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Home Mathematics Class 12 RD Sharma Solutions – Chapter 6 Determinants – Exercise 6.2 | Set 1

Class 12 RD Sharma Solutions – Chapter 6 Determinants – Exercise 6.2 | Set 1

Class 12 RD Sharma Solutions - Chapter 6 Determinants - Exercise 6.1

Class 12 RD Sharma Solutions - Chapter 6 Determinants - Exercise 6.2 | Set 2


Question 1. Evaluate the following determinant:

(i) 

Solution:

Considering the determinant, we have

As R1 and R2 are identical

Hence, β–³ = 0

(ii) 

Solution:

Considering the determinant, we have

C1β‡’C1 – 3C3

R3β‡’R3 + R2 and R1β‡’R1 + R2

R2β‡’R2 + 3R1

β–³ = 1(109 Γ— 40 – 119 Γ— 37)

Hence, β–³ = -43

(iii) 

Solution:

Considering the determinant, we have

β–³ = a(bc – f2) – h(hc – fg) + g(hf – gb)

β–³ = abc – af2 – h2c + fgh + fgh – g2b

Hence, β–³ = abc + 2fgh – af2 – ch2 – bg2

(iv) 

Solution:

Considering the determinant, we have

β–³ = 1(-2 – 10) + 3(8 – 6) + 2(20 + 3)

β–³ = 1(-12) + 3(2) + 2(23)

β–³ = -12 + 6 + 46

Hence, β–³ = 40

(v) 

Solution:

Considering the determinant, we have

β–³ = 1(225-256) + 4(100-144) + 9(64-81)

β–³ = 1(-31) – 4(-44) + 9(-17)

β–³ = -31 + 176 – 153

Hence, β–³ = -8

(vi) 

Solution:

Considering the determinant, we have

Taking -2 common from C1, C2 and C3

As C1 and C2 are identical

Hence, β–³ = 0

(vii) 

Solution:

Considering the determinant, we have

C1β‡’C1 + C2 + C3

C2β‡’C2 – C1

C3β‡’C3 – C1

C4β‡’C4 – C1

Taking 2, -2 and -2 common from C1, C2 and C3

Taking 4 common from R2 and R1β‡’R1+3R3

β–³ = (1 + 3 + 32 + 33)(4)(8)[40(9 – (-1))]

β–³ = (40)(4)(8)[40(9 + 1)]

β–³ = 40 Γ— 4 Γ— 8 Γ— 40 Γ— 10

Hence, β–³ = 512000

(viii) 

Solution:

Considering the determinant, we have

Taking 6 common from R1, we get

As R1 and R3 are identical

Hence, β–³ = 0

Question 2. Without expanding, show that the values of each of the following determinants are zero:

(i) 

Solution:

Considering the determinant, we have

Taking 4 common from C1, we get

As C1 and C2 are identical

Hence, β–³ = 0

(ii) 

Solution:

Considering the determinant, we have

Taking -2 common from C1, we get

As C1 and C2 are identical

Hence, β–³ = 0

(iii) 

Solution:

Considering the determinant, we have

R3β‡’R3 – R2

As R1 and R3 are identical

Hence, β–³ = 0

(iv) 

Solution:

Considering the determinant, we have

Multiplying and dividing β–³ by abc, we get

Multiplying R1, R2 and R3 by a, b and c respectively

Taking abc common from C3, we get

As C2 and C3 are identical

Hence, β–³ = 0

(v) 

Solution:

Considering the determinant, we have

C3β‡’C3 – C2 and C2β‡’C2 – C1

As C2 and C3 are identical

Hence, β–³ = 0

(vi) 

Solution:

Considering the determinant, we have

Splitting the determinant, we have

R2β‡’R2-R1 and R3β‡’R3-R1

Taking (b-a) and (c-a) common from R2 and R3, we have

β–³ = (b – a)(c – a)(c + a – (b + a)) – (b – a)(c – a)(-b – (-c))

β–³ = (b – a)(c – a)(c + a – b – a) – (b – a)(c – a)(-b + c)

β–³ = (b – a)(c – a)(c – b) – (b – a)(c – a)(c – b)

Hence, β–³ = 0

(vii) 

Solution:

Considering the determinant, we have

C1β‡’C1 – 8C3

As C1 and C2 are identical

Hence, β–³ = 0

(viii) 

Solution:

Considering the determinant, we have

Multiplying and dividing by xyz, we have

Multiplying C1, C2 and C3 by z, y and x respectively

Taking y, x and z common in R1, R2 and R3 respectively

C2β‡’C2 – C3

As C1 and C2 are identical

Hence, β–³ = 0

(ix) 

Solution:

Considering the determinant, we have

C2β‡’C2 – 7C3

As C1 and C2 are identical

Hence, β–³ = 0

(x) 

Solution:

Considering the determinant, we have

C3β‡’C3 – C2 and C4β‡’C4 – C1

Taking 3 common from C3, we get

As C3 and C4 are identical

Hence, β–³ = 0

(xi) 

Solution:

Considering the determinant, we have

R3β‡’R3 + R1 and R2β‡’R2 + R1

Taking 2 common from R2, we get

As R2 and R3 are identical

Hence, β–³ = 0

Question 3. 

