The system of equations $ k x+y+z=1, x+k y+z=k $ and $ x+y+z k=k^{2} $ has no solution if $ k $ is equal to:

Question

The system of equations $ k x+y+z=1, x+k y+z=k $ and $ x+y+z k=k^{2} $ has no solution if $ k $ is equal to:, $ 0 $, $ 1 $, $ -1 $, $ -2 $

MARKS App · Accepted Answer

k x + y + z = 1 x + k y + z = k x + y + z k = k 2 Δ = K 1 1 1 K 1 1 1 K = K K 2 - 1 - 1 K - 1 + 1 1 - K = K 3 - K - K + 1 + 1 - K = K 3 - 3 K + 2 = ( K - 1 ) 2 K + 2 For K = 1 Δ = Δ 1 = Δ 2 = Δ 3 = 0 But for K = - 2 , at least one out of Δ 1 , Δ 2 , Δ 3 are not zero Hence for no solution, K = - 2

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