# 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 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 problem-solving abilities of students, which are important from the exam point of view.

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## RD Sharma Solutions For Class 12 Chapter 6 Determinants Exercise 6.4

### Access other exercises of RD Sharma Solutions For Class 12 Chapter 6 – Determinants

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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 Cramerâ€™s rule:

1. x â€“ 2y = 4

-3x + 5y = -7

Solution:

Given x â€“ 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 1stÂ row

â‡’Â D = 5(1) â€“ (â€“ 3) (â€“ 2)

â‡’Â D = 5 â€“ 6

â‡’Â D = â€“ 1

Again,

Solving determinant, expanding along 1stÂ row

â‡’Â D1Â = 5(4) â€“ (â€“ 7) (â€“ 2)

â‡’Â D1Â = 20 â€“ 14

â‡’Â D1Â = 6

And

Solving determinant, expanding along 1stÂ row

â‡’Â D2Â = 1(â€“ 7) â€“ (â€“ 3) (4)

â‡’Â D2Â = â€“ 7 + 12

â‡’Â D2Â = 5

Thus by Cramerâ€™s Rule, we have

2. 2x â€“ y = 1

7x â€“ 2y = -7

Solution:

Given 2x â€“ y = 1 and

7x â€“ 2y = -7

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

Solving determinant, expanding along 1stÂ row

â‡’Â D1Â = 1(â€“ 2) â€“ (â€“ 7) (â€“ 1)

â‡’Â D1Â = â€“ 2 â€“ 7

â‡’Â D1Â = â€“ 9

And

Solving determinant, expanding along 1stÂ row

â‡’Â D2Â = 2(â€“ 7) â€“ (7) (1)

â‡’Â D2Â = â€“ 14 â€“ 7

â‡’Â D2Â = â€“ 21

Thus by Cramerâ€™s Rule, we have

3. 2x â€“ y = 17

3x + 5y = 6

Solution:

Given 2x â€“ 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 1stÂ row

â‡’Â D1Â = 17(5) â€“ (6) (â€“ 1)

â‡’Â D1Â = 85 + 6

â‡’Â D1Â = 91

Solving determinant, expanding along 1stÂ row

â‡’Â D2Â = 2(6) â€“ (17) (3)

â‡’Â D2Â = 12 â€“ 51

â‡’Â D2Â = â€“ 39

Thus by Cramerâ€™s Rule, we have

4. 3x + y = 19

3x â€“ y = 23

Solution:

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

Solving determinant, expanding along 1stÂ row

â‡’Â D = 3(â€“ 1) â€“ (3) (1)

â‡’Â D = â€“ 3 â€“ 3

â‡’Â D = â€“ 6

Again,

Solving determinant, expanding along 1stÂ row

â‡’Â D1Â = 19(â€“ 1) â€“ (23) (1)

â‡’Â D1Â = â€“ 19 â€“ 23

â‡’Â D1Â = â€“ 42

Solving determinant, expanding along 1stÂ row

â‡’Â D2Â = 3(23) â€“ (19) (3)

â‡’Â D2Â = 69 â€“ 57

â‡’Â D2Â = 12

Thus by Cramerâ€™s Rule, we have

5. 2x â€“ y = -2

3x + 4y = 3

Solution:

Given 2x â€“ 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 1stÂ row

â‡’Â D2Â = 3(2) â€“ (â€“ 2) (3)

â‡’Â D2Â = 6 + 6

â‡’Â D2Â = 12

Thus by Cramerâ€™s Rule, we have

6. 3x + ay = 4

2x + ay = 2, a â‰  0

Solution:

Given 3x + ay = 4 and

2x + ay = 2, a â‰  0

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

3x + ay = 4

2x + ay = 2, aâ‰ 0

So by comparing with the theorem, let’s find D, D1Â and D2

Solving determinant, expanding along 1stÂ row

â‡’Â D = 3(a) â€“ (2) (a)

â‡’Â D = 3a â€“ 2a

â‡’Â D = a

Again,

Solving determinant, expanding along 1stÂ row

â‡’Â D1Â = 4(a) â€“ (2) (a)

â‡’Â D = 4a â€“ 2a

â‡’Â D = 2a

Solving determinant, expanding along 1stÂ row

â‡’Â D2Â = 3(2) â€“ (2) (4)

