Question

If the system of equations $x+ky-z=0,3x-ky-z=0$ and $x-3y+z=0$, has non-zero solution, then k is equal to

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

Answer: (C)

$1$

The given system of equations has non- zero solution, if

$\left|\begin{array}{ccc}1& k& -1\\ 3& -k& -1\\ 1& -3& 1\end{array}\right|=0$

$⟹1(-k-3)-k(3+1)-1(-9+k)=0$

$⟹-6k+6=0$

$⟹k=1$

Hence, option C is correct.