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 $$k \neq 2, k \neq -2$$.
(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 $$k \neq -2$$.
Which of the following statements are correct?

The system of equations is $$x - 2y = 1$$, $$x - y + kz = -2$$, and $$ky + 4z = 6$$.

The coefficient matrix is $$\begin{pmatrix} 1 & -2 & 0 \\ 1 & -1 & k \\ 0 & k & 4 \end{pmatrix}$$. Computing the determinant by expanding along the first row: $$\Delta = 1 \cdot [(-1)(4) - k \cdot k] - (-2)[1 \cdot 4 - k \cdot 0] + 0 = (-4 - k^2) + 8 = 4 - k^2$$.

The determinant equals zero when $$k = \pm 2$$. When $$k \neq 2$$ and $$k \neq -2$$, the determinant is non-zero and the system has a unique solution. This confirms that statement (A) is correct.

For $$k = 2$$, the system becomes $$x - 2y = 1$$, $$x - y + 2z = -2$$, and $$2y + 4z = 6$$. Subtracting the first equation from the second gives $$y + 2z = -3$$. From the third equation, $$2y + 4z = 6$$ simplifies to $$y + 2z = 3$$. Since $$y + 2z$$ cannot simultaneously equal $$-3$$ and $$3$$, the system is inconsistent and has no solution when $$k = 2$$. This confirms that statement (D) is correct.

Statement (B) is incorrect because when $$k = -2$$, the determinant is zero, so there is no unique solution. Statement (C) is incorrect since $$k = 2$$ leads to no solution. Statement (E) is incorrect because the system does not have infinite solutions for all $$k \neq -2$$.

Therefore, the correct statements are (A) and (D) only.