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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
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)[c2 - b2 + a(c - b)]
△ = (a2 + b2 + c2)(a - b)(c - a)[(c - b)(c + b) + a(c - b)]
△ = (a2 + b2 + c2)(a - b)(c - a)(c - b)[c + b + a]
△ = (a2 + b2 + c2)(a + b + c)(a - b)(b - c)(c - a)
Hence proved
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
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
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
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
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
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
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 !!
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
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
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
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[c2((c2 + a2)(a2 + b2) - (b2)(c2)) + a2((b2)(c2) - (c2 + a2)(c2))]
△ = 2[c2((c2a2 + c2b2 + a4 + a2b2) - b2c2) + a2(b2c2 - (c4 + a2c2))]
△ = 2[c2(c2a2 + a2b2 + a4) + a2(b2c2 - c4 - a2c2)]
△ = 2[c4a2 + a2b2c2 +c2a]
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
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
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
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
Exercise 6.2 in Chapter 6 (Determinants) of RD Sharma's Class 12 mathematics textbook focuses on various properties and applications of determinants. This exercise covers topics such as:
The questions in this exercise help students develop a deeper understanding of determinants and their applications in solving linear equations and proving mathematical identities.
