1) If the system of equations $x+ky-z=0$ ,$3x-ky-z=0$ and $x-3y+z=0$ has non-zero solution , the k is equal to A) -1 B) 0 C) 1 D) 2 Answer: Option CExplanation:The system has non-zero solution. if $\begin{bmatrix}1 & k&-1 \\3 & -k&-1\\1&-3&1 \end{bmatrix}=0$ $\Rightarrow$ 1(-k-3)-k(3+1)-1(-9+k)=0 $\Rightarrow$ -6k+k=0 $\Rightarrow$ k=1
