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Q.

The system of linear equations $x + y + z = 1, x + 2 y + 4 z = λ; x + 4 y + 10 z = λ^{2}$ has a solution if it satisfies the following

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a

$λ = - 1$

b

$λ \neq 1$

c

$λ = 2$

d

$λ \in ϕ$

answer is C.

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Detailed Solution

$Δ = | \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{matrix} | = 0 Δ_{1} = | \begin{matrix} 1 & 1 & 1 \\ λ & 2 & 4 \\ λ^{2} & 4 & 10 \end{matrix} | = 2 (λ^{2} - 3 λ + 2) = 0 Δ_{2} = | \begin{matrix} 1 & 1 & 1 \\ 1 & λ & 4 \\ 1 & λ^{2} & 10 \end{matrix} | = 3 (λ^{2} - 3 λ + 2) = 0 Δ_{3} = | \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & λ \\ 1 & 4 & λ^{2} \end{matrix} | = λ^{2} - 3 λ + 2 = 0 λ = 1, 2$

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