RD Sharma Solutions for Class 12 Maths Exercise 6.4 Chapter 6 Determinants

RD Sharma Solutions for Class 12 Maths Exercise 6.4 Chapter 6 Determinants are available here. This exercise deals with the applications of determinants in solving a system of linear equations. It mainly consists of a set of examples before the exercise problems to make it possible for students to solve questions effortlessly. It improves the problem-solving abilities of students, which are important from the exam point of view.

Students can access exercise-wise solutions in PDF format to solve problems in RD Sharma textbook easily from the provided links. RD Sharma Solutions for Class 12 Maths Chapter 6 Exercise 6.4 are available here.

RD Sharma Solutions for Class 12 Chapter 6 Determinants Exercise 6.4

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Access answers to Maths RD Sharma Solutions for Class 12 Chapter 6 – Determinants Exercise 6.4

Exercise 6.4 Page No: 6.84

Solve the following system of linear equations by Cramer’s rule:

1. x – 2y = 4

-3x + 5y = -7

Solution:

Given x – 2y = 4

-3x + 5y = -7

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D = 5(1) – (– 3) (– 2)

⇒ D = 5 – 6

⇒ D = – 1

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 5(4) – (– 7) (– 2)

⇒ D₁ = 20 – 14

⇒ D₁ = 6

And

Solving determinant, expanding along 1^st row

⇒ D₂ = 1(– 7) – (– 3) (4)

⇒ D₂ = – 7 + 12

⇒ D₂ = 5

Thus by Cramer’s Rule, we have

2. 2x – y = 1

7x – 2y = -7

Solution:

Given 2x – y = 1 and

7x – 2y = -7

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D₁ = 1(– 2) – (– 7) (– 1)

⇒ D₁ = – 2 – 7

⇒ D₁ = – 9

And

Solving determinant, expanding along 1^st row

⇒ D₂ = 2(– 7) – (7) (1)

⇒ D₂ = – 14 – 7

⇒ D₂ = – 21

Thus by Cramer’s Rule, we have

3. 2x – y = 17

3x + 5y = 6

Solution:

Given 2x – y = 17 and

3x + 5y = 6

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D₁ = 17(5) – (6) (– 1)

⇒ D₁ = 85 + 6

⇒ D₁ = 91

Solving determinant, expanding along 1^st row

⇒ D₂ = 2(6) – (17) (3)

⇒ D₂ = 12 – 51

⇒ D₂ = – 39

Thus by Cramer’s Rule, we have

4. 3x + y = 19

3x – y = 23

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D = 3(– 1) – (3) (1)

⇒ D = – 3 – 3

⇒ D = – 6

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 19(– 1) – (23) (1)

⇒ D₁ = – 19 – 23

⇒ D₁ = – 42

Solving determinant, expanding along 1^st row

⇒ D₂ = 3(23) – (19) (3)

⇒ D₂ = 69 – 57

⇒ D₂ = 12

Thus by Cramer’s Rule, we have

5. 2x – y = -2

3x + 4y = 3

Solution:

Given 2x – y = -2 and

3x + 4y = 3

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D₂ = 3(2) – (– 2) (3)

⇒ D₂ = 6 + 6

⇒ D₂ = 12

Thus by Cramer’s Rule, we have

6. 3x + ay = 4

2x + ay = 2, a ≠ 0

Solution:

Given 3x + ay = 4 and

2x + ay = 2, a ≠ 0

Let there be a system of n simultaneous linear equations and with n unknown given by

3x + ay = 4

2x + ay = 2, a≠0

So by comparing with the theorem, let’s find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 3(a) – (2) (a)

⇒ D = 3a – 2a

⇒ D = a

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 4(a) – (2) (a)

⇒ D = 4a – 2a

⇒ D = 2a

Solving determinant, expanding along 1^st row

⇒ D₂ = 3(2) – (2) (4)

⇒ D = 6 – 8

⇒ D = – 2

Thus by Cramer’s Rule, we have

7. 2x + 3y = 10

x + 6y = 4

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D = 2 (6) – (3) (1)

