To solve the given system of linear equations for different values of \(k\) and determine when the system has no solutions, exactly one solution, or infinitely many solutions, we analyze the system of equations:

\(x + ky = 1\)

\(kx + y = -1\)

First, write the system of equations in matrix form:

To determine the conditions for no solution, exactly one solution, or infinitely many solutions, we calculate the determinant of the coefficient matrix. The system is invertible (has a unique solution) if the determinant is non-zero:

The determinant of the coefficient matrix is:

\(\text{det} = 1 \cdot 1 - k \cdot k = 1 - k^2\)

• If \(1 - k^2 \neq 0\), then the system has a unique solution.
• If \(1 - k^2 = 0\), then the determinant is zero, meaning the system could have no solution or infinitely many solutions. Thus, \(k^2 = 1\).

Solving \(k^2 = 1\), we find:

• \(k = 1\)
• \(k = -1\)

Let's analyze each case for \(k = 1\) and \(k = -1\):

Case 1: \(k = 1\)

Substituting \(k = 1\) in the original equations:

\(x + y = 1\)

\(x + y = -1\)

This leads to an inconsistent system (the lines are parallel and distinct), which has no solution.

Case 2: \(k = -1\)

Substituting \(k = -1\) in the original equations:

\(x - y = 1\)

\(-x + y = -1\)

This reduces to the same equation, implying the lines are coincident. Thus, there are infinitely many solutions.

Conclusion:

• There is exactly one value of \(k\) for which the system of equations has no solution: \(k = 1\).
• There exists exactly one value of \(k\) for which the system of equations has an infinite number of solutions: \(k = -1\).