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If the system of equations $x - 2 y + z = 4; x + y + λ z = 4; 2 x - y + 2 z = 2$ has no solution $' λ'$ lies in

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$(0, \frac{3}{4})$

$(2, 4)$

$(\frac{5}{4}, 2)$

$(\frac{3}{4}, \frac{5}{4})$

answer is B.

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Detailed Solution

$| \begin{matrix} 1 & - 2 & 1 \\ 1 & 1 & λ \\ 2 & - 1 & λ \end{matrix} | = 0 \Rightarrow 2 + λ + 2 (2 - 2 λ) + 1 (- 3) = 0$

$λ = 1 \Rightarrow λ = 1 \in (\frac{3}{4}, \frac{5}{4})$

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