If Δ_1 = beginvmatrix x & sinθ & cosθ \ -sinθ & -x & 1 \ cosθ & 1 & x endvmatrix and Δ_2 = beginvmatrix x & sin 2θ & cos 2θ \ -sin 2θ & -x & 1 \ cos 2θ & 1 & x endvmatrix, x ≠ 0, then for all θ ∈ (0, (π)/2 )

Δ_1 - Δ_2 = 2(cos 2θ - cos 4θ)

If Δ_1 = beginvmatrix x & sinθ & cosθ \ -sinθ & -x & 1 \… — JEE Main 2019 Mathematics PYQ

\Delta_1 = \begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}

and

\Delta_2 = \begin{vmatrix} x & \sin 2\theta & \cos 2\theta \\ -\sin 2\theta & -x & 1 \\ \cos 2\theta & 1 & x \end{vmatrix}

x \neq 0

, then for all

\theta \in \left(0, \dfrac{\pi}{2}\right)

A.
$\Delta_1 + \Delta_2 = -2(x^3 + x - 1)$
B.
$\Delta_1 - \Delta_2 = -2x^3$
C.
$\Delta_1 + \Delta_2 = -2x^3$
D.
$\Delta_1 - \Delta_2 = 2(\cos 2\theta - \cos 4\theta)$

Show answer and solution

Correct option: C

Solution

Expanding Δ₁ along the first row (or by cofactor expansion), the determinant of a matrix of the form shown gives:

Δ₁ = x(−x·x − 1·1) − sinθ(−sinθ·x − 1·cosθ) + cosθ(−sinθ·1 − (−x)cosθ)

= x(−x² − 1) − sinθ(−x sinθ − cosθ) + cosθ(−sinθ + x cosθ)

= −x³ − x + x sin²θ + sinθ cosθ − sinθ cosθ + x cos²θ

= −x³ − x + x(sin²θ + cos²θ)

= −x³ − x + x = −x³

The key observation is that the cross terms (sinθ cosθ) cancel and sin²θ + cos²θ = 1. Crucially, this result is independent of θ — the same algebra applies to Δ₂ with 2θ replacing θ, giving Δ₂ = −x³ as well.

Therefore Δ₁ + Δ₂ = −x³ + (−x³) = −2x³.

[JEE Main 2019 · 10 April · Shift II]

JEE Main 2019 Mathematics question on Matrices and Determinants