RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants

RD Sharma Solutions are primarily designed for CBSE students and are based on the latest syllabus prescribed as per the CCE guidelines by the CBSE Board. The practice is an essential task to learn and score well in Mathematics. Hence the solutions are designed by BYJU’S experts to boost confidence among students in understanding the concepts covered in this chapter. The PDF of RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants is provided here. The students are required to go through RD Sharma Solutions thoroughly before the final exams to score well and intensify their problem-solving abilities. This chapter consists of five exercises and explains the concepts of determinants and its properties.

Let us have a look at some of the important concepts that are discussed in this chapter.

  • Definition of determinants
  • The determinant of a square matrix of order 1, 2 and 3
  • The determinant of a square matrix of order 3 by using Sarrus diagram
  • Definition and meaning of singular matrix
  • Minors and cofactors of determinants
  • Properties of determinants
  • Evaluation of determinants
  • Evaluation of determinants by using the properties of determinants and proving identities
  • The solution of determinant equations
  • Addition of determinants
  • Evaluation of determinants by using factor theorem
  • Applications of determinants to coordinate geometry
    • Area of triangle
    • Conditions of collinearity of three points
    • Equation of a line passing through two given points
  • Applications of determinants in solving a system of linear equations
    • The solution of a non-homogeneous system of linear equations
    • Condition for consistency
    • Solutions of a homogeneous system of linear equations

RD Sharma Solutions For Class 12 Maths Chapter 6 Determinants:-Download PDF Here

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Access answers to Maths RD Sharma Solutions For Class 12 Chapter 6 – Determinants

Exercise 6.1 Page No: 6.10

1. Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case:

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 1
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image
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RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 6
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 7

Solution:

(i) Let Mij and Cij represents the minor and co–factor of an element, where i and j represent the row and column.The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, Cij = (–1)i+j × Mij

Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 8

From the given matrix we have,

M11 = –1

M21 = 20

C11 = (–1)1+1 × M11

= 1 × –1

= –1

C21 = (–1)2+1 × M21

= 20 × –1

= –20

Now expanding along the first column we get

|A| = a11 × C11 + a21× C21

= 5× (–1) + 0 × (–20)

= –5

(ii) Let Mij and Cij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, Cij = (–1)i+j × Mij

Given

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 9

From the above matrix we have

M11 = 3

M21 = 4

C11 = (–1)1+1 × M11

= 1 × 3

= 3

C21 = (–1)2+1 × 4

= –1 × 4

= –4

Now expanding along the first column we get

|A| = a11 × C11 + a21× C21

= –1× 3 + 2 × (–4)

= –11

(iii) Let Mij and Cij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, Cij = (–1)i+j × Mij

Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 10

M31 = –3 × 2 – (–1) × 2

M31 = –4

C11 = (–1)1+1 × M11

= 1 × –12

= –12

C21 = (–1)2+1 × M21

= –1 × –16

= 16

C31 = (–1)3+1 × M31

= 1 × –4

= –4

Now expanding along the first column we get

|A| = a11 × C11 + a21× C21+ a31× C31

= 1× (–12) + 4 × 16 + 3× (–4)

= –12 + 64 –12

= 40

(iv) Let Mij and Cij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, Cij = (–1)i+j × Mij

Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 11

M31 = a × c a – b × bc

M31 = a2c – b2c

C11 = (–1)1+1 × M11

= 1 × (ab2 – ac2)

= ab2 – ac2

C21 = (–1)2+1 × M21

= –1 × (a2b – c2b)

= c2b – a2b

C31 = (–1)3+1 × M31

= 1 × (a2c – b2c)

= a2c – b2c

Now expanding along the first column we get

|A| = a11 × C11 + a21× C21+ a31× C31

= 1× (ab2 – ac2) + 1 × (c2b – a2b) + 1× (a2c – b2c)

= ab2 – ac2 + c2b – a2b + a2c – b2c

(v) Let Mij and Cij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, Cij = (–1)i+j × Mij

Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 12

M31 = 2×0 – 5×6

M31 = –30

C11 = (–1)1+1 × M11

= 1 × 5

= 5

C21 = (–1)2+1 × M21

= –1 × –40

= 40

C31 = (–1)3+1 × M31

= 1 × –30

= –30

Now expanding along the first column we get

|A| = a11 × C11 + a21× C21+ a31× C31

= 0× 5 + 1 × 40 + 3× (–30)

= 0 + 40 – 90

= 50

(vi) Let Mij and Cij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, Cij = (–1)i+j × Mij

Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 13

M31 = h × f – b × g

M31 = hf – bg

C11 = (–1)1+1 × M11

= 1 × (bc– f2)

