Let the following system of equations
`{:(kx+y+z=1),(x+ky+z=k),(x+y+kz=k^2):}`
has no solution . Find |k|.

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the given system of equations for the value of \( |k| \) such that the system has no solution, we need to analyze the coefficient matrix and find its determinant.

The system of equations is:
1. \( kx + y + z = 1 \)
2. \( x + ky + z = k \)
3. \( x + y + kz = k^2 \)

### Step 1: Write the Coefficient Matrix
The coefficient matrix \( A \) for the system can be represented as:
\[ A = \begin{bmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{bmatrix} \]

### Step 2: Calculate the Determinant of the Coefficient Matrix
To find the values of \( k \) for which the system has no solution, we need to compute the determinant of matrix \( A \) and set it to zero:
\[ \text{det}(A) = \begin{vmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{vmatrix} \]

Using the determinant formula for a \( 3 \times 3 \) matrix, we expand the determinant:
\[ \text{det}(A) = k \begin{vmatrix} k & 1 \\ 1 & k \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & k \end{vmatrix} + 1 \begin{vmatrix} 1 & k \\ 1 & 1 \end{vmatrix} \]

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

### 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: Factor the Polynomial
To solve \( k^3 - 3k + 2 = 0 \), we can try to find rational roots using the Rational Root Theorem. Testing possible roots, we find:
1. \( k = 1 \):
\[ 1^3 - 3(1) + 2 = 1 - 3 + 2 = 0 \]
So, \( k = 1 \) is a root.

Now we can factor \( k - 1 \) out of \( k^3 - 3k + 2 \):
Using synthetic division or polynomial long division, we divide \( k^3 - 3k + 2 \) by \( k - 1 \):
\[ k^3 - 3k + 2 = (k - 1)(k^2 + k - 2) \]

### Step 5: Factor the Quadratic
Now we factor \( k^2 + k - 2 \):
\[ k^2 + k - 2 = (k - 1)(k + 2) \]

Thus, we have:
\[ k^3 - 3k + 2 = (k - 1)^2 (k + 2) = 0 \]

### Step 6: Solve for \( k \)
Setting each factor to zero gives:
1. \( k - 1 = 0 \) → \( k = 1 \)
2. \( k + 2 = 0 \) → \( k = -2 \)

### Step 7: Determine Valid Values of \( k \)
Since \( k = 1 \) leads to identical equations (no unique solution), we discard it. Thus, we only consider \( k = -2 \).

### Step 8: Find the Absolute Value of \( k \)
The absolute value of \( k \) is:
\[ |k| = |-2| = 2 \]

Thus, the final answer is:
\[ \boxed{2} \]