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JEE Advanced 2017 · Paper 1, Mathematics question 10

For a real number α\alpha, if the system

[1αα2α1αα2α1][xyz]=[1−11]\begin{bmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}

of linear equations, has infinitely many solutions, then 1+α+α2=1 + \alpha + \alpha^2 =

Answer and worked solution

Official answer: 1

Determinant.
det⁡=1(1−α2)−α(α−α3)+α2(α2−α2)=(1−α2)2.\det=1(1-\alpha^2)-\alpha(\alpha-\alpha^3)+\alpha^2(\alpha^2-\alpha^2)=(1-\alpha^2)^2.
Infinitely many solutions needs det⁡=0\det=0, so α=±1\alpha=\pm1.

Check consistency.

  • α=1\alpha=1: every row is (1,1,1)(1,1,1), but the right side is (1,−1,1)(1,-1,1). Inconsistent.
  • α=−1\alpha=-1: the rows are (1,−1,1)(1,-1,1), (−1,1,−1)(-1,1,-1), (1,−1,1)(1,-1,1) and the right side is (1,−1,1)(1,-1,1). Consistent, with infinitely many solutions.

So α=−1\alpha=-1 and 1+α+α2=11+\alpha+\alpha^2=1.

Answer: 1

Topics:Systems of linear equationsDeterminants and properties
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