RD Sharma Class 12 Ex 6.4 Solutions Chapter 6 Determinants

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Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	6
Exercise	6.4
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 6.1 Solutions Chapter 6 Determinants

Question 1. Solve the following system of linear equations by Cramer’s rule.

x – 2y = 4

−3x + 5y = −7

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 1 & -2 \\ -3 & 5 \end{array} \right|$

= 5 − 6

= −1

Also, we get,

$D_1=\left|\begin{array}{cc} 4 & -2 \\ -7 & 5 \end{array} \right|$

= 20 − 14

= 6

$D_2=\left|\begin{array}{cc} 1 & 4 \\ -3 & -7 \end{array} \right|$

= −7 + 12

= 5

So, x = D₁/D = 6/-1 = -6

And y = D₂/D = 5/-1 = -5

Therefore, x = −6 and y = −5.

Question 2. Solve the following system of linear equations by Cramer’s rule.

2x – y = 1

7x – 2y = −7

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 2 & -1 \\ 7 & -2 \end{array} \right|$

= −4 + 7

= 3

Also, we get,

$D_1=\left|\begin{array}{cc} 1 & -1 \\ -7 & -2 \end{array} \right|$

= −2 − 7

= −9

$D_2=\left|\begin{array}{cc} 2 & 1 \\ 7 & -7 \end{array} \right|$

= −14 − 7

= −21

So, x = D₁/D = -9/3 = -3

And y = D₂/D = -21/3 = -7

Therefore, x = −3 and y = −7.

Question 3. Solve the following system of linear equations by Cramer’s rule.

2x – y = 17

3x + 5y = 6

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 2 & -1 \\ 3 & 5 \end{array} \right|$

= 10 + 3

= 13

Also, we get,

$D_1=\left|\begin{array}{cc} 17 & -1 \\ 6 & 5 \end{array} \right|$

= 85 + 6

= 91

$D_2=\left|\begin{array}{cc} 2 & 17 \\ 3 & 6 \end{array} \right|$

= 12 − 51

= −39

So, x = D₁/D = 91/3 = 7

And y = D₂/D = -39/13 = -3

Therefore, x = 7 and y = −3.

Question 4. Solve the following system of linear equations by Cramer’s rule.

3x + y = 19

3x – y = 23

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 3 & 1 \\ 3 & -1 \end{array} \right|$

= −3 − 3

= −6

Also, we get,

$D_1=\left|\begin{array}{cc} 19 & 1 \\ 23 & -1 \end{array} \right|$

= −19 − 23

= −42

$D_2=\left|\begin{array}{cc} 3 & 19 \\ 3 & 23 \end{array} \right|$

= 69 − 57

= 12

So, x = D₁/D = -42/-6 = 7

And y = D₂/D = 12/-6 = -2

Therefore, x = 7 and y = −2.

Question 5. Solve the following system of linear equations by Cramer’s rule.

2x – y = –2

3x + 4y = 3

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 2 & -1 \\ 3 & 4 \end{array} \right|$

= 8 + 3

= 11

Also, we get,

$D_1=\left|\begin{array}{cc} -2 & -1 \\ 3 & 4 \end{array} \right|$

= −8 + 3

= −5

$D_2=\left|\begin{array}{cc} 2 & -2 \\ 3 & 3 \end{array} \right|$

= 6 + 6

= 12

So, x = D₁/D = -5/11

And y = D₂/D = 12/11

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

Question 6. Solve the following system of linear equations by Cramer’s rule.

3x + ay = 4

2x + ay = 2, a ≠ 0

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 3 & a \\ 2 & a \end{array} \right|$

= 3a − 2a

= a

Also, we get,

$D_1=\left|\begin{array}{cc} 4 & a \\ 2 & a \end{array} \right|$

= 4a − 2a

= 2a

$D_2=\left|\begin{array}{cc} 3 & 4 \\ 2 & 2 \end{array} \right|$

= 6 − 8

= −2

So, x = D₁/D = 2a/a = 2

And y = D₂/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,

$D=\left|\begin{array}{cc} 2 & 3 \\ 1 & 6 \end{array} \right|$

= 12 − 3

= 9

Also, we get,

$D_1=\left|\begin{array}{cc} 10 & 3 \\ 4 & 6 \end{array} \right|$

= 60 − 12

= 48

$D_2=\left|\begin{array}{cc} 2 & 10 \\ 1 & 4 \end{array} \right|$

= 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.

