Here we provide RD Sharma Class 12 Ex 6.2 Solutions Chapter 6 Determinants for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 6.2 Solutions Chapter 6 Determinants book pdf download. Now you will get step-by-step solutions to each question.
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 – 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
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, △ = 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)[(a2 – ac – ab + bc) – (cb – c2 – b2 + bc)]
△ = (a + b + c)[a2 – ac – ab + bc + c2 + b2 – 2bc]
Henfce, △ = (a + b + c)[a2 + b2 + c2 – 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.
= 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 – a2 – bc + ab) – (b2 – 2cb + c2)]
△ = (-1)(a + b + c)(ac – a2 – bc + ab – b2 + 2cb – c2)
△ = (a + b + c)(-ac + a2 – bc – ab + b2 + c2)
△ = (a + b + c)(a2 + b2 + c2 – 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 – a2 + ab – b2 – (ab – ac)]
△ = (a + b + c)[bc – c2 – bc + ab – ac + bc + ac – a2 + ab – b2 – ab + ac]
△ = (a + b + c)[bc – c2 + ab + ac – a2 – b2]
△ = (a + b + c)[bc + ab + ac – a2 – b2 – 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)[4b2 – 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(z2 + zy + y2) – 1(x2 + xy + y2))]
△ = (xyz)(x – y)(z – y)[z2 + zy + y2 – (x2 + xy + y2)]
△ = (xyz)(x – y)(z – y)[z2 + zy + y2 – x2 – xy – y2]
△ = (xyz)(x – y)(z – y)[z2 + zy – x2 – xy]
△ = (xyz)(x – y)(z – y)[z2 – x2 + 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 (a2 + b2 + 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
△ = (a2 + b2 + c2)(b – a)(c – a)[1((b + a)(-b) – (c + a)(-c))]
△ = (a2 + b2 + c2)(b – a)(c – a)[(b + a)(-b) + (c + a)c]
△ = (a2 + b2 + c2)(b – a)(c – a)[(-b2 – ab) + (c2 + ac)]
△ = (a2 + b2 + c2)(b – a)(c – a)
△ = (a2 + b2 + c2)(b – a)(c – a)[(c – b)(c + b) + a(c – b)]
△ = (a2 + b2 + c2)(b – a)(c – a)(c – b)
△ = (a2 + b2 + c2)(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)(a2 + b2 + c2)
Solution:
Considering the determinant, we have
C2⇢C2 – 2C1 – 2C3
Taking -(a2 + b2 + 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
△ = -(a2 + b2 + c2)(b – a)(c – a)[1((-b)(b + a) – (c + a)(-c))]
△ = (a2 + b2 + c2)(a – b)(c – a)[(-b)(b + a) + (c + a)c]
△ = (a2 + b2 + c2)(a – b)(c – a)[-b2 – ab + ac + c2]
△ = (a2 + b2 + c2)(a – b)(c – a)
△ = (a2 + b2 + c2)(a – b)(c – a)[(c – b)(c + b) + a(c – b)]
△ = (a2 + b2 + c2)(a – b)(c – a)(c – b)
△ = (a2 + b2 + 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)(c2 + a2 + ac) – (c + a – b)(b2 + a2 + 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((a2 – b2)(a2 – b2) – (b2 – 2ab)(2ab – a2))]
△ = (a + b)2 [(a2 – b2)2 + (b2 – 2ab)(a2 – 2ab)]
△ = (a + b)2 [(a2 + b2 – ab)2]
△ = (a3 + 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 (a2 + b2 + c2 + 1) common from R1, we get
C2⇢C2-C1 and C3⇢C3-C1
△ = (a2 + b2 + c2 + 1)[1((1)(1) – (0)(0))]
△ = (a2 + b2 + c2 + 1)[1]
△ = (a2 + b2 + c2 + 1)
Hence proved !!
Question 27.
= (a3 – 1)2
Solution:
Considering the determinant, we have
C1⇢C1 + C2 + C3
Taking (a2 + a + 1) common from C1, we get
R2⇢R2 – R1 and R3⇢R3 – R1
Taking (1 – a) common from R2 and R3, we get
△ = (a2 + a + 1)(1 – a)2[1((1 + a)(1) – (a)(-a))]
△ = (a2 + a + 1)(1 – a)2[(1 + a) + a2]
△ = (a2 + a + 1)(1 – a)2[1 + a + a2]
△ = ((a2 + a + 1)(1 – a))2
△ = (a3 – 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
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(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 (a2 + b2 + c2) common from R1 and R2, we get
C3⇢C3 + C1
△ = abc(a2 + b2 + c2)2[-1((-1)(a2 + c2 – b2) – (1)(2b2))]
△ = abc(a2 + b2 + c2)2[a2 + c2 – b2 + 2b2]
△ = abc(a2 + b2 + c2)2[a2 + c2 + b2]
△ = abc(a2 + b2 + 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]
△ = 3a2 + 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 – b2 – ca + cb – (c – a)2]
△ = 2(a + b + c)[ba – b2 – ca + cb – (c2 + a2 – 2ac)]
△ = 2(a + b + c)[ba – b2 – ca + cb – c2 – a2 + 2ac]
△ = 2(a + b + c)[ba + bc + ac – b2 – c2 – a2]
As, it is given that
△ = 0
2 (a + b + c)(ba + bc + ac – b2 – c2 – a2) = 0
(a + b + c)(ba + bc + ac – b2 – c2 – a2) = 0
So, either (a + b + c) = 0 or (ba + bc + ac – b2 – c2 – a2) = 0
As, ba + bc + ac – b2 – c2 – a2 = 0
On multiplying it by -2, we get
-2ba – 2bc – 2ac + 2b2 + 2c2 + 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
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)[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((c2 + x2 + cx)(1) – (b2 + x2 + bx)(1))]
△ = (b – x)(c – x)[(c2 + x2 + cx) – (b2 + x2 + 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
I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.
If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.