â‡’Â D = 6 â€“ 8

â‡’Â D = â€“ 2

Thus by Cramerâ€™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 1stÂ row

â‡’Â D = 2 (6) â€“ (3) (1)

â‡’Â D = 12 â€“ 3

â‡’Â D = 9

Again,

Solving determinant, expanding along 1stÂ row

â‡’Â D1Â = 10 (6) â€“ (3) (4)

â‡’Â D = 60 â€“ 12

â‡’Â D = 48

Solving determinant, expanding along 1stÂ row

â‡’Â D2Â = 2 (4) â€“ (10) (1)

â‡’Â D2Â = 8 â€“ 10

â‡’Â D2Â = â€“ 2

Thus by Cramerâ€™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 = â€“ 2

4x + 6y = â€“ 3

So by comparing with the theorem, let’s find D, D1Â and D2

Solving determinant, expanding along 1stÂ row

â‡’Â D = 5(6) â€“ (7) (4)

â‡’Â D = 30 â€“ 28

â‡’Â D = 2

Again,

Solving determinant, expanding along 1stÂ row

â‡’Â D1Â = â€“ 2(6) â€“ (7) (â€“ 3)

â‡’Â D1Â = â€“ 12 + 21

â‡’Â D1Â = 9

Solving determinant, expanding along 1stÂ row

â‡’Â D2Â = â€“ 3(5) â€“ (â€“ 2) (4)

â‡’Â D2Â = â€“ 15 + 8

â‡’Â D2Â = â€“ 7

Thus by Cramerâ€™s Rule, we have

9. 9x + 5y = 10

3y â€“ 2x = 8

Solution:

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

Solving determinant, expanding along 1stÂ row

â‡’Â D = 3(9) â€“ (5) (â€“ 2)

â‡’Â D = 27 + 10

â‡’Â D = 37

Again,

Solving determinant, expanding along 1stÂ row

â‡’Â D1Â = 10(3) â€“ (8) (5)

â‡’Â D1Â = 30 â€“ 40

â‡’Â D1Â = â€“ 10

Solving determinant, expanding along 1stÂ row

â‡’Â D2Â = 9(8) â€“ (10) (â€“ 2)

â‡’Â D2Â = 72 + 20

â‡’Â D2Â = 92

Thus by Cramerâ€™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 1stÂ row

â‡’Â D = 1(1) â€“ (3) (2)

â‡’Â D = 1 â€“ 6

â‡’Â D = â€“ 5

Again,

Solving determinant, expanding along 1stÂ row

â‡’Â D1Â = 1(1) â€“ (2) (4)

â‡’Â D1Â = 1 â€“ 8

â‡’Â D1Â = â€“ 7

Solving determinant, expanding along 1stÂ row

â‡’Â D2Â = 1(4) â€“ (1) (3)

â‡’Â D2Â = 4 â€“ 3

â‡’Â D2Â = 1

Thus by Cramerâ€™s Rule, we have

Solve the following system of linear equations by Cramerâ€™s rule:

11. 3x + y + z = 2

2x â€“ 4y + 3z = -1

4x + y â€“ 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 â€“ 4y + 3z = â€“ 1

4x + y â€“ 3z = â€“ 11

So by comparing with the theorem, let’s find D, D1, D2Â and D3

Solving determinant, expanding along 1stÂ row

â‡’Â D = 3[(â€“ 4) (â€“ 3) â€“ (3) (1)] â€“ 1[(2) (â€“ 3) â€“ 12] + 1[2 â€“ 4(â€“ 4)]

â‡’Â D = 3[12 â€“ 3] â€“ [â€“ 6 â€“ 12] + [2 + 16]

â‡’Â D = 27 + 18 + 18

â‡’Â D = 63

Again,

Solving determinant, expanding along 1stÂ row

â‡’Â D1Â = 2[( â€“ 4)( â€“ 3) â€“ (3)(1)] â€“ 1[( â€“ 1)( â€“ 3) â€“ ( â€“ 11)(3)] + 1[( â€“ 1) â€“ ( â€“ 4)( â€“ 11)]

â‡’Â D1Â = 2[12 â€“ 3] â€“ 1[3 + 33] + 1[â€“ 1 â€“ 44]

