The system of equations kx + y + z = 1, x + ky + z = k and x + y + zk = k2 has no solution…

Question

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 :

Accepted Answer

$$D = \left| {\begin{pmatrix} k & 1 & 1  \
   1 & k & 1  \
   1 & 1 & k  \

\end{pmatrix} } \right| = 0$$

$$ \Rightarrow k({k^2} - 1) - (k - 1) + (1 - k) = 0$$

$$ \Rightarrow (k - 1)({k^2} + k - 1 - 1) = 0$$

$$ \Rightarrow (k - 1)({k^2} + k - 2) = 0$$

$$ \Rightarrow (k - 1)(k - 1)(k + 2) = 0$$

$$ \Rightarrow k = 1,k = -2$$

for k = 1 equation identical so k = -2 for no solution.

Solutions · 1