The system of equations kx + y + z = 1, x + ky + z = k and x + y + zk = k² has no solution if k is equal to :

Solution

Maths sir · 3 steps

Find where the determinant vanishes, then add the equations to see which value gives a contradiction.

1. Coefficient determinant. A unique solution exists whenever D ≠ 0. So no solution is possible only at values of k that make D = 0.

   D = |k, 1, 1; 1, k, 1; 1, 1, k| = k³ - 3k + 2 = (k-1)²(k+2)

2. Two candidates. D = 0 only for k = 1 or k = -2. A zero determinant means either no solution or infinitely many, so test each one directly.

   k	D	What the system becomes
   0	2	D ≠ 0, unique solution
   -1	4	D ≠ 0, unique solution
   1	0	all three equations are x + y + z = 1, so infinitely many solutions
   -2	0	adding all three gives 0 = 1 - 2 + 4 = 3, a contradiction

3. So. At k = -2 the left sides add up to zero but the right sides add up to 3, so the equations cannot all hold. The answer is (c), k = -2.

The idea to keep. A zero determinant does not mean no solution by itself. It only means the solution is not unique. Plug the value back in and check whether the equations agree or contradict.

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