â‡’Â D1Â = 2[9] â€“ 36 â€“ 45

â‡’Â D1Â = 18 â€“ 36 â€“ 45

â‡’Â D1Â = â€“ 63

Again

Solving determinant, expanding along 1stÂ row

â‡’Â D2Â = 3[3 + 33] â€“ 2[â€“ 6 â€“ 12] + 1[â€“ 22 + 4]

â‡’Â D2Â = 3[36] â€“ 2(â€“ 18) â€“ 18

â‡’Â D2Â = 126

â‡’

Solving determinant, expanding along 1stÂ row

â‡’Â D3Â = 3[44 + 1] â€“ 1[â€“ 22 + 4] + 2[2 + 16]

â‡’Â D3Â = 3[45] â€“ 1(â€“ 18) + 2(18)

â‡’Â D3Â = 135 + 18 + 36

â‡’Â D3Â = 189

Thus by Cramerâ€™s Rule, we have

12. x â€“ 4y â€“ z = 11

2x â€“ 5y + 2z = 39

-3x + 2y + z = 1

Solution:

Given,

x â€“ 4y â€“ z = 11

2x â€“ 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 â€“ 4y â€“ z = 11

2x â€“ 5y + 2z = 39

â€“ 3x + 2y + z = 1

So by comparing with theorem, now we have to find D, D1Â and D2

Solving determinant, expanding along 1stÂ row

â‡’Â D = 1[(â€“ 5) (1) â€“ (2) (2)] + 4[(2) (1) + 6] â€“ 1[4 + 5(â€“ 3)]

â‡’Â D = 1[â€“ 5 â€“ 4] + 4[8] â€“ [â€“ 11]

â‡’Â D = â€“ 9 + 32 + 11

â‡’Â D = 34

Again,

Solving determinant, expanding along 1stÂ row

â‡’Â D1Â = 11[(â€“ 5) (1) â€“ (2) (2)] + 4[(39) (1) â€“ (2) (1)] â€“ 1[2 (39) â€“ (â€“ 5) (1)]

â‡’Â D1Â = 11[â€“ 5 â€“ 4] + 4[39 â€“ 2] â€“ 1[78 + 5]

â‡’Â D1Â = 11[â€“ 9] + 4(37) â€“ 83

â‡’Â D1Â = â€“ 99 â€“ 148 â€“ 45

â‡’Â D1Â = â€“ 34

Again

Solving determinant, expanding along 1stÂ row

â‡’Â D2Â = 1[39 â€“ 2] â€“ 11[2 + 6] â€“ 1[2 + 117]

â‡’Â D2Â = 1[37] â€“ 11(8) â€“ 119

â‡’Â D2Â = â€“ 170

And,

â‡’

Solving determinant, expanding along 1stÂ row

â‡’Â D3Â = 1[â€“ 5 â€“ (39) (2)] â€“ (â€“ 4) [2 â€“ (39) (â€“ 3)] + 11[4 â€“ (â€“ 5)(â€“ 3)]

â‡’Â D3Â = 1 [â€“ 5 â€“ 78] + 4 (2 + 117) + 11 (4 â€“ 15)

â‡’Â D3Â = â€“ 83 + 4(119) + 11(â€“ 11)

â‡’Â D3Â = 272

Thus by Cramerâ€™s Rule, we have

13. 6x + y â€“ 3z = 5

x + 3y â€“ 2z = 5

2x + y + 4z = 8

Solution:

Given

6x + y â€“ 3z = 5

x + 3y â€“ 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 â€“ 3z = 5

x + 3y â€“ 2z = 5

2x + y + 4z = 8

So by comparing with theorem, now we have to find D , D1Â and D2

Solving determinant, expanding along 1stÂ Row

â‡’Â D = 6[(4) (3) â€“ (1) (â€“ 2)] â€“ 1[(4) (1) + 4] â€“ 3[1 â€“ 3(2)]

â‡’Â D = 6[12 + 2] â€“ [8] â€“ 3[â€“ 5]

â‡’Â D = 84 â€“ 8 + 15

â‡’Â D = 91

Again, Solve D1Â formed by replacing 1stÂ column by B matrices

Here

Solving determinant, expanding along 1stÂ Row

â‡’Â D1Â = 5[(4) (3) â€“ (â€“ 2) (1)] â€“ 1[(5) (4) â€“ (â€“ 2) (8)] â€“ 3[(5) â€“ (3) (8)]

