 2k

This can be written as

(v) Given x + y + z = 6

x + 2y + 3z = 14

x + 4y + 7z = 30

|A| = 1(2) – 1(4) + 1(2)

= 2 – 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:

Cofactors of A are

C11 = (– 1)1 + 1 14 – 12 = 2

C21 = (– 1)2 + 1 7 – 4 = – 3

C31 = (– 1)3 + 1 3 – 2 = 1

C12 = (– 1)1 + 2 7 – 3 = – 4

C22 = (– 1)2 + 1 7 – 1 = 6

C32 = (– 1)3 + 1 3 – 1 = 2

C13 = (– 1)1 + 2 4 – 2 = 2

C23 = (– 1)2 + 1 4 – 1 = – 3

C33 = (– 1)3 + 1 2 – 1 = 1

Now, AX = B has infinite many solution

Let z = k

Then, x + y = 6 – k

x + 2y = 14 – 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) – 2(14) – 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:

Cofactors of A are:

C11 = (– 1)1 + 18 + 6 = 14

C21 = (– 1)2 + 1 4 + 12 = – 16

C31 = (– 1)3 + 1 – 2 + 8 = 6

C12 = (– 1)1 + 2 8 + 6 = – 14

C22 = (– 1)2 + 1 4 + 12 = 16

C32 = (– 1)3 + 1 – 2 + 8 = – 6

C13 = (– 1)1 + 2 24 – 24 = 0

C23 = (– 1)2 + 1 12 – 12 = 0

C33 = (– 1)3 + 1 8 – 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 – 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

Solution:

(i) Given 2x + 5y = 7

6x + 15y = 13

|A| = 30 – 30 = 0

So, A is singular,

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

C11 = (– 1)1 + 1 15 = 15

C12 = (– 1)1 + 2 6 = – 6

C21 = (– 1)2 + 1 5 = – 5

C22 = (– 1)2 + 2 2 = 2

(ii) Given 2x + 3y = 5

6x + 9y = 10

|A| = 18 – 18 = 0

So, A is singular,

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

C11 = (– 1)1 + 1 9 = 9

C12 = (– 1)1 + 2 6 = – 6

C21 = (– 1)2 + 1 3 = – 3

C22 = (– 1)2 + 2 2 = 2

(iii) Given 4x – 2y = 3

6x – 3y = 5

|A| = – 12 + 12 = 0

So, A is singular,

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

C11 = (– 1)1 + 1 – 3 = – 3

C12 = (– 1)1 + 2 6 = – 6

C21 = (– 1)2 + 1 – 2 = 2

C22 = (– 1)2 + 2 4 = 4

(iv) Given 4x – 5y – 2z = 2

5x – 4y + 2z = -2

2x + 2y + 8z = -1

|A| = 4(– 36) + 5(36) – 2(18)

|A| = 0

Cofactors of A are:

C11 = (– 1)1 + 1 – 32 – 4 = – 36

C21 = (– 1)2 + 1 – 40 + 4 = – 36

C31 = (– 1)3 + 1 – 10 – 8 = – 18

C12 = (– 1)1 + 2 40 – 4 = – 36

C22 = (– 1)2 + 1 32 + 4 = 36

C32 = (– 1)3 + 1 8 + 10 = – 18

C13 = (– 1)1 + 2 10 + 8 = 18

C23 = (– 1)2 + 1 8 + 10 = – 18

C33 = (– 1)3 + 1 – 16 + 25 = 9

(v) Given 3x – y – 2z = 2

2y – z = -1

3x – 5y = 3

|A| = 3(– 5) + 1(3) – 2(– 6)

|A| = 0

Cofactors of A are

C11 = (– 1)1 + 1 0 – 5 = – 5

C21 = (– 1)2 + 1 0 – 10 = 10

C31 = (– 1)3 + 1 1 + 4 = 5

C12 = (– 1)1 + 2 0 + 3 = – 3

C22 = (– 1)2 + 1 0 + 6 = 6

C32 = (– 1)3 + 1 – 3 – 0 = 3

C13 = (– 1)1 + 2 0 – 6 = – 6

C23 = (– 1)2 + 1 – 15 + 3 = 12

C33 = (– 1)3 + 1 6 – 0 = 6

(vi) Given x + y – 2z = 5

x – 2y + z = -2

-2x + y + z = 4

|A| = 1(– 3) – 1(3) – 2(– 3) = – 3 – 3 + 6

|A| = 0

Cofactors of A are:

