# RD Sharma Class 12 Ex 6.4 Solutions Chapter 6 Determinants

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## RD Sharma Class 12 Ex 6.1 Solutions Chapter 6 Determinants

### −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.

### 3x − y + 5z = −11

Solution:

Using Cramer’s Rule, we get,

Expanding along R1, we get,

= 2 (24) + 3 (−13) + 4 (-13)

= 48 − 21 – 52

= -25

Also, we get,

Expanding along R1, we get,

= 29 (24) + 3 (−64) + 4 (−40)

= 692 − 192 − 160

= 344

Expanding along R1, we get,

= 2 (−64) − 29 (−7) + 4 (23)

= −128 + 203 + 92

= 167

Expanding along R1, we get,

= 2 (40) + 3 (23) + 29 (−13)

= 80 + 69 − 377

= −228

So, x = D1/D = -344/25

y = D2/D = -167/25

z = D3/D = 228/25

Therefore, x = -344/25, y = -167/25, and z = 228/25.

### x − y − 2z = 3

Solution:

Using Cramer’s Rule, we get,

= 1(1) – 1(-3)

= 1 + 3

= 4

Also, we get,

Expanding along R1, we get,

= 1 (1) − 1 (9) + 0

= 1 − 9

= −8

Expanding along R1, we get,

= 1 (9) − 1 (−3)

= 9 + 3

= 12

Expanding along R1, we get,

= 1 (−6) − 1 (9) + 1 (−1)

= −6 − 9 − 1

= −16

So, x = D1/D = -8/4 = -2

y = D2/D = 12/4 = 3

z = D3/D = -16/4 = -4

Therefore, x = −2, y = 3 and z = −4.

### a2x + b2y + c2z + d2 = 0

Solution:

Using Cramer’s Rule, we get,

c2 -> c– c1, c3 -> c3 – c1

Taking common (b-a) from c2 and (c-a)c3

Expanding along R1, we get,

= (b – a)(c – a)(c + a – b – a)

= (b – a)(c – a)(c – b)

= (a – b)(b – c)(c – a)

= -(d – b)(b – c)(c – d)

= -(a – d)(d – c)(c – a)

= -(a – d)(b – d)(d – a)

So, x = D1/D = -(d – b)(b – c)(c – d)/(a – b)(b – c)(c – a)

y = D2/D = -(a – d)(d – c)(c – a)/(a – b)(b – c)(c – a)

z = D3/D = -(a – d)(b – d)(d – a)/(a – b)(b – c)(c – a)

### 3x − y + 3z − 3w = −3

Solution:

Using Cramer’s Rule, we get,  = −94

Also, we get,

So, x = D1/D = -188/94 = -2

y = D2/D = -282/-94 = 3

z = D3/D = -141/-94 = 3/2

w = D4/D = -47/94 = -1/2

Therefore, x = −2, y = 3 and z = 3/2 and w = -1/2.

### x + y + z = −1

Solution:

Using Cramer’s Rule, we get,

=  = −21

Also, we get,

So, x = D1/D = -21/-21 = 1

y = D2/D = -6/-21 = 2/7

z = D3/D = -6/-21 = 2/7

w = D4/D = -3/21 = -1/7

Therefore, x = 1, y = 2/7 and z = 2/7 and w = -1/7.

### 4x − 2y = 7

Solution:

Using Cramer’s Rule, we get,

= −4 + 4

= 0

Also, we get,

= − 10 + 7

= −3

= 14 − 20

= −6

Since D = 0 and D1 and D2 both are non-zero, the given system of equations is inconsistent.

Hence proved.

### −6x − 2y = 9

Solution:

Using Cramer’s Rule, we get,

= −6 + 6

= 0

Also, we get,

= −10 − 9

= −19

= 27 + 30

= 57

Since D = 0 and D1 and D2 both are non-zero, the given system of equations is inconsistent.

Hence proved.

### x − 2y − z = 1

Solution:

Using Cramer’s Rule, we get,

Expanding along R1, we get

= 3 (5) + 1 (−5) + 2 (−5)

= 15 − 5 − 10

= 0

Also, we get,

Expanding along R1, we get

= 3 (5) + 1 (−8) + 2 (−11)

= 15 − 8 − 22

= −15

Since D = 0 and D1 are non-zero, the given system of equations is inconsistent.

