Question

If the system of equation:- x – ky -z =0, kx – y – z =0 and x + y -z =0 has non-zero solutions, then the possible values of k are:-
A) $-1,2$
B) 1, 2
C) 0, 1
D) -1, 1

Answer 1

Hint:
Since, the system of equations has non-zero solutions, equate determinant of coefficient matrix to zero. Hence, by solving determinant find k.

Complete step by step solution:
The equation x – ky - z = 0
          ky - y – z = 0
          x + y – z = 0
Determinant of coefficient matrix should be zero so,
 $\left|\begin{array}{lll}1& -\mathrm{k}& -1\\ \mathrm{k}& -1& -1\\ 1& 1& -1\end{array}\right|=0$
 $\Rightarrow 1\left(1+1\right)+\mathrm{k}\left(-\mathrm{k}+1\right)-1\left(\mathrm{k}+1\right)=0$
 $\Rightarrow =2-{\mathrm{k}}^2+\mathrm{k}-\mathrm{k}-1=0$
 $\Rightarrow -{\mathrm{k}}^2+1=0$
 $\Rightarrow {\mathrm{k}}^2=1$
 $\Rightarrow \mathrm{k}=\pm 1$

So, k has values: -1, 1.
So, options (D) is correct.

Note:
Conditioning when a system of equations has non-zero solutions is important to solve this type of question. For a homogeneous system to be having non-zero solution, determinant of coefficient matrix should be zero.