Solution:

Considering the determinant, we have

C2β‡’C2+C1

Taking (a+b+c) common from C2, we get

R3β‡’R3-R1 and R2β‡’R2-R1

Taking (b – a) and (c – a) from R2 and R3, we have

β–³ = (a + b + c)(b – a)(c – a)[1(b + a – (c + a))]

β–³ = (a + b + c)(b – a)(c – a)(b + a – c – a)

Hence, β–³ = (a + b + c)(b – a)(c – a)(b – c)

Question 4. 

Solution:

Considering the determinant, we have

R3β‡’R3-R1 and R2β‡’R2-R1

Taking (b-a) and (c-a) from R2 and R3, we have

β–³ = (b – a)(c – a)[1((1)(-b) – (1)(-c))]

β–³ = (b – a)(c – a)[-b – (-c)]

β–³ = (b – a)(c – a)[-b + c]

Hence, β–³ = (a – b)(b – c)(c – a)

Question 5. 

Solution:

Considering the determinant, we have

C1β‡’C1+C2+C3

Taking (3x+Ξ») common from C1, we get

R3β‡’R3-R1 and R2β‡’R2-R1

β–³ = (3x + Ξ»)[Ξ»(Ξ»(1) – 0)]

β–³ = (3x + Ξ»)[Ξ»(Ξ»)]

Hence, β–³ = Ξ»2(3x + Ξ»)

Question 6. 

Solution:

Considering the determinant, we have

C1β‡’C1 + C2 + C3

Taking (a + b + c) common from C1, we get

R3β‡’R3 – R1 and R2β‡’R2 – R1

β–³ = (a + b + c)[1((a – b)(a – c) – (c – b)(b – c))]

β–³ = (a + b + c)[(a2 – ac – ab + bc) – (cb – c2 – b2 + bc)]

β–³ = (a + b + c)[a2 – ac – ab + bc + c2 + b2 – 2bc]

Henfce, β–³ = (a + b + c)[a2 + b2 + c2 – ac – ab – bc]

Question 7. 

Solution:

Considering the determinant, we have

C2β‡’C2 – C1

Using the trigonometric identity,

cos a cos b – sin a sin b =  cos (a + b)

As C2 and C3 are identical

Hence, β–³ = 0

Prove the following identities:

Question 8. = a3 + b2 + c3 – 3abc

Solution:

Considering the determinant, we have

R3β‡’R3 + R1 and R2β‡’R2 + R1

Taking (a + b + c) common from R3, we get

R2β‡’R2 – R1

Taking (-1) common from R2, we get

C1β‡’C1 – C2 and C2β‡’C2 – C3

β–³ = (-1)(a + b + c)[1((a – b)(c – a) – (b – c)(b – c))]

β–³ = (-1)(a + b + c)[(a – b)(c – a) – (b – c)2]

β–³ = (-1)(a + b + c)[(ac – a2 – bc + ab) – (b2 – 2cb + c2)]

β–³ = (-1)(a + b + c)(ac – a2 – bc + ab – b2 + 2cb – c2)

β–³ = (a + b + c)(-ac + a2 – bc – ab + b2 + c2)

β–³ = (a + b + c)(a2 + b2 + c2 – ac – ab – cb)

β–³ = a3 + b3 + c3 – 3abc

Hence proved 

Question 9. = 3abc – a3 – b2 – c3

Solution:

Considering the determinant, we have

C1β‡’C1 + C3

Taking (a + b + c) common from C1, we get

β–³ = (a + b + c)[1((b – c)c – b(c – a)) – 1((a – b)c – a(c – a)) + 1(b(a – b) – a(b – c))]

β–³ = (a + b + c)[(b – c)c – b(c – a) – (a – b)c + a(c – a) + b(a – b) – a(b – c)]

β–³ = (a + b + c)[(bc – c2-bc + ab) – (ac – bc) + ac – a2 + ab – b2 – (ab – ac)]

β–³ = (a + b + c)[bc – c2 – bc + ab – ac + bc + ac – a2 + ab – b2 – ab + ac]

β–³ = (a + b + c)[bc – c2 + ab + ac – a2 – b2]

β–³ = (a + b + c)[bc + ab + ac – a2 – b2 – c2]

β–³ = 3abc – a3 – b3 – c3

Hence proved 

Question 10. 

