For the system of linear equations :
x - 2y = 1, x - y + kz = -2, ky + 4z = 6, `k in R`, consider the following statements :
(A) The system has unique solution if `k ne 2, k ne - 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 ne 2`.
Which of the following statements are correct?

Options

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

Answer

To solve the system of linear equations given by:

1. \( x - 2y = 1 \)
2. \( x - y + kz = -2 \)
3. \( ky + 4z = 6 \)

we will analyze the conditions under which the system has a unique solution, no solution, or infinitely many solutions.

### Step 1: Formulate the Coefficient Matrix
The coefficient matrix \( A \) for the system can be represented as follows:

\[ A = \begin{bmatrix} 1 & -2 & 0 \\ 1 & -1 & k \\ 0 & k & 4 \end{bmatrix} \]

### Step 2: Calculate the Determinant of the Coefficient Matrix
To find the conditions for the existence of solutions, we need to calculate the determinant of the matrix \( A \):

\[ \text{det}(A) = \begin{vmatrix} 1 & -2 & 0 \\ 1 & -1 & k \\ 0 & k & 4 \end{vmatrix} \]

Using the method of cofactor expansion along the first row, we have:

\[ \text{det}(A) = 1 \cdot \begin{vmatrix} -1 & k \\ k & 4 \end{vmatrix} - (-2) \cdot \begin{vmatrix} 1 & k \\ 0 & 4 \end{vmatrix} + 0 \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} -1 & k \\ k & 4 \end{vmatrix} = (-1)(4) - (k)(k) = -4 - k^2 \)

2. \( \begin{vmatrix} 1 & k \\ 0 & 4 \end{vmatrix} = (1)(4) - (0)(k) = 4 \)

Thus, we have:

\[ \text{det}(A) = -4 - k^2 + 8 = 4 - k^2 \]

### Step 3: Analyze the Determinant
The determinant \( \text{det}(A) = 4 - k^2 \) gives us critical points:

- The determinant is zero when \( k = 2 \) or \( k = -2 \).
- If \( k \neq 2 \) and \( k \neq -2 \), the determinant is non-zero, indicating that the system has a unique solution.

### Step 4: Determine the Nature of Solutions
1. **Unique Solution**: The system has a unique solution if \( k \neq 2 \) and \( k \neq -2 \).
2. **No Solution or Infinite Solutions**: If \( k = 2 \) or \( k = -2 \), we need to analyze further.

### Step 5: Check for No Solution or Infinite Solutions
To check for no solution or infinite solutions when \( k = 2 \):

Substituting \( k = 2 \) into the equations, we have:

1. \( x - 2y = 1 \)
2. \( x - y + 2z = -2 \)
3. \( 2y + 4z = 6 \)

Rearranging the third equation gives us \( y + 2z = 3 \) or \( y = 3 - 2z \).

Substituting \( y = 3 - 2z \) into the first two equations leads to a contradiction, confirming that there is no solution for \( k = 2 \).

### Conclusion
- **Statement A**: Correct (The system has a unique solution if \( k \neq 2 \) and \( k \neq -2 \)).
- **Statement B**: Incorrect (The system does not have a unique solution when \( k = -2 \)).
- **Statement C**: Incorrect (The system does not have a unique solution when \( k = 2 \)).
- **Statement D**: Correct (The system has no solution if \( k = 2 \)).
- **Statement E**: Incorrect (The system does not have infinitely many solutions if \( k \neq 2 \)).

### Final Answer
The correct statements are A and D.

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