= bc– f2

C21 = (–1)2+1 × M21

= –1 × (hc – fg)

= fg – hc

C31 = (–1)3+1 × M31

= 1 × (hf – bg)

= hf – bg

Now expanding along the first column we get

|A| = a11 × C11 + a21× C21+ a31× C31

= a× (bc– f2) + h× (fg – hc) + g× (hf – bg)

= abc– af2 + hgf – h2c +ghf – bg2

(vii) Let Mij and Cij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, Cij = (–1)i+j × Mij

Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 14

M31 = –1(1 × 0 – 5 × (–2)) – 0(0 × 0 – (–1) × (–2)) + 1(0 × 5 – (–1) × 1)

M31 = –9

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 15

M41 = –1(1×1 – (–1) × (–2)) – 0(0 × 1 – 1 × (–2)) + 1(0 × (–1) – 1 × 1)

M41 = 0

C11 = (–1)1+1 × M11

= 1 × (–9)

= –9

C21 = (–1)2+1 × M21

= –1 × 9

= –9

C31 = (–1)3+1 × M31

= 1 × –9

= –9

C41 = (–1)4+1 × M41

= –1 × 0

= 0

Now expanding along the first column we get

|A| = a11 × C11 + a21× C21+ a31× C31 + a41× C41

= 2 × (–9) + (–3) × –9 + 1 × (–9) + 2 × 0

= – 18 + 27 –9

= 0

2. Evaluate the following determinants:

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 16

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 17

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 18

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 19

Solution:

(i) Given

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 20

⇒ |A| = x (5x + 1) – (–7) x

|A| = 5x2 + 8x

(ii) Given

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 21

⇒ |A| = cos θ × cos θ – (–sin θ) x sin θ

|A| = cos2θ + sin2θ

We know that cos2θ + sin2θ = 1

|A| = 1

(iii) Given

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 22

⇒ |A| = cos15° × cos75° + sin15° x sin75°

We know that cos (A – B) = cos A cos B + Sin A sin B

By substituting this we get, |A| = cos (75 – 15)°

|A| = cos60°

|A| = 0.5

(iv) Given

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 23

⇒ |A| = (a + ib) (a – ib) – (c + id) (–c + id)

= (a + ib) (a – ib) + (c + id) (c – id)

= a2 – i2 b2 + c2 – i2 d2

We know that i2 = -1

= a2 – (–1) b2 + c2 – (–1) d2

= a2 + b2 + c2 + d2

3. Evaluate:

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 24

Solution:

Since |AB|= |A||B|

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 25

= 2(17 × 12 – 5 × 20) – 3(13 × 12 – 5 × 15) + 7(13 × 20 – 15 × 17)

= 2 (204 – 100) – 3 (156 – 75) + 7 (260 – 255)

= 2×104 – 3×81 + 7×5

= 208 – 243 +35

= 0

Now |A|2 = |A|×|A|

|A|2= 0

4. Show that

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 26

Solution:

Given

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 27

Let the given determinant as A

Using sin (A+B) = sin A × cos B + cos A × sin B

⇒ |A| = sin 10° × cos 80° + cos 10° x sin 80°

|A| = sin (10 + 80)°

|A| = sin90°

|A| = 1

Hence Proved

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 28

Solution:

Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 29

= 2(1 × 1 – 4 × (–2)) – 3(7 × 1 – (–2) × (–3)) – 5(7 × 4 – 1 × (–3))

= 2(1 + 8) – 3(7 – 6) – 5(28 + 3)

= 2 × 9 – 3 × 1 – 5 × 31

= 18 – 3 – 155

= –140

Now by expanding along the second column

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 30

= 2(1 × 1 – 4 × (–2)) – 7(3 × 1 – 4 × (–5)) – 3(3 × (–2) – 1 × (–5))

= 2 (1 + 8) – 7 (3 + 20) – 3 (–6 + 5)

= 2 × 9 – 7 × 23 – 3 × (–1)

= 18 – 161 +3

= –140

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 31

Solution:

Given

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 32

⇒ |A| = 0 (0 – sinβ (–sinβ)) –sinα (–sinα × 0 – sinβ cosα) – cosα ((–sinα) (–sinβ) – 0 × cosα)

|A| = 0 + sinα sinβ cosα – cosα sinα sinβ

|A| = 0

Exercise 6.2 Page No: 6.57

1. Evaluate the following determinant:

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 33

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 34

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 35

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RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 38

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Solution:

(i) Given

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RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 42

(ii) Given

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 43

= 1[(109) (12) – (119) (11)]

