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← JEE Advanced 2015 · Paper 1
MathematicsMultiple correct+4 / -2Question 50 of 60

JEE Advanced 2015 · Paper 1, Mathematics question 50

Which of the following values of α\alpha satisfy the equation
∣(1+α)2(1+2α)2(1+3α)2(2+α)2(2+2α)2(2+3α)2(3+α)2(3+2α)2(3+3α)2∣=−648α ?\begin{vmatrix}(1+\alpha)^2&(1+2\alpha)^2&(1+3\alpha)^2\\(2+\alpha)^2&(2+2\alpha)^2&(2+3\alpha)^2\\(3+\alpha)^2&(3+2\alpha)^2&(3+3\alpha)^2\end{vmatrix}=-648\alpha\ ?

Choose one or more options
Answer and worked solution

Official answer: B, C

Factorise. Each entry is (i+jα)2=i2⋅1+2i⋅jα+1⋅j2α2(i+j\alpha)^2=i^2\cdot1+2i\cdot j\alpha+1\cdot j^2\alpha^2, so the matrix is the product UVUV with
U=[121441961] (rows i2,2i,1),V=[111α2α3αα24α29α2].U=\begin{bmatrix}1&2&1\\4&4&1\\9&6&1\end{bmatrix}\ (\text{rows }i^2,2i,1),\qquad V=\begin{bmatrix}1&1&1\\\alpha&2\alpha&3\alpha\\\alpha^2&4\alpha^2&9\alpha^2\end{bmatrix}.

Determinants. det⁡U=−4\det U=-4 and det⁡V=2α3\det V=2\alpha^3, so the determinant is −8α3-8\alpha^3.

Solve. −8α3=−648α-8\alpha^3=-648\alpha gives α2=81\alpha^2=81, so α=±9\alpha=\pm9.

(B), (C)

Topics:Determinants and properties
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