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

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Solution

The correct option is B $1$
The system of equations $k x + y + z = 1$ , $x + k y + z = k$ and $x + y + k z = k^{2}$ will have no solution when
$∣ ∣ ∣ ∣ \begin{matrix} k & 1 & 1 1 & k & 1 1 & 1 & k \end{matrix} ∣ ∣ ∣ ∣ = 0$
$\Rightarrow k^{3} - 3 k + 2 = 0$
$\Rightarrow k = 1$

Thus, the answer is $1$

Suggest Corrections

Similar questions

\begin{matrix} k x + y + z = 1 x + k y + z = k x + y + k z = k^{2} \end{matrix}

k x + y + z = 1, x + k y + z = k, x + y + k z = k^{2}

k = 1

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 k & k & 1 k^{2} & 1 & k \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

k = - 1

k x + (k + 1) y + (k - 1) z = 0

(k + 1) x + k y + (k + 2) z = 0

(k - 1) x + (k + 2) y + k z = 0

k x + y + z = 1, x + k y + z = k

x + y + z k = k^{2}

k

k

k x + y + z = k - 1, x + k y + z = k - 1, x + y + k z = k - 1

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