For the system of linear equations:
x - 2y = 1,x - y + kz = - 2,ky + 4z = 6,k ∈ R,
consider the following statements :
(A) The system has unique solution if k ≠ 2,k ≠ - 2.
(B) The system has unique solution if k = -2
(C) The system has unique solution if k = 2
(D) The system has no solution if k = 2
(E) The system has infinite number of solutions if k ≠ -2.
Which of the following statements are correct?

Solution

Maths sir · 4 steps

Find where the determinant vanishes, then test only those values of k by eliminating x.

1. Coefficient determinant. Expand along the first row. The system has a unique solution exactly when Δ ≠ 0.

   Δ = |1, -2, 0; 1, -1, k; 0, k, 4| = 1(-4 - k²) + 2(4 - 0) = 4 - k²

2. Unique case. Δ = 0 only when k = 2 or k = -2. For every other k the solution is unique, so (A) is true. This also rules out (C) and (E).
3. Eliminate x. Subtract the first equation from the second to get y + kz = -3. Compare it with the third equation, ky + 4z = 6, at each critical value of k.

   Value of k	From equations 1 and 2	Third equation	Result
   k = 2	y + 2z = -3	2y + 4z = 6, that is y + 2z = 3	contradiction, no solution
   k = -2	y - 2z = -3	-2y + 4z = 6, that is y - 2z = -3	same equation, infinitely many solutions

4. So. (A) and (D) are true, while (B), (C) and (E) are false. The answer is (d).

The idea to keep. Δ ≠ 0 settles the unique case at once. At each k that makes Δ = 0, reduce the equations and check whether they agree (infinitely many) or clash (none).

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