= 1308 – 1309

= – 1

So, Δ = – 1

(iii) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 44

= a (bc – f2) – h (hc – fg) + g (hf – bg)

= abc – af2 – ch2 + fgh + fgh – bg2

= abc + 2fgh – af2 – bg2 – ch2

So, Δ = abc + 2fgh – af2 – bg2 – ch2

(iv) Given

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 45

= 2[1(24 – 4)] = 40

So, Δ = 40

(v) Given

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 46
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 47

= 1[(– 7) (– 36) – (– 20) (– 13)]

= 252 – 260

= – 8

So, Δ = – 8

(vi) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 48

(vii) Given

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 49
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 50
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 51

(viii) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 52

2. Without expanding, show that the value of each of the following determinants is zero:

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 53

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 54

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RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 67

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Solution:

(i) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 70

(ii) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 71
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 72

(iii) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 73

(iv) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 74
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 75

(v) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 76

(vi) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 77
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 78

(vii) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 79

(viii) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 80

(ix) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 81

As, C1 = C2, hence determinant is zero

(x) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 82

(xi) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 83
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 84

(xii) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 85

(xiii) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 86

(xiv) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 87
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 88

(xv) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 89
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 90

(xvi) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 91
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 92

(xvii) Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 93

Hence proved.

Evaluate the following (3 – 9):

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 94

Solution:

Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 95

= (a + b + c) (b – a) (c – a) (b – c)

So, Δ = (a + b + c) (b – a) (c – a) (b – c)

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 96

Solution:

Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 97
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 98

Solution:

Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 99
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 100

Solution:

Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 101
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 102

Solution:

Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 103
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 104

Solution:

Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 105
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 106

Solution:

Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 107
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 108

= a [a (a + x + y) + az] + 0 + 0

= a2 (a + x + y + z)

So, Δ = a2 (a + x + y + z)

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 109

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 110

Prove the following identities (11 – 45):

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 111

Solution:

Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 112
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 113

Solution:

Consider,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 114
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 115

= – (a + b + c) [(b – c) (a + b – 2c) – (c – a) (c + a – 2b)]

= 3abc – a3 – b3 – c3

Therefore, L.H.S = R.H.S,

Hence the proof.

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 116

Solution:

Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 117
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 118
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 119

Solution:

Consider,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 120,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 121
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 122

Solution:

Consider,

L.H.S = 
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 123

Now by applying, R1→R1 + R2 + R3, we get,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 124
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 125

Solution:

Consider,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 126
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 127
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 128

Solution:

Consider,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 129
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 130
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 131

Solution:

Consider,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 132
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 133

Hence, the proof.

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 134

Solution:

Given,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 135

= – xyz(x – y) (z – y) [z2 + y2 + zy – x2 – y2 – xy]

= – xyz(x – y) (z – y) [(z – x) (z + x0 + y (z – x)]

= – xyz(x – y) (z – y) (z – x) (x + y + z)

= R.H.S

Hence, the proof.

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 136

Solution:

Consider,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 137
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 138

= (a2 + b2 + c2) (b – a) (c – a) [(b + a) (– b) – (– c) (c + a)]

= (a2 + b2 + c2) (a – b) (c – a) (b – c) (a + b + c)

= R.H.S

Hence, the proof.

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 139

Solution:

Consider,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 140

= [(2a + 4) (1) – (1) (2a + 6)]

= – 2

= R.H.S

Hence, the proof.

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 141

Solution:

Consider,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 142
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 143

= – (a2 + b2 + c2) (a – b) (c – a) [(– (b + a)) (– b) – (c) (c + a)]

= (a – b) (b – c) (c – a) (a + b + c) (a2 + b2 + c2)

= R.H.S

Hence, the proof.

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 144

Solution:

Consider,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 145

= R.H.S

Hence, the proof.

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 146

Solution:

Consider,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 147
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 148
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 149

Solution:

Consider,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 150
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 151
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 152

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 153
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 154
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 155

Expanding the determinant along R1, we have

Δ = 1[(1) (7) – (3) (2)] – 0 + 0

∴ Δ = 7 – 6 = 1

Thus, 
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 156

Hence the proof.

Exercise 6.3 Page No: 6.71

1. Find the area of the triangle with vertices at the points:

(i) (3, 8), (-4, 2) and (5, -1)

(ii) (2, 7), (1, 1) and (10, 8)

(iii) (-1, -8), (-2, -3) and (3, 2)

(iv) (0, 0), (6, 0) and (4, 3)

Solution:

(i) Given (3, 8), (-4, 2) and (5, -1) are the vertices of the triangle.

We know that, if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by:

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 157

(ii) Given (2, 7), (1, 1) and (10, 8) are the vertices of the triangle.

