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

Question

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

Accepted Answer

To determine why Option 2 is the correct answer, we need to analyze the system of equations:

1. $ x - ky + z = 0 $
2. $ kx + 3y - kz = 0 $
3. $ 3x + y - z = 0 $

The problem states that the trivial solution (i.e., $ x = 0, y = 0, z = 0 $) is the only solution. This implies that the determinant of the coefficient matrix of the system must be non-zero. The coefficient matrix is:

$$
\begin{bmatrix}
1 & -k & 1 \
k & 3 & -k \
3 & 1 & -1
\end{bmatrix}
$$

To find when the determinant is non-zero, we calculate the determinant of this matrix:

$$
\text{det} = \begin{vmatrix}
1 & -k & 1 \
k & 3 & -k \
3 & 1 & -1
\end{vmatrix}
$$

Expanding along the first row, the determinant is:

$$
= 1 \cdot \begin{vmatrix} 3 & -k \ 1 & -1 \end{vmatrix} - (-k) \cdot \begin{vmatrix} k & -k \ 3 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} k & 3 \ 3 & 1 \end{vmatrix}
$$

Calculating each of these 2x2 determinants:

1. $\begin{vmatrix} 3 & -k \ 1 & -1 \end{vmatrix} = (3)(-1) - (-k)(1) = -3 + k$
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 expression:

$$
= 1(-3 + k) + k(2k) + 1(k - 9)
$$
$$
= -3 + k + 2k^2 + k - 9
$$
$$
= 2k^2 + 2k - 12
$$

For the trivial solution to be the only solution, this determinant must be non-zero:

$$
2k^2 + 2k - 12 \neq 0
$$

Factoring the quadratic:

$$
2(k^2 + k - 6) = 2(k - 2)(k + 3)
$$

Thus, the determinant is zero when $ k = 2 $ or $ k = -3 $. Therefore, for the determinant to be non-zero, $ k $ must not be 2 or -3. Hence, the set of all values of $ k $ for which the trivial solution is the only solution is $ \mathbb{R} - \{2, -3\} $.

**Explanation of Options:**

1. **Option 1: {2, -3}** - This option suggests that $ k $ can only be 2 or -3, which is incorrect because these values make the determinant zero, allowing non-trivial solutions.

2. **Option 2: $ \mathbb{R} - \{2, -3\} $** - This is correct because it excludes the values of $ k $ that make the determinant zero, ensuring the trivial solution is the only solution.

3. **Option 3: $ \mathbb{R} - \{2, 3\} $** - This option incorrectly excludes 3 instead of -3. The value 3 does not make the determinant zero, so it should not be excluded.

4. **Option 4: {2, 3}** - This option is incorrect because it includes values that do not make the determinant zero, specifically 3, which does not affect the determinant being zero.

Thus, Option 2 is the correct answer.