RD Sharma Solutions For Class 12 Maths Exercise 6.1 Chapter 6 Determinants

The determinant of a square matrix of order 1, 2, 3 and 4, singular matrix, minors and cofactors of given determinants are the main topics which are explained under exercise 6.1. This exercise can be used as a model of reference by the students to improve their conceptual knowledge and understand the different ways used to solve the problems. The PDF of RD Sharma Solutions for Class 12 Maths Exercise 6.1 is available here. These solutions are formulated by BYJU’S experts in Maths to increase students confidence, which plays a crucial role in their board exams.

RD Sharma Solutions For Class 12 Chapter 6 Determinants Exercise 6.1Download PDF Here

RD Sharma Class 12 Maths Solutions Chapter 6 Determinants Exercise 6.1
RD Sharma Class 12 Maths Solutions Chapter 6 Determinants Exercise 6.1 1
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Access answers to Maths RD Sharma Solutions For Class 12 Chapter 6 – Determinants Exercise 6.1

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:

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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
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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:

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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:

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Solution:

Since |AB|= |A||B|

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= 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

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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

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