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

• Last Updated : 26 May, 2021

### âˆ’3x + 5y = âˆ’7

Solution:

Using Cramer’s Rule, we get,

= 5 âˆ’ 6

= âˆ’1

Also, we get,

= 20 âˆ’ 14

= 6

= âˆ’7 + 12

= 5

So, x = D1/D = 6/-1 = -6

And y = D2/D = 5/-1 = -5

Therefore, x = âˆ’6 and y = âˆ’5.

### 7x â€“ 2y = âˆ’7

Solution:

Using Cramer’s Rule, we get,

= âˆ’4 + 7

= 3

Also, we get,

= âˆ’2 âˆ’ 7

= âˆ’9

= âˆ’14 âˆ’ 7

= âˆ’21

So, x = D1/D = -9/3 = -3

And y = D2/D = -21/3 = -7

Therefore, x = âˆ’3 and y = âˆ’7.

### 3x + 5y = 6

Solution:

Using Cramer’s Rule, we get,

= 10 + 3

= 13

Also, we get,

= 85 + 6

= 91

= 12 âˆ’ 51

= âˆ’39

So, x = D1/D = 91/3 = 7

And y = D2/D = -39/13 = -3

Therefore, x = 7 and y = âˆ’3.

### 3x â€“ y = 23

Solution:

Using Cramer’s Rule, we get,

= âˆ’3 âˆ’ 3

= âˆ’6

Also, we get,

= âˆ’19 âˆ’ 23

= âˆ’42

= 69 âˆ’ 57

= 12

So, x = D1/D = -42/-6 = 7

And y = D2/D = 12/-6 = -2

Therefore, x = 7 and y = âˆ’2.

### 3x + 4y = 3

Solution:

Using Cramer’s Rule, we get,

= 8 + 3

= 11

Also, we get,

= âˆ’8 + 3

= âˆ’5

= 6 + 6

= 12

So, x = D1/D = -5/11

And y = D2/D = 12/11

Therefore, x = -5/11 and y = 12/11.

### 2x + ay = 2, a â‰  0

Solution:

Using Cramer’s Rule, we get,

= 3a âˆ’ 2a

= a

Also, we get,

= 4a âˆ’ 2a

= 2a

= 6 âˆ’ 8

= âˆ’2

So, x = D1/D = 2a/a = 2

And y = D2/D = -2/a

Therefore, x = a and y = -2/a.

### x + 6y = 4

Solution:

Using Cramer’s Rule, we get,

= 12 âˆ’ 3

= 9

Also, we get,

= 60 âˆ’ 12

= 48

= 8 âˆ’ 10

= âˆ’2

So, x = D1/D = 48/9 = 16/3

And y = D2/D = -2/9

Therefore, x = 4/3 and y = -2/9.

### 4x + 6y = âˆ’3

Solution:

Using Cramer’s Rule, we get,

= 30 âˆ’ 28

= 2

Also, we get,

= âˆ’12 + 21

= 9

= âˆ’15 + 8

= âˆ’7

So, x = D1/D = 9/2

And y = D2/D = -7/2

Therefore, x = 9/2 and y = -7/2.

### 3y â€“ 2x = 8

Solution:

Using Cramer’s Rule, we get,

= 27 + 10

= 37

Also, we get,

= 30 âˆ’ 40

= âˆ’10

= 72 + 20

= 92

So, x = D1/D = -10/37

And y = D2/D = 92/37

Therefore, x = -10/37 and y = 92/37.

### 3x + y = 4

Solution:

Using Cramer’s Rule, we get,

= 1 âˆ’ 6

= âˆ’5

Also, we get,

= 1 âˆ’ 8

= âˆ’7

= 4 âˆ’ 3

= 1

So, x = D1/D = -7/-5 = 7/5

And y = D2/D = -1/5

Therefore, x = 7/5 and y = -1/5.

### 4x + y â€“ 3z = âˆ’11

Solution:

Using Cramer’s Rule, we get,

Expanding along R1, we get,

= 3 (12 âˆ’ 3) + (âˆ’1) (âˆ’6 âˆ’ 12) + 1 (2 + 16)

= 27 + 18 + 18

= 63

Also, we get,

Expanding along R1, we get,

= 2 (12 âˆ’ 3) + (âˆ’1) (3 + 33) + 1 (âˆ’1 âˆ’ 44)

= 18 âˆ’ 36 âˆ’ 45

= âˆ’63

Expanding along R1, we get,

= 3 (3 + 33) + (âˆ’2) (âˆ’6 âˆ’ 12) + 1 (âˆ’22 + 4)

= 108 + 36 âˆ’ 18

= 126

Expanding along R1, we get,

= 3 (44 + 1) + (âˆ’1) (âˆ’22 + 4) + 2 (2 + 16)

= 135 + 18 + 36

= 189

So, x = D1/D = -63/63 = -1

y = D2/D = 126/63 = 2

z = D3/D = 189/63 = 3

Therefore, x = âˆ’1, y = 2 and z = 3.