We know that if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by:

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 158

(iii) Given (-1, -8), (-2, -3) and (3, 2) are the vertices of the triangle.

We know that if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by:

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 159
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 160

As we know area cannot be negative. Therefore, 15 square unit is the area

Thus area of triangle is 15 square units

(iv) Given (-1, -8), (-2, -3) and (3, 2) are the vertices of the triangle.

We know that if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by:

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 161

2. Using the determinants show that the following points are collinear:

(i) (5, 5), (-5, 1) and (10, 7)

(ii) (1, -1), (2, 1) and (10, 8)

(iii) (3, -2), (8, 8) and (5, 2)

(iv) (2, 3), (-1, -2) and (5, 8)

Solution:

(i) Given (5, 5), (-5, 1) and (10, 7)

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 162

(ii) Given (1, -1), (2, 1) and (10, 8)

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 163

(iii) Given (3, -2), (8, 8) and (5, 2)

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 164

Now, by substituting given value in above formula

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 165

Since, Area of triangle is zero

Hence, points are collinear.

(iv) Given (2, 3), (-1, -2) and (5, 8)

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 166
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 167

3. If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab

Solution:

Given (a, 0), (0, b) and (1, 1) are collinear

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 168

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 169

⇒ a + b = ab

Hence Proved

4. Using the determinants prove that the points (a, b), (a’, b’) and (a – a’, b – b) are collinear if a b’ = a’ b.

Solution:

Given (a, b), (a’, b’) and (a – a’, b – b) are collinear

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 170

⇒ a b’ = a’ b

Hence, the proof.

5. Find the value of λ so that the points (1, -5), (-4, 5) and (λ, 7) are collinear.

Solution:

Given (1, -5), (-4, 5) and (λ, 7) are collinear

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 171

⇒ – 50 – 10λ = 0

⇒ λ = – 5

6. Find the value of x if the area of ∆ is 35 square cms with vertices (x, 4), (2, -6) and (5, 4).

Solution:

Given (x, 4), (2, -6) and (5, 4) are the vertices of a triangle.

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 172
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 173

⇒ [x (– 10) – 4(– 3) + 1(8 – 30)] = ± 70

⇒ [– 10x + 12 + 38] = ± 70

⇒ ±70 = – 10x + 50

Taking positive sign, we get

⇒ + 70 = – 10x + 50

⇒ 10x = – 20

⇒ x = – 2

Taking –negative sign, we get

⇒ – 70 = – 10x + 50

⇒ 10x = 120

⇒ x = 12

Thus x = – 2, 12

Exercise 6.4 Page No: 6.84

Solve the following system of linear equations by Cramer’s rule:

1. x – 2y = 4

-3x + 5y = -7

Solution:

Given x – 2y = 4

-3x + 5y = -7

Let there be a system of n simultaneous linear equations and with n unknown given by

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 174
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 175

Solving determinant, expanding along 1st row

⇒ D = 5(1) – (– 3) (– 2)

⇒ D = 5 – 6

⇒ D = – 1

Again,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 176

Solving determinant, expanding along 1st row

⇒ D1 = 5(4) – (– 7) (– 2)

⇒ D1 = 20 – 14

⇒ D1 = 6

And

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 177

Solving determinant, expanding along 1st row

⇒ D2 = 1(– 7) – (– 3) (4)

⇒ D2 = – 7 + 12

⇒ D2 = 5

Thus by Cramer’s Rule, we have

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 178

2. 2x – y = 1

7x – 2y = -7

Solution:

Given 2x – y = 1 and

7x – 2y = -7

Let there be a system of n simultaneous linear equations and with n unknown given by

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 179

Solving determinant, expanding along 1st row

⇒ D1 = 1(– 2) – (– 7) (– 1)

⇒ D1 = – 2 – 7

⇒ D1 = – 9

And

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 180

Solving determinant, expanding along 1st row

⇒ D2 = 2(– 7) – (7) (1)

⇒ D2 = – 14 – 7

⇒ D2 = – 21

Thus by Cramer’s Rule, we have

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 181

3. 2x – y = 17

3x + 5y = 6

Solution:

Given 2x – y = 17 and

3x + 5y = 6

Let there be a system of n simultaneous linear equations and with n unknown given by

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 182
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 183

Solving determinant, expanding along 1st row

⇒ D1 = 17(5) – (6) (– 1)

⇒ D1 = 85 + 6

⇒ D1 = 91

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 184

Solving determinant, expanding along 1st row

⇒ D2 = 2(6) – (17) (3)