â‡’Â D1Â = 5[12 + 2] â€“ 1[20 + 16] â€“ 3[5 â€“ 24]

â‡’Â D1Â = 5[14] â€“ 36 â€“ 3(â€“ 19)

â‡’Â D1Â = 70 â€“ 36 + 57

â‡’Â D1Â = 91

Again, Solve D2Â formed by replacing 1stÂ column by B matrices

Here

Solving determinant

â‡’Â D2Â = 6[20 + 16] â€“ 5[4 â€“ 2(â€“ 2)] + (â€“ 3)[8 â€“ 10]

â‡’Â D2Â = 6[36] â€“ 5(8) + (â€“ 3) (â€“ 2)

â‡’Â D2Â = 182

And, Solve D3Â formed by replacing 1stÂ column by B matrices

Here

Solving determinant, expanding along 1stÂ Row

â‡’Â D3Â = 6[24 â€“ 5] â€“ 1[8 â€“ 10] + 5[1 â€“ 6]

â‡’Â D3Â = 6[19] â€“ 1(â€“ 2) + 5(â€“ 5)

â‡’Â D3Â = 114 + 2 â€“ 25

â‡’Â D3Â = 91

Thus by Cramerâ€™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 DjÂ be the determinant obtained from D after replacing the jthÂ 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, D1Â and D2

Solving determinant, expanding along 1stÂ Row

â‡’Â D = 1[1] â€“ 1[â€“ 1] + 0[â€“ 1]

â‡’Â D = 1 + 1 + 0

â‡’Â D = 2

Again, Solve D1Â formed by replacing 1stÂ column by B matrices

Here

Solving determinant, expanding along 1stÂ Row

â‡’Â D1Â = 5[1] â€“ 1[(3) (1) â€“ (4) (1)] + 0[0 â€“ (4) (1)]

â‡’Â D1Â = 5 â€“ 1[3 â€“ 4] + 0[â€“ 4]

â‡’Â D1Â = 5 â€“ 1[â€“ 1] + 0

â‡’Â D1Â = 5 + 1 + 0

â‡’Â D1Â = 6

Again, Solve D2Â formed by replacing 1stÂ column by B matrices

Here

Solving determinant

â‡’Â D2Â = 1[3 â€“ 4] â€“ 5[â€“ 1] + 0[0 â€“ 3]

â‡’Â D2Â = 1[â€“ 1] + 5 + 0

â‡’Â D2Â = 4

And, Solve D3Â formed by replacing 1stÂ column by B matrices

Here

Solving determinant, expanding along 1stÂ Row

â‡’Â D3Â = 1[4 â€“ 0] â€“ 1[0 â€“ 3] + 5[0 â€“ 1]

â‡’Â D3Â = 1[4] â€“ 1(â€“ 3) + 5(â€“ 1)

â‡’Â D3Â = 4 + 3 â€“ 5

â‡’Â D3Â = 2

Thus by Cramerâ€™s Rule, we have

15. 2y â€“ 3z = 0

x + 3y = -4

3x + 4y = 3

Solution:

Given

2y â€“ 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 â€“ 3z = 0

x + 3y = â€“ 4

3x + 4y = 3

So by comparing with theorem, now we have to find D, D1Â and D2

Solving determinant, expanding along 1stÂ Row

â‡’Â D = 0[0] â€“ 2[(0) (1) â€“ 0] â€“ 3[1 (4) â€“ 3 (3)]

â‡’Â D = 0 â€“ 0 â€“ 3[4 â€“ 9]

â‡’Â D = 0 â€“ 0 + 15

â‡’Â D = 15

Again, Solve D1Â formed by replacing 1stÂ column by B matrices

Here

Solving determinant, expanding along 1stÂ Row

â‡’Â D1Â = 0[0] â€“ 2[(0) (â€“ 4) â€“ 0] â€“ 3[4 (â€“ 4) â€“ 3(3)]

â‡’Â D1Â = 0 â€“ 0 â€“ 3[â€“ 16 â€“ 9]

â‡’Â D1Â = 0 â€“ 0 â€“ 3(â€“ 25)

