RD Sharma Class 12 Ex 6.4 Solutions Chapter 6 Determinants

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter6
Exercise6.4
CategoryRD 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 -> c– 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, Dand Dall 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.

MonthSale of unitsTotal Commission (in Rs)
ABC
Jan9010020              800
Feb1305040              900
Mar6010030              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
C1C2C3
S1234
S2112
S3321

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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