RD Sharma Solutions for Class 12 Maths Exercise 6.2 Chapter 6 Determinants

RD Sharma Solutions for Class 12 Maths Exercise 6.2 Chapter 6 Determinants is given here. This exercise deals with the properties of determinants. Exercise 6.2 of Chapter 6 consists of problems based on the evaluation of determinants with proving the identities. The solutions prepared by our faculty are in an interactive manner to make it easy for the students to understand the concepts.

Students can refer to the PDF of RD Sharma Solutions as a major study material to improve their speed in solving problems accurately. Students can access the RD Sharma Solutions for Class 12 Maths Exercise 6.2 of Chapter 6 Determinants.

RD Sharma Solutions for Class 12 Chapter 6 Determinants Exercise 6.2

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Access other exercises of RD Sharma Solutions For Class 12 Chapter 6 – Determinants

Exercise 6.1 Solutions

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Access answers to Maths RD Sharma Solutions For Class 12 Chapter 6 – Determinants Exercise 6.2

Exercise 6.2 Page No: 6.57

1. Evaluate the following determinant:

Solution:

(i) Given

(ii) Given

= 1[(109) (12) – (119) (11)]

= 1308 – 1309

= – 1

So, Δ = – 1

(iii) Given,

= 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

= 2[1(24 – 4)] = 40

So, Δ = 40

(v) Given

= 1[(– 7) (– 36) – (– 20) (– 13)]

= 252 – 260

= – 8

So, Δ = – 8

(vi) Given,

(vii) Given

(viii) Given,

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

Solution:

(i) Given,

(ii) Given,

(iii) Given,

(iv) Given,

(v) Given,

(vi) Given,

(vii) Given,

(viii) Given,

(ix) Given,

As, C₁ = C₂, hence determinant is zero

(x) Given,

(xi) Given,

(xii) Given,

(xiii) Given,

(xiv) Given,

(xv) Given,

(xvi) Given,

(xvii) Given,

Hence proved.

Evaluate the following (3 – 9):

Solution:

Given,

= (a + b + c) (b – a) (c – a) (b – c)

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

Solution:

Given,

Solution:

Given,

Solution:

Given,

Solution:

Given,

Solution:

Given,

Solution:

Given,

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

= a² (a + x + y + z)

So, Δ = a² (a + x + y + z)

Solution:

Prove the following identities (11 – 45):

Solution:

Given,

Solution:

Consider,

= – (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.

Solution:

Given,

Solution:

Consider,

,

Solution:

Consider,

L.H.S =

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

Solution:

Consider,

Solution:

Consider,

Solution:

Consider,

Hence, the proof.

Solution:

Given,

= – 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.

Solution:

Consider,

= (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.

Solution:

Consider,

= [(2a + 4) (1) – (1) (2a + 6)]

= – 2

= R.H.S

Hence, the proof.

Solution:

Consider,

= – (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.

Solution:

Consider,

= R.H.S

Hence, the proof.

Solution:

Consider,

Solution:

Consider,

Solution:

Expanding the determinant along R₁, we have

Δ = 1[(1) (7) – (3) (2)] – 0 + 0

∴ Δ = 7 – 6 = 1

Thus,

Hence the proof.