C11 = (– 1)1 + 1 – 2 – 1 = – 3

C21 = (– 1)2 + 1 1 + 2 = – 3

C31 = (– 1)3 + 1 1 – 4 = – 3

C12 = (– 1)1 + 2 1 + 2 = – 3

C22 = (– 1)2 + 1 1 – 4 = – 3

C32 = (– 1)3 + 1 1 + 2 = – 3

C13 = (– 1)1 + 2 1 – 4 = – 3

C23 = (– 1)2 + 1 1 + 2 = – 3

C33 = (– 1)3 + 1 – 2 – 1 = – 3

x – y = 3, 2x + 3y + 4z = 17, y + 2z = 7

Solution:

2x – 3y + 5z = 11, 3x + 2y – 4z = -5, x + y – 2z = -3.

Solution:

|A| = 2(0) + 3(– 2) + 5(1)

= – 1

Now, the cofactors of A

C11 = (– 1)1 + 1 – 4 + 4 = 0

C21 = (– 1)2 + 1 6 – 5 = – 1

C31 = (– 1)3 + 1 12 – 10 = 2

C12 = (– 1)1 + 2 – 6 + 4 = 2

C22 = (– 1)2 + 1 – 4 – 5 = – 9

C32 = (– 1)3 + 1 – 8 – 15 = 23

C13 = (– 1)1 + 2 3 – 2 = 1

C23 = (– 1)2 + 1 2 + 3 = – 5

C33 = (– 1)3 + 1 4 + 9 = 13

x + 2y + 5z = 10, x – y – z = -2, 2x + 3y – z = -11.

Solution:

Given

|A| = 1(1 + 3) + 2(– 1 + 2) + 5(3 + 2)

= 4 + 2 + 25

= 27

Now, the cofactors of A

C11 = (– 1)1 + 1 1 + 3 = 4

C21 = (– 1)2 + 1 – 2 – 15 = 17

C31 = (– 1)3 + 1 – 2 + 5 = 3

C12 = (– 1)1 + 2 – 1 + 2 = – 1

C22 = (– 1)2 + 1 – 1 – 10 = – 11

C32 = (– 1)3 + 1 – 1 – 5 = 6

C13 = (– 1)1 + 2 3 + 2 = 5

C23 = (– 1)2 + 1 3 – 4 = 1

C33 = (– 1)3 + 1 – 1 – 2 = – 3

Solution:

Given

|A| = 1(1 + 6) + 2(2 – 0) + 0

= 7 + 4

= 11

Now, the cofactors of A

C11 = (– 1)1 + 1 1 + 6 = 7

C21 = (– 1)2 + 1 – 2 – 0 = 2

C31 = (– 1)3 + 1 – 6 – 0 = – 6

C12 = (– 1)1 + 2 2 – 0 = – 2

C22 = (– 1)2 + 1 1 – 0 = 1

C32 = (– 1)3 + 1 3 – 0 = – 3

C13 = (– 1)1 + 2 – 4 – 0 = – 4

C23 = (– 1)2 + 1 – 2 – 0 = 2

C33 = (– 1)3 + 1 1 + 4 = 5

Solution:

Given

|A| = 3(3 – 0) + 4(2 – 5) + 2(0 – 3)

= 9 – 12 – 6

= – 9

Now, the cofactors of A

C11 = (– 1)1 + 1 3 – 0 = 3

C21 = (– 1)2 + 1 – 4 – 0 = 4

C31 = (– 1)3 + 1 – 20 – 6 = – 26

C12 = (– 1)1 + 2 2 – 5 = 3

C22 = (– 1)2 + 1 3 – 2 = 1

C32 = (– 1)3 + 1 15 – 4 = – 11

C13 = (– 1)1 + 2 0 – 3 = – 3

C23 = (– 1)2 + 1 0 + 4 = – 4

C33 = (– 1)3 + 1 9 + 8 = 17

Solution:

