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 \( k \) for which the system of equations has only the trivial solution. This occurs when the determinant of the coefficient matrix is non-zero. Let's proceed step by step.

### Step 1: Write the system of equations in matrix form
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 in matrix form as:
\[ \begin{bmatrix} 1 & -k & 1 \\ k & 3 & -k \\ 3 & 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix
We need to calculate the determinant of the matrix:
\[ A = \begin{bmatrix} 1 & -k & 1 \\ k & 3 & -k \\ 3 & 1 & -1 \end{bmatrix} \]

The determinant \( |A| \) can be calculated using the formula for a 3x3 matrix:
\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the second and third rows.

Substituting the values:
\[ |A| = 1 \cdot (3 \cdot (-1) - (-k) \cdot 1) - (-k) \cdot (k \cdot (-1) - (-k) \cdot 3) + 1 \cdot (k \cdot 1 - 3 \cdot (-k)) \]

Calculating each term:
1. First term: \( 1 \cdot (-3 + k) = -3 + k \)
2. Second term: \( k \cdot (k + 3k) = k \cdot 4k = 4k^2 \)
3. Third term: \( 1 \cdot (k + 3k) = 4k \)

Putting it all together:
\[ |A| = -3 + k + 4k^2 + 4k = 4k^2 + 5k - 3 \]

### Step 3: Set the determinant not equal to zero
For the system to have only the trivial solution, we need:
\[ 4k^2 + 5k - 3 \neq 0 \]

### Step 4: Find the roots of the equation
To find the values of \( k \) that make the determinant zero, we solve:
\[ 4k^2 + 5k - 3 = 0 \]

Using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 4, b = 5, c = -3 \):
\[ k = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4} \]
\[ k = \frac{-5 \pm \sqrt{25 + 48}}{8} \]
\[ k = \frac{-5 \pm \sqrt{73}}{8} \]

### Step 5: Determine the values of \( k \)
The roots are:
\[ k_1 = \frac{-5 + \sqrt{73}}{8}, \quad k_2 = \frac{-5 - \sqrt{73}}{8} \]

### Step 6: Conclusion
The system has only the trivial solution for all values of \( k \) except \( k_1 \) and \( k_2 \). Therefore, the set of all values of \( k \) is:
\[ k \in \mathbb{R} \setminus \left\{ \frac{-5 + \sqrt{73}}{8}, \frac{-5 - \sqrt{73}}{8} \right\} \]