Maths-
General
Easy

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. -1,2    
  2. 1, 2    
  3. (0,1)    
  4. -1,1    

hintHint:

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.
    open vertical bar table row 1 cell negative k end cell cell negative 1 end cell row k cell negative 1 end cell cell negative 1 end cell row 1 1 cell negative 1 end cell end table close vertical bar space equals space 0
1 left square bracket 1 space minus left parenthesis negative 1 right parenthesis right square bracket space minus left parenthesis negative k right parenthesis left square bracket negative k space minus left parenthesis negative 1 right parenthesis right square bracket plus left parenthesis negative 1 right parenthesis left square bracket k space plus space 1 right square bracket space equals space 0
1 left parenthesis space 1 space plus space 1 right parenthesis space plus space k left parenthesis negative k space plus space 1 right parenthesis space minus space 1 left parenthesis k space plus 1 right parenthesis space equals space 0
2 space minus k squared space plus space k space minus space k space minus 1 space equals space 0
minus k squared space plus space 1 space equals space 0
minus k squared space equals space minus 1
space space space k squared space equals space 1
T a k i n g space s q u a r e space r o o t
space space k space equals space plus-or-minus 1
    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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