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

### Question 18. = -2

Solution:

Considering the determinant, we have

R2â‡¢R2 – R1 and R3â‡¢R3 – R2

â–³ = 1[2(a + 2) – 2(a + 3)]

â–³ = (4a + 4 – (4a + 6))

â–³ = (4a + 4 – 4a – 6)

â–³ = -2

Hence proved

### Question 19. = (a – b)(b – c)(c – a)(a + b + c)(a2 + b2 + c2)

Solution:

Considering the determinant, we have

C2â‡¢C2 – 2C1 – 2C3

Taking -(a2 + b2 + c2) common from C2, we get

R2â‡¢R2 – R1 and R3â‡¢R3 – R1

Taking (b – a) and (c – a) common from R1 and R2, we get

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

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

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

â–³ = (a2 + b2 + c2)(a – b)(c – a)

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

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

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

Hence proved

### Question 20. = (a – b)(b – c)(c – a)(a2 + b2 + c2)

Solution:

Considering the determinant, we have

R2â‡¢R2 – R1 and R3â‡¢R3 – R1

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

â–³ = (b – a)(c – a)[1((b + a – c)(c2 + a2 + ac) – (c + a – b)(b2 + a2 + ab))]

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

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

Hence proved

### Question 21.  = 4a2b2c2

Solution:

Considering the determinant, we have

Taking a, b and c common from C1, C2 and C3 we get

C1â‡¢C1 + C2 + C3

Taking 2 common from C1, we get

C2â‡¢C2 – C1 and C3â‡¢C3 – C1

C1â‡¢C1 + C2 + C3

Taking c, a and b common from C1, C2 and C3 we get

R3â‡¢R3 – R1

â–³ = 2a2b2c2[1((-1)(-1) – (-1)(1))]

â–³ = 2a2b2c2[1 – (-1)]

â–³ = 2a2b2c2[1 + 1]

â–³ = 4a2b2c2

Hence proved

### Question 22. = 16(3x + 4)

Solution:

Considering the determinant, we have

C1â‡¢C1 + C2 + C3

Taking (3x + 4) common from C1, we get

R2â‡¢R2 – R1 and R3â‡¢R3 – R1

â–³ = (3x + 4)[1((4)(4) – (-4)(0))]

â–³ = (3x + 4)[16 – 0]

â–³ = 16(3x + 4)

Hence proved

### Question 23. = 1

Solution:

Considering the determinant, we have

C2â‡¢C2 – pC1 and C3â‡¢C3 – qC1

C3â‡¢C3 – pC2

C2â‡¢C2 – C1 and C3â‡¢C3 – C2

â–³ = 1[(1)(4) – (1)(3)]

â–³ = [4 – 3]

â–³ = 1

Hence proved

### Question 24. = (a + b – c)(b + c – a)(c + a – b)

Solution:

Considering the determinant, we have

R1â‡¢R1 – R2 – R3

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

C2â‡¢C2 + C1 and C3â‡¢C3 + C1

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

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

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

Hence proved

### Question 25.  = (a3 + b3)2

Solution:

Considering the determinant, we have

C1â‡¢C1 + C2 + C3

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

R2â‡¢R2 – R1 and R3â‡¢R3 – R1

R2â‡¢R2 – R3

â–³ = (a + b)2 [1((a2 – b2)(a2 – b2) – (b2 – 2ab)(2ab – a2))]

â–³ = (a + b)2 [(a2 – b2)2 + (b2 – 2ab)(a2 – 2ab)]

â–³ = (a + b)2 [(a2 + b2 – ab)2]

â–³ = (a3 + b3)2

Hence proved

### Question 26. = 1 + a2 + b2 + c2

Solution:

Considering the determinant, we have

Multiplying a, b and c to R1, R2 and R3 we get

Taking a, b and c common from C1, C2 and C3 we get

R1â‡¢R1 + R2 + R3

Taking (a2 + b2 + c2 + 1) common from R1, we get

C2â‡¢C2-C1 and C3â‡¢C3-C1

â–³ = (a2 + b2 + c2 + 1)[1((1)(1) – (0)(0))]

â–³ = (a2 + b2 + c2 + 1)[1]

â–³ = (a2 + b2 + c2 + 1)

Hence proved !!

