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

Prove the following identities:

Question 18. = -2

Solution:

Considering the determinant, we have

R2⇢R2 – R1 and R3⇢R3 – R2

△ = 1[2(a + 2) – 2(a + 3)]

△ = (4a + 4 – (4a + 6))

△ = (4a + 4 – 4a – 6)

△ = -2

Hence proved 

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

Solution:

Considering the determinant, we have

C2⇢C2 – 2C1 – 2C3

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

R2⇢R2 – R1 and R3⇢R3 – R1

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

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

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

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

△ = (a² + b² + c²)(a – b)(c – a)

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

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

△ = (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

R2⇢R2 – R1 and R3⇢R3 – R1

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

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

△ = (b – a)(c – a)(b – c)(a + b + c)

△ = -(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

C1⇢C1 + C2 + C3

Taking 2 common from C1, we get

C2⇢C2 – C1 and C3⇢C3 – C1

C1⇢C1 + C2 + C3

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

R3⇢R3 – R1

△ = 2a²b²c²[1((-1)(-1) – (-1)(1))]

△ = 2a²b²c²[1 – (-1)]

△ = 2a²b²c²[1 + 1]

△ = 4a²b²c²

Hence proved 

Question 22. = 16(3x + 4)

Solution:

Considering the determinant, we have

C1⇢C1 + C2 + C3

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

R2⇢R2 – R1 and R3⇢R3 – R1

△ = (3x + 4)[1((4)(4) – (-4)(0))]

△ = (3x + 4)[16 – 0]

△ = 16(3x + 4)

Hence proved 

Question 23. = 1

Solution:

Considering the determinant, we have

C2⇢C2 – pC1 and C3⇢C3 – qC1

C3⇢C3 – pC2

C2⇢C2 – C1 and C3⇢C3 – C2

△ = 1[(1)(4) – (1)(3)]

△ = [4 – 3]

△ = 1

Hence proved 

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

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

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

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

△ = (b + c – a)(b + a – c)(c + a – b)

Hence proved 

Question 25.  = (a³ + b³)²

Solution:

Considering the determinant, we have

C1⇢C1 + C2 + C3

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

R2⇢R2 – R1 and R3⇢R3 – R1

R2⇢R2 – R3

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

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

△ = (a + b)² [(a² + b² – ab)²]

△ = (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

R1⇢R1 + R2 + R3

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

C2⇢C2-C1 and C3⇢C3-C1

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

△ = (a² + b² + c² + 1)[1]

△ = (a² + b² + c² + 1)

Hence proved !!

Question 27. = (a³ – 1)²

Solution:

Considering the determinant, we have

C1⇢C1 + C2 + C3

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

R2⇢R2 – R1 and R3⇢R3 – R1

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

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

△ = (a² + a + 1)(1 – a)²[(1 + a) + a²]

△ = (a² + a + 1)(1 – a)²[1 + a + a²]

△ = ((a² + a + 1)(1 – a))²

△ = (a³ – 1)²

Hence proved 

Question 28. = 2(a + b)(b + c)(c + a)

Solution:

Considering the determinant, we have

C1⇢C1 + C3 and C2⇢C2 + C3

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

R2⇢R2 + R1 and R3⇢R3 + R2

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

△ = (c + a)(b + c)[0 + 2(a + b)]

△ = 2(a + b)(c + a)(b + c)

Hence proved 

Question 29. = 4abc

Solution:

Considering the determinant, we have

R1⇢R1 + R2 + R3

Taking 2 common from R1, we get

R2⇢R2 – R1 and R3⇢R3 – R1

R1⇢R1 + R2 + R3

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

△ = 2[-c(0 – ab) + b(ac – 0)]

△ = 2[abc + abc]

△ = 2[2abc]

△ = 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

R1⇢R1 + R2 + R3

Taking 2 common from R1, we get

R1⇢R1 – R2

△ = 2

△ = 2

△ = 2

△ = 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

C2⇢C2 – C3

△ = a³b³c³[1((1)(1) – (1)(-1))]

△ = a³b³c³[1 + 1]

△ = 2a³b³c³

Hence proved 

Question 32. = 4abc

Solution:

Considering the determinant, we have

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

R1⇢R1 – R2 – R3

Taking -2 common from R1, we get

R2⇢R2 – R1 and R3⇢R3 – R1

△ = 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

R1⇢R1 + R2 + R3

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

C1⇢C1 – C2 and C3⇢C3 – C2

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

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

△ = (ab + bc + ca)³ [-1(-1)]

△ = (ab + bc + ca)³

Hence proved 

Question 34. = (5x + 4)(4 -x)²

Solution:

Considering the determinant, we have

C1⇢C1 + C2 + C3

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

R2⇢R2 – R1 and R3⇢R3 – R1

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

△ = (5x + 4)[(4 – x)²]

△ = (5x + 4)(4 – x)²

Hence proved