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For the system \[ x-ky+z=0 \] \[ kx+3y-kz=0 \] \[ 3x+y-z=0 \] if the zero solution is the only solution, then the possible values of \(k\) are:

1. \(\mathbb{R}\setminus\{2,-3\}\)
2. \(\mathbb{R}\setminus\{2\}\)
3. \(\mathbb{R}\setminus\{-3\}\)
4. \(\{2,-3\}\)

Correct answer: \(\mathbb{R}\setminus\{2,-3\}\)

Solution

The zero solution being the only solution indicates that the system of equations must be consistent and have a unique solution, which occurs when the determinant of the coefficient matrix is non-zero. The values of k that make the determinant zero, specifically k = 2 and k = -3, must be excluded, leading to the conclusion that k can take any real number except these two.

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