â‡’Â D1Â = 0 â€“ 0 + 75

â‡’Â D1Â = 75

Again, Solve D2Â formed by replacing 2ndÂ column by B matrices

Here

Solving determinant

â‡’Â D2Â = 0[0] â€“ 0[(0) (1) â€“ 0] â€“ 3[1 (3) â€“ 3(â€“ 4)]

â‡’Â D2Â = 0 â€“ 0 + (â€“ 3) (3 + 12)

â‡’Â D2Â = â€“ 45

And, Solve D3Â formed by replacing 3rdÂ column by B matrices

Here

Solving determinant, expanding along 1stÂ Row

â‡’Â D3Â = 0[9 â€“ (â€“ 4) 4] â€“ 2[(3) (1) â€“ (â€“ 4) (3)] + 0[1 (4) â€“ 3 (3)]

â‡’Â D3Â = 0[25] â€“ 2(3 + 12) + 0(4 â€“ 9)

â‡’Â D3Â = 0 â€“ 30 + 0

â‡’Â D3Â = â€“ 30

Thus by Cramerâ€™s Rule, we have

16. 5x â€“ 7y + z = 11

6x â€“ 8y â€“ z = 15

3x + 2y â€“ 6z = 7

Solution:

Given

5x â€“ 7y + z = 11

6x â€“ 8y â€“ z = 15

3x + 2y â€“ 6z = 7

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

Now, here we have

5x â€“ 7y + z = 11

6x â€“ 8y â€“ z = 15

3x + 2y â€“ 6z = 7

So by comparing with theorem, now we have to find D, D1Â and D2

Solving determinant, expanding along 1stÂ Row

â‡’Â D = 5[(â€“ 8) (â€“ 6) â€“ (â€“ 1) (2)] â€“ 7[(â€“ 6) (6) â€“ 3(â€“ 1)] + 1[2(6) â€“ 3(â€“ 8)]

â‡’Â D = 5[48 + 2] â€“ 7[â€“ 36 + 3] + 1[12 + 24]

â‡’Â D = 250 â€“ 231 + 36

â‡’Â D = 55

Again, Solve D1Â formed by replacing 1stÂ column by B matrices

Here

Solving determinant, expanding along 1stÂ Row

â‡’Â D1Â = 11[(â€“ 8) (â€“ 6) â€“ (2) (â€“ 1)] â€“ (â€“ 7) [(15) (â€“ 6) â€“ (â€“ 1) (7)] + 1[(15)2 â€“ (7) (â€“ 8)]

â‡’Â D1Â = 11[48 + 2] + 7[â€“ 90 + 7] + 1[30 + 56]

â‡’Â D1Â = 11[50] + 7[â€“ 83] + 86

â‡’Â D1Â = 550 â€“ 581 + 86

â‡’Â D1Â = 55

Again, Solve D2Â formed by replacing 2ndÂ column by B matrices

Here

Solving determinant, expanding along 1stÂ Row

â‡’Â D2Â = 5[(15) (â€“ 6) â€“ (7) (â€“ 1)] â€“ 11 [(6) (â€“ 6) â€“ (â€“ 1) (3)] + 1[(6)7 â€“ (15) (3)]

â‡’Â D2 =Â 5[â€“ 90 + 7] â€“ 11[â€“ 36 + 3] + 1[42 â€“ 45]

â‡’Â D2Â = 5[â€“ 83] â€“ 11(â€“ 33) â€“ 3

â‡’Â D2Â = â€“ 415 + 363 â€“ 3

â‡’Â D2Â = â€“ 55

And, Solve D3Â formed by replacing 3rdÂ column by B matrices

Here

Solving determinant, expanding along 1stÂ Row

â‡’Â D3Â = 5[(â€“ 8) (7) â€“ (15) (2)] â€“ (â€“ 7) [(6) (7) â€“ (15) (3)] + 11[(6)2 â€“ (â€“ 8) (3)]

â‡’Â D3Â = 5[â€“ 56 â€“ 30] â€“ (â€“ 7) [42 â€“ 45] + 11[12 + 24]

â‡’Â D3Â = 5[â€“ 86] + 7[â€“ 3] + 11[36]

â‡’Â D3Â = â€“ 430 â€“ 21 + 396

â‡’Â D3Â = â€“ 55

Thus by Cramerâ€™s Rule, we have