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

5x + 7y = −2

4x + 6y = −3

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 5 & 7 \\ 4 & 6 \end{array} \right|$

= 30 − 28

= 2

Also, we get,

$D_1=\left|\begin{array}{cc} -2 & 7 \\ -3 & 6 \end{array} \right|$

= −12 + 21

= 9

$D_2=\left|\begin{array}{cc} 5 & -2 \\ 4 & -3 \end{array} \right|$

= −15 + 8

= −7

So, x = D₁/D = 9/2

And y = D₂/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 – 2x = 8

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 9 & 5 \\ -2 & 3 \end{array} \right|$

= 27 + 10

= 37

Also, we get,

$D_1=\left|\begin{array}{cc} 10 & 5 \\ 8 & 3 \end{array} \right|$

= 30 − 40

= −10

$D_2=\left|\begin{array}{cc} 9 & 10 \\ -2 & 8 \end{array} \right|$

= 72 + 20

= 92

So, x = D₁/D = -10/37

And y = D₂/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,

$D=\left|\begin{array}{cc} 1 & 2 \\ 3 & 1 \end{array} \right|$

= 1 − 6

= −5

Also, we get,

$D_1=\left|\begin{array}{cc} 1 & 2 \\ 4 & 1 \end{array} \right|$

= 1 − 8

= −7

$D_2=\left|\begin{array}{cc} 1 & 1 \\ 3 & 4 \end{array} \right|$

= 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.

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

3x + y + z = 2

2x – 4y + 3z = −1

4x + y – 3z = −11

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 3 & 1 & 1 \\ 2 & -4 & 3 \\ 4 & 1 & -3 \end{array} \right|$

Expanding along R₁, we get,

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

= 27 + 18 + 18

= 63

Also, we get,

$D_1=\left|\begin{array}{cc} 2 & 1 & 1 \\ -1 & -4 & 3 \\ -11 & 1 & -3 \end{array} \right|$

Expanding along R₁, we get,

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

= 18 − 36 − 45

= −63

$D_2=\left|\begin{array}{cc} 3 & 2 & 1 \\ 2 & -1 & 3 \\ 4 & 11 & -3 \end{array} \right|$

Expanding along R₁, we get,

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

= 108 + 36 − 18

= 126

$D_3=\left|\begin{array}{cc} 3 & 1 & 2 \\ 2 & -4 & -1 \\ 4 & 1 & 11 \end{array} \right|$

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.

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

x – 4y – z = 11

2x – 5y + 2z = 39

−3x + 2y + z = 1

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 1 & -4 & -1 \\ 2 & -5 & 2 \\ -3 & 2 & 1 \end{array} \right|$

Expanding along R₁, we get,

= 1 (−5 − 4) + 4 (2 + 6) − 1 (4 − 15)

= −9 + 32 + 11

= 34

Also, we get,

$D_1=\left|\begin{array}{cc} 11 & -4 & -1 \\ 39 & -5 & 2 \\ 1 & 2 & 1 \end{array} \right|$

Expanding along R₁, we get,

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

= −99 + 148 − 83

= −34

$D_2=\left|\begin{array}{cc} 1 & 11 & -1 \\ 2 & 39 & 2 \\ -3 & 1 & 1 \end{array} \right|$

Expanding along R₁, we get,

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

= 37 − 88 − 119

= −170

$D_3=\left|\begin{array}{cc} 1 & -4 & 11 \\ 2 & -5 & -39 \\ -3 & 2 & 1 \end{array} \right|$

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.

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

6x + y – 3z = 5

x + 3y – 2z = 5

2x + y + 4z = 8

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 6 & 1 & -3 \\ 1 & 3 & -2 \\ 2 & 1 & 4 \end{array} \right|$

Expanding along R₁, we get,

= 6 (12 + 2) − 1 (4 + 4) − 3 (1 − 6)

= 84 − 8 + 15

= 91

Also, we get,

$D_1=\left|\begin{array}{cc} 5 & 1 & -3 \\ 5 & 3 & -2 \\ 8 & 1 & 4 \end{array} \right|$

Expanding along R₁, we get,

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

= 70 − 36 + 57

= 91

$D_2=\left|\begin{array}{cc} 6 & 5 & -3 \\ 1 & 5 & -2 \\ 2 & 8 & 4 \end{array} \right|$

Expanding along R₁, we get,

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

= 216 − 40 + 6

= 182

$D_3=\left|\begin{array}{cc} 6 & 1 & 5 \\ 1 & 3 & 5 \\ 2 & 1 & 8 \end{array} \right|$

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.