⇒ D = 12 – 3

⇒ D = 9

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 10 (6) – (3) (4)

⇒ D = 60 – 12

⇒ D = 48

Solving determinant, expanding along 1^st row

⇒ D₂ = 2 (4) – (10) (1)

⇒ D₂ = 8 – 10

⇒ D₂ = – 2

Thus by Cramer’s Rule, we have

8. 5x + 7y = -2

4x + 6y = -3

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

5x + 7y = – 2

4x + 6y = – 3

So by comparing with the theorem, let’s find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 5(6) – (7) (4)

⇒ D = 30 – 28

⇒ D = 2

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = – 2(6) – (7) (– 3)

⇒ D₁ = – 12 + 21

⇒ D₁ = 9

Solving determinant, expanding along 1^st row

⇒ D₂ = – 3(5) – (– 2) (4)

⇒ D₂ = – 15 + 8

⇒ D₂ = – 7

Thus by Cramer’s Rule, we have

9. 9x + 5y = 10

3y – 2x = 8

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D = 3(9) – (5) (– 2)

⇒ D = 27 + 10

⇒ D = 37

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 10(3) – (8) (5)

⇒ D₁ = 30 – 40

⇒ D₁ = – 10

Solving determinant, expanding along 1^st row

⇒ D₂ = 9(8) – (10) (– 2)

⇒ D₂ = 72 + 20

⇒ D₂ = 92

Thus by Cramer’s Rule, we have

10. x + 2y = 1

3x + y = 4

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D = 1(1) – (3) (2)

⇒ D = 1 – 6

⇒ D = – 5

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 1(1) – (2) (4)

⇒ D₁ = 1 – 8

⇒ D₁ = – 7

Solving determinant, expanding along 1^st row

⇒ D₂ = 1(4) – (1) (3)

⇒ D₂ = 4 – 3

⇒ D₂ = 1

Thus by Cramer’s Rule, we have

Solve the following system of linear equations by Cramer’s rule:

11. 3x + y + z = 2

2x – 4y + 3z = -1

4x + y – 3z = -11

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

3x + y + z = 2

2x – 4y + 3z = – 1

4x + y – 3z = – 11

So by comparing with the theorem, let’s find D, D₁, D₂ and D₃

Solving determinant, expanding along 1^st row

⇒ D = 3[(– 4) (– 3) – (3) (1)] – 1[(2) (– 3) – 12] + 1[2 – 4(– 4)]

⇒ D = 3[12 – 3] – [– 6 – 12] + [2 + 16]

⇒ D = 27 + 18 + 18

⇒ D = 63

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 2[( – 4)( – 3) – (3)(1)] – 1[( – 1)( – 3) – ( – 11)(3)] + 1[( – 1) – ( – 4)( – 11)]

⇒ D₁ = 2[12 – 3] – 1[3 + 33] + 1[– 1 – 44]

⇒ D₁ = 2[9] – 36 – 45

⇒ D₁ = 18 – 36 – 45

⇒ D₁ = – 63

Again

Solving determinant, expanding along 1^st row

⇒ D₂ = 3[3 + 33] – 2[– 6 – 12] + 1[– 22 + 4]

⇒ D₂ = 3[36] – 2(– 18) – 18

⇒ D₂ = 126

⇒

Solving determinant, expanding along 1^st row

⇒ D₃ = 3[44 + 1] – 1[– 22 + 4] + 2[2 + 16]

⇒ D₃ = 3[45] – 1(– 18) + 2(18)

⇒ D₃ = 135 + 18 + 36

⇒ D₃ = 189

Thus by Cramer’s Rule, we have

12. x – 4y – z = 11

2x – 5y + 2z = 39

-3x + 2y + z = 1

Solution:

