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• RD Sharma Class 12 Solutions for Maths

# Class 12 RD Sharma Solutions – Chapter 6 Determinants – Exercise 6.2 | Set 3

### Question 35. = 4xyz

Solution:

Considering the determinant, we have

R1⇢R1 – R2 – R3

C2⇢C2 – C3

△ = [-2x((z)(-y) – (z)(y))]

△ = [-2x(-zy – zy)]

△ = [-2x(-2zy)]

△ = 4xyz

Hence proved

### Question 36. = abc(a2 + b2 + c2)3

Solution:

Considering the determinant, we have

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

R1⇢R1 – R3 and R2⇢R2 – R3

Taking (a2 + b2 + c2) common from R1 and R2, we get

C3⇢C3 + C1

△ = abc(a2 + b2 + c2)2[-1((-1)(a2 + c2 – b2) – (1)(2b2))]

△ = abc(a2 + b2 + c2)2[a2 + c2 – b2 + 2b2]

△ = abc(a2 + b2 + c2)2[a2 + c2 + b2]

△ = abc(a2 + b2 + c2)3

Hence proved

### Question 37. = a3 + 3a2

Solution:

Considering the determinant, we have

C1⇢C1 + C2 + C3

Taking (3 + a) common from C1, we get

R2⇢R2 – R1 and R3⇢R3 – R1

△ = (3 + a)[1((a)(a) – (0)(0))]

△ = (3 + a)[a2]

△ = 3a2 + a3

Hence proved

### Question 38. = (x + y + z)(x – z)2

Solution:

Considering the determinant, we have

R1⇢R1 + R2 + R3

Taking (x + y + z) common from R1, we get

C1⇢C1 – C2 – C3

△ = (x + y + z)[(x – z)(x – z)]

△ = (x + y + z)[(x – z)2]

△ = (x + y + z)(x – z)2

Hence proved

### Question 39. Without expanding, prove that

Solution:

Considering the determinant, we have

R1↔R2

R2↔R3

C1↔C2

R2↔R3

Taking transpose, we have

Hence proved

### Question 40. Show that  where a, b, c are in AP.

Solution:

Considering the determinant, we have

C1⇢C1 + C2 + C3

R2⇢R2 – R1 and R3⇢R3 – R2

As a, b and c are in AP

then, b – a = c – b = λ

As, R2 and R3 are identical

△ = 0

Hence proved

### Question 41. Show that  where α, β, γ are in AP.

Solution:

Considering the determinant, we have

C1⇢C1 + C2 + C3

R2⇢R2 – R1 and R3⇢R3 – R2

As α, β, γ are in AP

then, β – α = γ – β = λ

As, R2 and R3 are identical

△ = 0

Hence proved

### Question 42. Evaluate

Solution:

Considering the determinant, we have

C1⇢C1 + C2 + C3

Taking (x + 2) common from C1. we get

R2⇢R2 – R1 and R3⇢R3 – R2

△ = (x + 2)[1((x – 1)(x – 1) – (0)(0))]

△ = (x + 2)(x – 1)2

Hence proved

### Question 43. If a, b, c are real numbers such that  , then show that either a + b + c = 0 or,  a = b = c.

Solution:

Considering the determinant, we have

C1⇢C1 + C2 + C3

Taking 2(a + b + c) common from C1. We get

R2⇢R2 – R1 and R3⇢R3 – R2

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

△ = 2(a + b + c)[ba – b2 – ca + cb – (c – a)2]

△ = 2(a + b + c)[ba – b2 – ca + cb – (c2 + a2 – 2ac)]

△ = 2(a + b + c)[ba – b2 – ca + cb – c2 – a2 + 2ac]

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

As, it is given that

△ = 0

2 (a + b + c)(ba + bc + ac – b2 – c2 – a2) = 0

(a + b + c)(ba + bc + ac – b2 – c2 – a2) = 0

So, either (a + b + c) = 0 or (ba + bc + ac – b2 – c2 – a2) = 0

As, ba + bc + ac – b2 – c2 – a2 = 0

On multiplying it by -2, we get

-2ba – 2bc – 2ac + 2b2 + 2c2 + 2a2 = 0

(a – b)2 + (b – c)2 + (c – a)2= 0

As, square power is always positive

(a – b)2 = (b – c)2 = (c – a)2

(a – b) = (b – c) = (c – a)

a = b = c

Hence proved

### Question 44. Show that x=2 is a root of the equation  and solve it completely.

Solution:

