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JEE Advanced2018MathematicsDeterminantsActual

Let S be the of all column matrices b 1 b 2 b 3 such that b 1 , b 2 , b 3 ∈ ℝ and the system of equations (in real variables) - x + 2 y + 5 z = b 1 2 x - 4 y + 3 z = b 2 x - 2 y + 2 z = b 3 has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution of each b 1 b 2 b 3 ϵ S ?

Options

  1. Ax + 2 y + 3 z = b 1 , 4 y + 5 z = b 2 and x + 2 y + 6 z = b 3
  2. Bx + y + 3 z = b 1 ,   5 x + 2 y + 6 z = b 2 and - 2 x - y - 3 z = b 3
  3. C– x + 2 y - 5 z = b 1 , 2 x - 4 y + 10 z = b 2 and x - 2 y + 5 z = b 3
  4. Dx + 2 y + 5 z = b 1 , 2 x + 3 z = b 2 and x + 4 y - 5 z = b 3

Correct answer

A. x + 2 y + 3 z = b 1 , 4 y + 5 z = b 2 and x + 2 y + 6 z = b 3

Step-by-step solution

We find D = 0 & since no pair of planes are parallel, so there are an infinite number of solutions. So, the infinite solutions shall lie on a common line of intersection of these planes. Hence, we can write any plane as a linear combination of other two planes: α P 1 - λ P 2 = P 3 ⇒ P 1 + 7 P 2 = 13 P 3 (by comparing coefficients of x ,   y ,   z ) ⇒ b 1 + 7 b 2 = 13 b 3 (a) D ≠ 0 ⇒ unique solution for any b 1 ,   b 2 ,   b 3 (b) D = 0 but P 1 + 7 P 2 ≠ 13

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