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,

$D=\left|\begin{array}{cc} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{array} \right|$

Expanding along R₁, we get,

= 1 (1) − 1 (−1) + 0 (−1)

= 1 + 1

= 2

Also, we get,

$D_1=\left|\begin{array}{cc} 5 & 1 & 0 \\ 3 & 1 & 1 \\ 4 & 0 & 1 \end{array} \right|$

Expanding along R₁, we get,

= 5 (1) − 1 (−1) + 0 (−4)

= 5 + 1 + 0

= 6

$D_2=\left|\begin{array}{cc} 1 & 5 & 0 \\ 0 & 3 & 1 \\ 1 & 4 & 1 \end{array} \right|$

Expanding along R₁, we get,

= 1 (−1) − 5 (−1) + 0 (−3)

= −1 + 5 + 0

= 4

$D_3=\left|\begin{array}{cc} 1 & 1 & 5 \\ 0 & 1 & 3 \\ 1 & 0 & 4 \end{array} \right|$

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.

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

2y – 3z = 0

x + 3y = −4

3x + 4y = 3

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 0 & 2 & -3 \\ 1 & 3 & 0 \\ 3 & 4 & 0 \end{array} \right|$

Expanding along R₁, we get,

= 0 (0) − 2 (0) − 3 (−5)

= 0 − 0 + 15

= 15

Also, we get,

$D_1=\left|\begin{array}{cc} 0 & 2 & -3 \\ -4 & 3 & 0 \\ 3 & 4 & 0 \end{array} \right|$

Expanding along R₁, we get,

= 0 (0) − 2 (0) − 3 (−25)

= 0 − 0 + 75

= 75

$D_2=\left|\begin{array}{cc} 0 & 0 & -3 \\ 1 & -4 & 0 \\ 3 & 3 & 0 \end{array} \right|$

Expanding along R₁, we get,

= 0 (0) − 0 (0) − 3 (15)

= 0 − 0 − 45

= −45

$D_3=\left|\begin{array}{cc} 0 & 2 & 0 \\ 1 & 3 & -4 \\ 3 & 4 & 3 \end{array} \right|$

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.

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

5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 5 & -7 & 1 \\ 6 & -8 & -1 \\ 3 & 2 & -6 \end{array} \right|$

Expanding along R₁, we get,

= 5 (50) + 7 (−33) + 1 (36)

= 250 − 231 + 36

= 55

Also, we get,

$D_1=\left|\begin{array}{cc} 11 & -7 & 1 \\ 15 & -8 & -1 \\ 7 & 2 & -6 \end{array} \right|$

Expanding along R₁, we get,

= 11 (50) + 7 (−83) + 1 (86)

= 550 − 581 + 86

= 55

$D_2=\left|\begin{array}{cc} 5 & 11 & 1 \\ 6 & 15 & -1 \\ 3 & 7 & -6 \end{array} \right|$

Expanding along R₁, we get,

= 5 (−83) − 11 (−33) + 1 (−3)

= −415 + 363 − 3

= −55

$D_3=\left|\begin{array}{cc} 5 & -7 & 11 \\ 6 & -8 & 15 \\ 3 & 2 & 7 \end{array} \right|$

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.

Question 17. Solve the following system of the linear equations by Cramer’s rule.