Given,

x – 4y – z = 11

2x – 5y + 2z = 39

-3x + 2y + z = 1

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

x – 4y – z = 11

2x – 5y + 2z = 39

– 3x + 2y + z = 1

So by comparing with theorem, now we have to find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 1[(– 5) (1) – (2) (2)] + 4[(2) (1) + 6] – 1[4 + 5(– 3)]

⇒ D = 1[– 5 – 4] + 4[8] – [– 11]

⇒ D = – 9 + 32 + 11

⇒ D = 34

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 11[(– 5) (1) – (2) (2)] + 4[(39) (1) – (2) (1)] – 1[2 (39) – (– 5) (1)]

⇒ D₁ = 11[– 5 – 4] + 4[39 – 2] – 1[78 + 5]

⇒ D₁ = 11[– 9] + 4(37) – 83

⇒ D₁ = – 99 – 148 – 45

⇒ D₁ = – 34

Again

Solving determinant, expanding along 1^st row

⇒ D₂ = 1[39 – 2] – 11[2 + 6] – 1[2 + 117]

⇒ D₂ = 1[37] – 11(8) – 119

⇒ D₂ = – 170

And,

⇒

Solving determinant, expanding along 1^st row

⇒ D₃ = 1[– 5 – (39) (2)] – (– 4) [2 – (39) (– 3)] + 11[4 – (– 5)(– 3)]

⇒ D₃ = 1 [– 5 – 78] + 4 (2 + 117) + 11 (4 – 15)

⇒ D₃ = – 83 + 4(119) + 11(– 11)

⇒ D₃ = 272

Thus by Cramer’s Rule, we have

13. 6x + y – 3z = 5

x + 3y – 2z = 5

2x + y + 4z = 8

Solution:

Given

6x + y – 3z = 5

x + 3y – 2z = 5

2x + y + 4z = 8

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

6x + y – 3z = 5

x + 3y – 2z = 5

2x + y + 4z = 8

So by comparing with theorem, now we have to find D , D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 6[(4) (3) – (1) (– 2)] – 1[(4) (1) + 4] – 3[1 – 3(2)]

⇒ D = 6[12 + 2] – [8] – 3[– 5]

⇒ D = 84 – 8 + 15

⇒ D = 91

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 5[(4) (3) – (– 2) (1)] – 1[(5) (4) – (– 2) (8)] – 3[(5) – (3) (8)]

⇒ D₁ = 5[12 + 2] – 1[20 + 16] – 3[5 – 24]

⇒ D₁ = 5[14] – 36 – 3(– 19)

⇒ D₁ = 70 – 36 + 57

⇒ D₁ = 91

Again, Solve D₂ formed by replacing 1^st column by B matrices

Here

Solving determinant

⇒ D₂ = 6[20 + 16] – 5[4 – 2(– 2)] + (– 3)[8 – 10]

⇒ D₂ = 6[36] – 5(8) + (– 3) (– 2)

⇒ D₂ = 182

And, Solve D₃ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₃ = 6[24 – 5] – 1[8 – 10] + 5[1 – 6]

⇒ D₃ = 6[19] – 1(– 2) + 5(– 5)

⇒ D₃ = 114 + 2 – 25

⇒ D₃ = 91

Thus by Cramer’s Rule, we have

14. x + y = 5

y + z = 3

x + z = 4

Solution:

Given x + y = 5

y + z = 3

x + z = 4

Let there be a system of n simultaneous linear equations and with n unknown given by

Let D_j be the determinant obtained from D after replacing the j^th column by

Now, here we have

x + y = 5

y + z = 3

x + z = 4

So by comparing with theorem, now we have to find D, D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 1[1] – 1[– 1] + 0[– 1]

⇒ D = 1 + 1 + 0

⇒ D = 2

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 5[1] – 1[(3) (1) – (4) (1)] + 0[0 – (4) (1)]

⇒ D₁ = 5 – 1[3 – 4] + 0[– 4]

⇒ D₁ = 5 – 1[– 1] + 0

⇒ D₁ = 5 + 1 + 0

⇒ D₁ = 6

Again, Solve D₂ formed by replacing 1^st column by B matrices

Here

Solving determinant

⇒ D₂ = 1[3 – 4] – 5[– 1] + 0[0 – 3]