⇒ D2 = 12 – 51

⇒ D2 = – 39

Thus by Cramer’s Rule, we have

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 185

4. 3x + y = 19

3x – y = 23

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 186

Solving determinant, expanding along 1st row

⇒ D = 3(– 1) – (3) (1)

⇒ D = – 3 – 3

⇒ D = – 6

Again,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 187

Solving determinant, expanding along 1st row

⇒ D1 = 19(– 1) – (23) (1)

⇒ D1 = – 19 – 23

⇒ D1 = – 42

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 188

Solving determinant, expanding along 1st row

⇒ D2 = 3(23) – (19) (3)

⇒ D2 = 69 – 57

⇒ D2 = 12

Thus by Cramer’s Rule, we have

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 189

5. 2x – y = -2

3x + 4y = 3

Solution:

Given 2x – y = -2 and

3x + 4y = 3

Let there be a system of n simultaneous linear equations and with n unknown given by

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 190
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 191
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 192

Solving determinant, expanding along 1st row

⇒ D2 = 3(2) – (– 2) (3)

⇒ D2 = 6 + 6

⇒ D2 = 12

Thus by Cramer’s Rule, we have

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 193

6. 3x + ay = 4

2x + ay = 2, a ≠ 0

Solution:

Given 3x + ay = 4 and

2x + ay = 2, a ≠ 0

Let there be a system of n simultaneous linear equations and with n unknown given by

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 194
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 195

3x + ay = 4

2x + ay = 2, a≠0

So by comparing with the theorem, let’s find D, D1 and D2

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 196

Solving determinant, expanding along 1st row

⇒ D = 3(a) – (2) (a)

⇒ D = 3a – 2a

⇒ D = a

Again,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 197

Solving determinant, expanding along 1st row

⇒ D1 = 4(a) – (2) (a)

⇒ D = 4a – 2a

⇒ D = 2a

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 198

Solving determinant, expanding along 1st row

⇒ D2 = 3(2) – (2) (4)

⇒ D = 6 – 8

⇒ D = – 2

Thus by Cramer’s Rule, we have

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 199

7. 2x + 3y = 10

x + 6y = 4

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 200

Solving determinant, expanding along 1st row

⇒ D = 2 (6) – (3) (1)

⇒ D = 12 – 3

⇒ D = 9

Again,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 201

Solving determinant, expanding along 1st row

⇒ D1 = 10 (6) – (3) (4)

⇒ D = 60 – 12

⇒ D = 48

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 202

Solving determinant, expanding along 1st row

⇒ D2 = 2 (4) – (10) (1)

⇒ D2 = 8 – 10

⇒ D2 = – 2

Thus by Cramer’s Rule, we have

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 203

8. 5x + 7y = -2

4x + 6y = -3

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 204
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 205

Now, here we have

5x + 7y = – 2

4x + 6y = – 3

So by comparing with the theorem, let’s find D, D1 and D2

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 206

Solving determinant, expanding along 1st row

⇒ D = 5(6) – (7) (4)

⇒ D = 30 – 28

⇒ D = 2

Again,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 207

Solving determinant, expanding along 1st row

⇒ D1 = – 2(6) – (7) (– 3)

⇒ D1 = – 12 + 21

⇒ D1 = 9

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 208

Solving determinant, expanding along 1st row

⇒ D2 = – 3(5) – (– 2) (4)

⇒ D2 = – 15 + 8

⇒ D2 = – 7

Thus by Cramer’s Rule, we have

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 209
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 210

9. 9x + 5y = 10

3y – 2x = 8

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 211
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 212

Solving determinant, expanding along 1st row

⇒ D = 3(9) – (5) (– 2)

⇒ D = 27 + 10

⇒ D = 37

Again,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 213

Solving determinant, expanding along 1st row

⇒ D1 = 10(3) – (8) (5)

⇒ D1 = 30 – 40

⇒ D1 = – 10

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 214

Solving determinant, expanding along 1st row

⇒ D2 = 9(8) – (10) (– 2)

⇒ D2 = 72 + 20

⇒ D2 = 92

Thus by Cramer’s Rule, we have

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 215

10. x + 2y = 1

3x + y = 4

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 216

Solving determinant, expanding along 1st row

⇒ D = 1(1) – (3) (2)

⇒ D = 1 – 6

⇒ D = – 5

Again,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 217

Solving determinant, expanding along 1st row

⇒ D1 = 1(1) – (2) (4)

⇒ D1 = 1 – 8

⇒ D1 = – 7

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 218

Solving determinant, expanding along 1st row

⇒ D2 = 1(4) – (1) (3)