2x − 3y − 4z = 29

−2x + 5y − z = −15

3x − y + 5z = −11

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 2 & -3 & -4 \\ -2 & 5 & -1 \\ 3 & -1 & 5 \end{array} \right|$

Expanding along R₁, we get,

= 2 (24) + 3 (−13) + 4 (-13)

= 48 − 21 – 52

= -25

Also, we get,

$D_1=\left|\begin{array}{cc} 29 & -3 & -4 \\ -15 & 5 & -1 \\ 11 & -1 & 5 \end{array} \right|$

Expanding along R₁, we get,

= 29 (24) + 3 (−64) + 4 (−40)

= 692 − 192 − 160

= 344

$D_2=\left|\begin{array}{cc} 2 & 29 & -4 \\ -2 & -15 & -1 \\ 3 & 11 & 5 \end{array} \right|$

Expanding along R₁, we get,

= 2 (−64) − 29 (−7) + 4 (23)

= −128 + 203 + 92

= 167

$D_3=\left|\begin{array}{cc} 2 & -3 & 29 \\ -2 & 5 & -15 \\ 3 & -1 & 11 \end{array} \right|$

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.

Question 18. Solve the following system of the linear equations by Cramer’s rule.

x + y = 1

x + z = −6

x − y − 2z = 3

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & -1 & -2 \end{array} \right|$

= 1(1) – 1(-3)

= 1 + 3

= 4

Also, we get,

$D_1=\left|\begin{array}{cc} 1 & 1 & 0 \\ -6 & 0 & 1 \\ 3 & -1 & -2 \end{array} \right|$

Expanding along R₁, we get,

= 1 (1) − 1 (9) + 0

= 1 − 9

= −8

$D_2=\left|\begin{array}{cc} 1 & 1 & 0 \\ 1 & -6 & 1 \\ 1 & 3 & -2 \end{array} \right|$

Expanding along R₁, we get,

= 1 (9) − 1 (−3)

= 9 + 3

= 12

$D_3=\left|\begin{array}{cc} 1 & 1 & 1 \\ 1 & 0 & -6 \\ 1 & -1 & 3 \end{array} \right|$

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.

Question 19. Solve the following system of linear equations by Cramer’s rule.

x + y + z + 1 = 0

ax + by + cz + d = 0

a²x + b²y + c²z + d² = 0

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array} \right|$

c₂ -> c₂– c₁, c₃ -> c₃ – c₁

$D=\left|\begin{array}{cc} 1 & 1 & 1 \\ a & b-a & c-a \\ a^2 & b^2-a^2 & c^2-a^2 \end{array} \right|$

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

$D=(b-a)(c-a)\left|\begin{array}{cc} 1 & 1 & 1 \\ a & 1 & 1 \\ a^2 & b+a & c+a \end{array} \right|$

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)

$D_1=-\left|\begin{array}{cc} 1 & 1 & 1 \\ d & b & c \\ d^2 & b^2 & c^2 \end{array} \right|$

= -(d – b)(b – c)(c – d)

$D_2=-\left|\begin{array}{cc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array} \right|$

= -(a – d)(d – c)(c – a)

$D_3=-\left|\begin{array}{cc} 1 & 1 & 1 \\ a & b & d \\ a^2 & b^2 & d^2 \end{array} \right|$

= -(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)

Question 20. Solve the following system of linear equations by Cramer’s rule.

x + y + z + w = 2

x − 2y + 2z + 2w = −6

2x + y − 2z + 2w = −5

3x − y + 3z − 3w = −3

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 1 & 1 & 1 & 1 \\ 1 & -2 & 2 & 2 \\ 2 & 1 & -2 & 2 \\ 3 & -1 & 3 & -3 \end{array} \right|$

= $\left|\begin{array}{cc} 1 & 0 & 0 & 0 \\ 1 & -3 & 1 & 1 \\ 2 & -1 & -4 & 0 \\ 3 & -4 & 0 & -6 \end{array} \right|$

= $1\left|\begin{array}{cc} -3 & 1 & 1 \\ -1 & -4 & 0 \\ -4 & 0 & -6 \\ \end{array} \right|$

= −94

Also, we get,

$D_1=\left|\begin{array}{cc} 2 & 1 & 1 & 1 \\ -6 & -2 & 2 & 2 \\ -5 & 1 & -2 & 2 \\ -3 & -1 & 3 & -3 \end{array} \right| = 188$

$D_2=\left|\begin{array}{cc} 1 & 2 & 1 & 1 \\ 1 & -6 & 2 & 2 \\ 2 & -5 & -2 & 2 \\ 3 & -3 & 3 & -3 \end{array} \right| = −282$

$D_3=\left|\begin{array}{cc} 1 & 1 & 2 & 1 \\ 1 & -2 & -6 & 2 \\ 2 & 1 & -5 & 2 \\ 3 & -1 & -3 & -3 \end{array} \right| = −141$

$D_4=\left|\begin{array}{cc} 1 & 1 & 1 & 2 \\ 1 & -2 & 2 & -6 \\ 2 & 1 & -2 & -5 \\ 3 & -1 & 3 & -3 \end{array} \right| = 47$

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.