⇒ D₂ = 1[– 1] + 5 + 0

⇒ D₂ = 4

And, Solve D₃ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₃ = 1[4 – 0] – 1[0 – 3] + 5[0 – 1]

⇒ D₃ = 1[4] – 1(– 3) + 5(– 1)

⇒ D₃ = 4 + 3 – 5

⇒ D₃ = 2

Thus by Cramer’s Rule, we have

15. 2y – 3z = 0

x + 3y = -4

3x + 4y = 3

Solution:

Given

2y – 3z = 0

x + 3y = -4

3x + 4y = 3

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

2y – 3z = 0

x + 3y = – 4

3x + 4y = 3

So by comparing with theorem, now we have to find D, D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 0[0] – 2[(0) (1) – 0] – 3[1 (4) – 3 (3)]

⇒ D = 0 – 0 – 3[4 – 9]

⇒ D = 0 – 0 + 15

⇒ D = 15

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 0[0] – 2[(0) (– 4) – 0] – 3[4 (– 4) – 3(3)]

⇒ D₁ = 0 – 0 – 3[– 16 – 9]

⇒ D₁ = 0 – 0 – 3(– 25)

⇒ D₁ = 0 – 0 + 75

⇒ D₁ = 75

Again, Solve D₂ formed by replacing 2^nd column by B matrices

Here

Solving determinant

⇒ D₂ = 0[0] – 0[(0) (1) – 0] – 3[1 (3) – 3(– 4)]

⇒ D₂ = 0 – 0 + (– 3) (3 + 12)

⇒ D₂ = – 45

And, Solve D₃ formed by replacing 3^rd column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₃ = 0[9 – (– 4) 4] – 2[(3) (1) – (– 4) (3)] + 0[1 (4) – 3 (3)]

⇒ D₃ = 0[25] – 2(3 + 12) + 0(4 – 9)

⇒ D₃ = 0 – 30 + 0

⇒ D₃ = – 30

Thus by Cramer’s Rule, we have

16. 5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

Solution:

Given

5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

So by comparing with theorem, now we have to find D, D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 5[(– 8) (– 6) – (– 1) (2)] – 7[(– 6) (6) – 3(– 1)] + 1[2(6) – 3(– 8)]

⇒ D = 5[48 + 2] – 7[– 36 + 3] + 1[12 + 24]

⇒ D = 250 – 231 + 36

⇒ D = 55

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 11[(– 8) (– 6) – (2) (– 1)] – (– 7) [(15) (– 6) – (– 1) (7)] + 1[(15)2 – (7) (– 8)]

⇒ D₁ = 11[48 + 2] + 7[– 90 + 7] + 1[30 + 56]

⇒ D₁ = 11[50] + 7[– 83] + 86

⇒ D₁ = 550 – 581 + 86

⇒ D₁ = 55

Again, Solve D₂ formed by replacing 2^nd column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₂ = 5[(15) (– 6) – (7) (– 1)] – 11 [(6) (– 6) – (– 1) (3)] + 1[(6)7 – (15) (3)]

⇒ D₂ = 5[– 90 + 7] – 11[– 36 + 3] + 1[42 – 45]

⇒ D₂ = 5[– 83] – 11(– 33) – 3

⇒ D₂ = – 415 + 363 – 3

⇒ D₂ = – 55

And, Solve D₃ formed by replacing 3^rd column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₃ = 5[(– 8) (7) – (15) (2)] – (– 7) [(6) (7) – (15) (3)] + 11[(6)2 – (– 8) (3)]

⇒ D₃ = 5[– 56 – 30] – (– 7) [42 – 45] + 11[12 + 24]

⇒ D₃ = 5[– 86] + 7[– 3] + 11[36]

⇒ D₃ = – 430 – 21 + 396

⇒ D₃ = – 55

Thus by Cramer’s Rule, we have