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

Explanation

To determine the values of \( k \) for which the system of equations has no solution, we can analyze the given equations:

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

We can express this system in matrix form as:

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

Let \( A \) be the coefficient matrix:

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

For the system of equations to have no solution, the determinant of matrix \( A \) must be equal to zero:

\[ \text{det}(A) = 0 \]

We can calculate the determinant of \( A \):

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

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 expression:

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

Simplifying this expression:

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

Setting the determinant to zero for no solution:

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

To find the roots, we can factor the polynomial. Testing for possible rational roots, we find:

1. \( k = 1 \) is a root.
2. Using synthetic division or polynomial long division, we can factor \( k^3 - 3k + 2 \) as \( (k - 1)(k^2 + k - 2) \).

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

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

Thus, the complete factorization is:

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

Setting each factor to zero gives:

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

The values of \( k \) for which the system has no solution are \( k = 1 \) and \( k = -2 \).

However, since the question asks for the specific condition under which the system has no solution, we need to check the condition for \( k = 1 \):

When \( k = 1 \), the equations become identical, leading to infinitely many solutions rather than no solution. Thus, the only value of \( k \) for which the system has no solution is:

\[ \boxed{-2} \]