### âˆ’3x + 2y + z = 1

Solution:

Using Cramer’s Rule, we get,

Expanding along R1, we get,

= 1 (âˆ’5 âˆ’ 4) + 4 (2 + 6) âˆ’ 1 (4 âˆ’ 15)

= âˆ’9 + 32 + 11

= 34

Also, we get,

Expanding along R1, we get,

= 11 (âˆ’5 âˆ’ 4) + 4 (39 âˆ’ 2) âˆ’ 1 (78 + 5)

= âˆ’99 + 148 âˆ’ 83

= âˆ’34

Expanding along R1, we get,

= 1 (39 âˆ’ 2) âˆ’ 11 (2 + 6) âˆ’1 (2 + 117)

= 37 âˆ’ 88 âˆ’ 119

= âˆ’170

Expanding along R1, we get,

= 1 (âˆ’5 âˆ’ 78) + 4 (2 + 117) + 11 (4 âˆ’ 15)

= âˆ’83 + 476 âˆ’ 121

= 272

So, x = D1/D = -34/34 = -1

y = D2/D = -170/34 = -5

z = D3/D = 272/34 = 8

Therefore, x = âˆ’1, y = âˆ’5 and z = 8.

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

Solution:

Using Cramer’s Rule, we get,

Expanding along R1, we get,

= 6 (12 + 2) âˆ’ 1 (4 + 4) âˆ’ 3 (1 âˆ’ 6)

= 84 âˆ’ 8 + 15

= 91

Also, we get,

Expanding along R1, we get,

= 5 (12 + 2) âˆ’ 1 (20 + 16) âˆ’ 3 (5 âˆ’ 24)

= 70 âˆ’ 36 + 57

= 91

Expanding along R1, we get,

= 6 (20 + 16) âˆ’ 5 (4 + 4) âˆ’ 3 (8 âˆ’ 10)

= 216 âˆ’ 40 + 6

= 182

Expanding along R1, we get,

= 6 (24 âˆ’ 5) âˆ’ 1 (8 âˆ’ 10) + 5 (1 âˆ’ 6)

= 114 + 2 âˆ’ 25

= 91

So, x = D1/D = 91/91 = 1

y = D2/D = 182/91 = 2

z = D3/D = 92/92 =1

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

### x + z = 4

Solution:

Using Cramer’s Rule, we get,

Expanding along R1, we get,

= 1 (1) âˆ’ 1 (âˆ’1) + 0 (âˆ’1)

= 1 + 1

= 2

Also, we get,

Expanding along R1, we get,

= 5 (1) âˆ’ 1 (âˆ’1) + 0 (âˆ’4)

= 5 + 1 + 0

= 6

Expanding along R1, we get,

= 1 (âˆ’1) âˆ’ 5 (âˆ’1) + 0 (âˆ’3)

= âˆ’1 + 5 + 0

= 4

Expanding along R1, we get,

= 1 (4) âˆ’ 1 (âˆ’3) + 5 (âˆ’1)

= 4 + 3 âˆ’ 5

= 2

So, x = D1/D = 6/2 = 3

y = D2/D = 4/2 = 2

z = D3/D = 2/2 = 1

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

### 3x + 4y = 3

Solution:

Using Cramer’s Rule, we get,

Expanding along R1, we get,

= 0 (0) âˆ’ 2 (0) âˆ’ 3 (âˆ’5)

= 0 âˆ’ 0 + 15

= 15

Also, we get,

Expanding along R1, we get,

= 0 (0) âˆ’ 2 (0) âˆ’ 3 (âˆ’25)

= 0 âˆ’ 0 + 75

= 75

Expanding along R1, we get,

= 0 (0) âˆ’ 0 (0) âˆ’ 3 (15)

= 0 âˆ’ 0 âˆ’ 45

= âˆ’45

Expanding along R1, we get,

= 0 (25) âˆ’ 2 (15) + 0 (1)

= 0 âˆ’ 30 + 0

= âˆ’30

So, x = D1/D = 75/15 = 5

y = D2/D = -45/15 = -3

z = D3/D = -30/15 = -2

Therefore, x = 5, y = âˆ’3 and z = âˆ’2.

### 3x + 2y â€“ 6z = 7

Solution:

Using Cramer’s Rule, we get,

Expanding along R1, we get,

= 5 (50) + 7 (âˆ’33) + 1 (36)

= 250 âˆ’ 231 + 36

= 55

Also, we get,

Expanding along R1, we get,

= 11 (50) + 7 (âˆ’83) + 1 (86)

= 550 âˆ’ 581 + 86

= 55

Expanding along R1, we get,

= 5 (âˆ’83) âˆ’ 11 (âˆ’33) + 1 (âˆ’3)

= âˆ’415 + 363 âˆ’ 3

= âˆ’55

Expanding along R1, we get,

= 5 (âˆ’86) + 7 (âˆ’3) + 11 (36)

= âˆ’430 âˆ’ 21 + 396

= âˆ’55

So, x = D1/D = 55/55 = 1

y = D2/D = -55/55 = -1

z = D3/D = -55/55 = -1

Therefore, x = 1, y = âˆ’1 and z = âˆ’1.

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