Question 21. Solve the following system of linear equations by Cramer’s rule.

2x − 3z + w = 1

x − y + 2w = 1

−3y + z + w = 1

x + y + z = −1

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 2 & 0 & -3 & 1 \\ 1 & -1 & 0 & 2 \\ 0 & -3 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{array} \right|$

= $\left|\begin{array}{cc} 2 & -2 & -5 & 1 \\ 1 & -2 & -1 & 2 \\ 0 & -3 & 1 & 1 \\ 1 & 0 & 0 & 0 \end{array} \right|$

= $-1\left|\begin{array}{cc} -2 & -5 & 1 \\ -2 & -1 & 2 \\ -3 & 1 & 1 \\ \end{array} \right|$

= −21

Also, we get,

$D_1=\left|\begin{array}{cc} 1 & 0 & -3 & 1 \\ 1 & -1 & 0 & 2 \\ 1 & -3 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{array} \right| = -21$

$D_2=\left|\begin{array}{cc} 2 & 1 & -3 & 1 \\ 1 & 1 & 0 & 2 \\ 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{array} \right| = −6$

$D_3=\left|\begin{array}{cc} 2 & 0 & 1 & 1 \\ 1 & -1 & 1 & 2 \\ 0 & -3 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{array} \right| = −6$

$D_4=\left|\begin{array}{cc} 2 & 0 & -3 & 1 \\ 1 & -1 & 0 & 1 \\ 0 & -3 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{array} \right| = 3$

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.

Question 22. Show that the following system of linear equations is inconsistent.

2x − y = 5

4x − 2y = 7

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 2 & -1 \\ 4 & -2 \end{array} \right|$

= −4 + 4

= 0

Also, we get,

$D_1=\left|\begin{array}{cc} 5 & -1 \\ 7 & -2 \end{array} \right|$

= − 10 + 7

= −3

$D_2=\left|\begin{array}{cc} 2 & 5 \\ 4 & 7 \end{array} \right|$

= 14 − 20

= −6

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

Hence proved.

Question 23. Show that the following system of linear equations is inconsistent.

3x + y = 5

−6x − 2y = 9

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 3 & 1 \\ -6 & -2 \end{array} \right|$

= −6 + 6

= 0

Also, we get,

$D_1=\left|\begin{array}{cc} 5 & 1 \\ 9 & -2 \end{array} \right|$

= −10 − 9

= −19

$D_2=\left|\begin{array}{cc} 3 & 5 \\ -6 & 9 \end{array} \right|$

= 27 + 30

= 57

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

Hence proved.

Question 24. Show that the following system of linear equations is inconsistent.

3x − y + 2z = 3

2x + y + 3z = 5

x − 2y − z = 1

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 3 & -1 & 2 \\ 2 & 1 & 3 \\ 1 & -2 & 1 \end{array} \right|$

Expanding along R₁, we get

= 3 (5) + 1 (−5) + 2 (−5)

= 15 − 5 − 10

= 0

Also, we get,

$D_1=\left|\begin{array}{cc} 3 & -1 & 2 \\ 5 & 1 & 3 \\ 1 & -2 & 1 \end{array} \right|$

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.

Question 25. Show that the following system of linear equations is consistent and solve.

3x − y + 2z = 6

2x − y + z = 2

3x + 6y + 5z = 20

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 3 & -1 & 2 \\ 2 & -1 & 1 \\ 3 & 6 & 5 \end{array} \right|$

Expanding along R₁, we get

= 3 (−11) + 1 (7) + 2 (15)

= −33 + 7 + 30

= 4

Also, we get,

$D_1=\left|\begin{array}{cc} 6 & -1 & 2 \\ 2 & -1 & 1 \\ 20 & 6 & 5 \end{array} \right|$

Expanding along R₁, we get

= 6 (−11) + 1 (−10) + 2 (32)

= −66 − 10 + 64

= −12

$D_2=\left|\begin{array}{cc} 3 & 6 & 2 \\ 2 & 2 & 1 \\ 3 & 20 & 5 \end{array} \right|$

Expanding along R₁, we get

= 3 (−10) − 6 (7) + 2 (34)

= −30 − 42 + 68

= −4

$D_3=\left|\begin{array}{cc} 3 & -1 & 6 \\ 2 & -1 & 2 \\ 3 & 6 & 20 \end{array} \right|$

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.

