91Ó°ÊÓ

A system of equations is a set of two or more equations with the same variables. In the given exercise, we have the system of equations:
\(k x + y + z = 1\)
x + \(k y + z = k\)
x + y + \(k z = k^2\)
When solving a system of equations, the goal is to find values for the variables that satisfy all equations simultaneously. There are different methods to solve such systems: substitution, elimination, or matrix operations like using determinants.
When dealing with more complex systems, using determinants can simplify the process significantly, especially for larger systems involving three or more equations.

Determinants are numerical values computed from a square matrix. They are used in various algebraic operations, including solving systems of linear equations. For this exercise, we consider the coefficient matrix of the given system:
\(M = \begin{pmatrix} k & 1 & 1 \ 1 & k & 1 \ 1 & 1 & k\text{ \)
Besides straightforward calculation, determinants can help determine whether a system has a unique solution, no solution, or infinitely many solutions. For a 3x3 matrix like this one, the determinant is computed as follows: \(\text{det}(M) = k \times (k^2 - 1) - 1 \times (1 - k) + 1 \times (1 - k)\).
Simplifying this, we get: \(\text{det}(M) = k^3 - 2k + 1\).
Setting this determinant to zero helps identify critical values for k where the system’s behavior changes dramatically.

A system of equations yields infinite solutions if the equations describe the same plane or line in geometric terms. This happens when the determinant of the system’s coefficient matrix is zero, indicating that the system's equations are not independent.
Let's analyze this in the context of our problem:
We have already derived that the determinant of our coefficient matrix is: \(\text{det}(M) = k^3 - 2k + 1\).
Solving for when this determinant equals zero, we find: \(k = 1\) and \(k = -1\). These are the roots of the equation.
For these values, the system does not have a unique solution. Specifically, when \(k = 1\text{ \), all three rows of the matrix become identical, leading to infinite solutions. This is why Statement-1 is true, as it specifically notes \(k = 1\text{ \) as a case of infinite solutions. Statement-2 provides the mathematical basis for this conclusion, indicating that when the determinant is zero, it verifies the condition for infinite solutions.