JEE Mathematics Determinants MCQs Set 12

Answer: (b) \( \frac{-\Delta(b,c,d)}{\Delta(a,b,c)} \)

Explanation:
1. Write the system as a_ix + b_iy + c_iz = −d_i. By Cramer's rule x = Δ(−d, b, c)/Δ(a, b, c) = −Δ(d, b, c)/Δ(a, b, c).
2. The column order (d, b, c) is a cyclic permutation of (b, c, d), so Δ(d, b, c) = Δ(b, c, d).
3. Hence x = −Δ(b, c, d)/Δ(a, b, c).

Teacher's Note:
1. Move the constant terms to the right side before using Cramer's rule.

Answer: (b) 8cos³θ − 4cosθ + 6

Explanation:
1. Expanding along the first row: Δ = C(C² − 1) − 1(C − 6) = C³ − 2C + 6.
2. With C = 2cos θ: Δ = 8cos³θ − 4cosθ + 6.

Teacher's Note:
1. Expand along the first row.

Answer: (c) \( -2-\sqrt2 \)

Explanation:
1. The matrix is Hermitian with a₁₂ = e^iα, a₁₃ = e^iβ, a₂₃ = e^iγ, where α = π/3, β = π/4, γ = 2π/3.
2. Expanding, the determinant = 1 − 3 + 2cos(α + γ − β) = −2 + 2cos(π/3 + 2π/3 − π/4) = −2 + 2cos(3π/4) = −2 − √2.
3. So the value is −2 − √2.

Teacher's Note:
1. Expand and group the conjugate terms into cosines.

Answer: (c) 6

Explanation:
1. Apply R₁ → R₁ − R₂, R₂ → R₂ − R₃ and simplify; the determinant reduces to 2 + 4 sin 2x.
2. The maximum of sin 2x is 1, so the maximum value of f(x) is 6.

Teacher's Note:
1. Row operations make the determinant a simple trigonometric expression.

Answer: (d) 0

Explanation:
1. The matrix equals the sum of the all-ones matrix and the product of the column (x_i) with the row (y_j), so its rank is at most 2.
2. A 3 × 3 matrix of rank at most 2 has determinant 0.

Teacher's Note:
1. Write the matrix as a sum of two rank-one matrices.

Answer: (d) k ≠ 0

Explanation:
1. A unique solution requires the coefficient determinant to be non-zero.
2. det = 1(k + 2) − 1(2k + 3) + 1(4 − 3) = −k.
3. So k ≠ 0.

Teacher's Note:
1. Compute the coefficient determinant.

Answer: (c) Negative

Explanation:
1. Subtract R₃ from R₁ and R₂, then expand: the determinant equals (x − 3)(x − 1)².
2. For x < 1, x − 3 < 0 and (x − 1)² > 0, so the value is negative.

Teacher's Note:
1. Factorise the determinant.

Answer: (a) [z]

Explanation:
1. The determinant of this form equals 1 + [x] + [y] + [z].
2. Here [x] = −1, [y] = 0 and [z] = 1, so the value is 1 − 1 + 0 + 1 = 1 = [z].

Teacher's Note:
1. Evaluate [x], [y], [z] first.

Answer: (b) θ only

Explanation:
1. Expanding along the first row: x(−x² − 1) − sin θ(−x sin θ − cos θ) + cos θ(−sin θ + x cos θ).
2. Collecting terms gives −x³ + x(sin²θ + cos²θ) − x = −x³.
3. The value −x³ does not involve θ.

Teacher's Note:
1. Expand and use sin²θ + cos²θ = 1.

Answer: (d) \( -\frac12 \)

Explanation:
1. det(A⁻¹) = 1(2) − (−2)(−2) = 2 − 4 = −2.
2. det(A) = 1/det(A⁻¹) = −1/2.

Teacher's Note:
1. det A = 1/det A⁻¹.

Answer: (a) (14)⁴

Explanation:
1. det A = 1(1 + 2) − 2(−1 − 4) + (−1)(1 − 2) = 3 + 10 + 1 = 14.
2. For a 3 × 3 matrix, det(adj A) = (det A)² and det(adj(adj A)) = (det A)^(n−1)² = (det A)⁴.
3. So the value is 14⁴.

Teacher's Note:
1. Use |adj A| = |A|ⁿ⁻¹ twice.

Answer: (b) −2

Explanation:
1. Expand along the first row: f(x) = cos x(x² − 2x²) − x(2 sin x − 2x tan x) + (2x sin x − x² tan x) = −x² cos x − 2x sin x + 2x² tan x + 2x sin x − x² tan x = −x² cos x + x² tan x.
2. Near 0: f(x) ≈ −x² + x³ + ..., so f′(x) ≈ −2x + 3x² and f′(x)/x → −2.

Teacher's Note:
1. Expand the determinant and differentiate.