Question 26. Show that the following system of linear equations has infinite number of solutions.

x − y + z = 3

2x + y − z = 2

−x − 2y + 2z = 1

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & -2 & 2 \end{array} \right|$

= $\left|\begin{array}{cc} 0 & -1 & 0 \\ 3 & 1 & 0 \\ -3 & -2 & 0 \end{array} \right|$

= 0

Also, we get,

$D_1=\left|\begin{array}{cc} 3 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & -2 & 2 \end{array} \right|$

= $\left|\begin{array}{cc} 3 & -1 & 0 \\ 2 & 1 & 0 \\ 1 & -2 & 0 \end{array} \right|$

= 0

$D_2=\left|\begin{array}{cc} 1 & 3 & 1 \\ 2 & 2 & -1 \\ -1 & 1 & 2 \end{array} \right|$

= $\left|\begin{array}{cc} 0 & 0 & 0 \\ 2 & -4 & -3 \\ -1 & 4 & 3 \end{array} \right|$

= 0

$D_3=\left|\begin{array}{cc} 1 & -1 & 3 \\ 2 & 1 & 2 \\ -1 & -2 & 1 \end{array} \right|$

= $\left|\begin{array}{cc} 0 & 0 & 0 \\ 2 & 3 & -4 \\ -1 & -3 & 4 \end{array} \right|$

= 0

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

Hence proved.

Question 27. Show that the following system of linear equations has infinite number of solutions.

x + 2y = 5

3x + 2y = 15

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 1 & 2 \\ 3 & 6 \end{array} \right|$

= 6 − 6

= 0

Also, we get,

$D_1=\left|\begin{array}{cc} 5 & 2 \\ 15 & 6 \end{array} \right|$

= 30 − 30

= 0

$D_2=\left|\begin{array}{cc} 1 & 5 \\ 3 & 15 \end{array} \right|$

= 15 − 15

= 0

As D, D₁and D₂all are zero, the given system of equations has infinite number of solutions.

Hence proved.

Question 28. Show that the following system of linear equations has infinite number of solutions.

x + y − z = 0

x − 2y + z = 0

3x + 6y − 5z = 0

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 1 & 1 & -1 \\ 1 & -2 & 1 \\ 3 & 6 & -5 \end{array} \right|$

= $\left|\begin{array}{cc} 1 & 0 & 0 \\ 1 & -3 & 2 \\ 3 & 3 & -2 \end{array} \right|$

= 1 (6 − 6)

= 0

Also, we get,

$D_1=\left|\begin{array}{cc} 0 & 1 & -1 \\ 0 & -2 & 1 \\ 0 & 6 & -5 \end{array} \right|$

= 0

$D_2=\left|\begin{array}{cc} 1 & 0 & -1 \\ 1 & 0 & 1 \\ 3 & 0 & -5 \end{array} \right|$

= 0

$D_3=\left|\begin{array}{cc} 1 & 1 & 0 \\ 1 & -2 & 0 \\ 3 & 6 & 0 \end{array} \right|$

= 0

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

Hence proved.

Question 29. Show that the following system of linear equations has infinite number of solutions.

2x + y − 2z = 4

x − 2y + z = −2

5x − 5y + z = −2

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 2 & 1 & -2 \\ 1 & -2 & 1 \\ 5 & -5 & 1 \end{array} \right|$

= $\left|\begin{array}{cc} 12 & 9 & -2 \\ -4 & -3 & 1 \\ 0 & 0 & 1 \end{array} \right|$

= 1 (−36 + 36)

= 0

Also we get,

$D_1=\left|\begin{array}{cc} 4 & 1 & -2 \\ -2 & -2 & 1 \\ -2 & -5 & 1 \end{array} \right|$

= $\left|\begin{array}{cc} 0 & 1 & -2 \\ 0 & -2 & 1 \\ 0 & -5 & 1 \end{array} \right|$

