The system of equations
- kx + 3y - 14z = 25
- 15x + 4y - kz = 3
- 4x + y + 3z = 4
is consistent for all k in the set

Solution

Maths sir · 4 steps

Find where the coefficient determinant vanishes, then check whether those values of k really break the system.

1. Coefficient determinant. Expand along the first row: D = -k(12 + k) - 3(-45 - 4k) - 14(-15 + 16). The -12k and +12k terms cancel, which leaves a very clean result.

   D = |-k, 3, -14; -15, 4, -k; -4, 1, 3| = 121 - k²

2. When D is nonzero. D = 0 only when k = 11 or k = -11. For every other real k, D ≠ 0, so the system has a unique solution and is consistent.
3. Check k = ±11. Replace the x-column with the constants 25, 3, 4 to get D₁. For k = 11, D₁ = 575 - 159 + 182 = 598. For k = -11, D₁ = 25 + 105 + 182 = 312. In both cases D = 0 but D₁ ≠ 0, so the system has no solution.
4. So. The system is consistent exactly for k ∈ ℝ - {-11, 11}, which is option (d).

The idea to keep. Use Cramer's rule. If D ≠ 0, the system has a unique solution. If D = 0 and any of D₁, D₂, D₃ is nonzero, it has no solution. Always test the values that make D zero.

Answer from the NTA final answer key. Solution by PrepWiser.