# Class 12 RD Sharma Solutions – Chapter 6 Determinants – Exercise 6.4 | Set 1

### Question 1. Solve the following system of linear equations by Cramerβs rule.

### x β 2y = 4

### β3x + 5y = β7

**Solution:**

Using Cramer’s Rule, we get,

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= 5 β 6

= β1

Also, we get,

= 20 β 14

= 6

= β7 + 12

= 5

So, x = D

_{1}/D = 6/-1 = -6And y = D

_{2}/D = 5/-1 = -5

Therefore, x = β6 and y = β5.

### Question 2. Solve the following system of linear equations by Cramerβs rule.

### 2x β y = 1

### 7x β 2y = β7

**Solution:**

Using Cramer’s Rule, we get,

= β4 + 7

= 3

Also, we get,

= β2 β 7

= β9

= β14 β 7

= β21

So, x = D

_{1}/D = -9/3 = -3And y = D

_{2}/D = -21/3 = -7

Therefore, x = β3 and y = β7.

### Question 3. Solve the following system of linear equations by Cramerβs rule.

### 2x β y = 17

### 3x + 5y = 6

**Solution:**

Using Cramer’s Rule, we get,

= 10 + 3

= 13

Also, we get,

= 85 + 6

= 91

= 12 β 51

= β39

So, x = D

_{1}/D = 91/3 = 7And y = D

_{2}/D = -39/13 = -3

Therefore, x = 7 and y = β3.

### Question 4. Solve the following system of linear equations by Cramerβs rule.

### 3x + y = 19

### 3x β y = 23

**Solution:**

Using Cramer’s Rule, we get,

= β3 β 3

= β6

Also, we get,

= β19 β 23

= β42

= 69 β 57

= 12

So, x = D

_{1}/D = -42/-6 = 7And y = D

_{2}/D = 12/-6 = -2

Therefore, x = 7 and y = β2.

### Question 5. Solve the following system of linear equations by Cramerβs rule.

### 2x β y = β2

### 3x + 4y = 3

**Solution:**

Using Cramer’s Rule, we get,

= 8 + 3

= 11

Also, we get,

= β8 + 3

= β5

= 6 + 6

= 12

So, x = D

_{1}/D = -5/11And y = D

_{2}/D = 12/11

Therefore, x = -5/11 and y = 12/11.

### Question 6. Solve the following system of linear equations by Cramerβs rule.

### 3x + ay = 4

### 2x + ay = 2, a β 0

**Solution:**

Using Cramer’s Rule, we get,

= 3a β 2a

= a

Also, we get,

= 4a β 2a

= 2a

= 6 β 8

= β2

So, x = D

_{1}/D = 2a/a = 2And y = D

_{2}/D = -2/a

Therefore, x = a and y = -2/a.

### Question 7. Solve the following system of linear equation by Cramer’s rule.

### 2x + 3y = 10

### x + 6y = 4

**Solution:**

Using Cramer’s Rule, we get,

= 12 β 3

= 9

Also, we get,

= 60 β 12

= 48

= 8 β 10

= β2

So, x = D

_{1}/D = 48/9 = 16/3And y = D

_{2}/D = -2/9

Therefore, x = 4/3 and y = -2/9.

### Question 8. Solve the following system of linear equation by Cramer’s rule.

### 5x + 7y = β2

### 4x + 6y = β3

**Solution:**

Using Cramer’s Rule, we get,

= 30 β 28

= 2

Also, we get,

= β12 + 21

= 9

= β15 + 8

= β7

So, x = D

_{1}/D = 9/2And y = D

_{2}/D = -7/2

Therefore, x = 9/2 and y = -7/2.

### Question 9. Solve the following system of linear equation by Cramer’s rule.

### 9x + 5y = 10

### 3y β 2x = 8

**Solution:**

Using Cramer’s Rule, we get,

= 27 + 10

= 37

Also, we get,

= 30 β 40

= β10

= 72 + 20

= 92

So, x = D

_{1}/D = -10/37And y = D

_{2}/D = 92/37

Therefore, x = -10/37 and y = 92/37.

### Question 10. Solve the following system of linear equations by Cramer’s rule.

### x + 2y = 1

### 3x + y = 4

**Solution:**

Using Cramer’s Rule, we get,

= 1 β 6

= β5

Also, we get,

= 1 β 8

= β7

= 4 β 3

= 1

So, x = D

_{1}/D = -7/-5 = 7/5And y = D

_{2}/D = -1/5

Therefore, x = 7/5 and y = -1/5.

### Question 11. Solve the following system of linear equations by Cramer’s rule.

### 3x + y + z = 2

### 2x β 4y + 3z = β1

### 4x + y β 3z = β11

**Solution:**

Using Cramer’s Rule, we get,

Expanding along R

_{1}, we get,= 3 (12 β 3) + (β1) (β6 β 12) + 1 (2 + 16)

= 27 + 18 + 18

= 63

Also, we get,

Expanding along R

_{1}, we get,= 2 (12 β 3) + (β1) (3 + 33) + 1 (β1 β 44)

= 18 β 36 β 45

= β63

Expanding along R

_{1}, we get,= 3 (3 + 33) + (β2) (β6 β 12) + 1 (β22 + 4)

= 108 + 36 β 18

= 126

Expanding along R

_{1}, we get,= 3 (44 + 1) + (β1) (β22 + 4) + 2 (2 + 16)

= 135 + 18 + 36

= 189

So, x = D

_{1}/D = -63/63 = -1y = D

_{2}/D = 126/63 = 2z = D

_{3}/D = 189/63 = 3

Therefore, x = β1, y = 2 and z = 3.

