Question
If the system of equations x-ky-z=0,kx-y-z=0,x+y-z=0 has a non-zero solution then the possible values of k are
- -1,2
- 1, 2
- (0,1)
- -1,1
Hint:
We are given system of 3 equations. We are given that it has non zero solution. We have to find find the value of k so that it has non zero solution. We will use the determinant to solve the question.
The correct answer is: -1,1
The given equations are as follows:
x - ky - z = 0
kx - y - z = 0
x + y - z = 0
When the system of equations has non zero solution the determinant of the coefficients of the equations is zero.
We will use the property to solve the question.

So, the possible values of k are -1, 1.
For such questions, we should remember the requirement for non zero solution. We have to be careful when finding the determinant.
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