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

- 2

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Solution

The correct option is A $- 2$
Given system of equations
$k x + y + z = 1 x + k y + z = k x + y + z k = k^{2}$
Now,
$D = ∣ ∣ ∣ ∣ \begin{matrix} k & 1 & 1 1 & k & 1 1 & 1 & k \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow k (k^{2} - 1) - (k - 1) + (1 - k) = 0 \Rightarrow (k - 1) (k^{2} + k - 1 - 1) = 0 \Rightarrow (k - 1) (k^{2} + k - 2) \Rightarrow (k - 1) (k - 1) (k + 2) = 0 \Rightarrow k = 1, k = 2$
For $k = 1$ equations becomes identical, so $k = - 2$ for no solution.

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