Correct : a,d

Explanation:
1. Represent the System in Matrix Form:
The given system of linear equations can be written in matrix form as Ax = B:
    [ 1   k ] [ x ] = [  1 ]
    [ k   1 ] [ y ] = [ -1 ]

2. Calculate the Determinant of Coefficient Matrix A:
    |A| = (1 × 1) - (k × k) = 1 - k²

3. Case 1: Unique Solution (|A| ≠ 0)
• A unique (exactly one) solution exists when 1 - k² ≠ 0, which means k ≠ 1 and k ≠ -1.
• Since there are infinitely many real numbers other than 1 and -1, there are an infinite number of values of k that yield a unique solution. Therefore, Statement (c) is False.

4. Case 2: Determinant is Zero (|A| = 0)
This happens when 1 - k² = 0 → k = 1 or k = -1. Let's test both values to see if they yield no solution or infinitely many solutions:

• Sub-case 2a: If k = 1
    Equation 1: x + y = 1
    Equation 2: x + y = -1
These two lines are parallel and never intersect. This means there is no solution when k = 1. Since this is the only value of k that gives no solution, Statement (a) is True and Statement (b) is False.

• Sub-case 2b: If k = -1
    Equation 1: x - y = 1 → y = x - 1
    Equation 2: -x + y = -1 → y = x - 1
Both equations reduce to the exact same line. This means there are infinitely many solutions when k = -1. Since this is the single, unique value of k that causes this, Statement (d) is True.