= 0

$D_2=\left|\begin{array}{cc} 2 & 4 & -2 \\ 1 & -2 & 1 \\ 5 & -2 & 1 \end{array} \right|$

= $\left|\begin{array}{cc} 2 & 0 & -2 \\ 1 & 0 & 1 \\ 5 & 0 & 1 \end{array} \right|$

= 0

$D_3=\left|\begin{array}{cc} 2 & 1 & 4 \\ 1 & -2 & -2 \\ 5 & -5 & -2 \end{array} \right|$

= $\left|\begin{array}{cc} 4 & -3 & 0 \\ 1 & -2 & -2 \\ 4 & -3 & 0 \end{array} \right|$

= 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.

Question 30. Show that the following system of linear equations has infinite number of solutions.

x − y + 3z = 6

x + 3y − 3z = −4

5x + 3y + 3z = 10

Solution:

Using Cramer’s Rule, we get,

$D=\left|\begin{array}{cc} 1 & -1 & 3 \\ 1 & 3 & -3 \\ 5 & 3 & 3 \end{array} \right|$

= $\left|\begin{array}{cc} 2 & 2 & 0 \\ 1 & 3 & -3 \\ 6 & 6 & 0 \end{array} \right|$

= 3 (12 − 12)

= 0

Also we get,

$D_1=\left|\begin{array}{cc} 6 & -1 & 3 \\ -4 & 3 & -3 \\ 10 & 3 & 3 \end{array} \right|$

= $\left|\begin{array}{cc} 2 & 2 & 0 \\ -4 & 3 & -3 \\ 6 & 6 & 0 \end{array} \right|$

= 3 (12 − 12)

= 0

$D_2=\left|\begin{array}{cc} 1 & 6 & 3 \\ 1 & -4 & -3 \\ 5 & 10 & 3 \end{array} \right|$

= $\left|\begin{array}{cc} 2 & 2 & 0 \\ 1 & -4 & -3 \\ 6 & 6 & 0 \end{array} \right|$

= 3 (12 − 12)

= 0

$D_3=\left|\begin{array}{cc} 1 & -1 & 6 \\ 1 & 3 & -4 \\ 5 & 3 & 10 \end{array} \right|$

= $\left|\begin{array}{cc} 1 & -1 & 6 \\ 0 & 4 & -10 \\ 0 & 8 & -20 \end{array} \right|$

= 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.

Question 31. A salesman has the following record of sales during three months for three items A, B and C which have different rates of commission.

Month	Sale of units	Total Commission (in Rs)
A	B	C
Jan	90	100	20	800
Feb	130	50	40	900
Mar	60	100	30	850

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,

$D=\left|\begin{array}{cc} 90 & 100 & 20 \\ 130 & 50 & 40 \\ 60 & 100 & 30 \end{array} \right|$

= $\left|\begin{array}{cc} -170 & 0 & -60 \\ 130 & 50 & 40 \\ -200 & 0 & -50 \end{array} \right|$

= 50 (8500 − 12000)

= −175000

Also we get,

$D_1=\left|\begin{array}{cc} 800 & 100 & 20 \\ 900 & 50 & 40 \\ 850 & 100 & 30 \end{array} \right|$

= $\left|\begin{array}{cc} -1000 & 0 & -60 \\ 900 & 50 & 40 \\ -950 & 0 & -50 \end{array} \right|$

= 50 (50000 − 57000)

= −350000

$D_2=\left|\begin{array}{cc} 90 & 800 & 20 \\ 130 & 900 & 40 \\ 60 & 850 & 30 \end{array} \right|$

= $\left|\begin{array}{cc} 900 & 800 & 20 \\ -50 & -700 & 0 \\ -75 & -350 & 0 \end{array} \right|$

= 20 (17500 − 52500)

= −700000

$D_3=\left|\begin{array}{cc} 90 & 100 & 800 \\ 130 & 50 & 900 \\ 60 & 100 & 850 \end{array} \right|$

= $\left|\begin{array}{cc} -170 & 0 & -1000 \\ 130 & 50 & 900 \\ -200 & 0 & -950 \end{array} \right|$

= 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.

Question 32. An automobile company uses three types of steel S₁, S₂, and S₃ for producing three types of cars C₁, C₂ and C₃. Steel requirements (in tons) for each type of cars are given below.

