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Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 6 |

Exercise | 6.2 |

Category | RD Sharma Solutions |

**RD Sharma Class 12 Ex 6.2 Solutions Chapter 6 Determinants**

### Question 1. Evaluate the following determinant:

### (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 – f^{2}) – h(hc – fg) + g(hf – gb)

△ = abc – af^{2} – h^{2}c + fgh + fgh – g^{2}b

**Hence, △ = abc + 2fgh – af ^{2} – ch^{2} – bg^{2}**

### (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**

### Question 2. Without expanding, show that the values of each of the following determinants are zero:

### (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, △ = λ ^{2}(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)[(a^{2 }– ac – ab + bc) – (cb – c^{2 }– b^{2 }+ bc)]

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

**Henfce, △ = (a + b + c)[a ^{2 }+ b^{2 }+ c^{2 }– 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**

### Prove the following identities:

### Question 8. = a^{3 }+ b^{2 }+ c^{3 }– 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 – a^{2 }– bc + ab) – (b^{2 }– 2cb + c^{2})]

△ = (-1)(a + b + c)(ac – a^{2 }– bc + ab – b^{2 }+ 2cb – c^{2})

△ = (a + b + c)(-ac + a^{2 }– bc – ab + b^{2} + c^{2})

△ = (a + b + c)(a^{2 }+ b^{2 }+ c^{2 }– ac – ab – cb)

**△ = a ^{3} + b^{3} + c^{3} – 3abc**

Hence proved

### Question 9. = 3abc – a^{3 }– b^{2 }– c^{3}

**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 – c^{2}-bc + ab) – (ac – bc) + ac – a^{2 }+ ab – b^{2 }– (ab – ac)]

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

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

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

**△ = 3abc – a ^{3} – b^{3} – c^{3}**

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)b^{2}

**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)[4b^{2 }– b^{2}]

△ = 3(a + b)[3b^{2}]

**△ = 9(a + b)b ^{2}**

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(z^{2 }+ zy + y^{2}) – 1(x^{2 }+ xy + y^{2}))]

△ = (xyz)(x – y)(z – y)[z^{2 }+ zy + y^{2 }– (x^{2 }+ xy + y^{2})]

△ = (xyz)(x – y)(z – y)[z^{2 }+ zy + y^{2 }– x^{2 }– xy – y^{2}]

△ = (xyz)(x – y)(z – y)[z^{2 }+ zy – x^{2 }– xy]

△ = (xyz)(x – y)(z – y)[z^{2 }– x^{2 }+ 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)(a^{2 }+ b^{2 }+ c^{2})

**Solution:**

Considering the determinant, we have

C1⇢C1 + C2 – 2C3

Taking (a^{2 }+ b^{2 }+ c^{2}) 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

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

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

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

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

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

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

**△ = (a ^{2 }+ b^{2 }+ c^{2})(a + b + c)(a – b)(b – c)(c – a)**

Hence proved

### 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)(a^{2 }+ b^{2 }+ c^{2})

**Solution:**

Considering the determinant, we have

C2⇢C2 – 2C1 – 2C3

Taking -(a^{2 }+ b^{2 }+ c^{2}) 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

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

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

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

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

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

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

**△ = (a ^{2 }+ b^{2 }+ c^{2})(a + b + c)(a – b)(b – c)(c – a)**

Hence proved

### Question 20. = (a – b)(b – c)(c – a)(a^{2 }+ b^{2 }+ c^{2})

**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)(c^{2 }+ a^{2 }+ ac) – (c + a – b)(b^{2 }+ a^{2 }+ ab))]

△ = (b – a)(c – a)(b – c)(a + b + c)

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

Hence proved

### Question 21. = 4a^{2}b^{2}c^{2}

**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

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

△ = 2a^{2}b^{2}c^{2}[1 – (-1)]

△ = 2a^{2}b^{2}c^{2}[1 + 1]

**△ = 4a ^{2}b^{2}c^{2}**

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. = (a^{3 }+ b^{3})^{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((a^{2 }– b^{2})(a^{2 }– b^{2}) – (b^{2 }– 2ab)(2ab – a^{2}))]

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

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

**△ = (a ^{3 }+ b^{3})^{2}**

Hence proved

### Question 26. = 1 + a^{2 }+ b^{2 }+ c^{2}

**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 (a^{2 }+ b^{2 }+ c^{2 }+ 1) common from R1, we get

C2⇢C2-C1 and C3⇢C3-C1

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

△ = (a^{2 }+ b^{2 }+ c^{2 }+ 1)[1]

**△ = (a ^{2 }+ b^{2 }+ c^{2 }+ 1)**

Hence proved !!

### Question 27. = (a^{3 }– 1)^{2}

**Solution:**

Considering the determinant, we have

C1⇢C1 + C2 + C3

Taking (a^{2 }+ a + 1) common from C1, we get

R2⇢R2 – R1 and R3⇢R3 – R1

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

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

△ = (a^{2 }+ a + 1)(1 – a)^{2}[(1 + a) + a^{2}]

△ = (a^{2 }+ a + 1)(1 – a)^{2}[1 + a + a^{2}]

△ = ((a^{2 }+ a + 1)(1 – a))^{2}

△ = (a^{3 }– 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. = 4a^{2}b^{2}c^{2}

**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. = 2a^{3}b^{3}c^{3}

**Solution:**

Considering the determinant, we have

Taking a^{2}, b^{2} and c^{2} common from C1, C2 and C3. we get

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

C2⇢C2 – C3

△ = a^{3}b^{3}c^{3}[1((1)(1) – (1)(-1))]

△ = a^{3}b^{3}c^{3}[1 + 1]

**△ = 2a ^{3}b^{3}c^{3}**

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

**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)]

**△ = 4xyz**

Hence 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 get

C3⇢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

**△ = 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 – 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}) = 0

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

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

On 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}= 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**

### 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} = 0

x + 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} = 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((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**

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