If the system of equations \( x-ky-z=0,kx-y-z=0,x+y-z=0 \) has a non-zero solution, then possible values of \( k \) are
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1,-1 \)
A homogeneous system has a non-zero solution only when its coefficient determinant is 0.
\( \displaystyle \begin{vmatrix}1 & -k & -1\\k & -1 & -1\\1 & 1 & -1\end{vmatrix}=1(1+1)+k(-k+1)-1(k+1) \)
\( \displaystyle =2-k^2+k-k-1 \)
\( \displaystyle =1-k^2 \)
Setting it equal to 0:
\( \displaystyle 1-k^2=0 \)
\( \displaystyle k=\pm 1 \)