Consider the given system of linear equations for variables $x$ and $y$, where $k$ is a real-valued constant. Which of the following option(s) is/are CORRECT? $$x + ky = 1$$ $$kx + y = -1$$

To analyze the system of linear equations:

The determinant of the coefficient matrix $A$ is:

1. Unique solution: If $\det(A) \neq 0$, which means $k \neq 1$ and $k \neq -1$, the system has a unique solution. Since this holds for infinitely many real values of $k$, statement (C) is incorrect.

2. Case $k = 1$: The equations become:

   $$x + y = 1$$

   $$x + y = -1$$

   These two parallel lines never intersect, so there is no solution. Thus, $k = 1$ is the only value for which the system has no solution, making statement (A) correct, and statement (B) incorrect.

3. Case $k = -1$: The equations become:

   $$x - y = 1$$

   $$-x + y = -1 \implies x - y = 1$$

   Both equations represent the exact same line, resulting in infinitely many solutions. Thus, $k = -1$ is the only value for which the system has infinitely many solutions, making statement (D) correct.

Therefore, the correct options are A and D.