The system of equation kx + y + z = 1, x + ky + z = k and x + y + kz = `k^(2)` has no solution if k equals :

Options

A. A. 0
B. B. 1
C. C. -1
D. D. -2

Answer

To determine the value of \( k \) for which the system of equations has no solution, we will analyze the given equations step by step.

### Given Equations:
1. \( kx + y + z = 1 \) (Equation 1)
2. \( x + ky + z = k \) (Equation 2)
3. \( x + y + kz = k^2 \) (Equation 3)

### Step 1: Write the system in matrix form
We can represent the system of equations in matrix form as follows:

\[ \begin{bmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ k \\ k^2 \end{bmatrix} \]

### Step 2: Find the determinant of the coefficient matrix
To determine when the system has no solution, we need to find the determinant of the coefficient matrix and set it equal to zero.

The determinant \( D \) of the matrix is given by:

\[ D = \begin{vmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{vmatrix} \]

### Step 3: Calculate the determinant
Using the formula for the determinant of a 3x3 matrix:

\[ D = k \begin{vmatrix} k & 1 \\ 1 & k \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & k \end{vmatrix} + 1 \begin{vmatrix} 1 & k \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} k & 1 \\ 1 & k \end{vmatrix} = k^2 - 1 \)
2. \( \begin{vmatrix} 1 & 1 \\ 1 & k \end{vmatrix} = k - 1 \)
3. \( \begin{vmatrix} 1 & k \\ 1 & 1 \end{vmatrix} = 1 - k \)

Substituting these back into the determinant:

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

### Step 4: Set the determinant to zero
For the system to have no solution, we set the determinant equal to zero:

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

### Step 5: Solve the cubic equation
To find the values of \( k \), we can factor or use the Rational Root Theorem. Testing possible rational roots, we find:

1. \( k = 1 \) is a root.
2. Factoring \( k - 1 \) out of \( k^3 - 3k + 2 \):

Using synthetic division or polynomial long division, we find:

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

Now we factor \( k^2 + k - 2 \):

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

Thus, the complete factorization is:

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

### Step 6: Find the roots
Setting each factor to zero gives:

1. \( k - 1 = 0 \) → \( k = 1 \)
2. \( k + 2 = 0 \) → \( k = -2 \)

### Step 7: Determine which value leads to no solution
To check which value leads to no solution, we substitute back into the original equations:

- For \( k = 1 \):
- The equations become identical, leading to infinitely many solutions.

- For \( k = -2 \):
- The equations are distinct and do not intersect, leading to no solution.

### Conclusion
The value of \( k \) for which the system of equations has no solution is:

\[ \boxed{-2} \]