Hence proved.

### 3x + 6y + 5z = 20

Solution:

Using Cramer’s Rule, we get,

Expanding along R1, we get

= 3 (−11) + 1 (7) + 2 (15)

= −33 + 7 + 30

= 4

Also, we get,

Expanding along R1, we get

= 6 (−11) + 1 (−10) + 2 (32)

= −66 − 10 + 64

= −12

Expanding along R1, we get

= 3 (−10) − 6 (7) + 2 (34)

= −30 − 42 + 68

= −4

Expanding along R1, we get

= 3 (−32) + 1 (34) + 6 (15)

= −96 + 34 + 90

= 28

As D, D1, D2 and D3 all are non-zero, the given system of equations is consistent.

So, x = D1/D = -12/4 = -3

y = D2/D = -4/4 = -1

z = D3/D = 28/4 = 7

Therefore, x = −3, y = −1 and z = 7.

### −x − 2y + 2z = 1

Solution:

Using Cramer’s Rule, we get, = 0

Also, we get, = 0 = 0 = 0

As D, D1, D2 and D3 all are zero, the given system of equations has infinite number of solutions.

Hence proved.

### 3x + 2y = 15

Solution:

Using Cramer’s Rule, we get,

= 6 − 6

= 0

Also, we get,

= 30 − 30

= 0

= 15 − 15

= 0

As D, Dand Dall are zero, the given system of equations has infinite number of solutions.

Hence proved.

### 3x + 6y − 5z = 0

Solution:

Using Cramer’s Rule, we get, = 1 (6 − 6)

= 0

Also, we get,

= 0

= 0

= 0

As D, D1, D2 and D3 all are zero, the given system of equations has infinite number of solutions.

Hence proved.

### 5x − 5y + z = −2

Solution:

Using Cramer’s Rule, we get, = 1 (−36 + 36)

= 0

Also we get, = 0 = 0 = 2 (−12 + 12)

= 0

As D, D1, D2 and D3 all are zero, the given system of equations has infinite number of solutions.

Hence proved.

### 5x + 3y + 3z = 10

Solution:

Using Cramer’s Rule, we get, = 3 (12 − 12)

= 0

Also we get, = 3 (12 − 12)

= 0 = 3 (12 − 12)

= 0 = 1 (−80 + 80)

= 0

As D, D1, D2 and D3 all are zero, the given system of equations has infinite number of solutions.

Hence proved.

### Find out the rates of commission on items A, B and C by using the determinant method.

Solution:

Let the rates of commission on items A, B and C be x, y and z respectively.

According to the question, we have,

90x + 100y + 20z = 800

130x + 50y + 40z = 900

60x + 100y + 30z = 850

Using Cramer’s Rule, we get, = 50 (8500 − 12000)

= −175000

Also we get, = 50 (50000 − 57000)

= −350000 = 20 (17500 − 52500)

= −700000 = 50 (161500 − 200000)

= −1925000

So, x = D1/D = -350000/-175000 = 2

y = D2/D = -700000/-175000 = 4

z = D3/D = -1925000/-175000 = 11

Therefore, the rates of commission on items A, B and C are 2%, 4% and 11% respectively.

### Using Cramer’s Rule, find the number of cars of each type which can be produced using 29, 13 and 16 tons of steel of three types respectively.

Solution:

Let x, y and z be the number of cars C1, C2 and C3 produced respectively.

According to the question, we have,

2x + 3y + 4z = 29

x + y + 2z = 13

3x + 2y + z = 16

Using Cramer’s Rule, we get, = 1 (30 − 25)

= 5

Also we get, = 1 (105 − 95)

= 10 = 1 (190 − 175)

= 15 = −2 (16 − 26)

= 20

So, x = D1/D = 10/5 = 2

y = D2/D = 15/5 = 3

z = D3/D = 20/5 = 4

Therefore, the number of cars produced of type C1, C2 and C3 are 2, 3 and 4.

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