`kx+y+z=1 , x+ky+z=k , x+y+kz=k^2` be the system of equations with no solution , then k=

Answer

To solve the given system of equations for the value of \( k \) such that the system has no solution, we will follow these steps:

### Step 1: Write the system of equations
The given equations are:
1. \( kx + y + z = 1 \)
2. \( x + ky + z = k \)
3. \( x + y + kz = k^2 \)

### Step 2: Form the coefficient matrix
The coefficient matrix \( A \) for the system can be written as:
\[ A = \begin{bmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{bmatrix} \]

### Step 3: Calculate the determinant of the matrix
To find the determinant \( \Delta \) of matrix \( A \), we can use the formula for the determinant of a 3x3 matrix:
\[ \Delta = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix:
\[ \Delta = k \begin{vmatrix} k & 1 \\ 1 & k \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & k \end{vmatrix} + 1 \begin{vmatrix} 1 & k \\ 1 & 1 \end{vmatrix} \]

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

### Step 4: Set the determinant to zero for no solution
For the system to have no solution, the determinant must be equal to zero:
\[ k^3 - 3k + 2 = 0 \]

### Step 5: Factor the polynomial
To solve \( k^3 - 3k + 2 = 0 \), we can try to find rational roots using the Rational Root Theorem. Testing \( k = 1 \):
\[ 1^3 - 3(1) + 2 = 0 \]
Thus, \( k = 1 \) is a root. We can factor the polynomial as:
\[ (k - 1)(k^2 + k - 2) = 0 \]

### Step 6: Solve the quadratic equation
Now we solve \( k^2 + k - 2 = 0 \) using the quadratic formula:
\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4(1)(-2)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2} \]
This gives us:
\[ k = 1 \quad \text{or} \quad k = -2 \]

### Step 7: Determine the condition for no solution
We have two potential values for \( k \): \( k = 1 \) and \( k = -2 \).

- For \( k = 1 \), substituting back into the equations shows that they become dependent, leading to infinite solutions.
- For \( k = -2 \), we need to check if the equations yield a contradiction.

### Conclusion
Thus, the value of \( k \) for which the system has no solution is:
\[ \boxed{-2} \]