kx+y+z=1
x+ky+z=k
x+y+zk=k2
Δ=∣K111K111K∣=K(K2−1)−1(K−1)+1(1−K)
=K3−K−K+1+1−K
=K3−3K+2
=(K−1)2(K+2)
For K=1
Δ=Δ1=Δ2=Δ3=0
But for K=−2, at least one out of Δ1,Δ2,Δ3 are not zero
Hence for no solution, K=−2
The system of equations kx+y+z=1,x+ky+z=k and x+y+zk=k2 has no solution if k is equal to:
Held on 17 Mar 2021 · Verified 6 Jul 2026.
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