The system of equations kx+y+z=1, x+ky+z=k and x+y+kz=k^2 have no solution if k equals


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Question Number 88872 by Zainal Arifin last updated on 13/Apr/20

$The system of equations k x + y + z = 1,$ $x + k y + z = k and x + y + k z = k^{2} have$ $no solution if k equals$
Commented by john santu last updated on 13/Apr/20
$⇒k determinant (((k 1)),((1 k)))− determinant (((1 1)),((1 k)))+ determinant (((1 k)),((1 1)))=0 k(k^2 −1)−(k−1)+1−k = 0 k^3 −k−k+1+1−k = 0 k^3 −3k+2 = 0 (k−1)(k^2 +k−2) =0 (k−1)(k+2)(k−1) = 0 k = { (1),((−2)) :}$
$\Rightarrow k \| \begin{matrix} k 1 \\ 1 k \end{matrix} \| - \| \begin{matrix} 1 1 \\ 1 k \end{matrix} \| + \| \begin{matrix} 1 k \\ 1 1 \end{matrix} \| = 0$ $k (k^{2} - 1) - (k - 1) + 1 - k = 0$ $k^{3} - k - k + 1 + 1 - k = 0$ $k^{3} - 3 k + 2 = 0$ $(k - 1) (k^{2} + k - 2) = 0$ $(k - 1) (k + 2) (k - 1) = 0$ $k = {\begin{cases} 1 \\ - 2 \end{cases}$
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