Question

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

• (A) and (D) only
• (A) and (E) only
• (C) and (D) only
• (B) and (E) only

Hint:

Cramer's Rule: If $\Delta \neq 0$, unique solution. If $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \Delta_z \neq 0$, no solution. If all are 0, infinite or no solution.

Answer 1

The Correct Option is A

Step 1: Calculate determinant \( \Delta = \begin{vmatrix} 1 & -2 & 0 \\ 1 & -1 & k \\ 0 & k & 4 \end{vmatrix} \).
Step 2: \( \Delta = 1(-4 - k^2) + 2(4 - 0) = -k^2 - 4 + 8 = 4 - k^2 \).
Step 3: For a unique solution, \( \Delta \neq 0 \Rightarrow 4 - k^2 \neq 0 \Rightarrow k \neq \pm 2 \). Statement (A) is correct.
Step 4: If \( k = 2 \), then \( \Delta = 0 \). Check \( \Delta_x = \begin{vmatrix} 1 & -2 & 0 \\ -2 & -1 & 2 \\ 6 & 2 & 4 \end{vmatrix} = 1(-4 - 4) + 2(-8 - 12) = -8 - 40 = -48 \neq 0 \).
Step 5: Since \( \Delta = 0 \) and \( \Delta_x \neq 0 \), the system has no solution for \( k = 2 \). Statement (D) is correct.