⇒ D2 = 4 – 3

⇒ D2 = 1

Thus by Cramer’s Rule, we have

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 219

Solve the following system of linear equations by Cramer’s rule:

11. 3x + y + z = 2

2x – 4y + 3z = -1

4x + y – 3z = -11

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 220
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 221

Now, here we have

3x + y + z = 2

2x – 4y + 3z = – 1

4x + y – 3z = – 11

So by comparing with the theorem, let’s find D, D1, D2 and D3

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 222

Solving determinant, expanding along 1st row

⇒ D = 3[(– 4) (– 3) – (3) (1)] – 1[(2) (– 3) – 12] + 1[2 – 4(– 4)]

⇒ D = 3[12 – 3] – [– 6 – 12] + [2 + 16]

⇒ D = 27 + 18 + 18

⇒ D = 63

Again,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 223

Solving determinant, expanding along 1st row

⇒ D1 = 2[( – 4)( – 3) – (3)(1)] – 1[( – 1)( – 3) – ( – 11)(3)] + 1[( – 1) – ( – 4)( – 11)]

⇒ D1 = 2[12 – 3] – 1[3 + 33] + 1[– 1 – 44]

⇒ D1 = 2[9] – 36 – 45

⇒ D1 = 18 – 36 – 45

⇒ D1 = – 63

Again

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 224

Solving determinant, expanding along 1st row

⇒ D2 = 3[3 + 33] – 2[– 6 – 12] + 1[– 22 + 4]

⇒ D2 = 3[36] – 2(– 18) – 18

⇒ D2 = 126

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 225

Solving determinant, expanding along 1st row

⇒ D3 = 3[44 + 1] – 1[– 22 + 4] + 2[2 + 16]

⇒ D3 = 3[45] – 1(– 18) + 2(18)

⇒ D3 = 135 + 18 + 36

⇒ D3 = 189

Thus by Cramer’s Rule, we have

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 226

12. x – 4y – z = 11

2x – 5y + 2z = 39

-3x + 2y + z = 1

Solution:

Given,

x – 4y – z = 11

2x – 5y + 2z = 39

-3x + 2y + z = 1

Let there be a system of n simultaneous linear equations and with n unknown given by

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 227

Now, here we have

x – 4y – z = 11

2x – 5y + 2z = 39

– 3x + 2y + z = 1

So by comparing with theorem, now we have to find D, D1 and D2

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 228

Solving determinant, expanding along 1st row

⇒ D = 1[(– 5) (1) – (2) (2)] + 4[(2) (1) + 6] – 1[4 + 5(– 3)]

⇒ D = 1[– 5 – 4] + 4[8] – [– 11]

⇒ D = – 9 + 32 + 11

⇒ D = 34

Again,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 229

Solving determinant, expanding along 1st row

⇒ D1 = 11[(– 5) (1) – (2) (2)] + 4[(39) (1) – (2) (1)] – 1[2 (39) – (– 5) (1)]

⇒ D1 = 11[– 5 – 4] + 4[39 – 2] – 1[78 + 5]

⇒ D1 = 11[– 9] + 4(37) – 83

⇒ D1 = – 99 – 148 – 45

⇒ D1 = – 34

Again

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 230

Solving determinant, expanding along 1st row

⇒ D2 = 1[39 – 2] – 11[2 + 6] – 1[2 + 117]

⇒ D2 = 1[37] – 11(8) – 119

⇒ D2 = – 170

And,

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 231

Solving determinant, expanding along 1st row

⇒ D3 = 1[– 5 – (39) (2)] – (– 4) [2 – (39) (– 3)] + 11[4 – (– 5)(– 3)]

⇒ D3 = 1 [– 5 – 78] + 4 (2 + 117) + 11 (4 – 15)

⇒ D3 = – 83 + 4(119) + 11(– 11)

⇒ D3 = 272

Thus by Cramer’s Rule, we have

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 232

13. 6x + y – 3z = 5

x + 3y – 2z = 5

2x + y + 4z = 8

Solution:

Given

6x + y – 3z = 5

x + 3y – 2z = 5

2x + y + 4z = 8

Let there be a system of n simultaneous linear equations and with n unknown given by

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 233

Now, here we have

6x + y – 3z = 5

x + 3y – 2z = 5

2x + y + 4z = 8

So by comparing with theorem, now we have to find D , D1 and D2

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 234

Solving determinant, expanding along 1st Row

⇒ D = 6[(4) (3) – (1) (– 2)] – 1[(4) (1) + 4] – 3[1 – 3(2)]

⇒ D = 6[12 + 2] – [8] – 3[– 5]

