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NCERT Solutions for class 12 Maths | Chapter 4 - Determinants

(1) Find the minor of 6 and cofactor of 4 respectively in the determinant Δ=
[A] 6, 6
[B] 6, -6
[C] -6, -6
[D] -6, 6
Answer: -6, 6
(2) The value of the determinant
[A] (α + β)(β + γ)(γ + α)
[B] (α – β)(β – γ)(γ – α)(α + β + γ)
[C] (α + β + γ)2 (α – β – γ)2
[D] αβγ (α + β + γ)
Answer: (α – β)(β – γ)(γ – α)(α + β + γ)

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(3) The value of is independent of
[A] α
[B] β
[C] α, β
[D] none of these
Answer: α
(4) Find the area of the triangle whose vertices are (-2, 6), (3, -6) and (1, 5).
[A] 30 sq. units
[B] 35 sq. units
[C] 40 sq. units
[D] 15.5 sq. units
Answer: 15.5 sq. units
(5) If the points (2, -3), (k, -1) and (0, 4) are collinear, then find the value of 4k.
[A] 4
[B] 7/140
[C] 47
[D] 40/7
Answer: 40/7
(6) If the points (a1, b1), (a2, b2) and(a1 + a2, b1 + b2) are collinear, then
[A] a1b2 = a2b1
[B] a1 + a2 = b1 + b2
[C] a2b2 = a1b1
[D] a1 + b1 = a2 + b2
Answer: a1b2 = a2b1
(7) Find the area of the triangle with vertices P(4, 5), Q(4, -2) and R(-6, 2).
[A] 21 sq. units
[B] 35 sq. units
[C] 30 sq. units
[D] 40 sq. units
Answer: 35 sq. units
(8) If the system of equations x + ky – z = 0, 3x – ky – z = 0 & x – 3y + z = 0 has non-zero solution, then k is equal to
[A] -1
[B] 0
[C] 1
[D] 2
Answer: 1
(9) Solve the following system of equations x – y + z = 4, x – 2y + 2z = 9 and 2x + y + 3z = 1.
[A] x = -4, y = -3, z = 2
[B] x = -1, y = -3, z = 2
[C] x = 2, y = 4, z = 6
[D] x = 3, y = 6, z = 9
Answer: x = -1, y = -3, z = 2
(10) If the equations 2x + 3y + z = 0, 3x + y – 2z = 0 and ax + 2y – bz = 0 has non-trivial solution, then
[A] a – b = 2
[B] a + b + 1 = 0
[C] a + b = 3
[D] a – b – 8 = 0
Answer: a – b = 2
(11) If 4x + 3y + 6z = 25, x + 5y + 7z = 13, 2x + 9y + z = 1, then z = ________
[A] 1
[B] 3
[C] -2
[D] 2
Answer: 2
(12) A non-trivial solution of the system of equations x + λy + 2z = 0, 2x + λz = 0, 2λx – 2y + 3z = 0 is given by x : y : z =
[A] 1 : 2 : -2
[B] 1 : -2 : 2
[C] 2 : 1 : 2
[D] 2 : 1 : -2
Answer: 2 : 1 : -2
(13) If for the non-singular matrix A, A2 = I, then find A-1.
[A] A
[B] I
[C] O
[D] None of these
Answer: A
(14)
[A] -(a – b)(b – c)(c – a)(a2 + b2 + c2)
[B] (a – b)(b – c)(c – a)
[C] (a2 + b2 + c2)
[D] (a – b)(b – c)(c – a)(a2 + b2 + c2)
Answer: -(a – b)(b – c)(c – a)(a2 + b2 + c2)
(15) Using determinants, find the equation of the line joining the points (1, 2) and (3, 6).
[A] y = 2x
[B] x = 3y
[C] y = x
[D] 4x – y = 5
Answer: y = 2x
(16) The area of the triangle whose vertices are A(5, 4), B(-2, 4) and C(2, -6) is
[A] 15 square units.
[B] 25 square units.
[C] 35 square units.
[D] 45 square units.
Answer: 35 square units.
(17) Determinant of the 1 x 1 matrix [- 3] is
[A] 3
[B] - 3
[C] 0
[D] none of these
Answer: - 3
(18) If the equations (a + 1)3 x + (a + 2)3 y = (a + 3)3 (a + 1) x + (a + 2) y = a + 3 and x + y = 1 are consistent, then
[A] a = - 1
[B] a = - 2
[C] a = - 3
[D] none of these.
Answer: a = - 2
(19) If A is a square matrix of order 2, then det (adj A) is equal to
[A] 1
[B] det A
[C] (det A)2
[D] none of these
Answer: det A
(20) If the points (3, -2), (x, 2), (8, 8) are collinear, then find the value of x.
[A] 2
[B] 3
[C] 4
[D] 5
Answer: 5

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