If the trivial solution is the only solution of the system of equations `x-ky + z = 0, kx + 3y-kz=0, 3x + y-z = 0` Then the set of all values of `k` is:

Options

A. {2,-3}
B. R-{2,-3}
C. R-{2}
D. R-{-3}

Answer

To solve the problem, we need to determine the values of \( k \) such that the system of equations has only the trivial solution. The given system of equations is:

1. \( x - ky + z = 0 \)
2. \( kx + 3y - kz = 0 \)
3. \( 3x + y - z = 0 \)

We can express this system in matrix form as \( A \mathbf{X} = 0 \), where \( A \) is the coefficient matrix and \( \mathbf{X} \) is the vector of variables \( \begin{bmatrix} x \\ y \\ z \end{bmatrix} \).

### Step 1: Form the Coefficient Matrix
The coefficient matrix \( A \) can be written as:

\[ A = \begin{bmatrix} 1 & -k & 1 \\ k & 3 & -k \\ 3 & 1 & -1 \end{bmatrix} \]

### Step 2: Find the Determinant of the Matrix
To find the values of \( k \) for which the only solution is the trivial solution, we need to set the determinant of \( A \) equal to zero:

\[ \text{det}(A) = \begin{vmatrix} 1 & -k & 1 \\ k & 3 & -k \\ 3 & 1 & -1 \end{vmatrix} \]

Calculating the determinant using the formula for a 3x3 matrix:

\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 3 & -k \\ 1 & -1 \end{vmatrix} - (-k) \cdot \begin{vmatrix} k & -k \\ 3 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} k & 3 \\ 3 & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 3 & -k \\ 1 & -1 \end{vmatrix} = (3)(-1) - (-k)(1) = -3 + k = k - 3 \)
2. \( \begin{vmatrix} k & -k \\ 3 & -1 \end{vmatrix} = (k)(-1) - (-k)(3) = -k + 3k = 2k \)
3. \( \begin{vmatrix} k & 3 \\ 3 & 1 \end{vmatrix} = (k)(1) - (3)(3) = k - 9 \)

Substituting these back into the determinant:

\[ \text{det}(A) = 1(k - 3) + k(2k) + 1(k - 9) \]
\[ = k - 3 + 2k^2 + k - 9 \]
\[ = 2k^2 + 2k - 12 \]

### Step 3: Set the Determinant Equal to Zero
Now, we set the determinant equal to zero to find the values of \( k \):

\[ 2k^2 + 2k - 12 = 0 \]

Dividing the entire equation by 2:

\[ k^2 + k - 6 = 0 \]

### Step 4: Factor the Quadratic Equation
We can factor this quadratic equation:

\[ (k + 3)(k - 2) = 0 \]

### Step 5: Solve for \( k \)
Setting each factor equal to zero gives us:

1. \( k + 3 = 0 \) → \( k = -3 \)
2. \( k - 2 = 0 \) → \( k = 2 \)

### Final Answer
The set of all values of \( k \) for which the trivial solution is the only solution is:

\[ \{ -3, 2 \} \]