⇒ D = 84 – 8 + 15

⇒ D = 91

Again, Solve D1 formed by replacing 1st column by B matrices

Here

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 235

Solving determinant, expanding along 1st Row

⇒ D1 = 5[(4) (3) – (– 2) (1)] – 1[(5) (4) – (– 2) (8)] – 3[(5) – (3) (8)]

⇒ D1 = 5[12 + 2] – 1[20 + 16] – 3[5 – 24]

⇒ D1 = 5[14] – 36 – 3(– 19)

⇒ D1 = 70 – 36 + 57

⇒ D1 = 91

Again, Solve D2 formed by replacing 1st column by B matrices

Here

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 236

Solving determinant

⇒ D2 = 6[20 + 16] – 5[4 – 2(– 2)] + (– 3)[8 – 10]

⇒ D2 = 6[36] – 5(8) + (– 3) (– 2)

⇒ D2 = 182

And, Solve D3 formed by replacing 1st column by B matrices

Here

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 237

Solving determinant, expanding along 1st Row

⇒ D3 = 6[24 – 5] – 1[8 – 10] + 5[1 – 6]

⇒ D3 = 6[19] – 1(– 2) + 5(– 5)

⇒ D3 = 114 + 2 – 25

⇒ D3 = 91

Thus by Cramer’s Rule, we have

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 238

14. x + y = 5

y + z = 3

x + z = 4

Solution:

Given x + y = 5

y + z = 3

x + z = 4

Let there be a system of n simultaneous linear equations and with n unknown given by

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 239

Let Dj be the determinant obtained from D after replacing the jth column by

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 240

Now, here we have

x + y = 5

y + z = 3

x + z = 4

So by comparing with theorem, now we have to find D, D1 and D2

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 241

Solving determinant, expanding along 1st Row

⇒ D = 1[1] – 1[– 1] + 0[– 1]

⇒ D = 1 + 1 + 0

⇒ D = 2

Again, Solve D1 formed by replacing 1st column by B matrices

Here

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 242

Solving determinant, expanding along 1st Row

⇒ D1 = 5[1] – 1[(3) (1) – (4) (1)] + 0[0 – (4) (1)]

⇒ D1 = 5 – 1[3 – 4] + 0[– 4]

⇒ D1 = 5 – 1[– 1] + 0

⇒ D1 = 5 + 1 + 0

⇒ D1 = 6

Again, Solve D2 formed by replacing 1st column by B matrices

Here

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 243

Solving determinant

⇒ D2 = 1[3 – 4] – 5[– 1] + 0[0 – 3]

⇒ D2 = 1[– 1] + 5 + 0

⇒ D2 = 4

And, Solve D3 formed by replacing 1st column by B matrices

Here

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 244

Solving determinant, expanding along 1st Row

⇒ D3 = 1[4 – 0] – 1[0 – 3] + 5[0 – 1]

⇒ D3 = 1[4] – 1(– 3) + 5(– 1)

⇒ D3 = 4 + 3 – 5

⇒ D3 = 2

Thus by Cramer’s Rule, we have

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 245

15. 2y – 3z = 0

x + 3y = -4

3x + 4y = 3

Solution:

Given

2y – 3z = 0

x + 3y = -4

3x + 4y = 3

Let there be a system of n simultaneous linear equations and with n unknown given by

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 246

Now, here we have

2y – 3z = 0

x + 3y = – 4

3x + 4y = 3

So by comparing with theorem, now we have to find D, D1 and D2

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 247

Solving determinant, expanding along 1st Row

⇒ D = 0[0] – 2[(0) (1) – 0] – 3[1 (4) – 3 (3)]

⇒ D = 0 – 0 – 3[4 – 9]

⇒ D = 0 – 0 + 15

⇒ D = 15

Again, Solve D1 formed by replacing 1st column by B matrices

Here

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 248

Solving determinant, expanding along 1st Row

⇒ D1 = 0[0] – 2[(0) (– 4) – 0] – 3[4 (– 4) – 3(3)]

⇒ D1 = 0 – 0 – 3[– 16 – 9]

⇒ D1 = 0 – 0 – 3(– 25)

⇒ D1 = 0 – 0 + 75

⇒ D1 = 75

Again, Solve D2 formed by replacing 2nd column by B matrices

Here

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 249

Solving determinant

⇒ D2 = 0[0] – 0[(0) (1) – 0] – 3[1 (3) – 3(– 4)]

⇒ D2 = 0 – 0 + (– 3) (3 + 12)

⇒ D2 = – 45

And, Solve D3 formed by replacing 3rd column by B matrices

Here

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 250

Solving determinant, expanding along 1st Row

⇒ D3 = 0[9 – (– 4) 4] – 2[(3) (1) – (– 4) (3)] + 0[1 (4) – 3 (3)]

