Question
If the system of equations $$x - ky - z = 0, kx - y - z = 0, x + y - z = 0$$ has a non-zero solution, then the possible values of $$k$$ are
A.
$$- 1, 2$$
B.
$$1, 2$$
C.
$$0, 1$$
D.
$$- 1, 1$$
Answer :
$$- 1, 1$$
Solution :
For the given homogeneous system to have non-zero solution determinant of co-efficient matrix should be zero; i.e.,
\[ = \left| \begin{array}{l}
1\,\,\,\,\,\, - k\,\,\,\,\, - 1\\
k\,\,\,\,\, - 1\,\,\,\,\,\, - 1\\
1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\, - 1
\end{array} \right|\]
$$\eqalign{
& = 1\left( {1 + 1} \right) + k\left( { - k + 1} \right) - 1\left( {k + 1} \right) = 0 \cr
& \Rightarrow \,\,2 - {k^2} + k - k - 1 = 0 \cr
& \Rightarrow \,\,{k^2} = 1 \cr
& \Rightarrow \,\,k = \pm 1 \cr} $$