# RD Sharma Class 12 Ex 6.2 Solutions Chapter 6 Determinants

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

### (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 – f2) – h(hc – fg) + g(hf – gb)

△ = abc – af2 – h2c + fgh + fgh – g2b

Hence, △ = abc + 2fgh – af2 – ch2 – bg2

### (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

### (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, △ = λ2(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– ac – ab + bc) – (cb – c– b+ bc)]

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

Henfce, △ = (a + b + c)[a+ b+ c– 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

### Question 8. = a3 + b2 + c3 – 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– bc + ab) – (b– 2cb + c2)]

△ = (-1)(a + b + c)(ac – a– bc + ab – b+ 2cb – c2)

△ = (a + b + c)(-ac + a– bc – ab + b2 + c2)

△ = (a + b + c)(a+ b+ c– ac – ab – cb)

△ = a3 + b3 + c3 – 3abc

Hence proved

### Question 9. = 3abc – a3 – b2 – c3

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 – c2-bc + ab) – (ac – bc) + ac – a+ ab – b– (ab – ac)]

△ = (a + b + c)[bc – c– bc + ab – ac + bc + ac – a+ ab – b– ab + ac]

△ = (a + b + c)[bc – c+ ab + ac – a– b2]

△ = (a + b + c)[bc + ab + ac – a– b– c2]

△ = 3abc – a3 – b3 – c3

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)b2

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– b2]

△ = 3(a + b)[3b2]

△ = 9(a + b)b2

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+ zy + y2) – 1(x+ xy + y2))]

△ = (xyz)(x – y)(z – y)[z+ zy + y– (x+ xy + y2)]

△ = (xyz)(x – y)(z – y)[z+ zy + y– x– xy – y2]

△ = (xyz)(x – y)(z – y)[z+ zy – x– xy]

△ = (xyz)(x – y)(z – y)[z– x+ 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)(a2 + b2 + c2)

Solution:

Considering the determinant, we have

C1⇢C1 + C2 – 2C3

Taking (a+ b+ c2) 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+ b+ c2)(b – a)(c – a)[1((b + a)(-b) – (c + a)(-c))]

△ = (a+ b+ c2)(b – a)(c – a)[(b + a)(-b) + (c + a)c]

△ = (a+ b+ c2)(b – a)(c – a)[(-b– ab) + (c+ ac)]

△ = (a+ b+ c2)(b – a)(c – a)

△ = (a+ b+ c2)(b – a)(c – a)[(c – b)(c + b) + a(c – b)]

△ = (a+ b+ c2)(b – a)(c – a)(c – b)

△ = (a+ b+ c2)(a + b + c)(a – b)(b – c)(c – a)

Hence proved

### 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)(a2 + b2 + c2)

Solution:

Considering the determinant, we have

C2⇢C2 – 2C1 – 2C3

Taking -(a+ b+ c2) 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+ b+ c2)(b – a)(c – a)[1((-b)(b + a) – (c + a)(-c))]

△ = (a+ b+ c2)(a – b)(c – a)[(-b)(b + a) + (c + a)c]

△ = (a+ b+ c2)(a – b)(c – a)[-b– ab + ac + c2]

△ = (a+ b+ c2)(a – b)(c – a)

△ = (a+ b+ c2)(a – b)(c – a)[(c – b)(c + b) + a(c – b)]

△ = (a+ b+ c2)(a – b)(c – a)(c – b)

△ = (a+ b+ c2)(a + b + c)(a – b)(b – c)(c – a)

Hence proved

### Question 20. = (a – b)(b – c)(c – a)(a2 + b2 + c2)

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+ a+ ac) – (c + a – b)(b+ a+ ab))]

△ = (b – a)(c – a)(b – c)(a + b + c)

△ = -(a – b)(c – a)(b – c)(a + b + c)

Hence proved

### Question 21.  = 4a2b2c2

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

△ = 2a2b2c2[1((-1)(-1) – (-1)(1))]

△ = 2a2b2c2[1 – (-1)]

△ = 2a2b2c2[1 + 1]

△ = 4a2b2c2

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.  = (a3 + b3)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– b2)(a– b2) – (b– 2ab)(2ab – a2))]

△ = (a + b)2 [(a– b2)+ (b– 2ab)(a– 2ab)]

△ = (a + b)2 [(a+ b– ab)2]

△ = (a+ b3)2

Hence proved

### Question 26. = 1 + a2 + b2 + c2

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+ b+ c+ 1) common from R1, we get

C2⇢C2-C1 and C3⇢C3-C1

△ = (a+ b+ c+ 1)[1((1)(1) – (0)(0))]

△ = (a+ b+ c+ 1)[1]

△ = (a+ b+ c+ 1)

Hence proved !!

### Question 27. = (a3 – 1)2

Solution:

Considering the determinant, we have

C1⇢C1 + C2 + C3

Taking (a+ 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+ a + 1)(1 – a)2[1((1 + a)(1) – (a)(-a))]

△ = (a+ a + 1)(1 – a)2[(1 + a) + a2]

△ = (a+ a + 1)(1 – a)2[1 + a + a2]

△ = ((a+ a + 1)(1 – a))2

△ = (a– 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. = 4a2b2c2

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. = 2a3b3c3

Solution:

Considering the determinant, we have

Taking a2, b2 and c2 common from C1, C2 and C3. we get

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

C2⇢C2 – C3

△ = a3b3c3[1((1)(1) – (1)(-1))]

△ = a3b3c3[1 + 1]

△ = 2a3b3c3

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

### 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 (a+ b+ c2) common from R1 and R2, we get

C3⇢C3 + C1

△ = abc(a+ b+ c2)2[-1((-1)(a+ c– b2) – (1)(2b2))]

△ = abc(a+ b+ c2)2[a+ c– b+ 2b2]

△ = abc(a+ b+ c2)2[a+ c+ b2]

△ = abc(a+ b+ 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]

△ = 3a+ 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 – b– ca + cb – (c – a)2]

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

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

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

As, it is given that

△ = 0

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

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

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

As, ba + bc + ac – b– c– a2 = 0

On multiplying it by -2, we get

-2ba – 2bc – 2ac + 2b+ 2c+ 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((c+ x+ cx)(1) – (b+ x+ bx)(1))]

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