Steel	Cars
C₁	C₂	C₃
S₁	2	3	4
S₂	1	1	2
S₃	3	2	1

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,

$D=\left|\begin{array}{cc} 2 & 3 & 4 \\ 1 & 1 & 2 \\ 3 & 2 & 1 \end{array} \right|$

= $\left|\begin{array}{cc} -10 & -5 & 0 \\ -5 & -3 & 0 \\ 3 & 2 & 1 \end{array} \right|$

= 1 (30 − 25)

= 5

Also we get,

$D_1=\left|\begin{array}{cc} 29 & 3 & 4 \\ 13 & 1 & 2 \\ 16 & 2 & 1 \end{array} \right|$

= $\left|\begin{array}{cc} -35 & -5 & 0 \\ -19 & -3 & 0 \\ 16 & 2 & 1 \end{array} \right|$

= 1 (105 − 95)

= 10

$D_2=\left|\begin{array}{cc} 2 & 29 & 4 \\ 1 & 13 & 2 \\ 3 & 16 & 1 \end{array} \right|$

= $\left|\begin{array}{cc} -10 & -35 & 0 \\ -5 & -19 & 0 \\ 3 & 16 & 1 \end{array} \right|$

= 1 (190 − 175)

= 15

$D_3=\left|\begin{array}{cc} 2 & 3 & 29 \\ 1 & 1 & 13 \\ 3 & 2 & 16 \end{array} \right|$

= $\left|\begin{array}{cc} -2 & 0 & 0 \\ 1 & 1 & 13 \\ 3 & 2 & 16 \end{array} \right|$

= −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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RD Sharma Class 12 Ex 6.1 Solutions Chapter 6 Determinants

Question 1. Solve the following system of linear equations by Cramer’s rule.

x – 2y = 4

−3x + 5y = −7

Question 2. Solve the following system of linear equations by Cramer’s rule.

2x – y = 1

7x – 2y = −7

Question 3. Solve the following system of linear equations by Cramer’s rule.

2x – y = 17

3x + 5y = 6

Question 4. Solve the following system of linear equations by Cramer’s rule.

3x + y = 19

3x – y = 23

Question 5. Solve the following system of linear equations by Cramer’s rule.

2x – y = –2

3x + 4y = 3

Question 6. Solve the following system of linear equations by Cramer’s rule.

3x + ay = 4

2x + ay = 2, a ≠ 0

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

2x + 3y = 10

x + 6y = 4

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

5x + 7y = −2

4x + 6y = −3

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

9x + 5y = 10

3y – 2x = 8

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

x + 2y = 1

3x + y = 4

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

3x + y + z = 2

2x – 4y + 3z = −1

4x + y – 3z = −11

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

x – 4y – z = 11

2x – 5y + 2z = 39

−3x + 2y + z = 1

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

6x + y – 3z = 5

x + 3y – 2z = 5

2x + y + 4z = 8

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

x + y = 5

y + z = 3

x + z = 4

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

2y – 3z = 0

x + 3y = −4

3x + 4y = 3

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

5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

Question 17. Solve the following system of the linear equations by Cramer’s rule.

2x − 3y − 4z = 29

−2x + 5y − z = −15

3x − y + 5z = −11

Question 18. Solve the following system of the linear equations by Cramer’s rule.

x + y = 1

x + z = −6

x − y − 2z = 3

Question 19. Solve the following system of linear equations by Cramer’s rule.

x + y + z + 1 = 0

ax + by + cz + d = 0

a2x + b2y + c2z + d2 = 0

Question 20. Solve the following system of linear equations by Cramer’s rule.

x + y + z + w = 2

x − 2y + 2z + 2w = −6

2x + y − 2z + 2w = −5

3x − y + 3z − 3w = −3

Question 21. Solve the following system of linear equations by Cramer’s rule.

2x − 3z + w = 1

x − y + 2w = 1

−3y + z + w = 1

x + y + z = −1

Question 22. Show that the following system of linear equations is inconsistent.

2x − y = 5

4x − 2y = 7

a²x + b²y + c²z + d² = 0

Question 32. An automobile company uses three types of steel S₁, S₂, and S₃ for producing three types of cars C₁, C₂ and C₃. Steel requirements (in tons) for each type of cars are given below.