### Question 12. Solve the following system of linear equations by Cramer’s rule.

### x β 4y β z = 11

### 2x β 5y + 2z = 39

### β3x + 2y + z = 1

**Solution:**

Using Cramer’s Rule, we get,

Expanding along R

_{1}, we get,= 1 (β5 β 4) + 4 (2 + 6) β 1 (4 β 15)

= β9 + 32 + 11

= 34

Also, we get,

Expanding along R

_{1}, we get,= 11 (β5 β 4) + 4 (39 β 2) β 1 (78 + 5)

= β99 + 148 β 83

= β34

Expanding along R

_{1}, we get,= 1 (39 β 2) β 11 (2 + 6) β1 (2 + 117)

= 37 β 88 β 119

= β170

Expanding along R

_{1}, we get,= 1 (β5 β 78) + 4 (2 + 117) + 11 (4 β 15)

= β83 + 476 β 121

= 272

So, x = D

_{1}/D = -34/34 = -1y = D

_{2}/D = -170/34 = -5z = D

_{3}/D = 272/34 = 8

Therefore, x = β1, y = β5 and z = 8.

### Question 13. Solve the following system of linear equations by Cramer’s rule.

### 6x + y β 3z = 5

### x + 3y β 2z = 5

### 2x + y + 4z = 8

**Solution:**

Using Cramer’s Rule, we get,

Expanding along R

_{1}, we get,= 6 (12 + 2) β 1 (4 + 4) β 3 (1 β 6)

= 84 β 8 + 15

= 91

Also, we get,

Expanding along R

_{1}, we get,= 5 (12 + 2) β 1 (20 + 16) β 3 (5 β 24)

= 70 β 36 + 57

= 91

Expanding along R

_{1}, we get,= 6 (20 + 16) β 5 (4 + 4) β 3 (8 β 10)

= 216 β 40 + 6

= 182

Expanding along R

_{1}, we get,= 6 (24 β 5) β 1 (8 β 10) + 5 (1 β 6)

= 114 + 2 β 25

= 91

So, x = D

_{1}/D = 91/91 = 1y = D

_{2}/D = 182/91 = 2z = D

_{3}/D = 92/92 =1

Therefore, x = 1, y = 2 and z = 1.

### Question 14. Solve the following system of linear equations by Cramer’s rule.

### x + y = 5

### y + z = 3

### x + z = 4

**Solution:**

Using Cramer’s Rule, we get,

Expanding along R

_{1}, we get,= 1 (1) β 1 (β1) + 0 (β1)

= 1 + 1

= 2

Also, we get,

Expanding along R

_{1}, we get,= 5 (1) β 1 (β1) + 0 (β4)

= 5 + 1 + 0

= 6

Expanding along R

_{1}, we get,= 1 (β1) β 5 (β1) + 0 (β3)

= β1 + 5 + 0

= 4

Expanding along R

_{1}, we get,= 1 (4) β 1 (β3) + 5 (β1)

= 4 + 3 β 5

= 2

So, x = D

_{1}/D = 6/2 = 3y = D

_{2}/D = 4/2 = 2z = D

_{3}/D = 2/2 = 1

Therefore, x = 3, y = 2 and z = 1.

### Question 15. Solve the following system of linear equations by Cramer’s rule.

### 2y β 3z = 0

### x + 3y = β4

### 3x + 4y = 3

**Solution:**

Using Cramer’s Rule, we get,

Expanding along R

_{1}, we get,= 0 (0) β 2 (0) β 3 (β5)

= 0 β 0 + 15

= 15

Also, we get,

Expanding along R

_{1}, we get,= 0 (0) β 2 (0) β 3 (β25)

= 0 β 0 + 75

= 75

Expanding along R

_{1}, we get,= 0 (0) β 0 (0) β 3 (15)

= 0 β 0 β 45

= β45

Expanding along R

_{1}, we get,= 0 (25) β 2 (15) + 0 (1)

= 0 β 30 + 0

= β30

So, x = D

_{1}/D = 75/15 = 5y = D

_{2}/D = -45/15 = -3z = D

_{3}/D = -30/15 = -2

Therefore, x = 5, y = β3 and z = β2.

### Question 16. Solve the following system of linear equations by Cramer’s rule.

### 5x β 7y + z = 11

### 6x β 8y β z = 15

### 3x + 2y β 6z = 7

**Solution:**

Using Cramer’s Rule, we get,

Expanding along R

_{1}, we get,= 5 (50) + 7 (β33) + 1 (36)

= 250 β 231 + 36

= 55

Also, we get,

Expanding along R

_{1}, we get,= 11 (50) + 7 (β83) + 1 (86)

= 550 β 581 + 86

= 55

Expanding along R

_{1}, we get,= 5 (β83) β 11 (β33) + 1 (β3)

= β415 + 363 β 3

= β55

Expanding along R

_{1}, we get,= 5 (β86) + 7 (β3) + 11 (36)

= β430 β 21 + 396

= β55

So, x = D

_{1}/D = 55/55 = 1y = D

_{2}/D = -55/55 = -1z = D

_{3}/D = -55/55 = -1

Therefore, x = 1, y = β1 and z = β1.