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Determinants

Question
CBSEENMA12036010

The system of linear equations x+λy−z=0; λx−y−z=0; x+y−λz=0 has a non-trivial solution for

  • infinitely many values of λ.

  • exactly one value of λ.

  • exactly two values of λ.

  • exactly three values of λ.

Solution

D.

exactly three values of λ.

Given system of linear equations is 
x+λy−z=0;
λx−y−z=0;
x+y−λz=0 
Note that, given system will have a non-trivial solution only if the determinant of the coefficient matrix is zero, ie.
open vertical bar table row 1 straight lambda cell negative 1 end cell row straight lambda cell negative 1 end cell cell negative 1 end cell row 1 1 cell negative straight lambda end cell end table close vertical bar space equals 0

rightwards double arrow space 1 space left parenthesis straight lambda space plus 1 right parenthesis minus straight lambda left parenthesis negative straight lambda squared space plus 1 right parenthesis minus 1 left parenthesis straight lambda plus 1 right parenthesis space equals 0
rightwards double arrow space space straight lambda plus 1 space plus straight lambda squared minus straight lambda minus straight lambda minus 1 space equals 0
rightwards double arrow straight lambda cubed minus straight lambda space equals space 0
rightwards double arrow straight lambda left parenthesis straight lambda squared minus 1 right parenthesis space equals 0
straight lambda space equals space 0 space or space straight lambda space plus-or-minus 1
Hence, given system of linear equation has a non-trivial solution for exactly three values of λ.

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