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Matrices & Determinants: Value Determinant Vmatrix Vmatrix

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

Question
The value of the determinant (for n∈Nn \in \mathbb{N})
Δ=∣n!(n+1)!(n+2)!(n+1)!(n+2)!(n+3)!(n+2)!(n+3)!(n+4)!∣\Delta = \begin{vmatrix} n! & (n+1)! & (n+2)! \\ (n+1)! & (n+2)! & (n+3)! \\ (n+2)! & (n+3)! & (n+4)! \end{vmatrix}
 is:
A(n!)3(2n3+8n2+10n+4)(n!)^3(2n^3+8n^2+10n+4)correct
B(n!)3(2n2+8n+10)(n!)^3(2n^2+8n+10)
C(n!)2(2n3+8n2+10n+4)(n!)^2(2n^3+8n^2+10n+4)
DNone of these
Solution
Step 1: Factor out factorials from each row Factor n!n! from R1R_1, (n+1)!(n+1)! from R2R_2, (n+2)!(n+2)! from R3R_3:
Δ=(n!)3(n+1)2(n+2)∣1n+1(n+1)(n+2)1n+2(n+2)(n+3)1n+3(n+3)(n+4)∣\Delta = (n!)^3(n+1)^2(n+2)\begin{vmatrix} 1 & n+1 & (n+1)(n+2) \\ 1 & n+2 & (n+2)(n+3) \\ 1 & n+3 & (n+3)(n+4) \end{vmatrix}
 Step 2: Apply C2→C2−C1C_2 \to C_2 - C_1 and C3→C3−C1C_3 \to C_3 - C_1, then R2→R2−R1R_2 \to R_2-R_1, R3→R3−R2R_3 \to R_3-R_2 After row operations, the inner determinant simplifies to 22. Step 3: Combine
Δ=(n!)3⋅2(n+1)2(n+2)=(n!)3(2n3+8n2+10n+4)\Delta = (n!)^3 \cdot 2(n+1)^2(n+2) = (n!)^3(2n^3+8n^2+10n+4)
 Answer: (1)
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