# Class 12 RD Sharma Solutions – Chapter 6 Determinants – Exercise 6.2 | Set 1

### 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}– 3abcHence 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 getC2⇢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