Considering the determinant, we have

On putting x = 2, we get

As, R1 = R2

△ = 0

R3⇢R3 – R1

Taking (x + 3) common from R3, we get

R1⇢R1 – R2

Taking (x – 2) common from R1. We get

C3⇢C3 + C1

Now, taking (x – 1) common from C3. We get

△ = (x + 3)(x – 2)(x – 1)[1((1)(2) – (3)(-1))]

△ = (x + 3)(x – 2)(x – 1)[2 + 3]

△ = 5 (x + 3)(x – 2)(x – 1)

△ = 0

5 (x + 3)(x – 2)(x – 1) = 0

x = 2, 1, -3

### (i)

Solution:

Considering the determinant, we have

C1⇢C1 + C2 + C3

Now, taking (x + a + b + c) common from C1. We get

R2⇢R2 – R1 and R3⇢R3 – R1

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

△ = (x + a + b + c)[x2]

As △ = 0

(x + a + b + c) x2 = 0

x + a + b + c = 0 or x2 = 0

x = -(a + b + c) or x = 0

### (ii)

Solution:

Considering the determinant, we have

C1⇢C1 + C2 + C3

Now, taking (3x + a) common from C1. We get

R2⇢R2 – R1 and R3⇢R3 – R1

△ = (3x + a)[1((a)(a) – (0)(0))]

△ = (3x + a)[a2]

As △ = 0

(3x + a)[a2] = 0

x = -a/3

### (iii)

Solution:

Considering the determinant, we have

C1⇢C1 + C2 + C3

Now, taking (3x – 2) common from C1. We get

R2⇢R2 – R1 and R3⇢R3 – R1

△ = (3x – 2)[1((3x – 11)(3x – 11) – (0)(0))]

△ = (3x – 2)[(3x – 11)2]

As △ = 0

(3x – 2)(3x – 11)2 = 0

3x – 2 = 0 and 3x – 11 = 0

x = 2/3 and x = 11/3

### (iv)

Solution:

Considering the determinant, we have

R2⇢R2 – R1 and R3⇢R3 – R1

Now, taking (a – x) and (b – x) common from R2 and R3 respectively. We get

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

△ = (a – x)(b – x)[b + x – (a + x)]

△ = (a – x)(b – x)[b + x – a – x]

△ = (a – x)(b – x)[b – a]

As △ = 0

(a – x)(b – x)(b – a) = 0

a – x = 0 and b – x = 0

x = a and x = b

### (v)

Solution:

Considering the determinant, we have

C1⇢C1 + C2 + C3

Now, taking (x + 9) common from C1. We get

R2⇢R2 – R1 and R3⇢R3 – R1

△ = (x + 9)[1((x – 1)(x – 1) – (0)(0))]

△ = (x + 9)(x – 1)2

As △ = 0

(x + 9)(x – 1)2 = 0

x + 9 = 0 or (x – 1)2 = 0

x = -9 or x = 1

### (vi)

Solution:

Considering the determinant, we have

R2⇢R2 – R1 and R3⇢R3 – R1

Now, taking (b – x) and (c – x) common from R2 and R3 respectively. We get

△ = (b – x)(c – x)[1((c2 + x2 + cx)(1) – (b2 + x2 + bx)(1))]

△ = (b – x)(c – x)[(c2 + x2 + cx) – (b2 + x2 + bx)]

△ = (b – x)(c – x)

△ = (b – x)(c – x)

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

△ = (b – x)(c – x)(c – b)

As △ = 0

(b – x)(c – x)(c – b) = 0

b – x = 0 or c – x = 0 or c – b = 0 or c + b + x = 0

x = b or x = c or c = b or x = -(c + b)

### (vii)

Solution:

Considering the determinant, we have

R2⇢R2 – R3

R2⇢R2 – R1 and R1⇢R1 – R3

△ = -1[(8 – x)(6) – (-x – 4)(-3)]

△ = -1[(8 – x)(6) – (x + 4)(3)]

△ = [(x + 4)(3) – (8 – x)(6)]

△ = [3x + 12 – (48 – 6x)]

△ = [9x – 36]

As △ = 0

9x – 36 = 0

x = 4

### (viii)

Solution:

Considering the determinant, we have

R2⇢R2 – R1

Now, taking p common from R2. We get

R2⇢R2 – R1

Now, taking 1 – x common from R2. We get

△ = p(1 – x)[-1((x + 1)(1) – (3)(1))]

△ = p(x – 1)[x + 1 – 3]

△ = p(x – 1)[x – 2]

As △ = 0

p(x – 1)(x – 2) = 0

x – 1 = 0 or x – 2 = 0

x = 1 or x = 2

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