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NEST2026MathematicsDeterminants

Let S be the set of real values of k for which the system of equations kx + y + z = k x + ky + z = k x + y + kz = k has no solution. Then |S| is

Options

  1. A2
  2. B0
  3. C1

Correct answer

C. 1

Step-by-step solution

The given system of equations is: kx + y + z = k x + ky + z = k x + y + kz = k For the system to have no solution, the determinant of the coefficient matrix must be zero. = vmatrix k & 1 & 1 1 & k & 1 1 & 1 & k vmatrix = k(k^2 - 1) - 1(k - 1) + 1(1 - k) = (k - 1)[k(k + 1) - 1 - 1] = (k - 1)(k^2 + k - 2) = (k - 1)^2(k + 2) Setting = 0 gives k = 1 or k = -2 . For k = 1 , the system becomes: x + y + z = 1 x + y + z = 1 x + y + z = 1 This represents a single plane, so the system has infinitely many solutions. For k = -

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