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JEE Main 2020 · Mathematics · Matrices and Determinants · medium

The homogeneous system x+y+z=0, 3x+ky+kz=0, -3x+y+kz=0 has a non-trivial solution for exactly two real values of k. The sum of the squares of these…

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The homogeneous system x+y+z=0, 3x+ky+kz=0, −3x+y+kz=0x+y+z=0,\ 3x+ky+kz=0,\ -3x+y+kz=0 has a non-trivial solution for exactly two real values of kk. The sum of the squares of these values is
  1. A. 1616
  2. B. 1010
  3. C. 99
  4. D. 1313
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Answer: (B) 1010

A homogeneous system has non-trivial solutions exactly when the coefficient determinant vanishes: ∣1113kk−31k∣=1(k2−k)−1(3k+3k)+1(3+3k)=k2−4k+3.\begin{vmatrix}1&1&1\\3&k&k\\-3&1&k\end{vmatrix}=1(k^2-k)-1(3k+3k)+1(3+3k)=k^2-4k+3. Setting it to zero gives (k−1)(k−3)=0(k-1)(k-3)=0, so k=1k=1 or k=3k=3. The sum of squares is 12+32=101^2+3^2=10. Note that 1616 is the square of the sum and 99 is the square of the product, the common slips.
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