# 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)

△ = -2Hence 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 getR2⇢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]

△ = 1Hence 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 getR2⇢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 getC2⇢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 getR2⇢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]

△ = 4abcHence 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 getTaking 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

△ = 4abcHence 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

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