Question

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

• A. -1, 2
• B. 1, 2
• C. 0, 1
• D. -1, 1

Answer 1

Answer: (D)

Since, the given system has non-zero solution.
$\therefore \left|\begin{array}{ccc}1& -k& -1\\ k& -1& a-1\\ 1& 1& -1\end{array}\right|=0$
Applying $C_1\to C_1-C_2,C_2\to C_2+C_3$
$\Rightarrow \left|\begin{array}{ccc}1+k& -k-1& -1\\ 1+k& -2& -1\\ 0& 0& -1\end{array}\right|=0$
$\Rightarrow 2(k+1)-(k+1{)}^2=0$
$\Rightarrow (k+1)(2-k-1)=0\Rightarrow k=\pm 1$
NOTE There is a golden rule in determinant that $n$ one's $\Rightarrow (n-1)$ zero's or $n$ (constant) $\Rightarrow (n-1)$ zero's for all constant should be in a single row or a single column.