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]

Question

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]

Smart Achievers · Accepted Answer

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

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

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

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

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]

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Solution

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

Ans. (55) We have, x(α,1,2)+y(1,β,2)+z(2,3,γ)=(0,0,0) ⇒αx+y+2z=0; x+βy+3z=0; 2x+2y+γz=0 Since, xyz≠0 so the system of equations has non-trivial solution. Now, |α121β322γ|=0 ⇒α(βγ-6)-1(γ-6)+2(2-2β)=0 ⇒αβγ-6α-γ+6+4-4β=0 ⇒45+10-6α-γ-4β=0⇒6α+4β+γ=55

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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]