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,{\rm{\;}}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,
 $\left| {\begin{array}{*{20}{l}}
1&{ - {\rm{k}}}&{ - 1}\\
{\rm{k}}&{ - 1}&{ - 1}\\
1&1&{ - 1}
\end{array}} \right| = 0$
 $ \Rightarrow 1\left( {1 + 1} \right) + {\rm{k}}\left( { - {\rm{k}} + 1} \right) - 1\left( {{\rm{k}} + 1} \right) = 0$
 $ \Rightarrow = 2 - {{\rm{k}}^2} + {\rm{k}} - {\rm{k}} - 1 = 0$
 $ \Rightarrow - {{\rm{k}}^2} + 1 = 0$
 $ \Rightarrow {{\rm{k}}^2} = 1$
 $ \Rightarrow {\rm{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.