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Matrices & Determinants: Determinant Vmatrix Vmatrix Independent

JEE Maths question with a full step-by-step solution.

Question
The determinant Δ=∣−1+xsin⁡αcos⁡(x+α)sin⁡(x+α)13+xsin⁡βcos⁡(x+β)sin⁡(x+β)−12+xsin⁡γcos⁡(x+γ)sin⁡(x+γ)∣\Delta = \begin{vmatrix} -1+x\sin\alpha & \cos(x+\alpha) & \sin(x+\alpha) \\ 13+x\sin\beta & \cos(x+\beta) & \sin(x+\beta) \\ -12+x\sin\gamma & \cos(x+\gamma) & \sin(x+\gamma) \end{vmatrix} is independent of:
Axxcorrect
Bxx if x=α=β=γx = \alpha = \beta = \gamma
Cα,β,γ\alpha, \beta, \gamma
DNone of these
Solution
Step 1: Differentiate with respect to xx
dΔdx=∣sin⁡αcos⁡(x+α)sin⁡(x+α)sin⁡βcos⁡(x+β)sin⁡(x+β)sin⁡γcos⁡(x+γ)sin⁡(x+γ)∣+D2+D3\frac{d\Delta}{dx} = \begin{vmatrix}\sin\alpha&\cos(x+\alpha)&\sin(x+\alpha)\\\sin\beta&\cos(x+\beta)&\sin(x+\beta)\\\sin\gamma&\cos(x+\gamma)&\sin(x+\gamma)\end{vmatrix} + D_2 + D_3
 where D2D_2 and D3D_3 arise from differentiating columns 2 and 3. Step 2: Show each resulting determinant is zero D2D_2 has C2=−C3C_2 = -C_3 (since −sin⁡(x+⋅)=−sin⁡(x+⋅)-\sin(x+\cdot) = -\sin(x+\cdot)), so D2=0D_2 = 0. D3D_3 has C2=C3=cos⁡(x+⋅)C_2 = C_3 = \cos(x+\cdot), so D3=0D_3 = 0. For the first determinant, expanding using product-to-sum formulas yields four 3×33\times3 determinants each with two identical columns, each zero. Step 3: Conclude dΔdx=0\dfrac{d\Delta}{dx} = 0 for all xx, so Δ\Delta is constant, i.e., independent of xx. Note: The structure is possible because (−1)+13+(−12)=0(-1)+13+(-12) = 0 makes the constant terms in column 1 sum to zero. Answer: (1)
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