### Question 27. = (a3 – 1)2

Solution:

Considering the determinant, we have

C1â‡¢C1 + C2 + C3

Taking (a2 + a + 1) common from C1, we get

R2â‡¢R2 – R1 and R3â‡¢R3 – R1

Taking (1 – a) common from R2 and R3, we get

â–³ = (a2 + a + 1)(1 – a)2[1((1 + a)(1) – (a)(-a))]

â–³ = (a2 + a + 1)(1 – a)2[(1 + a) + a2]

â–³ = (a2 + a + 1)(1 – a)2[1 + a + a2]

â–³ = ((a2 + a + 1)(1 – a))2

â–³ = (a3 – 1)2

Hence proved

### Question 28. = 2(a + b)(b + c)(c + a)

Solution:

Considering the determinant, we have

C1â‡¢C1 + C3 and C2â‡¢C2 + C3

Taking (c + a) and (b + c) common from C1 and C2, we get

R2â‡¢R2 + R1 and R3â‡¢R3 + R2

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

â–³ = (c + a)(b + c)[0 + 2(a + b)]

â–³ = 2(a + b)(c + a)(b + c)

Hence proved

### Question 29. = 4abc

Solution:

Considering the determinant, we have

R1â‡¢R1 + R2 + R3

Taking 2 common from R1, we get

R2â‡¢R2 – R1 and R3â‡¢R3 – R1

R1â‡¢R1 + R2 + R3

â–³ = 2[-c((-c)(0) – (-a)(-b)) + b((-c)(-a) – (0)(-b))]

â–³ = 2[-c(0 – ab) + b(ac – 0)]

â–³ = 2[abc + abc]

â–³ = 2[2abc]

â–³ = 4abc

Hence proved

### Question 30. = 4a2b2c2

Solution:

Considering the determinant, we have

Multiplying a, b and c to R1, R2 and R3 respectively, we get

Taking common a, b and c to C1, C2 and C3 respectively, we get

R1â‡¢R1 + R2 + R3

Taking 2 common from R1, we get

R1â‡¢R1 – R2

â–³ = 2

â–³ = 2

â–³ = 2

â–³ = 2

### Question 31. = 2a3b3c3

Solution:

Considering the determinant, we have

Taking a2, b2 and c2 common from C1, C2 and C3. we get

Taking a, b and c common from R1, R2 and R3. we get

C2â‡¢C2 – C3

â–³ = a3b3c3[1((1)(1) – (1)(-1))]

â–³ = a3b3c3[1 + 1]

â–³ = 2a3b3c3

Hence proved

### Question 32. = 4abc

Solution:

Considering the determinant, we have

Multiplying c, a and b to R1, R2 and R3. We get

R1â‡¢R1 – R2 – R3

Taking -2 common from R1, we get

R2â‡¢R2 – R1 and R3â‡¢R3 – R1

â–³ = 4abc

Hence proved

### Question 33. = (ab + bc + ca)3

Solution:

Considering the determinant, we have

Multiplying a, b and c to R1, R2 and R3. We get

Taking a, b and c common from C1, C2 and C3. we get

R1â‡¢R1 + R2 + R3

Taking (ab + bc + ca) common from R1, we get

C1â‡¢C1 – C2 and C3â‡¢C3 – C2

Taking (ab + bc + ca) common from C1 and C2, we get

â–³ = (ab + bc + ca)3 [-1((1)(-1) – (1)(0))]

â–³ = (ab + bc + ca)3 [-1(-1)]

â–³ = (ab + bc + ca)3

Hence proved

### Question 34. = (5x + 4)(4 -x)2

Solution:

Considering the determinant, we have

C1â‡¢C1 + C2 + C3

Taking (5x + 4) common from C1, we get

R2â‡¢R2 – R1 and R3â‡¢R3 – R1

â–³ = (5x + 4)[1((4 – x)(4 – x) – (0)(0))]

â–³ = (5x + 4)[(4 – x)2]

â–³ = (5x + 4)(4 – x)2

Hence proved

Previous
Next