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

**Prove the following identities:**

### 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)]

△ = 4xyzHence proved

### Question 36. = abc(a^{2 }+ b^{2 }+ c^{2})^{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 (a

^{2 }+ b^{2 }+ c^{2}) common from R1 and R2, we getC3⇢C3 + C1

△ = abc(a

^{2 }+ b^{2 }+ c^{2})^{2}[-1((-1)(a^{2 }+ c^{2 }– b^{2}) – (1)(2b^{2}))]△ = abc(a

^{2 }+ b^{2 }+ c^{2})^{2}[a^{2 }+ c^{2 }– b^{2 }+ 2b^{2}]△ = abc(a

^{2 }+ b^{2 }+ c^{2})^{2}[a^{2 }+ c^{2 }+ b^{2}]

△ = abc(a^{2 }+ b^{2 }+ c^{2})^{3}Hence proved

### Question 37. = a^{3 }+ 3a^{2}

**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)[a

^{2}]

△ = 3a^{2 }+ a^{3}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

△ = 0Hence 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

△ = 0Hence 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 – b

^{2 }– ca + cb – (c – a)^{2}]△ = 2(a + b + c)[ba – b

^{2 }– ca + cb – (c^{2 }+ a^{2 }– 2ac)]△ = 2(a + b + c)[ba – b

^{2 }– ca + cb – c^{2 }– a^{2 }+ 2ac]△ = 2(a + b + c)[ba + bc + ac – b

^{2 }– c^{2 }– a^{2}]As, it is given that

△ = 0

2 (a + b + c)(ba + bc + ac – b

^{2 }– c^{2 }– a^{2}) = 0(a + b + c)(ba + bc + ac – b

^{2 }– c^{2 }– a^{2}) = 0So, either (a + b + c) = 0 or (ba + bc + ac – b

^{2 }– c^{2 }– a^{2}) = 0As, ba + bc + ac – b

^{2 }– c^{2 }– a^{2}= 0On multiplying it by -2, we get

-2ba – 2bc – 2ac + 2b

^{2 }+ 2c^{2 }+ 2a^{2}= 0(a – b)

^{2}+ (b – c)^{2}+ (c – a)^{2}= 0As, square power is always positive

(a – b)

^{2}= (b – c)^{2}= (c – a)^{2}(a – b) = (b – c) = (c – a)

a = b = cHence 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

△ = 0R3⇢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

### Question 45. Solve the following determinant equations:

### (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)[x

^{2}]As △ = 0

(x + a + b + c) x

^{2}= 0x + a + b + c = 0 or x

^{2}= 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)[a

^{2}]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}= 03x – 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}= 0x + 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((c

^{2 }+ x^{2 }+ cx)(1) – (b^{2 }+ x^{2 }+ bx)(1))]△ = (b – x)(c – x)[(c

^{2 }+ x^{2 }+ cx) – (b^{2 }+ x^{2 }+ 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