Question

Consider the system of the equations \(k x+y+z=1, x+k y+z=k\) and \(x+y+k z=k^{2}\) Statement- 1: System of equations has infinite solutions for \(k=1\). Statement- \(2:\) : The determinant \(\left|\begin{array}{lll}1 & 1 & 1 \\ k & k & 1 \\ k^{2} & 1 & k\end{array}\right|=0\) then \(k=1\) or \(-1\) (1) Statement \(-1\) is True, Statement \(-2\) is True ; Statement \(-2\) is a correct explanation for Statement \(-1\) (2) Statement- 1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation for Statement-1 (3) Statement \(-1\) is True, Statement \(-2\) is False (4) Statement \(-1\) is False, Statement \(-2\) is True

Short answer

Option (1) is correct: Statement-1 is True, Statement-2 is True, and Statement-2 is a correct explanation for Statement-1.

Step-by-step solution

Set Up the Determinant

Consider the system's coefficient matrix and form the determinant. The matrix is: \[M = \begin{pmatrix} k & 1 & 1 \ 1 & k & 1 \ 1 & 1 & k \end{pmatrix} \]The determinant of this matrix should be calculated as follows: \[\text{det}(M) = \left| \begin{array}{ccc} k & 1 & 1 \ 1 & k & 1 \ 1 & 1 & k \end{array} \right| \]

Calculate the Determinant

Using the formula for the determinant of a 3x3 matrix, compute: \[\text{det}(M) = k \cdot (k^2 - 1) - 1 \cdot (1 - k) + 1 \cdot (1 - k) \]This simplifies to: \[k^3 - k - 1 + k + 1 - k \]Finally, simplifying further: \[k^3 - 2k + 1 \]

Solve Determinant for Zero

Set the determinant to zero to find the values of \(k\): \[k^3 - 2k + 1 = 0 \]Factor this equation to find the roots: \[k = 1, -1\]

Verify Statements

Compare the values found with the given statements:1. For \(k = 1\), the determinant is zero and the system has infinite solutions.2. The determinant equation \(k^3 - 2k + 1 = 0\) is true for \(k = 1\) and \(k = -1\).Both statements are true. Statement-2 is the correct explanation for Statement-1 since the determinant being zero confirms infinite solutions.

Key concepts

System of Equations

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

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.

Infinite Solutions

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.