For what value of 'K', the system of equations `kx+y+z=1, x+ky+z=k" and "x+y+kz=K^(2)` has no solution ?

Options

A. `0`
B. `-1`
C. `1`
D. None of these

Answer

To determine the value of 'K' for which the system of equations has no solution, we need to analyze the 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 express the system of equations in matrix form \( A \mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the variable vector, and \( \mathbf{b} \) is the constant vector.

The coefficient matrix \( A \) is:
\[ A = \begin{bmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{bmatrix} \]

The variable vector \( \mathbf{x} \) is:
\[ \mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \]

The constant vector \( \mathbf{b} \) is:
\[ \mathbf{b} = \begin{bmatrix} 1 \\ k \\ K^2 \end{bmatrix} \]

### Step 2: Find the determinant of matrix \( A \)
For the system to have no solution, the determinant of matrix \( A \) must be equal to zero. We calculate the determinant of \( A \):

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

Using the determinant formula for a 3x3 matrix, we expand it:

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

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 calculation:

\[ \text{det}(A) = k(k^2 - 1) - (k - 1) + (1 - k) \]
\[ = k^3 - k - k + 1 + 1 - k \]
\[ = k^3 - 3k + 2 \]

### Step 3: 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 4: Solve the cubic equation
To solve the cubic equation, we can try to find rational roots using the Rational Root Theorem. Testing \( k = 1 \):

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

Thus, \( k - 1 \) is a factor. We can factor the cubic polynomial:

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

Next, we factor \( k^2 + k - 2 \):

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

So, the complete factorization is:

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

### Step 5: Find the roots
Setting the factors equal to zero gives us:

1. \( k - 1 = 0 \) → \( k = 1 \) (infinite solutions)
2. \( k + 2 = 0 \) → \( k = -2 \) (no solution)

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

\[ \boxed{-2} \]