A) \[-1\]
B) \[0\]
C) \[1\]
D) \[2\]
Correct Answer: C
Solution :
Key Idea: If the system of equations has non-zero solution, then value of determinant should be zero. Given, system of equations are\[x+ky-z=0\],\[3x-ky-z=0\] and \[x-3y+z=0\] has non-zero solution. \[\therefore \] \[\left| \begin{matrix} 1 & k & -1 \\ 3 & -k & -1 \\ 1 & -3 & 1 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[1(-k-3)-k(3+1)-1(-9+k)=0\] \[\Rightarrow \] \[-6k+6=0\] \[\Rightarrow \] \[k=1\] Note: If the system of equations has zero solution, then the value of determinant should not be zero.
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