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Matrices & Determinants: Vmatrix Vmatrix Vmatrix Vmatrix

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

Question
If Δ1=∣xbbaxbaax∣\Delta_1 = \begin{vmatrix} x & b & b \\ a & x & b \\ a & a & x \end{vmatrix} and Δ2=∣xbax∣\Delta_2 = \begin{vmatrix} x & b \\ a & x \end{vmatrix}, then:
AΔ1=3Δ22\Delta_1 = 3\Delta_2^2
Bddx(Δ1)=3Δ22\dfrac{d}{dx}(\Delta_1) = 3\Delta_2^2
Cddx(Δ1)=3Δ2\dfrac{d}{dx}(\Delta_1) = 3\Delta_2correct
DNone of these
Solution
Step 1: Differentiate Δ1\Delta_1 using the row-differentiation formula
ddx(Δ1)=∣1bb0xb0ax∣+∣x0ba1ba0x∣+∣xb0ax0aa1∣\frac{d}{dx}(\Delta_1) = \begin{vmatrix} 1 & b & b \\ 0 & x & b \\ 0 & a & x \end{vmatrix} + \begin{vmatrix} x & 0 & b \\ a & 1 & b \\ a & 0 & x \end{vmatrix} + \begin{vmatrix} x & b & 0 \\ a & x & 0 \\ a & a & 1 \end{vmatrix}
 Step 2: Evaluate each 3×33\times3 determinant Each reduces to (x2−ab)=Δ2(x^2 - ab) = \Delta_2:
ddx(Δ1)=Δ2+Δ2+Δ2=3Δ2\frac{d}{dx}(\Delta_1) = \Delta_2 + \Delta_2 + \Delta_2 = 3\Delta_2
 Verification: Δ1=x3−3abx+2ab2\Delta_1 = x^3 - 3abx + 2ab^2, so ddx(Δ1)=3x2−3ab=3(x2−ab)=3Δ2\dfrac{d}{dx}(\Delta_1) = 3x^2-3ab = 3(x^2-ab) = 3\Delta_2. ✓\checkmark Answer: (3)
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