Let = 45. If x ( 1 , 2 ) + y ( 1 , 2 ) + z ( 2 , 3 ) = ( 0 , 0 , 0 ) for some x , y , z ? R , ? x y z ? 0 , then 6 + 4 is equal to _______. [2024]

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Solution

Q.
Let $α β γ = 45; α, β, γ \in R .$ If $x (α, 1, 2) + y (1, β, 2) + z (2, 3, γ) = (0, 0, 0)$ for some $x, y, z \in R, x y z \neq 0,$ then $6 α + 4 β + γ$ is equal to _______. [2024]

Ans.
(55)

We have, $x (α, 1, 2) + y (1, β, 2) + z (2, 3, γ) = (0, 0, 0)$

$\Rightarrow α x + y + 2 z = 0; x + β y + 3 z = 0; 2 x + 2 y + γ z = 0$

Since, $x y z \neq 0$ so the system of equations has non-trivial solution.

Now, $| \begin{matrix} α & 1 & 2 \\ 1 & β & 3 \\ 2 & 2 & γ \end{matrix} |$ $= 0$

$\Rightarrow α (β γ - 6) - 1 (γ - 6) + 2 (2 - 2 β) = 0$

$\Rightarrow α β γ - 6 α - γ + 6 + 4 - 4 β = 0$

$\Rightarrow 45 + 10 - 6 α - γ - 4 β = 0 \Rightarrow 6 α + 4 β + γ = 55$

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