Finding Conditions for a Unique Solution in Linear Equations

A system of linear equations has a unique solution if and only if the determinant of its coefficient matrix is non-zero. We are given the system of linear equations:

1. \(kx + y + z = 1\)
2. \(x + ky + z = 1\)
3. \(x + y + kz = 1\)

The coefficient matrix, let's call it A, for this system is:

\[ A = \begin{pmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{pmatrix} \]

To find the condition for a unique solution, we need to calculate the determinant of this matrix, \( \text{det}(A) \), and find the values of \( k \) for which \( \text{det}(A) \ne 0 \).

Calculating the Determinant of the Coefficient Matrix

We can calculate the determinant using the cofactor expansion along the first row:

\[ \text{det}(A) = k \begin{vmatrix} k & 1 \\ 1 & k \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & k \end{vmatrix} + 1 \begin{vmatrix} 1 & k \\ 1 & 1 \end{vmatrix} \]

\[ \text{det}(A) = k(k \cdot k - 1 \cdot 1) - 1(1 \cdot k - 1 \cdot 1) + 1(1 \cdot 1 - k \cdot 1) \]

\[ \text{det}(A) = k(k^2 - 1) - (k - 1) + (1 - k) \]

\[ \text{det}(A) = k^3 - k - k + 1 + 1 - k \]

\[ \text{det}(A) = k^3 - 3k + 2 \]

Finding Values of k for Non-Zero Determinant

For the system to have a unique solution, the determinant must be non-zero:

\[ k^3 - 3k + 2 \ne 0 \]

First, let's find the values of \( k \) for which \( k^3 - 3k + 2 = 0 \). We can try simple integer values. If we test \( k=1 \):

\[ (1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0 \]

So, \( k=1 \) is a root of the polynomial, which means \( (k-1) \) is a factor. We can perform polynomial division or synthetic division to find the other factors.

Using synthetic division with root 1:

The quotient is \( k^2 + k - 2 \). So, the polynomial can be factored as:

\[ k^3 - 3k + 2 = (k-1)(k^2 + k - 2) \]

Now, we factor the quadratic term \( k^2 + k - 2 \). We look for two numbers that multiply to -2 and add to 1. These numbers are 2 and -1. So, \( k^2 + k - 2 = (k+2)(k-1) \).

Substituting this back into the determinant expression:

\[ \text{det}(A) = (k-1)(k+2)(k-1) = (k-1)^2 (k+2) \]

For a unique solution, we need \( \text{det}(A) \ne 0 \):

\[ (k-1)^2 (k+2) \ne 0 \]

This inequality holds if and only if both factors are non-zero:

• \( (k-1)^2 \ne 0 \implies k-1 \ne 0 \implies k \ne 1 \)
• \( k+2 \ne 0 \implies k \ne -2 \)

Therefore, the system of linear equations has a unique solution if and only if \( k \ne 1 \) and \( k \ne -2 \).

Comparing with Given Options

We compare our derived condition with the provided options:

• Option 1: \( k \ne 1 \) and \( k \ne -2 \)
• Option 2: \( k \ne 1 \) and \( k \ne 2 \)
• Option 3: \( k \ne -1 \) and \( k \ne -2 \)
• Option 4: \( k \ne -1 \) and \( k \ne 2 \)

Our condition matches Option 1.

Revision Table: System of Linear Equations

Additional Information: Determinant and System Solutions

The determinant of the coefficient matrix is a fundamental tool in analyzing systems of linear equations. For a square system (number of equations equals number of variables), the determinant provides quick insight into the nature of solutions:

• Unique Solution: A non-zero determinant indicates that the matrix is invertible, and the system has exactly one solution. This is often solved using Cramer's rule or matrix inversion.
• Determinant is Zero: If the determinant is zero, the matrix is singular (not invertible). In this case, the system either has no solution (inconsistent) or infinitely many solutions (consistent and dependent). To distinguish between these two cases when the determinant is zero, one would typically analyze the augmented matrix (coefficient matrix combined with the constant terms) using methods like Gaussian elimination or comparing ranks of the coefficient matrix and the augmented matrix.

In this specific problem, we only needed the condition for a unique solution, which solely depends on the coefficient matrix's determinant being non-zero.