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← JEE Main 2014 · 12 Apr · Shift 1
MathematicsSingle correct+4 / -1Question 64 of 90

JEE Main 2014 · 12 Apr · Shift 1, Mathematics question 64

If
∣a2b2c2(a+λ)2(b+λ)2(c+λ)2(a−λ)2(b−λ)2(c−λ)2∣=kλ∣a2b2c2abc111∣, λ≠0,\begin{vmatrix}a^2&b^2&c^2\\(a+\lambda)^2&(b+\lambda)^2&(c+\lambda)^2\\(a-\lambda)^2&(b-\lambda)^2&(c-\lambda)^2\end{vmatrix}=k\lambda\begin{vmatrix}a^2&b^2&c^2\\a&b&c\\1&1&1\end{vmatrix},\ \lambda\ne0,
then k is equal to :

Choose one option
Answer and worked solution

Official answer: C

R2→R2−R3R_2\to R_2-R_3 gives 4λ(a,b,c)4\lambda(a,b,c). Then R3→R3−R1+12R2R_3\to R_3-R_1+\frac12R_2 leaves (λ2,λ2,λ2)(\lambda^2,\lambda^2,\lambda^2). So the left side is
4λ⋅λ2∣a2b2c2abc111∣,4\lambda\cdot\lambda^2\begin{vmatrix}a^2&b^2&c^2\\a&b&c\\1&1&1\end{vmatrix},
and kλ=4λ3k\lambda=4\lambda^3 gives k=4λ2k=4\lambda^2.

Topics:Determinants and properties
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