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• RD Sharma Class 12 Solutions for Maths

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

(i)

Solution:

Considering the determinant, we have

As R1 and R2 are identical

Hence, △ = 0

(ii)

Solution:

Considering the determinant, we have

C1⇢C1 – 3C3

R3⇢R3 + R2 and R1⇢R1 + R2

R2⇢R2 + 3R1

△ = 1(109 × 40 – 119 × 37)

Hence, △ = -43

(iii)

Solution:

Considering the determinant, we have

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

△ = abc – af2 – h2c + fgh + fgh – g2b

Hence, △ = abc + 2fgh – af2 – ch2 – bg2

(iv)

Solution:

Considering the determinant, we have

△ = 1(-2 – 10) + 3(8 – 6) + 2(20 + 3)

△ = 1(-12) + 3(2) + 2(23)

△ = -12 + 6 + 46

Hence, △ = 40

(v)

Solution:

Considering the determinant, we have

△ = 1(225-256) + 4(100-144) + 9(64-81)

△ = 1(-31) – 4(-44) + 9(-17)

△ = -31 + 176 – 153

Hence, △ = -8

(vi)

Solution:

Considering the determinant, we have

Taking -2 common from C1, C2 and C3

As C1 and C2 are identical

Hence, △ = 0

(vii)

Solution:

Considering the determinant, we have

C1⇢C1 + C2 + C3

C2⇢C2 – C1

C3⇢C3 – C1

C4⇢C4 – C1

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

Taking 4 common from R2 and R1⇢R1+3R3

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

△ = (40)(4)(8)[40(9 + 1)]

△ = 40 × 4 × 8 × 40 × 10

Hence, △ = 512000

(viii)

Solution:

Considering the determinant, we have

Taking 6 common from R1, we get

As R1 and R3 are identical

Hence, △ = 0

(i)

Solution:

Considering the determinant, we have

Taking 4 common from C1, we get

As C1 and C2 are identical

Hence, △ = 0

(ii)

Solution:

Considering the determinant, we have

Taking -2 common from C1, we get

As C1 and C2 are identical

Hence, △ = 0

(iii)

Solution:

Considering the determinant, we have

R3⇢R3 – R2

As R1 and R3 are identical

Hence, △ = 0

(iv)

Solution:

Considering the determinant, we have

Multiplying and dividing △ 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, △ = 0

(v)

Solution:

Considering the determinant, we have

C3⇢C3 – C2 and C2⇢C2 – C1

As C2 and C3 are identical

Hence, △ = 0

(vi)

Solution:

Considering the determinant, we have

Splitting the determinant, we have

R2⇢R2-R1 and R3⇢R3-R1

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

△ = (b – a)(c – a)(c + a – (b + a)) – (b – a)(c – a)(-b – (-c))

△ = (b – a)(c – a)(c + a – b – a) – (b – a)(c – a)(-b + c)

△ = (b – a)(c – a)(c – b) – (b – a)(c – a)(c – b)

Hence, △ = 0

(vii)

Solution:

Considering the determinant, we have

C1⇢C1 – 8C3

As C1 and C2 are identical

Hence, △ = 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

C2⇢C2 – C3

As C1 and C2 are identical

Hence, △ = 0

(ix)

Solution:

Considering the determinant, we have

C2⇢C2 – 7C3

As C1 and C2 are identical

Hence, △ = 0

(x)

Solution:

Considering the determinant, we have

C3⇢C3 – C2 and C4⇢C4 – C1

Taking 3 common from C3, we get

As C3 and C4 are identical

Hence, △ = 0

(xi)

Solution:

Considering the determinant, we have

R3⇢R3 + R1 and R2⇢R2 + R1

Taking 2 common from R2, we get

As R2 and R3 are identical

Hence, △ = 0

Question 3.

Solution:

Considering the determinant, we have

C2⇢C2+C1

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

R3⇢R3-R1 and R2⇢R2-R1

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

△ = (a + b + c)(b – a)(c – a)[1(b + a – (c + a))]

△ = (a + b + c)(b – a)(c – a)(b + a – c – a)

Hence, △ = (a + b + c)(b – a)(c – a)(b – c)

Question 4.