⇒ D3 = 0[25] – 2(3 + 12) + 0(4 – 9)

⇒ D3 = 0 – 30 + 0

⇒ D3 = – 30

Thus by Cramer’s Rule, we have

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 251

16. 5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

Solution:

Given

5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

Let there be a system of n simultaneous linear equations and with n unknown given by

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 252

Now, here we have

5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

So by comparing with theorem, now we have to find D, D1 and D2

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 253

Solving determinant, expanding along 1st Row

⇒ D = 5[(– 8) (– 6) – (– 1) (2)] – 7[(– 6) (6) – 3(– 1)] + 1[2(6) – 3(– 8)]

⇒ D = 5[48 + 2] – 7[– 36 + 3] + 1[12 + 24]

⇒ D = 250 – 231 + 36

⇒ D = 55

Again, Solve D1 formed by replacing 1st column by B matrices

Here

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 254
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 255

Solving determinant, expanding along 1st Row

⇒ D1 = 11[(– 8) (– 6) – (2) (– 1)] – (– 7) [(15) (– 6) – (– 1) (7)] + 1[(15)2 – (7) (– 8)]

⇒ D1 = 11[48 + 2] + 7[– 90 + 7] + 1[30 + 56]

⇒ D1 = 11[50] + 7[– 83] + 86

⇒ D1 = 550 – 581 + 86

⇒ D1 = 55

Again, Solve D2 formed by replacing 2nd column by B matrices

Here

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 256

Solving determinant, expanding along 1st Row

⇒ D2 = 5[(15) (– 6) – (7) (– 1)] – 11 [(6) (– 6) – (– 1) (3)] + 1[(6)7 – (15) (3)]

⇒ D2 = 5[– 90 + 7] – 11[– 36 + 3] + 1[42 – 45]

⇒ D2 = 5[– 83] – 11(– 33) – 3

⇒ D2 = – 415 + 363 – 3

⇒ D2 = – 55

And, Solve D3 formed by replacing 3rd column by B matrices

Here

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 257

Solving determinant, expanding along 1st Row

⇒ D3 = 5[(– 8) (7) – (15) (2)] – (– 7) [(6) (7) – (15) (3)] + 11[(6)2 – (– 8) (3)]

⇒ D3 = 5[– 56 – 30] – (– 7) [42 – 45] + 11[12 + 24]

⇒ D3 = 5[– 86] + 7[– 3] + 11[36]

⇒ D3 = – 430 – 21 + 396

⇒ D3 = – 55

Thus by Cramer’s Rule, we have

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 258

Exercise 6.5 Page No: 6.89

Solve each of the following system of homogeneous linear equations:

1. x + y – 2z = 0

2x + y – 3z =0

5x + 4y – 9z = 0

Solution:

Given x + y – 2z = 0

2x + y – 3z =0

5x + 4y – 9z = 0

Any system of equation can be written in matrix form as AX = B

Now finding the Determinant of these set of equations,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 259

= 1(1 × (– 9) – 4 × (– 3)) – 1(2 × (– 9) – 5 × (– 3)) – 2(4 × 2 – 5 × 1)

= 1(– 9 + 12) – 1(– 18 + 15) – 2(8 – 5)

= 1 × 3 –1 × (– 3) – 2 × 3

= 3 + 3 – 6

= 0

Since D = 0, so the system of equation has infinite solution.

Now let z = k

⇒ x + y = 2k

And 2x + y = 3k

Now using the Cramer’s rule

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 260
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 261

2. 2x + 3y + 4z = 0

x + y + z = 0

2x + 5y – 2z = 0

Solution:

Given

2x + 3y + 4z = 0

x + y + z = 0

2x + 5y – 2z = 0

Any system of equation can be written in matrix form as AX = B

Now finding the Determinant of these set of equations,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 262
RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 263

= 2(1 × (– 2) – 1 × 5) – 3(1 × (– 2) – 2 × 1) + 4(1 × 5 – 2 × 1)

= 2(– 2 – 5) – 3(– 2 – 2) + 4(5 – 2)

= 1 × (– 7) – 3 × (– 4) + 4 × 3

= – 7 + 12 + 12

= 17

Since D ≠ 0, so the system of equation has infinite solution.

Therefore the system of equation has only solution as x = y = z = 0.

Also, access RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants

Exercise 6.1 Solutions

Exercise 6.2 Solutions

Exercise 6.3 Solutions

Exercise 6.4 Solutions

Exercise 6.5 Solutions

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