Solution:

Given

|A| = 1(– 1 – 1) – 2(– 2 – 0) + 0

= – 2 + 4

= 2

Now, the cofactors of A

C11 = (– 1)1 + 1 – 1 – 1 = – 2

C21 = (– 1)2 + 1 2 – 0 = 2

C31 = (– 1)3 + 1 – 2 – 0 = – 2

C12 = (– 1)1 + 2 2 – 0 = – 2

C22 = (– 1)2 + 1 1 – 0 = 1

C32 = (– 1)3 + 1 – 1 – 0 = 1

C13 = (– 1)1 + 2 – 2 – 0 = – 2

C23 = (– 1)2 + 1 – 1 – 0 = 1

C33 = (– 1)3 + 1 – 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
…… (i)

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

x + 2y + z = 1 …… (ii)

Again,

x + z + 5(x) = 6

5x + y + z = 6 …… (iii)

A X = B

|A| = 1(1) – 1(– 4) + 1(– 9)

= 1 + 4 – 9

= – 4

Hence, the unique solution given by x = A – 1B

C11 = (– 1)1 + 1 (2 – 1) = 1

C12 = (– 1)1 + 2 (1 – 5) = 4

C13 = (– 1)1 + 3 (1 – 10) = – 9

C21 = (– 1)2 + 1 (1 – 1) = 0

C22 = (– 1)2 + 2 (1 – 5) = – 4

C23 = (– 1)2 + 3 (1 – 5) = 4

C31 = (– 1)3 + 1 (1 – 2) = – 1

C32 = (– 1)3 + 2 (1 – 1) = 0

C33 = (– 1)3 + 3 (2 – 1) = 1

10. An amount of ₹10,000 is put into three investments at the rate of 10, 12 and 15% per annum. The combined incomes are ₹1310 and the combined income of first and second investment is ₹ 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 …… (i)

Also,

0.1x + 0.12y + 0.15z = 1310 …… (ii)

Again,

0.1x + 0.12y – 0.15z = – 190 …… (iii)

A X = B

|A| = 1(– 0.036) – 1(– 0.03) + 1(0)

= – 0.006

Hence, the unique solution given by x = A – 1B

C11 = – 0.036

C12 = 0.27

C13 = 0

C21 = 0.27

C22 = – 0.25

C23 = – 0.02

C31 = 0.03

C32 = – 0.05

C33 = 0.02

Exercise 8.2 Page No: 8.20

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

1. 2x – y + z = 0

3x + 2y – z = 0

x + 4y + 3z = 0

Solution:

Given

2x – y + z = 0

3x + 2y – 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 – 2)

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

|A| = 40 ≠ 0

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

2. 2x – y + 2z = 0
5x + 3y – z = 0
X + 5y – 5z = 0

Solution:

Given 2x – y + 2z = 0

5x + 3y – z = 0

X + 5y – 5z = 0

A X = 0

Now, |A| = 2(– 15 + 5) + 1(– 25 + 1) + 2(25 – 3)

|A| = – 20 – 24 + 44

|A| = 0

Hence, the system has infinite solutions

Let z = k

2x – y = – 2k

5x + 3y = k

3. 3x – y + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z = 0

Solution:

Given 3x – y + 2z = 0

4x + 3y + 3z = 0

5x + 7y + 4z = 0

A X = 0

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

|A| = – 27 + 1 + 26

|A| = 0

Hence, the system has infinite solutions

Let z = k

3x – y = – 2k

4x + 3y = – 3k

4. x + y – 6z = 0
x – y + 2z = 0
– 3x + y + 2z = 0

Solution:

Given x + y – 6z = 0

x – y + 2z = 0

– 3x + y + 2z = 0

A X = 0

Now, |A| = 1(– 2 – 2) – 1(2 + 6) – 6(1 – 3)

|A| = – 4 – 8 + 12

|A| = 0

Hence, the system has infinite solutions

Let z = k

x + y = 6k

x – y = – 2k

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

Exercise 8.1 Solutions

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