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• RD Sharma Class 12 Solutions for Maths

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

### −3x + 5y = −7

Solution:

Using Cramer’s Rule, we get, = 5 − 6

= −1

Also, we get, = 20 − 14

= 6 = −7 + 12

= 5

So, x = D1/D = 6/-1 = -6

And y = D2/D = 5/-1 = -5

Therefore, x = −6 and y = −5.

### 7x – 2y = −7

Solution:

Using Cramer’s Rule, we get, = −4 + 7

= 3

Also, we get, = −2 − 7

= −9 = −14 − 7

= −21

So, x = D1/D = -9/3 = -3

And y = D2/D = -21/3 = -7

Therefore, x = −3 and y = −7.

### 3x + 5y = 6

Solution:

Using Cramer’s Rule, we get, = 10 + 3

= 13

Also, we get, = 85 + 6

= 91 = 12 − 51

= −39

So, x = D1/D = 91/3 = 7

And y = D2/D = -39/13 = -3

Therefore, x = 7 and y = −3.

### 3x – y = 23

Solution:

Using Cramer’s Rule, we get, = −3 − 3

= −6

Also, we get, = −19 − 23

= −42 = 69 − 57

= 12

So, x = D1/D = -42/-6 = 7

And y = D2/D = 12/-6 = -2

Therefore, x = 7 and y = −2.

### 3x + 4y = 3

Solution:

Using Cramer’s Rule, we get, = 8 + 3

= 11

Also, we get, = −8 + 3

= −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 ≠ 0

Solution:

Using Cramer’s Rule, we get, = 3a − 2a

= a

Also, we get, = 4a − 2a

= 2a = 6 − 8

= −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 − 3

= 9

Also, we get, = 60 − 12

= 48 = 8 − 10

= −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 = −3

Solution:

Using Cramer’s Rule, we get, = 30 − 28

= 2

Also, we get, = −12 + 21

= 9 = −15 + 8

= −7

So, x = D1/D = 9/2

And y = D2/D = -7/2

Therefore, x = 9/2 and y = -7/2.

### 3y – 2x = 8

Solution:

Using Cramer’s Rule, we get, = 27 + 10

= 37

Also, we get, = 30 − 40

= −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 − 6

= −5

Also, we get, = 1 − 8

= −7 = 4 − 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 – 3z = −11

Solution:

Using Cramer’s Rule, we get, Expanding along R1, we get,

= 3 (12 − 3) + (−1) (−6 − 12) + 1 (2 + 16)

= 27 + 18 + 18

= 63

Also, we get, Expanding along R1, we get,

= 2 (12 − 3) + (−1) (3 + 33) + 1 (−1 − 44)

= 18 − 36 − 45

= −63 Expanding along R1, we get,

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

= 108 + 36 − 18

= 126 Expanding along R1, we get,

= 3 (44 + 1) + (−1) (−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 = −1, y = 2 and z = 3.

### −3x + 2y + z = 1

Solution:

Using Cramer’s Rule, we get, Expanding along R1, we get,

= 1 (−5 − 4) + 4 (2 + 6) − 1 (4 − 15)

= −9 + 32 + 11

= 34

Also, we get, Expanding along R1, we get,

= 11 (−5 − 4) + 4 (39 − 2) − 1 (78 + 5)

= −99 + 148 − 83

= −34 Expanding along R1, we get,

= 1 (39 − 2) − 11 (2 + 6) −1 (2 + 117)

= 37 − 88 − 119

= −170 Expanding along R1, we get,

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

= −83 + 476 − 121

= 272

So, x = D1/D = -34/34 = -1

y = D2/D = -170/34 = -5

z = D3/D = 272/34 = 8

Therefore, x = −1, y = −5 and z = 8.

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

Solution:

Using Cramer’s Rule, we get, Expanding along R1, we get,

= 6 (12 + 2) − 1 (4 + 4) − 3 (1 − 6)

= 84 − 8 + 15

= 91

Also, we get, Expanding along R1, we get,

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

= 70 − 36 + 57

= 91 Expanding along R1, we get,

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

= 216 − 40 + 6

= 182 Expanding along R1, we get,

= 6 (24 − 5) − 1 (8 − 10) + 5 (1 − 6)

= 114 + 2 − 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) − 1 (−1) + 0 (−1)

= 1 + 1

= 2

Also, we get, Expanding along R1, we get,

= 5 (1) − 1 (−1) + 0 (−4)

= 5 + 1 + 0

= 6 Expanding along R1, we get,

= 1 (−1) − 5 (−1) + 0 (−3)

= −1 + 5 + 0

= 4 Expanding along R1, we get,

= 1 (4) − 1 (−3) + 5 (−1)

= 4 + 3 − 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) − 2 (0) − 3 (−5)

= 0 − 0 + 15

= 15

Also, we get, Expanding along R1, we get,

= 0 (0) − 2 (0) − 3 (−25)

= 0 − 0 + 75

= 75 Expanding along R1, we get,

= 0 (0) − 0 (0) − 3 (15)

= 0 − 0 − 45

= −45 Expanding along R1, we get,

= 0 (25) − 2 (15) + 0 (1)

= 0 − 30 + 0

= −30

So, x = D1/D = 75/15 = 5

y = D2/D = -45/15 = -3

z = D3/D = -30/15 = -2

Therefore, x = 5, y = −3 and z = −2.

### 3x + 2y – 6z = 7

Solution:

Using Cramer’s Rule, we get, Expanding along R1, we get,

= 5 (50) + 7 (−33) + 1 (36)

= 250 − 231 + 36

= 55

Also, we get, Expanding along R1, we get,

= 11 (50) + 7 (−83) + 1 (86)

= 550 − 581 + 86

= 55 Expanding along R1, we get,

= 5 (−83) − 11 (−33) + 1 (−3)

= −415 + 363 − 3

= −55 Expanding along R1, we get,

= 5 (−86) + 7 (−3) + 11 (36)

= −430 − 21 + 396

= −55

So, x = D1/D = 55/55 = 1

y = D2/D = -55/55 = -1

z = D3/D = -55/55 = -1

Therefore, x = 1, y = −1 and z = −1.

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