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 If the system of equations x=ky+z,y=kxzandz=x+yhas a non-zero solution, then the possible values of ‘k’ are 

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a
-1,2
b
1,2
c
0,1
d
-1,1

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detailed solution

Correct option is D

The system has non-zero solution⇒1-k-1k-1-111-1=0⇒1(1+1)+k(-k+1)-1(k+1)=0⇒-k2+1=0⇒k2=1⇒k=-1,1


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