Solution:

Considering the determinant, we have

R3⇢R3-R1 and R2⇢R2-R1

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

△ = (b – a)(c – a)[1((1)(-b) – (1)(-c))]

△ = (b – a)(c – a)[-b – (-c)]

△ = (b – a)(c – a)[-b + c]

Hence, △ = (a – b)(b – c)(c – a)

Question 5.

Solution:

Considering the determinant, we have

C1⇢C1+C2+C3

Taking (3x+λ) common from C1, we get

R3⇢R3-R1 and R2⇢R2-R1

△ = (3x + λ)[λ(λ(1) – 0)]

△ = (3x + λ)[λ(λ)]

Hence, △ = λ2(3x + λ)

Question 6.

Solution:

Considering the determinant, we have

C1⇢C1 + C2 + C3

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

R3⇢R3 – R1 and R2⇢R2 – R1

△ = (a + b + c)[1((a – b)(a – c) – (c – b)(b – c))]

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

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

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

Question 7.

Solution:

Considering the determinant, we have

C2⇢C2 – C1

Using the trigonometric identity,

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

As C2 and C3 are identical

Hence, △ = 0

Question 8. = a3 + b2 + c3 – 3abc

Solution:

Considering the determinant, we have

R3⇢R3 + R1 and R2⇢R2 + R1

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

R2⇢R2 – R1

Taking (-1) common from R2, we get

C1⇢C1 – C2 and C2⇢C2 – C3

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

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

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

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

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

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

△ = a3 + b3 + c3 – 3abc

Hence proved

Question 9. = 3abc – a3 – b2 – c3

Solution:

Considering the determinant, we have

C1⇢C1 + C3

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

△ = (a + b + c)[1((b – c)c – b(c – a)) – 1((a – b)c – a(c – a)) + 1(b(a – b) – a(b – c))]

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

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

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

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

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

△ = 3abc – a3 – b3 – c3

Hence proved

Question 10.

Solution:

Considering the determinant, we have

C1⇢C1 + C2 + C3

Taking 2 common from C1, we get

C2⇢C2 – C1 and C3⇢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

C1⇢C1 + C2 + C3

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

R2⇢R2 – R1 and R3⇢R3 – R1

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

△ = 2(a + b + c)[(a + b + b)2]

△ = 2(a + b + c)3

Hence proved

Question 12. = (a + b + c)3

Solution:

Considering the determinant, we have

R1⇢R1 + R2 + R3

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

C2⇢C2 – C1 and C3⇢C3 – C1

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

△ = (a + b + c)[(b + c + a)(b + c + a)]

△ = (a + b + c)[(b + c + a)2]

△ = (a + b + c)3

Hence proved

Question 13.  = (a – b)(b – c)(c – a)

Solution:

Considering the determinant, we have

R2⇢R2 – R1 and R3⇢R3 – R1

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

△ = (a – b)(a – c)[1(1(a + c) – 1(a + b))]

△ = (a – b)(a – c)[(a + c) – (a + b)]

△ = (a – b)(a – c)[a + c – a – b]

△ = (a – b)(a – c)

△ = (a – b)(a – c)(c – b)

△ = (a – b)(b – c)(c – a)

Hence proved

Question 14. = 9(a + b)b2

Solution:

Considering the determinant, we have

C1⇢C1 + C2 + C3

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

C2⇢C2 – C1 and C3⇢C3 – C2

△ = 3(a + b)[1((-2b)(-2b) – b(b))]

△ = 3(a + b)[4b2 – b2]

△ = 3(a + b)[3b2]

△ = 9(a + b)b2

Hence proved

Question 15.

Solution:

Considering the determinant, we have

R1⇢aR1, R2⇢bR2 and R3⇢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

C1⇢C1 – C2 and C3⇢C3 – C2

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

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

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

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

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

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

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

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

△ = (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

C1⇢C1 + C2 – 2C3

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

C2⇢C2-C1 and C3⇢C3-C1

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

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

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

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

△ = (a2 + b2 + c2)(b – a)(c – a)

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

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

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

Hence proved

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