Solution:

Considering the determinant, we have

C1β‡’C1 + C2 + C3

Taking 2 common from C1, we get

C2β‡’C2 – C1 and C3β‡’C3 – C1

Taking (-1) and (-1) common from C2 and C3,

By splitting the determinant, we get

Hence proved 

Question 11.  = 2(a + b + c)3

Considering the determinant, we have

C1β‡’C1 + C2 + C3

Taking (2a + 2b + 2c) common from C1, we get

R2β‡’R2 – R1 and R3β‡’R3 – R1

β–³ = 2(a + b + c)[1((a + b + b)(a + b + c) – 0)]

β–³ = 2(a + b + c)[(a + b + b)2]

β–³ = 2(a + b + c)3

Hence proved 

Question 12. = (a + b + c)3

Solution:

Considering the determinant, we have

R1β‡’R1 + R2 + R3

Taking (a + b + c) common from R1, we get

C2β‡’C2 – C1 and C3β‡’C3 – C1

β–³ = (a + b + c)[1((-b – c – a)(-b – c – a) – 0)]

β–³ = (a + b + c)[(b + c + a)(b + c + a)]

β–³ = (a + b + c)[(b + c + a)2]

β–³ = (a + b + c)3

Hence proved 

Question 13.  = (a – b)(b – c)(c – a)

Solution:

Considering the determinant, we have

R2β‡’R2 – R1 and R3β‡’R3 – R1

Taking (a – b) and (a – c) common from R2 and R3 respectively, we get

β–³ = (a – b)(a – c)[1(1(a + c) – 1(a + b))]

β–³ = (a – b)(a – c)[(a + c) – (a + b)]

β–³ = (a – b)(a – c)[a + c – a – b]

β–³ = (a – b)(a – c)

β–³ = (a – b)(a – c)(c – b)

β–³ = (a – b)(b – c)(c – a)

Hence proved 

Question 14. = 9(a + b)b2

Solution:

Considering the determinant, we have

C1β‡’C1 + C2 + C3

Taking (3a + 3b) common from R1, we get

C2β‡’C2 – C1 and C3β‡’C3 – C2

β–³ = 3(a + b)[1((-2b)(-2b) – b(b))]

β–³ = 3(a + b)[4b2 – b2]

β–³ = 3(a + b)[3b2]

β–³ = 9(a + b)b2

Hence proved 

Question 15. 

Solution:

Considering the determinant, we have

R1β‡’aR1, R2β‡’bR2 and R3β‡’cR3

Taking (abc) common from C3, we get

Hence proved 

Question 16. = xyz(x – y)(y – z)(z – x)(x + y + z)

Solution:

Considering the determinant, we have

C1↔C2 and then 

C2↔C3

R1↔R2

R2↔R3

Taking,

Taking x, y and z common from C1, C2 and C3 respectively

C1β‡’C1 – C2 and C3β‡’C3 – C2

Taking (x – y) and (z – y) common from C1 and C3 respectively, we get

β–³ = (xyz)(x – y)(z – y)[1(1(z2 + zy + y2) – 1(x2 + xy + y2))]

β–³ = (xyz)(x – y)(z – y)[z2 + zy + y2 – (x2 + xy + y2)]

β–³ = (xyz)(x – y)(z – y)[z2 + zy + y2 – x2 – xy – y2]

β–³ = (xyz)(x – y)(z – y)[z2 + zy – x2 – xy]

β–³ = (xyz)(x – y)(z – y)[z2 – x2 + zy – xy]

β–³ = (xyz)(x – y)(z – y)[(z – x)(z + x) + y(z – x)]

β–³ = (xyz)(x – y)(z – y)(z – x)[z + x + y]

β–³ = (xyz)(x – y)(z – y)(z – x)(x + y + z)

Hence proved 

Question 17. = (a – b)(b – c)(c – a)(a + b + c)(a2 + b2 + c2)

Solution:

Considering the determinant, we have

C1β‡’C1 + C2 – 2C3

Taking (a2 + b2 + c2) common from C1, we get

C2β‡’C2-C1 and C3β‡’C3-C1

Taking (b – a) and (c – a) common from R2 and R3, we get

β–³ = (a2 + b2 + c2)(b – a)(c – a)[1((b + a)(-b) – (c + a)(-c))]

β–³ = (a2 + b2 + c2)(b – a)(c – a)[(b + a)(-b) + (c + a)c]

β–³ = (a2 + b2 + c2)(b – a)(c – a)[(-b2 – ab) + (c2 + ac)]

β–³ = (a2 + b2 + c2)(b – a)(c – a)

β–³ = (a2 + b2 + c2)(b – a)(c – a)[(c – b)(c + b) + a(c – b)]

β–³ = (a2 + b2 + c2)(b – a)(c – a)(c – b)

β–³ = (a2 + b2 + c2)(a + b + c)(a – b)(b – c)(c – a)

Hence proved 



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#Maths-Class-12 #RD Sharma Class-12 #Class 12 #Mathematics #RD Sharma Solutions #School Learning

Class 12 RD Sharma Solutions - Chapter 6 Determinants - Exercise 6.1

Class 12 RD Sharma Solutions - Chapter 6 Determinants - Exercise 6.2 | Set 2